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[ "If we list the rows by iterations, then we get\n$1,2,3,4$\n$6,7,8,9,10$\n$13,14,15,16,17,18$ etc.\nso that the $500,000$ th number is the $506$ th number on the $997$ th row because $4+5+6+7......+999 = 499,494$ . The last number of the $996$ th row (when including the numbers skipped) is $499,494 + (1+2+3+4.....+996)= 996,000$ , (we add the $1-996$ because of the numbers we skip) so our answer is $996,000 + 506 = \\boxed{996,506}$", "One may also note we simply need to add the number of skipped numbers to $500,000$ to get our answer. The number of skipped numbers is $\\frac{996\\cdot 997}{2}$ which has a units digit of $6$ . Looking at the answer choices, it becomes apparent that the answer is $\\boxed{996,506}$", "Let's start with natural numbers, with no skips in between.\n$1,2,3,4,5,...,500,000$\nAll we need to do is count how many numbers are skipped, $n$ , and \"push\" (add on to) $500,000$ according to however many numbers are skipped.\nClearly, $\\frac{999(1000)}{2}<500,000<\\frac{1000(1001)}{2}$ . This means that there are $999-3=996$ skipped number \"blocks\" in the sequence because we started counting from 4.\nTherefore $n=\\frac{996(997)}{2}=496,506$ , and the answer is $496,506+500000=\\boxed{996,506}$", "First, we group the numbers together in the following way: ${1, 2, 3, 4, (5)}; {6, 7, 8, 9, 10, (11, 12)}; ...$ We quickly realize that the number of terms in the curly braces follow a pattern: $5, 7, 9, 11, ... , n$ (where $n$ is the $n^\\text{th}$ block. Now, we can tell that the last number in a curly brace will be the number of terms in the set added to the number of terms in all the previous sets. Luckily for us, odd numbers are easy to add. If we pretend that there was a $1, 3$ at the beginning, then the sum of all of the numbers before and including $n$ is $(n+2)^2$ . However, we have to subtract $1+3$ which results in $n^2+4n$ . The amount of numbers in the parenthesis are the $n^\\text{th}$ triangular number or $\\frac{n(n+1)}{2}$ . Next, we want to find the greatest $n$ , where $(n^2+4n) - \\frac{n(n+1)}{2}<500000$ . Simplifying, we get $n^2+7n<1000000$ . We realize that $n=1000$ results in a number just $7000$ greater than our target. Next, we square $999$ $(1000 - 1)^2 = 1000000 - 2001$ . As we decrease $n$ by $1$ , we decrease the result of the equation by approximately $2000$ . In order to decrease by at least $7000$ , we have to decrease $4$ times leading to $n=1000-4=996$ . We plug it in to $n^2+4n$ getting $996\\cdot1000=996000$ . This is the last number in the $996^\\text{th}$ set. The number of terms used is $\\frac{n\\cdot(n+7)}{2}=\\frac{996\\cdot1003}{2}=498\\cdot1003=498000+1494=499494$ . We need to add $500000-499494=506$ terms to get an answer of $\\boxed{996,506}$" ]

[ "List them out, adding the three previous numbers to get the next number,\n\\[1,2,3,6,11,20,37,\\boxed{68}\\]" ]

[ "The first square has a sidelength of $1$ , the second square $2$ , and so on. The seventh square has $7$ and is made of $7^2=49$ unit tiles. The sixth square has $6$ and is made of $6^2=36$ unit tiles. The seventh square has $49-36=\\boxed{13}$ more tiles than the sixth square.", "The edge of each square is one tile length longer than the edge of the previous square, which means that each square has $2*$ edge length $- 1$ more tiles than the previous square, because each square is just one edge added on the top and on the right to the previous square, with one overlapping tile. Then the seventh square has $2*7-1=13$ more tiles than the sixth square, which is $\\boxed{13}$", "We see a pattern of 1, 4, and 9, all of which are the squares of 1, 2,and 3 respectively. So, the 6th and 7th squares will also follow the same pattern. Via the difference of squares, we see that $7^2 - 6^2$ . Now we can see that the seventh square has $(7-6)(7+6) =$ $\\boxed{13}$ more tiles than the sixth square." ]

[ "At first, $S=\\{0,10\\}$\n\\[\\begin{tabular}{r c l c l} \\(10x+10\\) & has root & \\(x=-1\\) & so now & \\(S=\\{-1,0,10\\}\\) \\\\ \\(-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10\\) & has root & \\(x=1\\) & so now & \\(S=\\{-1,0,1,10\\}\\) \\\\ \\(x+10\\) & has root & \\(x=-10\\) & so now & \\(S=\\{-10,-1,0,1,10\\}\\) \\\\ \\(x^3+x-10\\) & has root & \\(x=2\\) & so now & \\(S=\\{-10,-1,0,1,2,10\\}\\) \\\\ \\(x+2\\) & has root & \\(x=-2\\) & so now & \\(S=\\{-10,-2,-1,0,1,2,10\\}\\) \\\\ \\(2x-10\\) & has root & \\(x=5\\) & so now & \\(S=\\{-10,-2,-1,0,1,2,5,10\\}\\) \\\\ \\(x+5\\) & has root & \\(x=-5\\) & so now & \\(S=\\{-10,-5,-2,-1,0,1,2,5,10\\}\\) \\end{tabular}\\]\nAt this point, no more elements can be added to $S$ . To see this, let\n\\begin{align*} a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0 &= 0 \\\\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 &= 0 \\\\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) &= -a_0 \\end{align*}\nwith each $a_i$ in $S$ $x$ is a factor of $a_0$ , and $a_0$ is in $S$ , so $x$ has to be a factor of some element in $S$ . There are no such integers left, so there can be no more additional elements. $\\{-10,-5,-2,-1,0,1,2,5,10\\}$ has $9$ elements $\\to \\boxed{9}$" ]

[ "If $(2,3)$ is in $S$ , then $(3,2)$ is also, and quickly we see that every point of the form $(\\pm 2, \\pm 3)$ or $(\\pm 3, \\pm 2)$ must be in $S$ . Now note that these $8$ points satisfy all of the symmetry conditions. Thus the answer is $\\boxed{8}$" ]

[ "Define $T$ to be the set of all integral triples $(a, b, c)$ such that $a \\ge b \\ge c$ $b+c > a$ , and $a, b, c < 5$ . Now we enumerate the elements of $T$\n$(4, 4, 4)$\n$(4, 4, 3)$\n$(4, 4, 2)$\n$(4, 4, 1)$\n$(4, 3, 3)$\n$(4, 3, 2)$\n$(3, 3, 3)$\n$(3, 3, 2)$\n$(3, 3, 1)$\n$(3, 2, 2)$\n$(2, 2, 2)$\n$(2, 2, 1)$\n$(1, 1, 1)$\nIt should be clear that $|S|$ is simply $|T|$ minus the larger \"duplicates\" (e.g. $(2, 2, 2)$ is a larger duplicate of $(1, 1, 1)$ ). Since $|T|$ is $13$ and the number of higher duplicates is $4$ , the answer is $13 - 4$ or $\\boxed{9}$" ]

[ "Let these four numbers be $a$ $b$ $c$ , and $d$ , where $a>b>c>d$ $x+y$ needs to be maximized, so let $x=a+b$ and $y=a+c$ because these are the two largest pairwise sums. Now $x+y=2a+b+c$ needs to be maximized. Notice that $2a+b+c=3(a+b+c+d)-(a+2b+2c+3d)=3((a+c)+(b+d))-((a+d)+(b+c)+(b+d)+(c+d))$ . No matter how the numbers $189$ $320$ $287$ , and $234$ are assigned to the values $a+d$ $b+c$ $b+d$ , and $c+d$ , the sum $(a+d)+(b+c)+(b+d)+(c+d)$ will always be $189+320+287+234$ . Therefore we need to maximize $3((a+c)+(b+d))-(189+320+287+234)$ . The maximum value of $(a+c)+(b+d)$ is achieved when we let $a+c$ and $b+d$ be $320$ and $287$ because these are the two largest pairwise sums besides $x$ and $y$ . Therefore, the maximum possible value of $x+y=3(320+287)-(189+320+287+234)=\\boxed{791}$", "Let the four numbers be $a$ $b$ $c$ , and $d$ , in no particular order. Adding the pairwise sums, we have $3a+3b+3c+3d=1030+x+y$ , so $x+y=3(a+b+c+d)-1030$ . Since we want to maximize $x+y$ , we must maximize $a+b+c+d$\nOf the four sums whose values we know, there must be two sums that add to $a+b+c+d$ . To maximize this value, we choose the highest pairwise sums, $320$ and $287$ . Therefore, $a+b+c+d=320+287=607$\nWe can substitute this value into the earlier equation to find that $x+y=3(607)-1030=1821-1030=\\boxed{791}$", "There are two cases we can consider. Let the elements of our set be denoted $a,b,c,d$ , and say that the largest sums $x$ and $y$ will be consisted of $b+d$ and $c+d$ . Thus, we want to maximize $b+c+2d$ , which means $d$ has to be as large as possible, and $a$ has to be as small as possible to maximize $b$ and $c$ . So, the two cases we look at are:\nCase 1:\n\\[a+d = 287\\] \\[b+c = 320\\] \\[a+b = 234\\] \\[a+c = 189\\]\nCase 2:\n\\[a+d = 320\\] \\[b+c = 189\\] \\[a+b = 234\\] \\[a+c = 287\\]\nNote we have determined these cases by maximizing the value of $a+d$ determined by our previous conditions. So, the answers for each ( after some simple substitution ) will be:\nCase 1:\n\\[(a,b,c,d) = (\\frac{103}{2},\\frac{365}{2},\\frac{275}{2},\\frac{471}{2})\\]\nCase 2:\n\\[(a,b,c,d) = (166,68,121,154)\\]\nSee the first case has our largest $d$ , so our answer will be $471+\\frac{640}{2} = \\boxed{791}$" ]

[ "There are $25$ ways to choose the first square. The four remaining squares in its row and column and the square you chose exclude nine squares from being chosen next time.\nThere are $16$ remaining blocks to be chosen for the second square. The three remaining spaces in its row and column and the square you chose must be excluded from being chosen next time.\nFinally, the last square has $9$ remaining choices.\nThe number of ways to choose $3$ squares is $25 \\cdot 16 \\cdot 9,$ but the order in which you chose the squares does not matter as the blocks are indistinguishable, so we divide by $3!$\n\\[\\frac{25 \\cdot 16 \\cdot 9}{3 \\cdot 2 \\cdot 1} = 25 \\cdot 8 \\cdot 3 = 100 \\cdot 6 = \\boxed{600}\\]", "Once we choose our three squares, we will have occupied three separate columns $(A, B, C)$ and three separate rows. There are ${5 \\choose 3} \\times {5 \\choose 3}$ ways to choose these rows and columns.\nThere are $3$ ways to assign the square in column $A$ to a row, $2$ ways to assign the square in column $B$ to one of the remaining two rows, and poor square in column $C$ doesn't get to choose. $:($\nIn total, we have \\[{5 \\choose 3} \\times {5 \\choose 3} \\times 3!\\] which totals out to $\\boxed{600}$", "We know that there are $\\binom{25}{3}=2300$ ways to choose three blocks. However, the restriction clearly limits the number of ways we can choose our blocks. Hence, only $\\text{(A)}$ $\\text{(B)}$ , or $\\text{(C)}$ could be the correct answer. Clearly, there are more than $125$ ways, thus yielding $\\boxed{600}$ ways." ]

[ "Let there be $T$ teams. For each team, there are ${n-5\\choose 4}$ different subsets of $9$ players that includes a given full team, so the total number of team-(group of 9) pairs is\n\\[T{n-5\\choose 4}.\\]\nThus, the expected value of the number of full teams in a random set of $9$ players is\n\\[\\frac{T{n-5\\choose 4}}{{n\\choose 9}}.\\]\nSimilarly, the expected value of the number of full teams in a random set of $8$ players is\n\\[\\frac{T{n-5\\choose 3}}{{n\\choose 8}}.\\]\nThe condition is thus equivalent to the existence of a positive integer $T$ such that\n\\[\\frac{T{n-5\\choose 4}}{{n\\choose 9}}\\frac{T{n-5\\choose 3}}{{n\\choose 8}} = 1.\\]\n\\[T^2\\frac{(n-5)!(n-5)!8!9!(n-8)!(n-9)!}{n!n!(n-8)!(n-9)!3!4!} = 1\\]\n\\[T^2 = \\big((n)(n-1)(n-2)(n-3)(n-4)\\big)^2 \\frac{3!4!}{8!9!}\\]\n\\[T^2 = \\big((n)(n-1)(n-2)(n-3)(n-4)\\big)^2 \\frac{144}{7!7!8\\cdot8\\cdot9}\\]\n\\[T^2 = \\big((n)(n-1)(n-2)(n-3)(n-4)\\big)^2 \\frac{1}{4\\cdot7!7!}\\]\n\\[T = \\frac{(n)(n-1)(n-2)(n-3)(n-4)}{2^5\\cdot3^2\\cdot5\\cdot7}\\]\nNote that this is always less than ${n\\choose 5}$ , so as long as $T$ is integral, $n$ is a possibility. Thus, we have that this is equivalent to\n\\[2^5\\cdot3^2\\cdot5\\cdot7\\big|(n)(n-1)(n-2)(n-3)(n-4).\\]\nIt is obvious that $5$ divides the RHS, and that $7$ does iff $n\\equiv 0,1,2,3,4\\mod 7$ . Also, $3^2$ divides it iff $n\\not\\equiv 5,8\\mod 9$ . One can also bash out that $2^5$ divides it in $16$ out of the $32$ possible residues $\\mod 32$\nNote that $2016 = 7*9*32$ so by using all numbers from $2$ to $2017$ , inclusive, it is clear that each possible residue $\\mod 7,9,32$ is reached an equal number of times, so the total number of working $n$ in that range is $5\\cdot 7\\cdot 16 = 560$ . However, we must subtract the number of \"working\" $2\\leq n\\leq 8$ , which is $3$ . Thus, the answer is $\\boxed{557}$" ]

[ "There are $10 \\cdot 2+1=21$ teams. Any of the $\\tbinom{21}3=1330$ sets of three teams must either be a fork (in which one team beat both the others) or a cycle:\n But we know that every team beat exactly $10$ other teams, so for each possible $A$ at the head of a fork, there are always exactly $\\tbinom{10}2$ choices for $B$ and $C$ as $A$ beat exactly 10 teams and we are choosing 2 of them. Therefore there are $21\\cdot\\tbinom{10}2=945$ forks, and all the rest must be cycles.\nThus the answer is $1330-945=385$ which is $\\boxed{385}$", "Since there are $21$ teams and for each set of three teams there is a cycle, there are a total of $\\tbinom{21}3=1330$ cycles of three teams. Because about $1/4$ of the cycles $\\{A, B, C\\}$ satisfy the conditions of the problems, our answer is close to $1/4 \\cdot 1330=332.5$ . Looking at the answer choices, we find that $332.5$ is closer to $385$ than any other answer choices, and that the next closest is $665$ which is twice of $332.5$ , so our answer is $385$ which is $\\boxed{385}$", "Let's arrange all the teams in a circle and assume that each team won against the first 10 teams clockwise of themselves, and lost against the first 10 teams counter-clockwise of themselves.\nConsider a working set of $3$ teams: moving clockwise, we know that the first team beat the team clockwise of itself, that team beat the next team clockwise of itself, and the final team beat the first team, which would be clockwise of itself. However, we also must remember that if the first team beat the second team, moving clockwise, the second team cannot be more than $10$ teams away from the first team; the same applies to the second and third team, and the third and first team.\nLet's say, WLOG, that the first team, team $A$ , is at position $0$ out of $20$ on the circle.\nIf team $A$ is to beat team $B$ , since we are assuming each team beats the 10 members clockwise of themselves, there are $10$ places on the circle that team $B$ could be: positions $1$ through $10$ . Also, if team $C$ is to beat team $A$ , team $C$ must be located from positions $11$ to $20$\nIf Team $B$ is in position $n$ $C$ must be located from position $n+1$ to position $n+10$ , since $B$ beats $C$ . We also just found that $C$ 's position must be between $11$ and $20$ inclusive. So, when $B$ is in position $1$ $C$ can only be in position $11$ . When $B$ is in position $2$ $C$ can be in position $11$ or $12$ . In general, when $B$ is in position $n$ , there are $n$ choices for where $C$ can be. So, there are $1+2+3+ \\dots + 10 = 55$ ways to place $B$ and $C$ . There are $21$ players to choose as player $A$ , and each working set will be counted $3$ times, so our final answer is $55 \\cdot 21 \\div 3 = \\boxed{385,}$", "This is a Graph Theory problem with directed graph. There are $21$ teams in total. WLOG, pick team $A$ , there are $10$ teams that lost to $A$ and $10$ teams that won over $A$ . Call the group of teams that lost to $A$ group $L$ , and the group of teams that won over $A$ group $W$\n\nAny team from group $L$ that won a team from group $W$ will form a cycle with $A$ . Now we need to count how many teams in group $L$ won over a team from group $W$ . The total number of wins in group $L$ is $10 \\cdot 10 =100$ . There are $\\tbinom{10}2=45$ wins among the teams inside group $L$ . So group $L$ has $100-45=55$ wins over group $W$\n\\[\\frac{21 \\cdot 55}{3}=\\boxed{385,}\\]" ]

[ "We use complementary counting. First, because each team played $20$ other teams, there are $21$ teams total. All sets that do not have $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A$ have one team that beats both the other teams. Thus we must count the number of sets of three teams such that one team beats the two other teams and subtract that number from the total number of ways to choose three teams.\nThere are $21$ ways to choose the team that beat the two other teams, and $\\binom{10}{2} = 45$ to choose two teams that the first team both beat. This is $21 * 45 = 945$ sets. There are $\\binom{21}{3} = 1330$ sets of three teams total. Subtracting, we obtain $1330 - 945 = \\boxed{385})$ is our answer.", "Note that there are $21$ teams total and $\\binom{21}{3}=1330$ ways to pick ${A,B,C}.$ The possible arrangements are one team beats the other two or they each win/lose equally (we want the second case). Approximately $\\frac{1}{4}$ of all the arrangements satisfy the second case, and $\\frac{1330}{4}=332.5,$ which is by far the closest to $\\boxed{385}.$" ]

[ "As above, note that there are 21 teams, and call them A, B, C, ... T, U. WLOG, assume that A beat teams B-L and lost to teams M-U. We will count the number of sets satisfying the “cycle-win” condition—e.g. here, A beats a team in X which beats a team in Y which beats A. The first and third part of the condition are already met by our wlog, so we just need to count of number of ways the second condition is true (a team in X beats a team in Y). These are the number of cycle-wins that include A, then multiply by 21 (for each team) and divide by 3 (since every set will be counted by each of the 3 teams that are a part of that set).\nTo do this, let X $=\\{B, ..., L\\}$ and Y $=\\{M, ..., U\\}$ . Since a total of $10*10=100$ losses total were suffered by teams in Y and $\\binom{10}{2}=45^{*}$ losses were suffered by teams in Y from teams in Y, we have $100-45=55$ losses suffered by teams in Y from teams in X. Hence, for each of these $55$ losses, there is exactly one set of three teams that includes A that satisfies the problem conditions. Thus, the answer is $\\frac{55\\cdot 21}{3}=\\boxed{385}$" ]

[ "The pattern is quite simple to see after listing a couple of terms.\n\\[\\begin{tabular}{|r|r|r|} \\hline \\#&\\text{Removed}&\\text{Left}\\\\ \\hline 1&10&90\\\\ 2&9&81\\\\ 3&9&72\\\\ 4&8&64\\\\ 5&8&56\\\\ 6&7&49\\\\ 7&7&42\\\\ 8&6&36\\\\ 9&6&30\\\\ 10&5&25\\\\ 11&5&20\\\\ 12&4&16\\\\ 13&4&12\\\\ 14&3&9\\\\ 15&3&6\\\\ 16&2&4\\\\ 17&2&2\\\\ 18&1&1\\\\ \\hline \\end{tabular}\\]\nThus, the answer is $\\boxed{18}$", "Given $n^2$ tiles, a step removes $n$ tiles, leaving $n^2 - n$ tiles behind. Now, $(n-1)^2 = n^2 - n + (1-n) < n^2 - n < n^2$ , so in the next step $n-1$ tiles are removed. This gives $(n^2 - n) - (n-1) = n^2 - 2n + 1 = (n-1)^2$ , another perfect square.\nThus each two steps we cycle down a perfect square, and in $(10-1)\\times 2 = 18$ steps, we are left with $1$ tile, hence our answer is $\\boxed{18}$" ]

[ "We start of with $100 = 10 \\cdot 10$ numbers. When we use the certain operation, call if $P(x)$ , have $100 - 10 = 90 = 10 \\cdot 9$ .\nThen we do $P(x)$ again, to subtract $9$ numbers and get $9 \\cdot 9$ . In the end, we will want $1 = 1 \\cdot 1$ . We can say we have to use $P(x)$ once to make $n \\cdot n$ into $n \\cdot (n-1)$ . Thus we must use it twice to get from $n \\cdot n$ to $(n-1)(n-1)$ . For example, it takes us $2$ of $P(x)$ to get from $10 \\cdot 10$ to $9 \\cdot 9$ . Then $2$ of $P(x)$ to get from $9 \\cdot 9$ to $8 \\cdot 8$ . You should try this with $7$ and $6$ , and see it works. This means we can have $n$ be the number we start with, and $1$ be the number we want. Then we would use $P(x)$ $2(n - 1)$ times to get $1 \\cdot 1$ . Substituting $n$ for $10$ we get $2(10-1) = 2 \\cdot 9 = \\boxed{18}$ .\n- Wiselion" ]

[ "Let $C_1$ be the point the ship would reach if it turned $45^\\circ$ , and $C_2$ the point it would reach if it turned $60^\\circ$ . Obviously, $C_1$ is the furthest possible point from $A$ , and $C_2$ is the closest possible point to $A$\nHence the interval of possible values for $AC^2$ is $[AC_2^2,AC_1^2]$\n\nWe can find $AC_1^2$ and $AC_2^2$ as follows:\nLet $D_1$ and $D_2$ be the feet of the heights from $C_1$ and $C_2$ onto $AB$ . The angles in the triangle $BD_1C_1$ are $45^\\circ$ $45^\\circ$ , and $90^\\circ$ , hence $BD_1 = D_1C_1 = BC_1 / \\sqrt 2$ . Similarly, the angles in the triangle $BD_2C_2$ are $30^\\circ$ $60^\\circ$ , and $90^\\circ$ , hence $BD_2 = BC_2 / 2$ and $D_2C_2 = BC_2 \\sqrt 3 / 2$\n\n\nHence we get:\n\\[AC_2^2 = AD_2^2 + D_2C_2^2 = 20^2 + (10\\sqrt 3)^2 = 400 + 300 = 700\\]\n\\[AC_1^2 = AD_1^2 + D_1C_1^2 = (10 + 20/\\sqrt 2)^2 + (20\\sqrt 2)^2 = 100 + 400/\\sqrt 2 + 200 + 200 = 500 + 200\\sqrt 2 < 500 + 200\\cdot 1.5 = 800\\]\nTherefore for any valid $C$ the value $AC^2$ is surely in the interval $\\boxed{700,800}$" ]

[ "Let the original price of an item be $x$\nFirst, everything is half-off, so the price is now $\\frac{x}{2} = 0.5x$\nNext, the extra coupon applies 20% off on the sale price , so the price after this discount will be $100\\% - 20\\% = 80\\%$ of what it was before. (Notice how this is not applied to the original price; if it were, the solution would be applying 50% + 20 % = 70% off the original price.)\n$80\\% \\cdot 0.5 x = \\frac{4}{5} \\cdot 0.5 x = 0.4x$\nThe price of the item after all discounts have been applied is $0.4x = 40\\% \\cdot x$ . However, we need to find the percentage off the original price, not the current percentage of the original price. We then subtract $40\\% x$ from $100\\% x$ (the original price of the item), to find the answer, $\\boxed{60}$", "Since the problem implies that the percentage off the original price will be the same for every item in the store, fakesolving is applicable here. Say we are buying an item worth 10 dollars, a convenient number to work with. First, it is clear that we'll get 50% off, which makes the price then 5 dollars. Taking 20% off of 5 dollars gives us 4 dollars. Therefore, we have saved a total of $\\frac{10-4}{10} = \\frac{6}{10} = \\frac{60}{100} = \\boxed{60} \\%$", "The price is $50\\%$ off, which means the price would be $100\\% - 50\\% = 50\\%$ of what it was before. Then, you get $20\\%$ off. Sidenote: A common mistake is finding $20\\%$ of the price, but forgetting to subtract it. $20\\%$ off the price is $100\\% - 20\\% = 80\\%.$ Now, we can multiply the two discounts as decimals. $0.5 \\times 0.8 = 0.4 = 40\\%.$ Our final answer is $100\\% - 40\\% = \\boxed{60}$" ]

[ "After the $20\\%$ sale, the coat costs $100(0.8)=80$ dollars. Then $5$ dollars are taken off for a cost of $80-5=75$ . Adding on the sales tax, the final amount is $(75)(1.08)=\\boxed{81.00}$" ]

[ "Let the listed price be $(100 + p)$ , where $p > 0$\nCoupon A saves us: $0.15(100+p) = (0.15p + 15)$\nCoupon B saves us: $30$\nCoupon C saves us: $0.25p$\nNow, the condition is that A has to be greater than or equal to either B or C which gives us the following inequalities:\n$A \\geq B \\Rightarrow 0.15p + 15 \\geq 30 \\Rightarrow p \\geq 100$\n$A \\geq C \\Rightarrow 0.15p + 15 \\geq 0.25p \\Rightarrow p \\leq 150$\nWe see here that the greatest possible value for $p$ is $150$ , thus $y = 100 + 150 = 250$ and the smallest value for $p$ is $100$ so $x = 100 + 100 = 200$\nThe difference between $y$ and $x$ is $y - x = 250 - 200 = \\boxed{50}$" ]

[ "50% off the price of half a pound of fish is $3, so 100%, the regular price, of a half pound of fish is $6. If half a pound of fish costs $6, then a whole pound of fish is $\\boxed{12}$ dollars.", "Suppose a full pound at normal price costs $x$ dollars. Then, with the 50% off deal, the full pound would cost $x/2$ dollars. A half pound of this would cost $x/4$ dollars (including the 50% off). 50% off half a pound is 3, so we form the equation $x/4=3\\Rightarrow x=\\boxed{12}$" ]

[ "The least common multiple of $7$ and $11$ is $77$ . Therefore, there must be $77$ adults and $77$ children. The total number of benches is $\\frac{77}{7}+\\frac{77}{11}=11+7=\\boxed{18}$ .~taarunganesh", "Let $x$ denote how many adults there are. Since the number of adults is equal to the number of children we can write $N$ as $\\frac{x}{7}+\\frac{x}{11}=N$ . Simplifying we get $\\frac{18x}{77} = N$ Since both $n$ and $x$ have to be positive integers, $x$ has to equal $77$ . Therefore, $N=\\boxed{18}$ is our final answer." ]

[ "Since there are 6 players, a total of $\\frac{6(6-1)}{2}=15$ games are played. So far, $4+3+2+2+2=13$ games finished (one person won from each game), so Monica needs to win $15-13 = \\boxed{2}$" ]

[ "To find how many small bottles we need, we can simply divide $500$ by $35$ . This simplifies to $\\frac{100}{7}=14 \\frac{2}{7}.$ Since the answer must be an integer greater than $14$ , we have to round up to $15$ bottles, or $\\boxed{15}$", "We double $35$ to get $70.$ We see that $70\\cdot7=490,$ which is very close to $500.$ Thus, $2\\cdot7+1=\\boxed{15}$ bottles are enough. \n~Technodoggo" ]

[ "To find how many small bottles we need, we can simply divide $500$ by $35$ . This simplifies to $\\frac{100}{7}=14 \\frac{2}{7}.$ Since the answer must be an integer greater than $14$ , we have to round up to $15$ bottles, or $\\boxed{15}$", "We double $35$ to get $70.$ We see that $70\\cdot7=490,$ which is very close to $500.$ Thus, $2\\cdot7+1=\\boxed{15}$ bottles are enough. \n~Technodoggo" ]

[ "The lines passing through $A$ and $C$ divide the square into three parts, two right triangles and a parallelogram . Using the smaller side of the parallelogram, $1/n$ , as the base, where the height is 1, we find that the area of the parallelogram is $A = \\frac{1}{n}$ . By the Pythagorean Theorem , the longer base of the parallelogram has length $l = \\sqrt{1^2 + \\left(\\frac{n - 1}{n}\\right)^2} = \\frac{1}{n}\\sqrt{2n^2 - 2n + 1}$ , so the parallelogram has height $h = \\frac{A}{l} = \\frac{1}{\\sqrt{2n^2 - 2n + 1}}$ . But the height of the parallelogram is the side of the little square, so $2n^2 - 2n + 1 = 1985$ . Solving this quadratic equation gives $n = \\boxed{32}$", "Aime.png\nSurrounding the square with area $\\frac{1}{1985}$ are $4$ right triangles with hypotenuse $1$ (sides of the large square). Thus, $X + \\frac{1}{1985} = 1$ , where $X$ is the area of the of the 4 triangles.\nWe can thus use proportions to solve this problem. \\begin{eqnarray*} \\frac{GF}{BE}=\\frac{CG}{CB}\\implies \\frac{\\frac{1}{\\sqrt{1985}}}{BE}=\\frac{\\frac{1}{n}}{1}\\implies BE=\\frac{n\\sqrt{1985}}{1985} \\end{eqnarray*} Also, \\begin{eqnarray*} \\frac{BE}{1}=\\frac{EC}{\\frac{n-1}{n}}\\implies EC=\\frac{\\sqrt{1985}}{1985}(n-1) \\end{eqnarray*} Thus, \\begin{eqnarray*} 2(BE)(EC)+\\frac{1}{1985}=1\\\\ 2n^{2}-2n+1=1985\\\\ n(n-1)=992 \\end{eqnarray*} Simple factorization and guess and check gives us $\\boxed{32}$", "AIME 1985 Problem 4 Solution 3 Diagram.png\nLine Segment $DE = \\frac{1}{n}$ , so $EC = 1 - \\frac{1}{n} = \\frac{n-1}{n}$ . Draw line segment $HE$ parallel to the corresponding sides of the small square, $HE$ has length $\\frac{1}{\\sqrt{1985}}$ , as it is the same length as the sides of the square. Notice that $\\triangle CEL$ is similar to $\\triangle HDE$ by $AA$ similarity. Thus, $\\frac{LC}{HE} = \\frac{EC}{DE} = n-1$ , so $LC = \\frac{n-1}{\\sqrt{1985}}$ . Notice that $\\triangle CEL$ is also similar to $\\triangle CDF$ by $AA$ similarity. Thus, $\\frac{FC}{EC} = \\frac{DC}{LC}$ , and the expression simplifies into a quadratic equation $n^2 - n - 992 = 0$ . Solving this quadratic equation yields $n =\\boxed{32}$" ]

[ "There are $0-3$ substitutions. The number of ways to sub any number of times must be multiplied by the previous number. This is defined recursively. The case for $0$ subs is $1$ , and the ways to reorganize after $n$ subs is the product of the number of new subs ( $12-n$ ) and the players that can be ejected ( $11$ ). The formula for $n$ subs is then $a_n=11(12-n)a_{n-1}$ with $a_0=1$\nSumming from $0$ to $3$ gives $1+11^2+11^{3}\\cdot 10+11^{4}\\cdot 10\\cdot 9$ . Notice that $10+9\\cdot11\\cdot10=10+990=1000$ . Then, rearrange it into $1+11^2+11^3\\cdot (10+11\\cdot10\\cdot9)= 1+11^2+11^3\\cdot (1000)$ . When taking modulo $1000$ , the last term goes away. What is left is $1+11^2=\\boxed{122}$" ]

[ "We know that, for the games where they scored an odd number of runs, they cannot have scored twice as many runs as their opponents, as odd numbers are not divisible by $2$ . Thus, from this, we know that the five games where they lost by one run were when they scored $1$ $3$ $5$ $7$ , and $9$ runs, and the others are where they scored twice as many runs. We can make the following chart:\n$\\begin{tabular}{|l|l|} \\hline Them & Opponent \\\\ \\hline 1 & 2 \\\\ 2 & 1 \\\\ 3 & 4 \\\\ 4 & 2 \\\\ 5 & 6 \\\\ 6 & 3 \\\\ 7 & 8 \\\\ 8 & 4 \\\\ 9 & 10 \\\\ 10 & 5 \\\\ \\hline \\end{tabular}$\nThe sum of their opponent's scores is $2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = \\boxed{45}$" ]

[ "The volume of the original box is $15\\cdot10\\cdot8=1200.$\nThe volume of each cube that is removed is $3\\cdot3\\cdot3=27.$\nSince there are $8$ corners on the box, $8$ cubes are removed.\nSo the total volume removed is $8\\cdot27=216$\nTherefore, the desired percentage is $\\frac{216}{1200}\\cdot100 = \\boxed{18}.$" ]

[ "The volume of the original box is $15\\cdot10\\cdot8=1200.$\nThe volume of each cube that is removed is $3\\cdot3\\cdot3=27.$\nSince there are $8$ corners on the box, $8$ cubes are removed.\nSo the total volume removed is $8\\cdot27=216$\nTherefore, the desired percentage is $\\frac{216}{1200}\\cdot100 = \\boxed{18}.$" ]

[ "Imagine making the cuts one at a time. The first cut removes a box $2\\times 2\\times 3$ . The second cut removes two boxes, each of dimensions $2\\times 2\\times 0.5$ , and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is $12 + 4 + 4 = 20$\nTherefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \\boxed{7}$", "We can use Principle of Inclusion-Exclusion (PIE) to find the final volume of the cube.\nThere are 3 \"cuts\" through the cube that go from one end to the other. Each of these \"cuts\" has $2 \\times 2 \\times 3=12$ cubic inches. However, we can not just sum their volumes, as\nthe central $2\\times 2\\times 2$ cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.\nHence the total volume of the cuts is $3(2 \\times 2 \\times 3) - 2(2\\times 2\\times 2) = 36 - 16 = 20$\nTherefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \\boxed{7}$", "We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.\nEach edge can be seen as a $2\\times 0.5\\times 0.5$ box, and each corner can be seen as a $0.5\\times 0.5\\times 0.5$ box.\n$12\\cdot{\\frac{1}{2}} + 8\\cdot{\\frac{1}{8}} = 6+1 = \\boxed{7}$" ]

[ "Imagine making the cuts one at a time. The first cut removes a box $2\\times 2\\times 3$ . The second cut removes two boxes, each of dimensions $2\\times 2\\times 0.5$ , and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is $12 + 4 + 4 = 20$\nTherefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \\boxed{7}$", "We can use Principle of Inclusion-Exclusion (PIE) to find the final volume of the cube.\nThere are 3 \"cuts\" through the cube that go from one end to the other. Each of these \"cuts\" has $2 \\times 2 \\times 3=12$ cubic inches. However, we can not just sum their volumes, as\nthe central $2\\times 2\\times 2$ cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.\nHence the total volume of the cuts is $3(2 \\times 2 \\times 3) - 2(2\\times 2\\times 2) = 36 - 16 = 20$\nTherefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \\boxed{7}$", "We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.\nEach edge can be seen as a $2\\times 0.5\\times 0.5$ box, and each corner can be seen as a $0.5\\times 0.5\\times 0.5$ box.\n$12\\cdot{\\frac{1}{2}} + 8\\cdot{\\frac{1}{8}} = 6+1 = \\boxed{7}$" ]

[ "We can use Euler's polyhedron formula that says that $F+V=E+2$ . We know that there are originally $6$ faces on the cube, and each corner cube creates $3$ more. $6+8(3) = 30$ . In addition, each cube creates $7$ new vertices while taking away the original $8$ , yielding $8(7) = 56$ vertices. Thus $E+2=56+30$ , so $E=\\boxed{84}$", "The removal of each cube adds nine additional edges to the solid. Since a cube initially has $12$ edges and there are eight vertices, the number of edges will be $12 + 9 \\times 8 = \\boxed{84}$" ]

[ "The $231$ cubes which are not visible must lie below exactly one layer of cubes. Thus, they form a rectangular solid which is one unit shorter in each dimension. If the original block has dimensions $l \\times m \\times n$ , we must have $(l - 1)\\times(m-1) \\times(n - 1) = 231$ . The prime factorization of $231 = 3\\cdot7\\cdot11$ , so we have a variety of possibilities; for instance, $l - 1 = 1$ and $m - 1 = 11$ and $n - 1 = 3 \\cdot 7$ , among others. However, it should be fairly clear that the way to minimize $l\\cdot m\\cdot n$ is to make $l$ and $m$ and $n$ as close together as possible, which occurs when the smaller block is $3 \\times 7 \\times 11$ . Then the extra layer makes the entire block $4\\times8\\times12$ , and $N= \\boxed{384}$" ]

[ "Let $P_k$ be the probability of emptying the bag when it has $k$ pairs in it. Let's consider the possible draws for the first three cards:\nTherefore, we obtain the recursion $P_k = \\frac {3}{2k - 1}P_{k - 1}$ . Iterating this for $k = 6,5,4,3,2$ (obviously $P_1 = 1$ ), we get $\\frac {3^5}{11*9*7*5*3} = \\frac {9}{385}$ , and $p+q=\\boxed{394}$", "Call the case that we begin with [ABCDEF]. It doesn't matter what letter we choose at first, so WLOG assume we choose A. Now there is BCDEFABCDEF remaining in the bag. We have two cases to consider here.\n1. We pick the other A. There's a $\\frac{1}{11}$ chance for this to happen. We remain with the case [BCDEF] if this is the case.\n2. We pick any other letter that is not an A. There's a $\\frac{10}{11}$ chance for this to happen. WLOG, assume we pick the letter B. Now in order for us to continue the game, we must choose either the other A or B. There's a $\\frac{2}{10}$ chance for this to happen. WLOG, assume we choose A. Now we have BCDEFCDEF left.\nNotice however that in the first case, the probability of emptying the bag with [BCDEF] is the same thing as with BCDEFCDEF, as the only difference is you've removed one of the letters (and it doesn't matter which you chose).\nHence for this case, there is a $\\frac{1}{11} + \\frac{10}{11}*\\frac{2}{10} = \\frac{3}{11}$ * [BCDEF] chance to empty the bag.\nContinuing this process, we get that:\n[BCDEF] = $\\frac{3}{9}$ * [CDEF]\n[CDEF] = $\\frac{3}{7}$ * [DEF]\n[DEF] = $\\frac{3}{5}$ * [EF]\n[EF] = 1 (clearly, since if we are only left with EFEF then we are going to empty the bag).\nAnd hence [ABCDEF] = $1*\\frac{3}{5}*\\frac{3}{7}*\\frac{3}{9}*\\frac{3}{11} = \\frac{9}{385}$ so our answer is $9+385=\\boxed{394}$" ]

[ "Assume that there are $100$ people in the audience.\n$20$ people heard $60$ minutes of the talk, for a total of $20\\cdot 60 = 1200$ minutes heard.\n$10$ people heard $0$ minutes.\n$\\frac{70}{2} = 35$ people heard $20$ minutes of the talk, for a total of $35\\cdot 20 = 700$ minutes.\n$35$ people heard $40$ minutes of the talk, for a total of $1400$ minutes.\nAltogether, there were $1200 + 0 + 700 + 1400 = 3300$ minutes heard among $100$ people.\nThus, the average is $\\frac{3300}{100} = 33$ minutes, and the answer is $\\boxed{33}$" ]

[ "Without loss of generality, assume that the $8$ numbers on Sharon's cards are $1$ $1$ $2$ $3$ $4$ $5$ $6$ , and $7$ , in that order, and assume the $8$ colors are red, red, and six different arbitrary colors. There are ${8\\choose2}-1$ ways of assigning the two red cards to the $8$ numbers; we subtract $1$ because we cannot assign the two reds to the two $1$ 's.\nIn order for Sharon to be able to remove at least one card and still have at least one card of each color, one of the reds have to be assigned with one of the $1$ s. The number of ways for this to happen is $2 \\cdot 6 = 12$ (the first card she draws has to be a $1$ (2 choices), while the second card can be any card but the remaining card with a $1$ (6 choices)). Each of these assignments is equally likely, so desired probability is $\\frac{12}{{8\\choose2}-1}=\\frac{4}{9} \\implies 4 + 9 = 13 = \\boxed{013}$", "First note that out of the $8$ selected cards, one pair of cards have to share the same number and another pair of cards have to share the same color. Now, these $2$ pairs of cards can't be the same or else there will be $2$ cards which are completely same. Then, WLOG let the numbers be $1,1,2,3,4,5,6,$ and $7$ and the colors be $a,a,b,c,d,e,f,$ and $g$ . We therefore obtain only $2$ cases:\nCase One: $1a,1b,2a,3c,4d,5e,6f,$ and $7g$ In this case, we can discard $1a$ .\nThere are $2*6=12$ situations in this case.\nCase Two: $1b,1c,2a,3a,4d,5e,6f,$ and $7g$ In this case, we can't discard.\nThere are $\\dbinom{6}{2}=15$ situations in this case.\nSo the probability is $\\frac{12}{12+15}=\\frac{4}{9}$ , giving us the answer of $4+9=\\boxed{013}$", "There are $7!$ ways to choose a set of 7 cards that have all the numbers from 1-7 and all 7 colors. There are then $42$ cards remaining. Thus, there are $7!(42)$ desired sets.\nNow, the next thing to find is the number of ways to choose 8 cards where there is not a set of 7 such cards. In this case, one color must have 2 cards and one number must have 2 cards, and they can't be the same number/color card. The number of ways to pick this is equal to a multiplication of $\\binom{7}{2}$ ways to pick 2 numbers, $7$ colors to assign them to, $\\binom{6}{2}$ ways to pick 2 nonchosen colors, $5$ ways to pick a number to assign them to, and $4!$ ways to assign the rest.\nThus, the answer is $\\frac{7!(42)}{7!(42) + 21(15)(7)(5!)}$ . Dividing out $5!$ yields $\\frac{42(42)}{42(42) + 21(15)(7)}$ which is equal to $\\frac{2(42)}{2(42) + 15(7)}$ which is equal to $\\frac{12}{12 + 15}$ which is equal to $\\frac{4}{9}$ giving a final answer of $\\boxed{013}$", "We can rewrite the problem as \"What is the probability that, given 8 cards with numbers a, b, c, d, e, f, g, g and some assortment of seven colors A, B, C, D, E, F, G, G where G is repeated, one of the cards is Gg?\" Note that this is a valid restatement because Sharon has to have two of one number and two of one color. She needs to be able to take away one of the cards with the duplicate number, but this also has to have the duplicate color. There are two cases.\nCase I: One of the cards is Gg. This implies that the other card with color G can be placed in $6$ ways, and the rest of the colors can be paired with cards in $6!$ ways.\nCase II: None of the cards are Gg. This implies that the cards with color G can be chosen in $\\dbinom{6}{2}=15$ ways, and the rest of the colors can be paired with cards in $\\frac{6!}{2}$ ways, with the divide by 2 because of the double-g.\nNote that there is no case with Gg, Gg because all 49 cards are unique!\nTherefore our answer is $\\frac{6\\times6!}{\\frac{15}{2}\\times6! + 6\\times6!}=\\frac{4}{9}$ , so $\\boxed{013}$", "We count the two entities: The number of sets of eight cards that contain all seven numbers and all seven colors; and The number of sets of eight cards that contain all seven numbers and colors, and one card can be removed and the property still holds.\nFor the first: It is equivalent to counting:\nHow many ways can (colors) $A, A, B, C, D, E, F, G$ be matched to (numbers) $1, 1, 2, 3, 4, 5, 6, 7$ such that the two $A's$ cannot both be matched to the two ones. (This would mean we've chosen two identical cards from the deck, both with color $A$ and number one.)\nWe consider two cases:\n1) Neither $A$ is matched with a one. Then there are $\\binom{6}{2}$ ways to choose which two numbers are matched with the $A's,$ and $6! / 2!$ ways to order the remaining numbers. (There are two one's, so we divide by two.)\n2) One of the $A$ 's is matched with a one. There are $6 \\cdot 6!$ ways to arrange the remaining color\nSince there are seven ways to choose the extra color and likewise for the extra number, \nwe have in total $7 \\cdot 7 \\cdot \\left( 6C2 \\cdot \\frac{6!}{2!} + 6 \\cdot 6! \\right)$ ways to choose eight cards satisfying the first condition.\nFor the second: View the cards as rooks on a seven by seven chess board. Seven cards that contain all colors and numbers is can be represented as placing seven rooks on the board so that no two rooks are attacking each other. There are $7!$ ways to do this. For each arrangement, we can choose any of the other 42 positions on the board (cards) to add an extra rook. Thus, there are $7! \\cdot 42$ good sets of eight cards. Note that we are not overcounting.\nThus, the probability of getting a good set given the first condition is $\\frac{ 7! \\cdot 42 }{ 7 \\cdot 7 \\cdot \\left( 6C2 \\cdot \\frac{6!}{2!} + 6 \\cdot 6! \\right) } = \\frac{4}{9},$ and the answer is $\\boxed{013}.$", "We recast this problem to coloring 8 cells in a 7x7 grid, such that each row and column has at least one colored cell. If a cell in row a and column b is colored, that means we drew a card with color a and number b.\nBy pigeonhole principle, there exists a row that has 2 colored cells, and intuitively, that row is unique. There also exists a column with 2 colored cells. Call the unique row, \"Row A\", and Call the unique column, \"Column B\".\nIf we are able to remove a card so that we still have a card of each color and number, that means, we can erase a cell, so each row still has at least one colored cell, and so does each column. If we erase a cell not in Row A, then the erased cell's row will be empty. If we erase a cell not in Column B, the erased cell's column will be empty. Thus the cell that we erase must be in both Row A and Column B. There is only one cell in both Row A and Column B, and thus that cell must be colored.\nWe first count the number of successful outcomes. We must chose which row to be \"Row A\", and which column to be \"Column B\". There are 7 ways to chose \"Row A\", and 7 ways to chose \"Column B\". Because of the logic used above, the intersection of Row A and Column B must be colored. However, Row A needs to have 2 colored cells. There are 6 other uncolored cells to chose from. Similarly, there are 6 uncolored cells in Column B.\nNote that we can ignore all the rows and columns that contain colored cells. This is because Row A already contains 2 colored cells, and thus we can't color any more cells in Row A, similarly for Column B. The other row that contains a colored cell is not Row A, so it can only contain 1 colored cell, which it has, thus we can't color anymore cells in it. Similarly for the other column.\nWe can actually remove these rows/columns. This leaves a 5x5 grid, full of uncolored cells. Each row can have 1 cell, and so can each column. We count the # of colorings from the perspective of the rows. We look at the first row, it has 5 choices. We can look at the 2nd row, it has 4 choices, as it's selected cell can't occupy the same column. We continue down the line, and see there are $5! = 120$ ways to color.\nThus, the # of successful outcomes is $5! \\cdot 6 \\cdot 6 \\cdot 7 \\cdot 7$\nWe can then count the # of possible outcomes.\nWe can count the # of unsuccessful outcomes. We select \"Row A\" and \"Column B\" as usual, with $7 \\cdot 7$ ways. We don't color the intersection, so we have to color 2 cells in each. There are 6 other cells that can be colored in each, so we have ${6 \\choose 2} = 15$ ways to color each. Then, we can get rid of rows and columns using the method above and we are left with a 4x4 grid. There are $4! = 24$ ways to color.\nThus, the # of unsuccessful outcomes is $4! \\cdot 15 \\cdot 15 \\cdot 7 \\cdot 7$\nAfter some calculation, we get $\\frac{successful}{possible} = \\frac{successful}{successful+unsuccessful} = \\frac{5! \\cdot 6^2 \\cdot 7^2}{7^2 \\cdot(5! \\cdot 6^2 + 4! \\cdot 15^2} = \\frac{5! \\cdot 6^2}{5! \\cdot 6^2 +4! \\cdot 15^2} = \\frac{5 \\cdot 6^2}{5 \\cdot 6^2 + 15^2} = \\frac{180}{180+225} = \\frac{4}{9} => 4+9 = \\boxed{13}$" ]

[ "The perimeter of the triangle is $6.1+8.2+9.7=24$ cm. A square's perimeter is four times its sidelength, since all its sidelengths are equal. If the square's perimeter is $24$ , the sidelength is $24/4=6$ , and the area is $6^2=\\boxed{36}$" ]

[ "The diagram can be quartered as shown: and reassembled into two smaller squares of side $k$ , each of which looks like this: The border in this figure is the former cross, which still occupies 36% of the area. Therefore the inner square occupies 64% of the area, from which we deduce that it is $0.8k \\times 0.8k$ , and that one blue square must be $0.1k\\times 0.1k=0.01k^2$ or 1% each. Thus the blue area is $\\boxed{2}$ of the total." ]

[ "Let the points be $A=(x_1,0)$ $B=(x_2,1)$ $C=(x_3,5)$ , and $D=(x_4,4)$\nNote that the difference in $y$ value of $B$ and $C$ is $4$ . By rotational symmetry of the square, the difference in $x$ value of $A$ and $B$ is also $4$ . Note that the difference in $y$ value of $A$ and $B$ is $1$ . We now know that $AB$ , the side length of the square, is equal to $\\sqrt{1^2+4^2}=\\sqrt{17}$ , so the area is $\\boxed{17}$", "By translation, we can move the square with point $A$ at the origin. Then, $A=(0,0), B=(x_1,1), C=(x_2,5), D=(x_3,4)$ . We will use the relationship among the 4 sides of being perpendicular and equal.\nThe slope of $AB$ is $\\frac{1-0}{x_1-0}=\\frac{1}{x_1}$\nBecause $BC$ is perpendicular to $AB$ , the slope of $BC=-x_1$ .\nFrom the information above we could have the equation:\nBecause $CD$ is perpendicular to $BC$ , the slope of $CD=\\frac{1}{x_1}$ .\nFrom the information above we could have the equation:\nBecause $AD=AB,$\nNote that the square with $x_1=-4$ is just the reflection of square with $x_1=4$ over the origin. I will use $x_1=4$ $B=(4,1), AB=\\sqrt{17}, [ABCD]=\\boxed{17}$", "In this solution, we will use the fact that the diagonals of a square bisect each other, they are perpendicular to each other, and they are equal in length.\nUsing the fact that the diagonals bisect each other, we get the equation:\nNow we use the fact that the diagonals are perpendicular to each other:\nUsing the fact that the diagonals are equal in length, we get the equation:\nNow we have 3 equations with 3 variables:\n$\\begin{cases} x_2=x_1+x_3 \\\\ x_2(x_1-x_3)=15 \\\\ (x_3-x_1)^2-x_2^2=16 \\end{cases}$\nWe substitute $x_2$ into the 2 other equations:\nNow we have 2 equations of $x_1$ and $x_3$\n$\\begin{cases} x_1^2-x_3^2=15 \\\\ x_1x_3=-4 \\end{cases}$\nThis is the same equation as solution $2$ . So $x_1= \\pm 4, AB=\\sqrt{17}, [ABCD]=\\boxed{17}$", "Draw it out, by inspection the coordinates are $(-1, 4)$ $(0, 0)$ $(4, 1)$ , and $(3, 5)$ . The side length is $\\sqrt{17}, [ABCD]=\\boxed{17}$ ~JH. L" ]

[ "We let $c, e,$ and $m$ be the probability of reaching a corner before an edge when starting at an \"inside corner\" (e.g. $(1, 1)$ ), an \"inside edge\" (e.g. $(1, 0)$ ), and the middle respectively.\nStarting in the middle, there is a $\\frac{4}{8}$ chance of moving to an inside edge and a $\\frac{4}{8}$ chance of moving to an inside corner, so\n\\[m = \\frac{1}{2}e + \\frac{1}{2}c.\\]\nStarting at an inside edge, there is a $\\frac{2}{8}$ chance of moving to another inside edge, a $\\frac{2}{8}$ chance of moving to an inside corner, a $\\frac{1}{8}$ chance of moving into the middle, and a $\\frac{3}{8}$ chance of reaching an outside edge and stopping. Therefore,\n\\[e = \\frac{1}{4}e + \\frac{1}{4}c + \\frac{1}{8}m + \\frac{3}{8}\\cdot 0 = \\frac{1}{4}e + \\frac{1}{4}c + \\frac{1}{8}m.\\]\nStarting at an inside corner, there is a $\\frac{2}{8}$ chance of moving to an inside edge, a $\\frac{1}{8}$ chance of moving into the middle, a $\\frac{4}{8}$ chance of moving to an outside edge and stopping, and finally a $\\frac{1}{8}$ chance of reaching that elusive outside corner. This gives\n\\[c = \\frac{1}{4}e + \\frac{1}{8}m + \\frac{1}{2}0 + \\frac{1}{8}\\cdot 1 = \\frac{1}{4}e + \\frac{1}{8}m + \\frac{1}{8}.\\]\nSolving this system of equations gives\n\\[m = \\frac{4}{35},\\] \\[e = \\frac{1}{14},\\] \\[c = \\frac{11}{70}.\\]\nSince the particle starts at $(0, 0),$ it is $m$ we are looking for, so the final answer is\n\\[4 + 35 = \\boxed{39}.\\]" ]

[ "Since the area of the square is $40$ , the length of a side is $\\sqrt{40}=2\\sqrt{10}$ . The distance between the center of the semicircle and one of the bottom vertices of the square is half the length of the side, which is $\\sqrt{10}$\nUsing the Pythagorean Theorem to find the radius $r$ of the semicircle, $r^2 = (2\\sqrt{10})^2 + (\\sqrt{10})^2 = 50$ . So, the area of the semicircle is $\\frac{1}{2}\\cdot \\pi \\cdot 50 = \\boxed{25}$" ]

[ "We can set the point on $CD$ where the fold occurs as point $F$ . Then, we can set $FD$ as $x$ , and $CF$ as $1-x$ because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for $x$ , we get,\n\\[x^2 + \\left(\\frac{1}{3}\\right)^2 = (1-x)^2 \\rightarrow x^2 + \\frac{1}{9} = x^2 - 2x + 1 \\rightarrow x=\\frac{4}{9}\\]\nWe know this is a 3-4-5 triangle because the side lengths are $\\frac{3}{9}, \\frac{4}{9}, \\frac{5}{9}$ . We also know that $EAC'$ is similar to $C'DF$ because angle $EC'F$ is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of $C'DF \\times \\frac{AC'}{DF}$ . That's just $\\frac{4}{3} \\times \\frac{\\frac{2}{3}}{\\frac{4}{9}} = \\frac{4}{3} \\times \\frac{3}{2} = 2$ . Therefore, the final answer is $\\boxed{2}$", "Let the line we're reflecting over be $\\ell$ , and let the points where it hits $AB$ and $CD$ , be $M$ and $N$ , respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line $\\ell$ . The segment $CC'$ has slope $\\frac{0 - 1}{1 - 2/3} = -3$ , implying line $\\ell$ has a slope of $\\frac{1}{3}$ . Also, the midpoint of segment $CC'$ is $\\left( \\frac{5}{6}, \\frac{1}{2} \\right)$ , so line $\\ell$ passes through this point. Then, we get the equation of line $\\ell$ is simply $y = \\frac{1}{3} x + \\frac{2}{9}$ . Then, if the point where $B$ is reflected over line $\\ell$ is $B'$ , then we get $BB'$ is the line $y = -3x$ . The intersection of $\\ell$ and segment $BB'$ is $\\left( - \\frac{1}{15}, \\frac{1}{5} \\right)$ . So, we get $B' = \\left(- \\frac{2}{15}, \\frac{2}{5} \\right)$ . Then, line segment $B'C'$ has equation $y = \\frac{3}{4} x + \\frac{1}{2}$ , so the point $E$ is the $y$ -intercept, or $\\left(0, \\frac{1}{2} \\right)$ . This implies that $AE = \\frac{1}{2}, AC' = \\frac{2}{3}$ , and by the Pythagorean Theorem, $EC' = \\frac{5}{6}$ (or you could notice $\\triangle AEC'$ is a $3-4-5$ right triangle). Then, the perimeter is $\\frac{1}{2} + \\frac{2}{3} + \\frac{5}{6} = 2$ , so our answer is $\\boxed{2}$ . ~rocketsri", "Assume that E is the midpoint of $\\overline{AB}$ . Then, $\\overline{AE}=\\frac{1}{2}$ and since $C'D=\\frac{1}{3}$ $\\overline{AC'}=\\frac{2}{3}$ . By the Pythagorean Theorem, $\\overline{EC'}=\\frac{5}{6}$ . It easily follows that our desired perimeter is $\\boxed{2}$ ~samrocksnature", "As described in Solution 1, we can find that $DF=\\frac{4}{9}$ , and $C'F = \\frac{5}{9}.$\nThen, we can find we can find the length of $\\overline{AE}$ by expressing the length of $\\overline{EF}$ in two different ways, in terms of $AE$ . If let $AE = a$ , by the Pythagorean Theorem we have that $EC = \\sqrt{a^2 + \\left(\\frac{2}{3}\\right)^2} = \\sqrt{a^2 + \\frac{4}{9}}.$ Therefore, since we know that $\\angle EC'F$ is right, by Pythagoras again we have that $EF = \\sqrt{\\left(\\sqrt{a^2+\\frac{4}{9}}\\right)^2 + \\left(\\frac{5}{9}\\right)^2} = \\sqrt{a^2 + \\frac{61}{81}}.$\nAnother way we can express $EF$ is by using Pythagoras on $\\triangle XEF$ , where $X$ is the foot of the perpendicular from $F$ to $\\overline{AE}.$ We see that $ADFX$ is a rectangle, so we know that $AD = 1 = FX$ . Secondly, since $FD = \\frac{4}{9}, EX = a - \\frac{4}{9}$ . Therefore, through the Pythagorean Theorem, we find that $EF = \\sqrt{\\left(a-\\frac{4}{9}\\right)^2 + 1^2} = \\sqrt{a^2 - \\frac{8}{9}a +\\frac{97}{81}}.$\nSince we have found two expressions for the same length, we have the equation $\\sqrt{a^2 + \\frac{61}{81}} = \\sqrt{a^2 - \\frac{8}{9}a +\\frac{97}{81}}.$ Solving this, we find that $a=\\frac{1}{2}$\nFinally, we see that the perimeter of $\\triangle AEC'$ is $\\frac{1}{2} + \\frac{2}{3} + \\sqrt{\\left(\\frac{1}{2}\\right)^2 + \\frac{4}{9}},$ which we can simplify to be $2$ . Thus, the answer is $\\boxed{2}.$ ~laffytaffy", "Draw a perpendicular line from $\\overline{AB}$ at $E$ , and let it intersect $\\overline{DC}$ at $E'$ . The angle between $\\overline{AB}$ and $\\overline{EE'}$ is $2\\theta$ , where $\\theta$ is the angle between the fold and a line perpendicular to $\\overline{AD}$ . The slope of the fold is $\\frac{1}{3}$ because it is perpendicular to $\\overline{CC'}$ $\\overline{CC'}$ has a slope of $-3$ using points $C'$ and $C$ , and perpendicular lines have slopes negative inverses of each other). Using tangent double angle formula, the slope of $\\overline{EC'}$ is $\\frac{3}{4}$ , which implies $\\overline{AE}$ $\\frac{1}{2}$ . By the Pythagorean Theorem, $\\overline{EC'}=\\frac{5}{6}$ . It easily follows that our desired perimeter is $\\boxed{2}$ ~forrestc", "It is easy to prove that the ratio of the sum of the larger leg and hypotenuse to the smaller leg depends monotonically on the angle of a right triangle, which means: \\[C' F + DF = CF + DF = CD = AD = 3 C'D \\implies C' D : DF : C' F = 3 : 4 : 5.\\]\nFor a similar triangle, the ratio of the perimeter to the larger leg is $\\frac {3 + 4 +5}{4} = 3.$\n$\\triangle AEC' \\sim \\triangle DC'F \\implies$ the perimeter of triangle $\\bigtriangleup AEC'$ is $3 AC' = \\boxed{2}.$" ]

[ "In the original picture, let $P$ be the corner, and $M$ and $N$ be the two points whose distance is $\\sqrt{17}$ from $P$ . Also, let $R$ be the point where the two cuts intersect.\nUsing $\\triangle{MNP}$ (a 45-45-90 triangle), $MN=MP\\sqrt{2}\\quad\\Longrightarrow\\quad MN=\\sqrt{34}$ $\\triangle{MNR}$ is equilateral , so $MR = NR = \\sqrt{34}$ . (Alternatively, we could find this by the Law of Sines .)\nThe length of the perpendicular from $P$ to $MN$ in $\\triangle{MNP}$ is $\\frac{\\sqrt{17}}{\\sqrt{2}}$ , and the length of the perpendicular from $R$ to $MN$ in $\\triangle{MNR}$ is $\\frac{\\sqrt{51}}{\\sqrt{2}}$ . Adding those two lengths, $PR=\\frac{\\sqrt{17}+\\sqrt{51}}{\\sqrt{2}}$ . (Alternatively, we could have used that $\\sin 75^{\\circ} = \\sin (30+45) = \\frac{\\sqrt{6}+\\sqrt{2}}{4}$ .)\nDrop a perpendicular from $R$ to the side of the square containing $M$ and let the intersection be $G$\n\\begin{align*}PG&=\\frac{PR}{\\sqrt{2}}=\\frac{\\sqrt{17}+\\sqrt{51}}{2}\\\\ MG=PG-PM&=\\frac{\\sqrt{17}+\\sqrt{51}}{2}-\\sqrt{17}=\\frac{\\sqrt{51}-\\sqrt{17}}{2}\\end{align*}\nLet $A'B'C'D'$ be the smaller square base of the tray and let $ABCD$ be the larger square, such that $AA'$ , etc, are edges. Let $F$ be the foot of the perpendicular from $A$ to plane $A'B'C'D'$\nWe know $AA'=MR=\\sqrt{34}$ and $A'F=MG\\sqrt{2}=\\frac{\\sqrt{51}-\\sqrt{17}}{\\sqrt{2}}$ . Now, use the Pythagorean Theorem on triangle $AFA'$ to find $AF$\n\\begin{align*}\\left(\\frac{\\sqrt{51}-\\sqrt{17}}{\\sqrt{2}}\\right)^2+AF^2&=\\left(\\sqrt{34}\\right)^2\\\\ \\frac{51-34\\sqrt{3}+17}{2}+AF^2&=34\\\\AF&=\\sqrt{34-\\frac{68-34\\sqrt{3}}{2}}\\\\AF&=\\sqrt{\\frac{34\\sqrt{3}}{2}}\\\\AF&=\\sqrt[4]{867}\\end{align*}\nThe answer is $867 + 4 = \\boxed{871}$", "In the final pyramid, let $ABCD$ be the smaller square and let $A'B'C'D'$ be the larger square such that $AA'$ , etc. are edges.\nIt is obvious from the diagram that $\\angle A'AB = \\angle A'AD = 105^\\circ$\nLet $AB$ and $AD$ be the positive $x$ and $y$ axes in a 3-d coordinate system such that $A'$ has a positive $z$ coordinate. Let $\\alpha$ be the angle made with the positive $x$ axis . Define $\\beta$ and $\\gamma$ analogously.\nIt is easy to see that if $P: = (x,y,z)$ , then $x = AA'\\cdot \\cos\\alpha$ . Furthermore, this means that $\\frac {x^2}{AA'^2} + \\frac {y^2}{AA'^2} + \\frac {z^2}{AA'^2} = \\cos^2\\alpha + \\cos^2\\beta + \\cos^2\\gamma = 1$\nWe have that $\\alpha = \\beta = 105^\\circ$ , so $\\cos^2 105^\\circ + \\cos^2105^\\circ + \\cos^2\\gamma = 1\\implies \\cos\\gamma = \\sqrt [4]{\\frac {3}{4}}$\nIt is easy to see from the Law of Sines that $\\frac {AA'}{\\sin 45^\\circ} = \\frac {\\sqrt {17}}{\\sin 30^\\circ}\\implies AA' = \\sqrt {34}$\nNow, $z = AA'\\cdot \\cos\\gamma = \\sqrt [4]{34^2\\cdot \\frac {3}{4}} = \\sqrt [4]{867}$\nIt follows that the answer is $867 + 4 = \\boxed{871}$" ]

[ "Note first that the intersection is a pentagon\nUse 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. $A(-2,2,0),\\ B(2,2,0),\\ C(2,-2,0),\\ D(-2,-2,0),\\ E(0,2\\sqrt{2})$ . Using the coordinates of the three points of intersection $(-1,1,\\sqrt{2}),\\ (2,0,0),\\ (0,-2,0)$ , it is possible to determine the equation of the plane. The equation of a plane resembles $ax + by + cz = d$ , and using the points we find that $2a = d \\Longrightarrow d = \\frac{a}{2}$ $-2b = d \\Longrightarrow d = \\frac{-b}{2}$ , and $-a + b + \\sqrt{2}c = d \\Longrightarrow -\\frac{d}{2} - \\frac{d}{2} + \\sqrt{2}c = d \\Longrightarrow c = d\\sqrt{2}$ . It is then $x - y + 2\\sqrt{2}z = 2$\nWrite the equation of the lines and substitute to find that the other two points of intersection on $\\overline{BE}$ $\\overline{DE}$ are $\\left(\\frac{\\pm 3}{2},\\frac{\\pm 3}{2},\\frac{\\sqrt{2}}{2}\\right)$ . To find the area of the pentagon, break it up into pieces (an isosceles triangle on the top, an isosceles trapezoid on the bottom). Using the distance formula $\\sqrt{a^2 + b^2 + c^2}$ ), it is possible to find that the area of the triangle is $\\frac{1}{2}bh \\Longrightarrow \\frac{1}{2} 3\\sqrt{2} \\cdot \\sqrt{\\frac 52} = \\frac{3\\sqrt{5}}{2}$ . The trapezoid has area $\\frac{1}{2}h(b_1 + b_2) \\Longrightarrow \\frac 12\\sqrt{\\frac 52}\\left(2\\sqrt{2} + 3\\sqrt{2}\\right) = \\frac{5\\sqrt{5}}{2}$ . In total, the area is $4\\sqrt{5} = \\sqrt{80}$ , and the solution is $\\boxed{080}$" ]

[ "The first answer choice ${\\textbf{(A)}\\ 3}$ , can be eliminated since there must be $10$ squares with integer side lengths. We then test the next smallest sidelength which is $4$ . The square with area $16$ can be partitioned into $8$ squares with area $1$ and two squares with area $4$ , which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is $\\boxed{4}$" ]

[ "To find the probability, we look at the $\\frac{\\text{success region}}{\\text{total possible region}}$ . For the coin to be completely contained within the square, we must have the distance from the center of the coin to a side of the square to be at least $\\frac{1}{2}$ , as it's the radius of the coin. This implies the $\\text{total possible region}$ is a square with side length $8 - \\frac{1}{2} - \\frac{1}{2} = 7$ , with an area of $49$ . Now, we consider cases where needs to land to partially cover a black region.\nNear The Center Square\nWe can have the center of the coin land within $\\frac{1}{2}$ outside of the center square, or inside of the center square. So, we have a region with $\\frac{1}{2}$ emanating from every point on the exterior of the square, forming four quarter circles and four rectangles. The four quarter circles combine to make a full circle of radius $\\frac{1}{2}$ , so the area is $\\frac{\\pi}{4}$ . The area of a rectangle is $2 \\sqrt 2 \\cdot \\frac{1}{2} = \\sqrt 2$ , so $4$ of them combine to an area of $4 \\sqrt 2$ . The area of the black square is simply $\\left(2\\sqrt 2\\right)^2 = 8$ . So, for this case, we have a combined total of $8 + 4\\sqrt 2 + \\frac{\\pi}{4}$ . Onto the second (and last) case.\nNear A Triangle\nWe can also have the coin land within $\\frac{1}{2}$ outside of one of the triangles. By symmetry, we can just find the successful region for one of them, then multiply by $4$ . Consider the above diagram. We can draw an altitude from the bottom corner of the square to hit the hypotenuse of the green triangle. The length of this when passing through the black region is $\\sqrt 2$ , and when passing through the white region (while being contained in the green triangle) is $\\frac{1}{2}$ . However, we have to subtract off when it doesn't pass through the red square. Then, it's the hypotenuse of a small isosceles right triangle with side lengths of $\\dfrac{1}{2}$ which is $\\dfrac{\\sqrt{2}}{2}.$ So, the altitude of the green triangle is $\\sqrt 2 + \\frac{1}{2} - \\frac{\\sqrt 2}{2} = \\frac{\\sqrt 2 + 1}{2}$ . Then, recall, the area of an isosceles right triangle is $h^2$ , where $h$ is the altitude from the right angle. So, squaring this, we get $\\frac{3 + 2\\sqrt 2}{4}$ . Now, we have to multiply this by $4$ to account for all of the black triangles, to get $3 + 2\\sqrt 2$ as the final area for this case.\nFinishing\nThen, to have the coin touching a black region, we add up the area of our successful regions, or $8 + 4\\sqrt 2 + \\frac{\\pi}{4} + 3 + 2\\sqrt 2 = 11 + 6\\sqrt 2 + \\frac{\\pi}{4} = \\frac{44 + 24\\sqrt 2 + \\pi}{4}$ . The total region is $49$ , so our probability is $\\frac{\\frac{44 + 24\\sqrt 2 + \\pi}{4}}{49} = \\frac{44 + 24\\sqrt 2 + \\pi}{196}$ , which implies $a+b = 44+24 = 68$ . This corresponds to answer choice $\\boxed{68}$" ]

[ "Let the region within the circle and square be $a$ . In other words, it is the area inside the circle $\\textbf{and}$ the square. Let $r$ be the radius. We know that the area of the circle minus $a$ is equal to the area of the square, minus $a$\nWe get:\n$\\pi r^2 -a=4-a$\n$r^2=\\frac{4}{\\pi}$\n$r=\\frac{2}{\\sqrt{\\pi}}$\nSo the answer is $\\boxed{2}$", "We realize that since the areas of the regions outside of the circle and the square are equal to each other, the area of the circle must be equal to the area of the square.\n$\\pi r^2=4$\n$r^2=\\frac{4}{\\pi}$\n$r=\\frac{2}{\\sqrt{\\pi}}$\nSo the answer is $\\boxed{2}$" ]

[ "The area of the trapezoid is $\\frac{1}{3}$ , and the shorter base and height are both $\\frac{1}{2}$ . Therefore, \\[\\frac{1}{3}=\\frac{1}{2}\\cdot \\frac{1}{2}\\cdot \\left(\\frac{1}{2}+x\\right) \\Rightarrow x=\\frac{5}{6}\\rightarrow \\boxed{56}\\]", "Divide the pentagon into 2 small congruent trapezoids by extending the common shorter base of the 2 larger trapezoids.\nSince each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base $\\frac{1}{2}$ , we have \\[b_{large}+\\frac{1}{2}=2(b_{small}+ \\frac{1}{2})\\] \\[x+\\frac{1}{2}=2((1-x)+\\frac{1}{2})\\] \\[x=\\frac{5}{6}\\boxed{56}\\]" ]

[ "Since the number of tiles lying on both diagonals is $37$ , counting one tile twice, there are $37=2x-1\\implies x=19$ tiles on each side. Therefore, our answer is $19^2=361=\\boxed{361}$", "Visualize it as 4 separate diagonals connecting to one square in the middle. Each square on the diagonal corresponds to one square of horizontal/vertical distance (because it's a square). So, we figure out the length of each separate diagonal, multiply by two, and then add 1. (Realize that we can just join two of the separate diagonals on opposite sides together to save some time in calculations.) Therefore, the edge length is: \\[\\frac{37-1}{4} \\cdot 2 + 1 = 19\\] Thus, our solution is $19^2 = 361 = \\boxed{361}$" ]

[ "We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, $2(1024 - 1000) = 48$ , to be exact. Once these cards go through, 1999 will be the $512 - 48 = 464^\\text{th}$ card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position $464 \\times 2 = 928$ , meaning that there were $\\boxed{927}$ cards are above the one labeled $1999$", "To simplify matters, we want a power of $2$ . Hence, we will add $48$ 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position $1024$ in a $2048$ card stack, where the fake cards towards the front.\nLet the fake cards have positions $1, 3, 5, \\cdots, 95$ . Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the $2000$ card case, where all of them are below $1999$ . From this, we know that the cards from positions $1$ to $96$ alternate in fake-real-fake-real, where we have the correct order of cards once the first $96$ have moved and we can start putting real cards on the table. Hence, $1999$ is in position $1024 - 96 = 928$ , so $\\boxed{927}$ cards are above it. - Spacesam", "We work backwards. To reverse the process, we must move the bottom card to the top, and add a new number to the top. Let $d_n$ equal the number of cards below 1999 after $n$ process reversals. Reversing one process, our deck only has $2000$ , so we reverse again to obtain $1999, 2000$ . So, $d_2 = 1$ . When $d_{n-1} > 0$ , after a process reversal, the bottom card is moved to the top (it goes above $1999$ ), so we subtract by one to account for this (the addition of the new card doesn't matter since it goes above $1999$ ), giving us $d_n = d_{n-1} - 1$ . So, $d_3 = 0$ . Then, when $d_{n-1} = 0$ , after a process reversal, $1999$ moves to the top and one more card is added above $1999$ . Since at $d_{n}$ $n$ cards have been added, there must be $n - 2$ cards below $1999$ . So, $d_4 = 2$ . Then, consider all $d_{n-1} = 0$ . Then, $d_n = n-2$ as previously stated. After $n-2$ process reversals, we go back to $d_{2n - 2} = n - 2 - (n - 2) = 0$ . Then, $d_{2n-1} = 2n - 3$ . Next, we can simply calculate $2 \\cdot 4 - 1 = 7$ $2 \\cdot 7 - 1 = 13$ , and so on (which is a very simple bash, as $2000$ is a small number. If you don't want to do this, define sequence $a_n = 2a_{n-1} - 1$ , and solve for the closed form, which is very easy). Consequently, we derive $d_{1537} = 1537 - 2 = 1535$ $2000 - 1537 = 463$ steps remain. So, $d_{2000} = d_{1537} - 463 = 1072$ . Finally, there are $2000 - 1 - 1072 = \\boxed{927}$ numbers above $1999$", "Let us treat each run through the deck as a separate \"round\". For example, in round one, you would go through all of the $2000$ cards initially in the deck once, in round two, you would go through all $1000$ cards initially in the deck once, so on and so forth. For each round, let us record what the initial and final actions are ( $r$ for moving the card to the right, $b$ for moving the card to the bottom), the number of cards moved to the right, the number of cards left, and what the position of the cards moved to the right were in the original $2000$ card deck (as $n = a + ck$ where $n$ is the position, $c$ is the spacing of the cards moved, $a$ is an integer such that the correct first card is moved, and $k$ is an integer greater than or equal to $1$ which represents which card out of all the cards moved to the right you are finding the position of). Then,\nRound 1: $r$ to $b$ $1000$ to right, $1000$ left in deck, $n = -1 + 2k$\nRound 2: $r$ to $b$ $500$ to right, $500$ left in deck, $n = -2 + 4k$\nRound 3: $r$ to $b$ $250$ to right, $250$ left in deck, $n = -4 + 8k$\nRound 4: $r$ to $b$ $125$ to right, $125$ left in deck, $n = -8 + 16k$\nLet us treat the remaining deck of $125$ cards as a totally independent deck, note that the positions of card in this deck are $n = 16k$ . Also note that the first action of a new round is never the same action as the last action of the previous round because actions alternate. Also note that for every new round, the spacing between cards moved doubles, and that the cards remaining in the beginning of a new round have position $n = a + c/2 + ck$ for the values $a, c$ of the previous round. Also note that if there are an odd number of cards in an initial deck, the first and last actions are the same. Then,\nRound 5: $r$ to $r$ $63$ to right, $62$ left in deck, $n = -1 + 2k$\nRound 6: $b$ to $r$ $31$ to right, $31$ left in deck, $n = 4k$ , because $n = 2k$ positioned cards are left at the beginning of this round and the first card is sent to the bottom, only every second card is sent to the right, and because spacing doubles every round,\nRound 7: $b$ to $b$ $15$ to right, $16$ left in deck, $n = -2 + 8k$ , because $n = 2 + 4k$ positioned cards are left at the beginning and only every second card is sent to the right, and because spacing doubles every round,\nRound 8: $r$ to $b$ $8$ to right, $8$ left in deck, $n = -14 + 16k$ , by similar reasoning, since the cards $n = 2 + 8k$ are left and spacing doubles every round, from here on things get real easy,\nRound 9: $r$ to $b$ $4$ to right, $4$ left in deck, $n = -22 + 32k$\nRound 10: $r$ to $b$ $2$ to right, $2$ left in deck, $n = -38 + 64k$\nRound 11: $r$ to $b$ $1$ to right, $1$ left in deck, $n = 58$ , since $58 = -38 + 64/2 + 64$\nThis card must be labeled 1999 since it is second to last. Then, since it is the $58th$ in the deck of $125$ , it must be the $58 * 8 = 928th$ card in the original deck, and because we've been numbering from the top of the deck to the bottom (also because AIME answers are 000-999), there were $928 - 1 = \\boxed{927}$ cards above it in the deck - Romulus and minimaxweii", "Similar to Solution 5, we treat each run-through of the deck from the lowest-indexed card to the highest-indexed card as a separate round. Notice that after each round, approximately half the deck will remain, with the other half having been cast aside to the right in sorted order. Then, we can model each round as if we are running a \"Sieve\" through the deck with powers of 2s, filtering out the cards to put to the right and the cards to be pushed to the bottom.\nAt the beginning, when all $2000$ cards are still in the deck, notice that the first run-through of the deck consists of removing all the cards such that their index $I\\equiv 1\\pmod{2}$ , while we leave behind all the cards such that their index $I\\equiv 2\\pmod{2}$ . (Since the cards are not $0$ -indexed, using $2$ instead of $0\\pmod{2}$ is simpler to keep track of the cards. This will appear again later.) The remaining $1000$ cards, then, will all share the attribute that their index numbers are even. To split this further for round 2, we look more specifically at each index, categorizing them further as $I\\equiv 2\\pmod 4$ or $I\\equiv 4\\pmod 4$ in that order (The card with the smaller remainder will always be the card on the top of the new deck). Then, in Round 2, we will remove all the cards with index values that leave a remainder of $2$ when divided by $4$ , while leaving the multiples of $4$ behind. Notice that each time, in each new deck, we are removing all the cards with an odd position in that specific deck, and leaving behind all the cards with an even position in that specific deck.\nThis process will continue until we reach Round 5, where the number of cards that remain ( $125$ ) is no longer a multiple of 2. Therefore, when we remove all the cards with indices $I\\equiv 16\\pmod{32}$ and leave behind all the cards with indices $I\\equiv 32\\pmod{32}$ , the last card in the deck will be removed instead of being placed to the bottom, with $125-63=62$ cards remaining. Because of this, when Round 6 begins, we cannot follow our normal procedure, since the top card of the Round 6 deck would have been moved to the bottom! Therefore, when we are using our sieve with $I\\equiv 32\\pmod{64}$ and $I\\equiv 64\\pmod{64}$ , the order at which the remainders are presented have flipped, so all the cards with indices that are multiples of 64 will be removed, while all the cards with indices that are 32 more than a multiple of 64 will remain. The $62$ cards from Round 6 will be reduced to $31$ , and the last card in the deck will be removed just like in Round 5, which means Round 7 will be affected the same way. Of the choices of $I\\equiv 32\\pmod{128}$ and $I\\equiv 32+64=96\\pmod{128}$ , we will remove the latter, leaving behind the cards with indices that are 32 more than a multiple of 128. However, since 31 is an odd number, the last card of the deck will NOT be removed, which means our sieving process will return to normal after Round 7, with $31-15=16$ cards remaining. Since $16$ is a perfect power of $2$ , we will not need to worry about this scenario again in the following rounds.\nRound 8: $I\\equiv 32\\pmod{256}$ removed; $I\\equiv 32+128=160\\pmod{256}$ remains; $16-8=8$ cards left.\nRound 9: $I\\equiv 160\\pmod{512}$ removed; $I\\equiv 160+256=416\\pmod{512}$ remains; $8-4=4$ cards left.\nRound 10: $I\\equiv 416\\pmod{1024}$ removed; $I\\equiv 416+512=928\\pmod{1024}$ remains; $4-2=2$ cards left.\nRound 11: $I\\equiv 928\\pmod{2048}$ removed; $I\\equiv 928+1024=1952\\pmod{2048}$ remains; $2-1=1$ card left.\nRound 12: The last card in position $1952$ is sorted into place (This is card # $2000$ !).\nIt is obvious that the single card put into place in Round 11 is the second-to-last card in the deck, which is our target # $1999$ . Then, its position in the original deck leaves a remainder of $928$ when divided by $2056$ . It is easy to see that $928$ is the only index that satisfies this, so there will be $928-1=\\boxed{927}$ cards above card # $1999$" ]

[ "The product of all six numbers is $6!=720$ . The products of numbers that can be visible are $720/1$ $720/2$ , ..., $720/6$ .\nThe answer to this problem is their greatest common divisor -- which is $720/L$ , where $L$ is the least common multiple of $\\{1,2,3,4,5,6\\}$ .\nClearly $L=60$ and the answer is $720/60=\\boxed{12}$", "The product P can be one of the following six numbers excluding the number that is hidden under, so we have:\n\\begin{align*}\n2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 = 2^4 \\cdot 3^2 \\cdot 5 \\\\\n1 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 = 2^3 \\cdot 3^2 \\cdot 5 \\\\\n1 \\cdot 2 \\cdot 4 \\cdot 5 \\cdot 6 = 2^4 \\cdot 3 \\cdot 5 \\\\\n1 \\cdot 2 \\cdot 3 \\cdot 5 \\cdot 6 = 2^2 \\cdot 3^2 \\cdot 5 \\\\\n1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 6 = 2^4 \\cdot 3^2 \\\\\n1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 = 2^3 \\cdot 3 \\cdot 5\n\\end{align*}\nThe largest number that is certain to divide product P is basically GCD of all the above 6 products which is $2^2 \\cdot 3$\nHence $P=3\\cdot2^2=\\boxed{12}$" ]

[ "Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are $6^4 = 1296$ outcomes).\nCase 1 (easy): Four 5's are rolled. This has probability $\\frac{1}{6^4}$ of occurring.\nCase 2: Two 5's are rolled.\nCase 3: No 5's are rolled.\nTo find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For $n \\ge 1$ , let $a_n$ equal the number of outcomes after rolling the die $n$ times, with the property that the product is a square. Thus, $a_1 = 2$ as 1 and 4 are the only possibilities.\nTo find $a_{n+1}$ given $a_n$ (where $n \\ge 1$ ), we observe that if the first $n$ rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives $2a_n$ outcomes. Otherwise, the first $n$ rolls do not multiply to a perfect square ( $5^n - a_n$ outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first $n$ rolls is $2^x 3^y$ where $x$ and $y$ are not both even, then we observe that if $x$ and $y$ are both odd, then the last roll must be 6; if only $x$ is odd, the last roll must be 2, and if only $y$ is odd, the last roll must be 3. Thus, we have $5^n - a_n$ outcomes in this case, and $a_{n+1} = 2a_n + (5^n - a_n) = 5^n + a_n$\nComputing $a_2$ $a_3$ $a_4$ gives $a_2 = 7$ $a_3 = 32$ , and $a_4 = 157$ . Thus for Case 3, there are 157 outcomes. For case 2, we multiply $a_2$ by $\\binom{4}{2} = 6$ to distribute the two 5's among four rolls. Thus the probability is\n\\[\\frac{1 + 6 \\cdot 7 + 157}{6^4} = \\frac{200}{6^4} = \\frac{25}{162} \\implies m+n = \\boxed{187}\\]", "We can solve this without finding the amount of cases. Notice when we roll a 1 or a 4, it does not affect whether or not the product is a square number. We have a 1/3 chance of rolling either a 1 or 4. We have a 2/3 chance of rolling a 2,3,5 or 6. Let's call rolling 1 or 4 rolling a dud (a perfect square).\nProbability of rolling 4 duds: $\\left(\\frac{1}{3}\\right)^4$\nProbability of rolling 3 duds: $4 * \\left(\\frac{1}{3}\\right)^3 * \\frac{2}{3}$\nProbability of rolling 2 duds: $6 * \\left(\\frac{1}{3}\\right)^2 * \\left(\\frac{2}{3}\\right)^2$\nProbability of rolling 1 dud: $4 * \\frac{1}{3} * \\left(\\frac{2}{3}\\right)^3$\nProbability of rolling 0 duds: $\\left(\\frac{2}{3}\\right)^4$\nNow we will find the probability of a square product given we have rolled each amount of duds\nProbability of getting a square product given 4 duds: 1\nProbability of getting a square product given 3 duds: 0 (you will have 1 non-dud and that's never going to be square)\nProbability of getting a square product given 2 duds: $\\frac{1}{4}$ (as long as our two non-duds are the same, our product will be square)\nProbability of getting a square product given 1 dud: $\\frac{3!}{4^3}$ $\\frac{3}{32}$ (the only way to have a square product is rolling a 2,3 and 6. There are 3! ways of doing that and a total of $4^3$ ways to roll 3 non-duds).\nProbability of getting a square product given 0 duds: $\\frac{40}{4^4}$ $\\frac{5}{32}$ (We can have any two non-duds twice. For example, 2,2,5,5. There are $\\binom{4}{2} = 6$ ways of choosing which two non-duds to use and $\\binom{4}{2} = 6$ ways of choosing how to arrange those 4 numbers. That gives us 6*6=36 combinations. We can also have 2,2,2,2 or 3,3,3,3 or 5,5,5,5 or 6,6,6,6. This gives us a total of 40 combinations).\nWe multiply each probability of rolling k duds with the probability of getting a square product given k duds and then sum all the values. \\[\\left(\\frac{1}{3}\\right)^4 * 1 + 4 * \\left(\\frac{1}{3}\\right)^3 * \\frac{2}{3} * 0 + 6 * \\left(\\frac{1}{3}\\right)^2 * \\left(\\frac{2}{3}\\right)^2 * \\frac{1}{4} + 4 * \\frac{1}{3} * \\left(\\frac{2}{3}\\right)^3 * \\frac{3}{32} + \\left(\\frac{2}{3}\\right)^4 * \\frac{5}{32} = \\frac{25}{162}.\\]\n$25+162$ $\\boxed{187}$", "Note that rolling a 1/4 will not affect whether or not the product is a perfect square. This means that in order for the product to be a perfect square, all non 1/4 numbers rolled must come in pairs, with the only exception being the triplet 2,3, 6. Now we can do casework:\nIf there are four 1/4's, then there are $2^4=16$ combinations. \nIf there are three 1/4's, then there are 0 combinations, because the fourth number isn't a square. \nIf there are two 1/4's, there are $2^2=4$ ways to choose the two 1/4's, 4 ways to choose the remaining pair of numbers, and $\\frac{4!}{2!2!}=6$ ways to arrange, so there are $4\\cdot 4\\cdot 6=96$ combinations for this case. \nIf there is one 1/4, then there are 2 ways to choose whether it is a 1 or 4, and the remaining three numbers must be 2, 3, and 6, so there are $4!$ ways to order, meaning there are $2\\cdot 4!=48$ combinations for this case. \nOur final case is if there are no 1/4's, in which case we must have two pairs. If the two pairs are of different numbers, then there $\\binom{4}{2}$ to choose the numbers and $\\frac{4!}{2!2!}=6$ ways to arrange them, so $6\\cdot 6=36$ . If all four numbers are the same there are $4$ combinations, so there are $4+36=40$ combinations for this case.\nHence there are $16+0+96+48+40=200$ combinations where the product of the dice is a perfect square, and there are $6^4=1296$ total combinations, so the desired probability is $\\frac{200}{1296}=\\frac{25}{162}$ , yielding an answer of $25+162=\\boxed{187}$", "Another way to solve this problem is to do casework on all the perfect squares from $1^2$ to $36^2$ , and how many ways they can be ordered $1^2$ $1,1,1,1$ $1$ way. $2^2$ $4,1,1,1$ or $2,2,1,1$ $\\binom{4}{2}+4=10$ ways. $3^2$ $3,3,1,1$ $\\binom{4}{2}=6$ ways. $4^2$ $4,4,1,1$ $2,2,2,2$ , or $2,2,4,1$ $\\binom{4}{2}+1+12=19$ ways. $5^2$ $5,5,1,1$ $\\binom{4}{2}=6$ ways. $6^2$ $6,6,1,1$ $1,2,3,6$ $2,3,2,3$ $3,3,4,1$ $2*\\binom{4}{2}+4!+12=48$ ways. $7^2$ - Since there is a prime greater than 6 in its prime factorization there are $0$ ways. $8^2$ $4,4,4,1$ or $2,4,2,4$ $\\binom{4}{2}+4=10$ ways. $9^2$ $3,3,3,3$ $1$ way. $10^2$ $2,2,5,5$ or $1,4,5,5$ $6+12=18$ ways. $11^2$ $0$ ways for the same reason as $7^2$ $12^2$ $6,6,2,2$ $4,4,3,3$ $2,3,4,6$ , or $1,4,6,6$ $2*\\binom{4}{2}+4!+12=48$ ways. $13^2$ $0$ ways. $14^2$ $0$ ways. $15^2$ $3,3,5,5$ $\\binom{4}{2}=6$ ways. $16^2$ $4,4,4,4$ $1$ way. $17^2$ $0$ ways. $18^2$ $3,3,6,6$ $\\binom{4}{2}=6$ ways. $19^2$ $0$ ways. $20^2$ $4,4,5,5$ $\\binom{4}{2}=6$ ways. $21^2$ $0$ ways. $22^2$ $0$ ways. $23^2$ $0$ ways. $24^2$ $4,4,6,6$ $\\binom{4}{2}=6$ ways. $25^2$ $5,5,5,5$ $1$ way. $26^2$ $0$ ways. $27^2$ $0$ ways. $28^2$ $0$ ways. $29^2$ $0$ ways. $30^2$ $5,5,6,6$ $\\binom{4}{2}$ ways. $31^2$ $0$ ways. $32^2$ $0$ ways. $33^2$ $0$ ways. $34^2$ $0$ ways. $35^2$ $0$ ways. $36^2$ $6,6,6,6$ $1$ way.\nThere are $6^4=1296$ ways that the dice can land. Summing up the ways, it is easy to see that there are $200$ ways. \nThis results in a probability of $\\frac{200}{1296}=\\frac{25}{162}\\implies\\boxed{187}$ -superninja2000", "We can do recursion on the number of rolls to find the number of ways we can get $4$ rolls to multiply to a square.\nAfter $n$ rolls, let us say that the product is $p = 2^a3^b5^c$\nDefine the following:\n$A_{n} =$ the number of ways to have a product after $n$ rolls where $a$ is odd, and $b$ $c$ are even\n$B_{n} =$ the number of ways to have a product after $n$ rolls where $b$ is odd, and $a$ $c$ are even\n$C_{n} =$ the number of ways to have a product after $n$ rolls where $c$ is odd, and $a$ $b$ are even\n$D_{n} =$ the number of ways to have a product after $n$ rolls where $c$ is even, and $a$ $b$ are odd\n$E_{n} =$ the number of ways to have a product after $n$ rolls where $b$ is even, and $a$ $c$ are odd\n$F_{n} =$ the number of ways to have a product after $n$ rolls where $a$ is even, and $b$ $c$ are odd\n$G_{n} =$ the number of ways to have a product after $n$ rolls where $a, b,$ and $c$ are all odd\n$S_{n} =$ the number of ways to have a product after $n$ rolls where $a, b,$ and $c$ are all even (square!)\nWe have the following equations after considering the possible values of the nth roll:\n\\[A_{n} = S_{n-1}+B_{n-1}+D_{n-1}+E_{n-1}+2A_{n-1}\\]\n\\[B_{n} = A_{n-1}+D_{n-1}+F_{n-1}+S_{n-1}+2B_{n-1}\\]\n\\[C_{n} = S_{n-1}+E_{n-1}+F_{n-1}+G_{n-1}+2C_{n-1}\\]\n\\[D_{n} = S_{n-1}+A_{n-1}+B_{n-1}+G_{n-1}+2D_{n-1}\\]\n\\[E_{n} = A_{n-1}+C_{n-1}+F_{n-1}+G_{n-1}+2E_{n-1}\\]\n\\[F_{n} = B_{n-1}+E_{n-1}+C_{n-1}+G_{n-1}+2F_{n-1}\\]\n\\[G_{n} = C_{n-1}+D_{n-1}+F_{n-1}+E_{n-1}+2G_{n-1}\\]\n\\[S_{n} = A_{n-1}+C_{n-1}+B_{n-1}+D_{n-1}+2S_{n-1}\\]\nWe have the following values after considering the possible values of the 1st roll:\n\\[A_1 = B_1 = C_1 = D_1 = 1; E_1 = F_1 = G_1 = 0; S_1 = 2\\]\nAfter applying recursion twice, we get:\n\\[A_2 = B_2 = D_2 = 6, C_2 = 4, E_2 = F_2 = G_2 = 2, S_2 = 8\\]\n\\[A_3 = B_3 = D_3 = 34, C_3 = 22, E_3 = F_3 = G_3 = 18, S_3 = 38\\]\nFinally, we have $S_4 = 200$ $\\frac{m}{n} = \\frac{200}{1296} = \\frac{25}{162}$ meaning our answer is $\\boxed{187}$", "Consider all the distinct \"fundamental\" groups of integers from $1$ to $6$ whose product is a perfect square. A \"fundamental\" group is one that cannot be broken into two smaller groups that each have a perfect square product. For example, $\\{2,2\\}$ is a fundamental group, while $\\{3,3,4\\}$ is not, because it can be broken up into $\\{3,3\\}$ and $\\{4\\}$\n$1$ and $4$ are already perfect squares, so they each form a \"fundamental\" group and cannot belong in any other group. Pairs of the other $4$ numbers ( $\\{2,2\\}$ $\\{3,3\\}$ , etc. ) form \"fundamental\" groups as well. The last \"fundamental\" group is $\\{2,3,6\\}$ . It can be easily seen that no more groups exist.\nThus, we have the \"fundamental\" groups $\\{1\\}$ $\\{4\\}$ $\\{2,2\\}$ $\\{3,3\\}$ $\\{5,5\\}$ $\\{6,6\\}$ , and $\\{2,3,6\\}$\nWe now consider the ways to use these groups to form a sequence of $4$ numbers whose product is a perfect square. To form a set, we can simply select zero to two groups of size $2$ or $3$ and fill in any remaining spots with $1$ s and $4$ s. We can do this in one of $5$ ways: Using only $1$ s and $4$ s, using one group of size $2$ , using one group of size $3$ , using two different groups of size $2$ , and using the same group of size $2$ twice.\nIf we only use $1$ s and $4$ s, each of the $4$ slots can be filled with one of the $2$ numbers, so there are $2^4=16$ possibilities.\nIf we use one group of size $2$ , there are $4$ options for the group to use, $\\binom{4}{2}$ ways to place the two numbers (since they are identical), and $2^2$ ways to fill in the remaining slots with $1$ s and $4$ s, so there are $4\\cdot\\binom{4}{2}\\cdot2^2=96$ possibilities.\nIf we use one group of size $3$ , there is only $1$ option for the group to use, $4\\cdot3\\cdot2$ ways to place the three numbers (since they are distinct), and $2$ ways to fill in the remaining slot, so there are $4\\cdot3\\cdot2\\cdot2=48$ possibilities.\nIf we use two different groups of size $2$ , there are $\\binom{4}{2}$ options for the groups to use and $\\binom{4}{2}$ ways to place the four numbers (since there are $2$ groups of identical numbers, and one group's placement uniquely determines the other's), so there are $\\binom{4}{2}\\cdot\\binom{4}{2}=36$ possibilities.\nIf we use the same group of size $2$ twice, there are $4$ options for the group to use and $1$ way to place the four numbers (since they are all identical), so there are $4=4$ possibilities.\nThis gives us a total of $16+96+48+36+4=200$ possibilities, and since there are $6^4=1296$ total sequences that can be rolled, the probability is equal to $\\frac{200}{1296}=\\frac{25}{162}$ , so the answer is $25+162=\\boxed{187}$ . ~ emerald_block", "There are a total of $2^4=1296$ possible die rolls.\nWe use casework:\nCase 1 : All 4 numbers are the same.\nThere are obviously $6$ ways.\nCase 2 : Sets of 2 different numbers.\nA set of two different numbers is basically $(x,x,y,y)$ . There are a total of $\\frac{4!}{2!\\cdot 2!}=6$ ways to arrange the numbers.\nBy listing these cases, we quickly see a pattern:\n$(1,1,2,2)$\n$(1,1,3,3)$\n$...$\n$(1,1,6,6)$\n$(2,2,3,3)$\n$...$\n$(2,2,6,6)$\n$...$\n$(5,5,6,6)$\nThere are a total of $5+4+3+2+1=15$ cases. Multiplying this by $6$ yields $15\\cdot 6=90$ ways.\nCase 3 : Sets of numbers in the form of $(x,x,1,4)$\nA special case must be made for the number $4$ because $4$ itself is a perfect square.\n$(1,1,1,4)$ $4$\n$(2,2,1,4)$ $12$\n$(3,3,1,4)$ $12$\n$(4,4,1,4)$ $4$\n$(5,5,1,4)$ $12$\n$(6,6,1,4)$ $12$\nSumming these up yields a total of $4+12+12+4+12+12=56$ ways.\nCase 4 : Sets with all 4 numbers different\nNote that the sets\n$(1,2,3,6)$\n$(2,3,4,6)$\nMultiply to perfect square. The total of these cases are $24+24=48$\nAdding all these cases together yields $6+90+56+48=200$ ways that the product of the values of the die can be a perfect square.\nTherefore the probability is\n$\\frac{200}{1296}=\\frac{25}{162}$\n$m+n = 25+162 = \\boxed{187}$" ]

[ "Suppose the fraction of discount is $x$ . That means $(1-x)(1+x)=0.84$ ; so, $1-x^{2}=0.84$ , and $(x^{2})=0.16$ , obtaining $x=0.4$ . Therefore, the price was increased and decreased by $40$ %, or $\\boxed{40}$", "After the first increase by $p$ percent, the shirt price became $(1+p)$ times greater than the original. Upon the decrease in p percent on this price, the shirt price became $(1-p)$ times less than $(1+p)$ , or $(1-p)(1+p)$ . We know that this price is $84$ percent of the original, so $(1-p)(1+p) = 0.84$\nFrom here, we can list the factors of $0.84$ and see which are equidistant from $1$ . We see that $0.6$ and $1.4$ are both $0.4$ from $1$ , so $p = 0.4 = 40 \\%$ , or choice $\\boxed{40}$", "We can try out every option and see which one works. By this method, we get $\\boxed{40}$", "Let our original cost be $$ 100$ , so we are looking for a whole number of $$ 84$ . Also, we can see that (A), (C), and (D) give us answers with decimals while we know that (B) and (E) give us whole numbers. Therefore, we only need to try these two: (B) $$100$ increased by 20% = $$120$ , and $$120$ decreased by 20% = $$96$ , a whole number, and (E) $$100$ increased by 40% = $$140$ , and $$140$ decreased by 40% = $$84$ , a whole number.\nThus, $40$ % or $\\boxed{40}$ is the answer." ]

[ "Let our original cost be $$ 100.$ We are looking for a result of $$ 84,$ then. We try 16% and see it gets us higher than 84. We try 20% and see it gets us lower than 16 but still higher than 84. We know that the higher the percent, the less the value. We try 36, as we are not progressing much, and we are close! We try $\\boxed{40}$ , and we have the answer; it worked.\n(OR: try (C) first to eliminate 2 answer choices)" ]

[ "The store's offer means that every $5$ th window is free.\nDave would get $\\left\\lfloor\\frac{7}{5}\\right\\rfloor=1$ free window.\nDoug would get $\\left\\lfloor\\frac{8}{5}\\right\\rfloor=1$ free window.\nThis is a total of $2$ free windows.\nTogether, they would get $\\left\\lfloor\\frac{8+7}{5}\\right\\rfloor = \\left\\lfloor\\frac{15}{5}\\right\\rfloor=3$ free windows.\nSo they get $3-2=1$ additional window if they purchase the windows together.\nTherefore they save $1\\cdot100= \\boxed{100}$" ]

[ "$1500\\times 0.1=150$ , so the store owner is $$150$ below profit. Therefore he needs to sell $150+100= 250$ dollars worth of pencils. Selling them at $$0.25$ each gives $250/0.25= \\boxed{1000}$" ]

[ "This is a $1$ yard by $20$ yard by $1/12$ yard sidewalk, so its volume in yards is \\[1\\times 20\\times \\frac{1}{12} = 1.\\overline{6}.\\] Since concrete must be ordered in a whole number of cubic yards, we need $2\\rightarrow \\boxed{2}$" ]

[ "Define $m$ as the number of minutes they swim for.\nLet their meeting point be $A$ . Melanie is swimming against the current, so she must aim upstream from point $A$ , to compensate for this; in particular, since she is swimming for $m$ minutes, the current will push her $14m$ meters downstream in that time, so she must aim for a point $B$ that is $14m$ meters upstream from point $A$ . Similarly, Sherry is swimming downstream for $m$ minutes, so she must also aim at point $B$ to compensate for the flow of the current.\nIf Melanie and Sherry were to both aim at point $B$ in a currentless river with the same dimensions, they would still both meet at that point simultaneously. Since there is no current in this scenario, the distances that Melanie and Sherry travel, respectively, are $80m$ and $60m$ meters. We can draw out this new scenario, with the dimensions that we have: (While it is indeed true that the triangle above with side lengths $60m$ $80m$ and $D$ is a right triangle, we do not know this yet, so we cannot assume this based on the diagram.)\nBy the Pythagorean Theorem, we have \\begin{align*} 264^{2} + \\left( \\frac{D}{2} - 14m \\right) ^{2} &= 3600m^{2} \\\\ 264^{2} + \\left( \\frac{D}{2} + 14m \\right) ^{2} &= 6400m^{2}. \\end{align*}\nSubtracting the first equation from the second gives us $28Dm = 2800m^{2}$ , so $D = 100m$ . Substituting this into our first equation, we have that \\begin{align*}264^{2} + 36^{2} m^{2} &= 60^{2}m^{2} \\\\ 264^{2} &= 96 \\cdot 24 \\cdot m^{2} \\\\ 11^{2} &= 4 \\cdot m^{2} \\\\ m &= \\frac{11}{2}. \\end{align*}\nSo $D = 100m = \\boxed{550}$", "Claim\nMedian $AM$ and altitude $AH$ are drawn in triangle $ABC$ $AB = c, AC = b < c, BC = a$ are known. Let's denote $MH = x$\nProve that \\begin{align*}2ax = c^{2} - b^{2}\\end{align*}\nProof \\[BH + CH = a,\\] \\begin{align*} BH^{2} - CH^{2} = c^{2} - b^{2}\\implies BH - CH &= \\frac{c^{2} - b^{2}} {a},\\end{align*} \\[BH = \\frac{c^{2} - b^{2}}{2a} + \\frac{a}{2},\\] \\begin{align*}MH = BH - BM &= \\frac{c^{2} - b^{2}} {2a}.\\end{align*}\nSolution\nIn the coordinate system associated with water, the movement is described by the scheme in the form of a triangle, the side on which Melanie floats is $80t$ , where t is the time of Melanie's movement, the side along which Sherry floats is $60t$\nThe meeting point floated away at a distance of $14t$ from the midpoint between the starting points of Melanie and Sherry.\nIn the notation of the Claim \\begin{align*} c = 80t, b = 60t, x = 14t \\implies a = \\frac{(80t)^2-(60t)^2}{2 \\cdot 14t}=\\frac{20^2}{4}\\cdot \\frac{16-9}{7}t = 100t.\\end{align*} Hence, \\begin{align*} AH = \\sqrt{BC^2-BH^2}= \\sqrt{(80t)^2-(50t+14t)^2}=16t \\cdot \\sqrt{5^2-4^2}= 48t = 264 \\implies t = 5.5.\\end{align*} \\[D = a = 100t = \\boxed{550}\\] [email protected], vvsss", "We have the following diagram: Since Melanie and Sherry swim for the same distance and the same amount of time, they swim at the same net speed.\nLet $x$ and $y$ be some positive numbers. We have the following table: \\[\\begin{array}{c||c|c|c} & \\textbf{Net Velocity Vector (m/min)} & \\textbf{Natural Velocity Vector (m/min)} & \\textbf{Natural Speed (m/min)} \\\\ \\hline \\hline &&& \\\\ [-2.25ex] \\textbf{Melanie} & \\langle -x,y\\rangle & \\langle -x-14,y\\rangle & 80 \\\\ \\hline &&& \\\\ [-2.25ex] \\textbf{Sherry} & \\langle x,y\\rangle & \\langle x-14,y\\rangle & 60 \\end{array}\\] Recall that $|\\text{velocity}|=\\text{speed},$ so \\begin{align*} (-x-14)^2 + y^2 &= 80^2, &&(1) \\\\ (x-14)^2 + y^2 &= 60^2. &&(2) \\end{align*} We subtract $(2)$ from $(1)$ to get $56x=2800,$ from which $x=50.$ Substituting this into either equation, we have $y=48.$\nIt follows that Melanie and Sherry both swim for $264\\div y=5.5$ minutes. Therefore, the answer is \\[D=2x\\cdot5.5=\\boxed{550}.\\] ~MRENTHUSIASM", "We can break down movement into two components: the $x$ -component and the $y$ -component. Suppose that Melanie travels a distance of $a$ in the $x$ -direction and a distance of $c$ in the $y$ -direction in one minute when there is no current. Similarly, suppose that Sherry travels a distance of $a$ in the $x$ -direction but a distance of $b$ in the $y$ -direction in one minute when there is no current. The current only affects the $x$ -components because it goes in the $x$ -direction.\n\nNow, note that $a^2 + b^2 = 60^2$ because Sherry travels 60 meters in a minute. Thus, $a^2 + c^2 = 80^2$ because Melanie travels 80 meters in a minute. Also, the distance they travel with the current must be the same in one minute because they reach the point equidistant from them at the same time. That means $b + 14 = c - 14$ or $b = c - 28$ . So now we can plug that into the two equations to get: \\begin{align*} a^2 + c^2 &= 80^2, \\\\ a^2 + (c-28)^2 &= 60^2. \\end{align*} We can solve the system of equations to get $a = 48$ and $c = 64$ . From this, we can figure out that it must've taken them $5.5$ minutes to get to the other side, because $264/48 = 5.5$ . This means that there are $5.5$ lengths of $48$ in each person's travel. Also, $D$ must be equal to $11(b+14) = 11(c-14)$ because there are $(5.5)2 = 11$ lengths of $b-14$ between them, $5.5$ on each person's side. Since $c = 64$ , we have $c-14 = 50$ , so the answer is \\[D=11\\cdot50=\\boxed{550}.\\] ~Curious_crow" ]

[ "Drawing the problem out, we see we get a parallelogram with a height of $40$ and a base of $15$ , giving an area of $600$\nIf we look at it the other way, we see the distance between the stripes is the height and the base is $50$\n\nThe area is still the same, so the distance between the stripes is $600/50 = \\boxed{12}$", "Alternatively, we could use similar triangles--the $30-40-50$ triangle (created by the length of the bordering stripe and the difference between the two curbs) is similar to the $x-y-15$ triangle, where we are trying to find $y$ (the shortest distance between the two stripes). Therefore, $y$ would have to be $\\boxed{12}$" ]

[ "Instead of setting $a_1$ equal to something and $a_2$ equal to something, note that it is rather easier to set $a_1=x^2$ and $a_3=y^2$ so that $a_2=xy,a_4=y(2y-x),a_5=(2y-x)^2$ and so on until you reached $a_{13}=(6y-5x)^2$ (Or simply notice the pattern), so $6y-5x=\\sqrt{2016}=12\\sqrt{14}$ . Note that since each of the terms has degree 2 so if you multiply $x$ and $y$ by $\\sqrt{14}$ you multiply each term by $14$ so each term is still a integer if the terms are already integers before you multiply $x$ and $y$ by $\\sqrt{14}$ , so let $w=\\frac{x}{\\sqrt{14}}$ and $z=\\frac{y}{\\sqrt{14}}$ so $6z-5w=12$ . Then, for the sequence to be strictly increasing positive integers we have $(w,z)=(6,7)$ so $x=6\\sqrt{14}$ and $a_1=x^2=6^2 \\cdot 14=\\boxed{504}$ Ddk001", "The thirteenth term of the sequence is $2016$ , which makes that fourteenth term of the sequence $2016+r$ and the $15^{\\text{th}}$ term $\\frac{(2016+r)^2}{2016}$ . We note that $r$ is an integer so that means $\\frac{r^2}{2016}$ is an integer. Thus, we assume the smallest value of $r$ , which is $168$ . We bash all the way back to the first term and get our answer of $\\boxed{504}$", "Let $a_{2k-1}=s$ where $k=1$ . Then, $a_{2k}=sr \\Longrightarrow a_{2k+1}=sr^2$ . Continuing on, we get $a_{2k+2}=sr(r-1)+sr^2=sr(2r-1) \\Longrightarrow a_{2k+3}=sr^2(\\frac{2r-1}{r})^2=s(2r-1)^2$ . Moreover, $a_{2k+4}=s(2r-1)(r-1)+s(2r-1)^2=s(2r-1)(3r-2) \\Longrightarrow a_{2k+5}=s(2r-1)^2(\\frac{3r-2}{2r-1})^2=s(3r-2)^2$\nIt is clear now that $a_{2k+2c}=s(cr-(c-1))((c+1)r-c)$ and $a_{2k+2c-1}=s(cr-(c-1))^2$ . Plugging in $c=6$ $a_{13}=s(6r-5)^2=2016$ . The prime factorization of $2016=2^5\\cdot3^2\\cdot7$ so we look for perfect squares.\n$6r-5\\equiv (6r-5)^2\\equiv 1\\pmod{6}$ if $r$ is an integer, and $\\frac{\\omega+5}{6}=r \\Longrightarrow 6\\mid{s}$ if $r$ is not an integer and $\\omega$ is rational, so $6\\mid{s}$ . This forces $s=2\\cdot3^2\\cdot7\\cdot{N}$ . Assuming $(6r-5)$ is an integer, it can only be $2^x$ $x\\in{1,2}$\nIf $6r-5=2^1$ $r=\\frac{7}{6}$ . If $6r-5=2^2$ $r=\\frac{3}{2}$ . Note that the latter cannot work since $a_{2k+1}=s(\\frac{9}{4}) \\Longrightarrow 4\\mid{s}$ but $N=1 \\Longrightarrow s=2\\cdot3^2\\cdot7$ in this scenario. Therefore, $r=\\frac{7}{6} \\Longrightarrow s=\\frac{2016}{2^2}=504$ . Plugging back $k=1$ $a_1=s=\\boxed{504}$" ]

[ "Let the number of students on the council be $x$ . To select a two-person committee, we can select a \"first person\" and a \"second person.\" There are $x$ choices to select a first person; subsequently, there are $x-1$ choices for the second person. This gives a preliminary count of $x(x-1)$ ways to choose a two-person committee. However, this accounts for the order of committees. To understand this, suppose that Alice and Bob are two students in the council. If we choose Alice and then Bob, that is the same as choosing Bob and then Alice and so latter and former arrangements would be considered the same. Therefore, we have to divide by $2$ to account for overcounting. Thus, there are $\\dfrac{x(x-1)} 2=10$ ways to choose the two-person committee. Solving this equation, we find that $5$ and $-4$ are integer solutions. $-4$ is a ridiculous situation, so there are $5$ people on the student council. The solution is $\\dbinom 5 3=10\\implies \\boxed{10}$", "To choose $2$ people from $n$ total people and get $10$ as a result, we can establish the equation $\\binom{n}{2}=10$ which we can easily see $n=5$ , so there are $5$ people. The question asks how many ways to choose $3$ people from the $5$ , so there are $\\binom{5}{3}=\\boxed{10}$ ways." ]

[ "Let us split this up into two cases.\nCase $1$ : The student chooses both algebra and geometry.\nThis means that $3$ courses have already been chosen. We have $3$ more options for the last course, so there are $3$ possibilities here.\nCase $2$ : The student chooses one or the other.\nHere, we simply count how many ways we can do one, multiply by $2$ , and then add to the previous.\nAssume the mathematics course is algebra. This means that we can choose $2$ of History, Art, and Latin, which is simply $\\dbinom{3}{2} = 3$ . If it is geometry, we have another $3$ options, so we have a total of $6$ options if only one mathematics course is chosen.\nThus, overall, we can choose a program in $6 + 3 = \\boxed{9}$ ways", "We can use complementary counting. Since there must be an English class, we will add that to our list of classes leaving $3$ remaining spots for rest of classes. We are also told that there needs to be at least one math class. This calls for complementary counting. The total number of ways of choosing $3$ classes out of the $5$ is $\\binom{5}3$ . The total number of ways of choosing only non-mathematical classes is $\\binom{3}3$ . Therefore the amount of ways you can pick classes with at least one math class is $\\binom{5}3-\\binom{3}3=10-1=\\boxed{9}$ ways.", "Similar to Solution 1, note that for Case 1 of solution the answer is simply $\\binom{4}2$ and for the second case it is $\\binom{3}2$ hence $\\boxed{9}$ ways" ]

[ "Note that $12.5\\% = \\frac{1}{8}$ $25\\% = \\frac{1}{4}$ , and $50\\% = \\frac{1}{2}$ . Thus, since the frequencies must be integers, $N$ must be divisible by $2$ $4$ , and $8$ (so that $\\frac{N}{8}$ etc. are integers), or in other words, $N$ is divisible by $8$ . Thus the smallest possible value of $N$ is the smallest positive multiple of $8$ , which is $8$ itself, or $\\boxed{8}$" ]

[ "To find the largest number of \"repeated\" crates necessary, we must account for all the possibilities of the number of apples in each crate. Since each crate contains a minimum of $120$ apples and a maximum of $144$ apples, there are $144 - 120 + 1 = 25$ different amounts possible for the number of apples per crate.\nNow, we have to count for the worst case scenario: the $25$ amounts are repeated as many times as possible.\n$25$ can go into $128$ exactly $5$ times because $5 \\cdot 25 = 125$ , which is less than $128$ . This leaves a remainder of $3$ crates.\nThe worst case scenario would be that these $3$ crates have a different number of apples each. It doesn't actually matter how many apples are in these $3$ crates because any of the $25$ values would be repeated again anyway. So, the answer is $5 + 1 = \\boxed{6}$ jiang147369" ]

[ "There are five miles which need to be traveled. The cost of these five miles is $(0.25\\cdot5) = 1.25$ . Adding this to $1.50$ , we get $\\boxed{2.75}$" ]

[ "We can assume there are $10$ people in the class. Then there will be $1$ junior and $9$ seniors. The sum of everyone's scores is $10 \\cdot 84 = 840$ . Since the average score of the seniors was $83$ , the sum of all the senior's scores is $9 \\cdot 83 = 747$ . The only score that has not been added to that is the junior's score, which is $840 - 747 = \\boxed{93}$", "Let the average score of the juniors be $j$ . The problem states the average score of the seniors is $83$ . The equation for the average score of the class (juniors and seniors combined) is $\\frac{j}{10} + \\frac{83 \\cdot 9}{10} = 84$ . Simplifying this equation yields $j = \\boxed{93}$" ]

[ "We can assume there are $10$ people in the class. Then there will be $1$ junior and $9$ seniors. The sum of everyone's scores is $10 \\cdot 84 = 840$ . Since the average score of the seniors was $83$ , the sum of all the senior's scores is $9 \\cdot 83 = 747$ . The only score that has not been added to that is the junior's score, which is $840 - 747 = \\boxed{93}$", "Let the average score of the juniors be $j$ . The problem states the average score of the seniors is $83$ . The equation for the average score of the class (juniors and seniors combined) is $\\frac{j}{10} + \\frac{83 \\cdot 9}{10} = 84$ . Simplifying this equation yields $j = \\boxed{93}$" ]

[ "Ben adds $1$ to $6$ , getting $1 + 6 = 7$ . Then Ben doubles it, to get $7\\cdot 2 = 14$ . He gives this number to Sue. Then Sue subtracts $1$ from the number, to get $14 - 1 = 13$ . Finally, Sue doubles the result to get $13\\cdot 2 = 26$ , and the answer is $\\boxed{26}$" ]

[ "Note that $\\operatorname{lcm}(6,7)=42.$ It is clear that $42\\not\\in S$ and $84\\not\\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$\nIn the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple, we obtain cycles: If $n$ is a possible maximum value of $S,$ then $n+42$ must be another possible maximum value of $S,$ and vice versa. By observations, we circle all possible maximum values of $S.$ From the second row of the table above, we perform casework on the possible maximum value of $S:$ \\[\\begin{array}{c||c|c|l} & & & \\\\ [-2.5ex] \\textbf{Max Value} & \\boldsymbol{S} & \\textbf{Valid?} & \\hspace{16.25mm}\\textbf{Reasoning/Conclusion} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 49 & \\{46,47,48,49\\} & & \\text{The student who gets } 46 \\text{ will reply yes.} \\\\ 50 & \\{47,48,49,50\\} & \\checkmark & \\text{Another possibility is } S=\\{89,90,91,92\\}. \\\\ 51 & \\{48,49,50,51\\} & & \\text{The student who gets } 51 \\text{ will reply yes.} \\\\ 56 & \\{53,54,55,56\\} & & \\text{The student who gets } 53 \\text{ will reply yes.} \\\\ 57 & \\{54,55,56,57\\} & & \\text{The student who gets } 57 \\text{ will reply yes.} \\\\ 63 & \\{60,61,62,63\\} & & \\text{The students who get } 60,61,62 \\text{ will reply yes.} \\\\ 66 & \\{63,64,65,66\\} & & \\text{The students who get } 64,65,66 \\text{ will reply yes.} \\\\ 72 & \\{69,70,71,72\\} & & \\text{The student who gets } 69 \\text{ will reply yes.} \\\\ 73 & \\{70,71,72,73\\} & & \\text{The student who gets } 73 \\text{ will reply yes.} \\\\ 78 & \\{75,76,77,78\\} & & \\text{The student who gets } 75 \\text{ will reply yes.} \\\\ 79 & \\{76,77,78,79\\} & \\checkmark & \\text{Another possibility is } S=\\{34,35,36,37\\}. \\\\ 80 & \\{77,78,79,80\\} & & \\text{The student who gets } 80 \\text{ will reply yes.} \\end{array}\\] Finally, all possibilities for $S$ are $\\{34,35,36,37\\}, \\{47,48,49,50\\}, \\{76,77,78,79\\},$ and $\\{89,90,91,92\\},$ from which the answer is $37+50+79+92=\\boxed{258}.$", "We know right away that $42\\not\\in S$ and $84\\not\\in S$ as stated in Solution 1.\nTo get a feel for the problem, let’s write out some possible values of $S$ based on the teacher’s remarks. The first multiple of 7 that is two-digit is 14. The closest multiple of six from 14 is 12, and therefore there are two possible sets of four consecutive integers containing 12 and 14; $\\{11,12,13,14\\}$ and $\\{12,13,14,15\\}$ . Here we get our first crucial idea; that if the multiples of 6 and 7 differ by two, there will be 2 possible sets of $S$ without any student input. Similarly, if they differ by 3, there will be only 1 possible set, and if they differ by 1, 3 possible sets.\nNow we read the student input. Each student says they can’t figure out what $S$ is just based on the teacher’s information, which means each student has to have a number that would be in 2 or 3 of the possible sets (This is based off of the first line of student input). However, now that each student knows that all of them have numbers that fit into more than one possible set, this means that S cannot have two possible sets because otherwise, when shifting from one set to the other, one of the end numbers would not be in the shifted set, but we know each number has to fall in two or more possible sets. For example, take $\\{11,12,13,14\\}$ and $\\{12,13,14,15\\}$ . The numbers at the end, 11 and 15, only fall in one set, but each number must fall in at least two sets. This means that there must be three possible sets of S, in which case the actual S would be the middle S.\nTake for example $\\{33,34,35,36\\}$ $\\{34,35,36,37\\}$ , and $\\{35,36,37,38\\}$ . 37 and 34 fall in two sets while 35 and 36 fall in all three sets, so the condition is met. Now, this means that the multiple of 6 and 7 must differ by 1. \nSince 42 means the difference is 0, when you add/subtract 6 and 7, you will obtain the desired difference of 1, and similarly adding/subtracting 6 or 7 from 84 will also obtain the difference of 1. Thus there are four possible sets of $S$ $\\{34,35,36,37\\}$ $\\{47,48,49,50\\}$ $\\{76,77,78,79\\}$ and $\\{89,90,91,92.\\}$ . Therefore the sum of the greatest elements of the possible sets $S$ is $37+50+79+92=\\boxed{258}$", "In a solution that satisfies these constraints, the multiple of 6 must be adjacent to multiple of 7. The other two numbers must be on either side.\nWLOG assume the set is $\\{a,6j,7k,b\\}$ . The student with numbers $a$ $6j$ , and $7k$ can think the set is $\\{a-1, a,6j,7k\\}$ or $\\{a,6j,7k,b\\}$ , and the students with number $6j$ $7k$ , and $b$ can think the set is $\\{a,6j,7k,b\\}$ or $\\{6j,7k,b, b+1\\}$ . Therefore, none of the students know the set for sure.\nPlaying around with the arrangement of the multiple of 6 and multiple of 7 shows that this is the only configuration viewed as ambiguous to all the students. (Therefore when they hear nobody else knows either, they can find out it is this configuration)\nConsidering $S$ as $\\{a,6j,7k,b\\}$ ,b is 2 mod 6 and 1 mod 7, so $b$ is 8 mod 42. (since it is all 2-digit, the values are either 50 or 92).\nSimilarly, considering $S$ as $\\{a,7j,6k,b\\}$ $b$ is 1 mod 6 and 2 mod 7, so $b$ is 37 mod 42. The values that satisfy that are 37 and 79.\nThe total sum of all these values is therefore $50+92+37+89=\\boxed{258}$", "Consider the tuple $(a, a+1, a+2, a+3)$ as a possible $S$ . If one of the values in $S$ is $3$ or $4 \\pmod{7}$ , observe the student will be able to deduce $S$ with no additional information. This is because, if a value is $b = 3 \\pmod{7}$ and $S$ contains a $0 \\pmod{7}$ , then the values of $S$ must be $(b-3, b-2, b-1, b)$ . Similarly, if we are given a $b \\equiv 4 \\pmod{7}$ and we know that $0 \\pmod{7}$ is in $S$ $S$ must be $(b, b+1, b+2, b+3).$ Hence, the only possibility for $a$ is $5, 6 \\pmod{7}.$\nIn either case, we are guaranteed there is a $6, 0, 1 \\pmod{7}$ value in $7$ . The difference comes down to if there is a $5 \\pmod{7}$ value or a $1 \\pmod{7}$ value. The person receiving such value will be able to determine all of $S$ but the $6, 0, 1 \\pmod{7}$ people will not be able to differentiate the two cases ... yet.\nNow consider which value among the consecutive integers is $c \\equiv 3 \\pmod{6}$ , if any. The person will know that $S$ is either $(c-3, c-2, c-1, c)$ or $(c, c+1, c+2, c+3)$ to have a $0 \\pmod{6}$ value in $S$ . Neither the $6, 1 \\pmod{7}$ person can be $3 \\pmod{7}$ , else they can decipher what $S$ is right off the bat by considering which set has $0 \\pmod{7}.$ This translates to the possible $5 \\pmod{7}$ or $1 \\pmod{7}$ person cannot be $0 \\pmod{6}$ . We are given that $0 \\pmod{7}$ and $0 \\pmod{6}$ cannot be the same person. Hence we conclude one of the $6 \\pmod{7}$ or $1 \\pmod{7}$ must be the $0 \\pmod{6}$ person.\nLet the $6 \\pmod{7}$ person be $0 \\pmod{6}.$ Then--hypothetical-- $c \\equiv 2 \\pmod{7}$ person is $3 \\pmod{6}.$ After the first round, the $6, 0, 1 \\pmod{7}$ people realize that $2 \\pmod{7}$ is not in $S$ else they would have deduced $S$ by noting $S$ was either $(c-3, c-2, c-1, c)$ or $(c, c+1, c+2, c+3)$ to have a $0 \\pmod{6}$ and choosing the former based on where $0 \\pmod{7}$ is. Hence they figure out $S$ by knowing $5 \\pmod{7}.$ So $a \\equiv 5 \\pmod{7}$ and $a \\equiv 5 \\pmod{6}$ (from $6 \\pmod{7}$ person being $0 \\pmod{6}$ ).\nSimilarly, if $1 \\pmod{7}$ person is $0 \\pmod{6}$ we find that $a \\equiv 6 \\pmod{7}$ (so a $2 \\pmod{7}$ is in $S$ ) and $a \\equiv 4 \\pmod{6}.$\nBy CRT, the possibilities are $a \\equiv 5, 34 \\pmod{42}.$ The sum of the greatest values of $S$ are the sum of $a + 3$ and so we get $(34 + 3) + (47 + 3) + (76 + 3) + (89 + 3) = \\boxed{258}.$" ]

[ "Noting that 70% is the same as $\\frac{70}{100}=\\frac{7}{10}$ , and that, when x is the amount of wins in the last 40 games, the fraction of games won is $\\frac{40+x}{50+40}=\\frac{40+x}{90}$ , all we have to do is set them equal: \\[\\frac{40+x}{90}=\\frac{7}{10}\\] \\[40+x=63\\] \\[x=\\boxed{23}\\]", "Alternatively we can note that they will play a total of $40+50=90$ games and must win $0.7(90)=63$ games. Since they won $40$ games already they need $63-40=\\boxed{23}$ more games." ]

[ "Let $n$ be the number of matches won, so that $\\frac{n}{2n}=\\frac{1}{2}$ , and $\\frac{n+3}{2n+4}>\\frac{503}{1000}$\nCross multiplying $1000n+3000>1006n+2012$ , so $n<\\frac{988}{6}=164 \\dfrac {4}{6}=164 \\dfrac{2}{3}$ . Thus, the answer is $\\boxed{164}$", "Let $n$ be the number of matches she won before the weekend began. Since her win ratio started at exactly . $500 = \\tfrac{1}{2},$ she must have played exactly $2n$ games total before the weekend began. After the weekend, she would have won $n+3$ games out of $2n+4$ total. Therefore, her win ratio would be $(n+3)/(2n+4).$ This means that \\[\\frac{n+3}{2n+4} > .503 = \\frac{503}{1000}.\\] Cross-multiplying, we get $1000(n+3) > 503(2n+4),$ which is equivalent to $n < \\frac{988}{6} = 164.\\overline{6}.$ Since $n$ must be an integer, the largest possible value for $n$ is $\\boxed{164}.$" ]

[ "Let the side length of the tetrahedron be $1$\nThe altitude of the equilateral triangle base is $\\frac{\\sqrt{3}}{2}$ . Thus, the distance from the center of the equilateral triangle to its vertex is $\\frac{\\sqrt{3}}{2} \\cdot \\frac 23 = \\frac{\\sqrt{3}}{3}$ . Therefore, the altitude of the tetrahedron is $\\sqrt{1^2-(\\frac{\\sqrt{3}}{3})^2} = \\frac{\\sqrt{6}}{3}$\nLet the radius of the large sphere be $R$ \\[(\\frac{\\sqrt{3}}{3})^2 + (\\frac{\\sqrt{6}}{3} - R)^2 = R^2\\] \\[R = \\frac{\\sqrt{6}}{4}\\]\nThe distance from the center of the tetrahedron to the center of one of the bases is $\\frac{\\sqrt{6}}{3} - R = \\frac{\\sqrt{6}}{3} - \\frac{\\sqrt{6}}{4} = \\frac{\\sqrt{6}}{12}$\nThe distance from the center of the tetrahedron to the center of one of the bases is also the same as the radius of the small sphere in the center of the tetrahedron.\nThe radius of the smaller spheres tangent to the large sphere and the tetrahedron is $\\frac{\\frac{\\sqrt{6}}{4}-\\frac{\\sqrt{6}}{12}}{2} = \\frac{\\sqrt{6}}{12}$\nTherefore, the probability that $P$ lies inside one of the five small spheres is $\\frac{5 \\cdot (\\frac{\\sqrt{6}}{12})^3 \\cdot \\frac43 \\cdot \\pi }{(\\frac{\\sqrt{6}}{4})^3 \\cdot \\frac43 \\cdot \\pi} = \\frac{5}{27} \\approx \\boxed{0.2}$" ]

[ "We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use\n$\\left(\\frac{3}{5}\\right)^2=\\frac{12}{x}$\nWe can then solve the equation to get $x=\\frac{100}{3}$ which is closest to $\\boxed{33.3}$", "Also recall that the area of an equilateral triangle is $\\frac{a^2\\sqrt3}{4}$ so we can give a ratio as follows:\n$\\frac{\\frac{9\\sqrt3}{4}}{12}$ $=$ $\\frac{\\frac{25\\sqrt3}{4}}{x}$\nCross multiplying and simplifying, we get $12 \\cdot \\frac{25}{9}$\nWhich is $33.\\overline{3}$ $\\approx$ $\\boxed{33.3}$", "Note that the ratio of the two triangle's weights is equal to the ratio of their areas, as the height is the same. The ratio of their areas is equal to the square of the ratio of their sides. So if $x$ denotes the weight of the second triangle, we have \\[\\frac{x}{12}=\\frac{5^2}{3^2}=\\frac{25}{9}\\] Solving gives us $x \\approx 33.33$ so the answer is $\\boxed{33.3}$" ]

[ "By: dragonfly\nWe can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use\n$\\left(\\frac{3}{5}\\right)^2=\\frac{12}{x}$\nWe can then solve the equation to get $x=\\frac{100}{3}$ which is closest to $\\boxed{33.3}$", "Another approach to this problem, very similar to the previous one but perhaps explained more thoroughly, is to use proportions. First, since the thickness and density are the same, we can set up a proportion based on the principle that $d=\\frac{m}{V}$ , thus $dV=m$\nHowever, since density and thickness are the same and $A$ is proportional to $s^2$ (recognizing that the area of an equilateral triangle is $\\frac{(s)^2\\sqrt{3}}{4}$ ), we can say that $m$ is proportional to $s^2$\nThen, by increasing s by a factor of $\\frac{5}{3}$ $s^2$ is increased by a factor of $\\frac{25}{9}$ , thus $m=12*\\frac{25}{9}$ or $\\boxed{33.3}$" ]

[ "The three digit numbers are $135,153,351,315,513,531$ . The numbers that end in $5$ are divisible are $5$ , and the probability of choosing those numbers is $\\boxed{13}$", "The number is divisible by 5 if and only if the number ends in $5$ (also $0$ , but that case can be ignored, as none of the digits are $0$ )\nIf we randomly arrange the three digits, the probability of the last digit being $5$ is $\\boxed{13}$" ]

[ "Perform the coordinate transformation $(x, y)\\rightarrow (x+y, x-y)$ . Then we can see that a movement up, right, left, or down in the old coordinates adds the vectors $\\langle 1, -1 \\rangle$ $\\langle 1, 1 \\rangle$ $\\langle -1, -1 \\rangle$ $\\langle -1, 1 \\rangle$ respectively. Moreover, the transformation takes the equation $|y| = |x|$ to the union of the x and y axis. Exactly half of the moves go up in the new coordinates, and half of them go down. In order to end up on the x axis, we need to go up thrice and down thrice. The number of possible sequences of up and down moves is the number of permutations of $UUUDDD$ , which is just $\\binom63 = 20$ . The probability of any of these sequences happening is $\\left(\\frac12\\right)^6$ . Thus, the probability of ending on the x axis is $\\frac{20}{2^6}$ . Similarly, the probability of ending on the y axis is the same.\nHowever, we overcount exactly one case: ending at $(0, 0)$ . Since ending on the x axis and ending on the y axis are independent events, the probability of both is simply $\\left(\\frac{20}{2^6}\\right)^2 = \\frac{25}{256}$ . Using PIE, the total probability is $\\frac{20}{64} + \\frac{20}{64} - \\frac{25}{256} = \\frac{135}{256}$ , giving an answer of $\\boxed{391}$", "We have 4 possible moves: U, D, R, and L. The total number of paths that could be taken is $4^6$ , or $4096$ . There are 4 possible cases that land along the line $y = x$ $x,y = \\pm 1; x,y = \\pm 2; x,y = \\pm 3;$ or $x = y = 0$ . We will count the number of ways to end up at $(1,1), (2,2),$ and $(3,3)$ , multiply them by 4 to account for the other quadrants, and add this to the number of ways to end up at $(0,0)$\nThus, the total number of ways to end up at $(1,1)$ is $300$\nThus, the total number of ways to end up at $(0,0)$ is $400$\nAdding these cases together, we get that the total number of ways to end up on $y = x$ is $4(20 + 120 + 300) + 400 = 2160$ . Thus, our probability is $\\frac{2160}{4096}$ . When this fraction is fully reduced, it is $\\frac{135}{256}$ , so our answer is $135 + 256 = \\boxed{391}.$", "We split this into cases by making a chart. In each row, the entries $(\\pm1)$ before the dividing line represent the right or left steps (without regard to order), and the entries after the dividing line represent the up or down steps (again, without regard to order). This table only represents the cases where the ending position $(x,y)$ satisfies $x=y$ \\[\\begin{array}{ccccccccccccl} \\multicolumn{5}{c}{R (+)\\qquad L (-)}& |&\\multicolumn{5}{c}{U (+)\\qquad D (-)}\\\\ +1&& +1&& +1&| & +1&& +1&& +1\\\\ +1&& +1&& -1& | & +1&& +1&& -1\\\\ +1&& -1&& -1& | & +1&& -1&& -1\\\\ -1 && -1&& -1& | & -1&& -1&& -1\\\\ \\\\ +1&& +1&& +1&& -1 &|& +1&& +1\\\\ +1&& +1&& -1 && -1 &|& +1 && -1\\\\ +1&& -1&& -1 && -1 &|& -1 && -1 &&(\\times 2 \\text{ for symmetry by swapping }R-L\\text{ and }U-D)\\\\ \\\\ +1&& +1 &&+1 &&-1&& -1& |& +1\\\\ +1&& +1 &&-1&& -1&& -1 &|& -1&& (\\times 2\\text{ symmetry})\\\\ \\\\ +1&& +1 &&+1&& -1&& -1 &&-1&| & (\\times2 \\text{ symmetry})\\\\ \\end{array}\\] Note that to account for the cases when $x=-y$ , we can simply multiply the $U-D$ steps by $-1$ , so for example, the first row would become \\[+1 \\qquad+1\\qquad +1 \\ \\ \\ \\ |\\ \\ \\ -1\\qquad -1\\qquad -1.\\] Therefore, we must multiply the number of possibilities in each case by $2$ , except for when $x=y=0$\nNow, we compute the number of possibilities for each case. In particular, we must compute the number of $RLUD$ words, where $R$ represents $+1$ to the left of $|$ $L$ represents $-1$ to the left of $|$ $U$ represents $+1$ to the right of $|$ , and $D$ represents $-1$ to the right of $|$ . Using multinomials, we compute the following numbers of possibilities for each case. \\[{6\\choose 3}\\cdot 2+ \\frac{6!}{2!2!}\\cdot 2 + \\frac{6!}{2!2!} \\cdot 2 + {6\\choose 3} \\cdot 2 = 2(20 + 180 + 180 + 20) = 800\\] \\[\\frac{6!}{3!2!}\\cdot 2 + \\frac{6!}{2!2!} + \\frac{6!}{3!2!}\\cdot 2 = 120 + 180 + 120 = 420\\ (\\times2\\text{ for symmetry})\\] \\[\\frac{6!}{3!2!} \\cdot 2 + \\frac{6!}{3!2!} \\cdot 2 = 120 + 120 = 240\\ (\\times2\\text{ for symmetry})\\] \\[{6\\choose 3} = 20\\ (\\times 2\\text{ for symmetry})\\]\nThus, there are $800 + 840 + 480 + 40 = 2160$ possibilities where $|x|=|y|$ . Because there are $4^6$ total possibilities, the probability is $\\frac{2160}{4^6} = \\frac{135}{256}$ , so the answer is $\\boxed{391}.$", "Denote $(x, y)_n$ the probability that starting from point $(x, y)$ , the token reaches a point on the graph of $|y| = |x|$ in exactly $n$ moves. The problem asks us to find $(0, 0)_6$ . We start by breaking this down: \\[(0, 0)_6 = \\frac14 \\cdot ((0, 1)_5 + (0, -1)_5 + (1, 0)_5 + (-1, 0)_5)\\] Notice that by symmetry, $(0, 1)_5 = (0, -1)_5 = (1, 0)_5 = (-1, 0)_5$ , so the equation simplifies to \\[(0, 0)_6 = (0, 1)_5\\] We now expand $(0, 1)_5$ \\[(0, 1)_5 = \\frac14 \\cdot ((0, 0)_4 + (0, 2)_4 + 2(1, 1)_4)\\] First, we find $(0, 0)_4$ \\[(0, 0)_4 = (0, 1)_3\\] \\[(0, 1)_3 = \\frac14 \\cdot ((0, 0)_2 + (0, 2)_2 + 2(1, 1)_2)\\] At this point, we can just count the possibilities to find $(0, 0)_2 = \\frac34$ $(0, 2)_2 = \\frac{7}{16}$ , and $(1, 1)_2 = \\frac58$ . Therefore, \\[(0, 1)_3 = \\frac14 \\cdot (\\frac34 + \\frac{7}{16} + 2 \\cdot \\frac58)\\] \\[(0, 1)_3 = \\frac{39}{64}\\] Next, we find $(0, 2)_4$ \\[(0, 2)_4 = \\frac14 \\cdot ((0, 1)_3 + (0, 3)_3 + 2(1, 2)_3)\\] We already calculated $(0, 1)_3$ , so we just need to find $(0, 3)_3$ and $(1, 2)_3$ \\[(0, 3)_3 = \\frac14 \\cdot ((0, 2)_2 + (0, 4)_2 + 2(1, 3)_2)\\] \\[(0, 3)_3 = \\frac14 \\cdot (\\frac{7}{16} + 0 + 2 \\cdot \\frac{1}{4})\\] \\[(0, 3)_3 = \\frac{15}{64}\\] \\[(1, 2)_3 = \\frac14 \\cdot ((1, 3)_2 + (1, 1)_2 + (0, 2)_2 + (2, 2)_2)\\] \\[(1, 2)_3 = \\frac14 \\cdot (\\frac14 + \\frac58 + \\frac{7}{16} + \\frac12)\\] \\[(1, 2)_3 = \\frac{29}{64}\\] Therefore, \\[(0, 2)_4 = \\frac14 \\cdot (\\frac{39}{64} + \\frac{15}{64} + 2 \\cdot \\frac{29}{64})\\] \\[(0, 2)_4 = \\frac{7}{16}\\] Finally, we find $(1, 1)_4$ \\[(1, 1)_4 = \\frac12 \\cdot ((0, 1)_3 + (1, 2)_3)\\] \\[(1, 1)_4 = \\frac12 \\cdot (\\frac{39}{64} + \\frac{29}{64})\\] \\[(1, 1)_4 = \\frac{17}{32}\\] Putting it all together, \\[(0, 0)_6 = (0, 1)_5 =\\frac14 \\cdot (\\frac{39}{64} + \\frac{7}{16} + 2 \\cdot \\frac{17}{32})\\] \\[(0, 0)_6 = \\frac{135}{256}\\] Thus, the answer is $135 + 256 = \\boxed{391}$" ]

[ "Let the speed of the train be $x$ miles per hour, and let $D$ miles be the total distance of the trip, where $x$ and $D$ are unit-less quantities. Then for the trip that actually occurred, the train travelled 1 hour before the crash, and then travelled $D-x$ miles after the crash. In other words, the train travelled for $\\frac{(D-x)\\text{mi}}{\\frac{3x}{4}\\frac{\\text{mi}}{\\text{hr}}}=\\frac{D-x}{3x/4}\\text{hr}$ after the crash. So, not including the half hour detention after the crash, the entire trip took \\[\\left(1+\\frac{D-x}{3x/4}\\right)\\text{hours}.\\] Now let us consider the alternative trip, where the accident happened $90$ miles farther along the line. Originally, the accident happened $x\\frac{\\text{mi}}{\\text{hr}}\\cdot(1\\text{ hr})=x\\text{ mi}$ down the line, but in this situation it happens $x+90$ miles from the line. Therefore the accident happens $\\frac{(x+90)\\text{ mi}}{x\\frac{\\text{mi}}{\\text{hr}}}=\\frac{x+90}{x}\\text{hr}$ into the trip. The length of the remaining part of the trip is now $D-x-90$ , so the remaining trip takes $\\frac{(D-x-90)\\text{mi}}{\\frac{3x}{4}\\frac{\\text{mi}}{\\text{hr}}}=\\frac{D-x-90}{3x/4}\\text{hr}$ . So, not including the half hour detention after the crash, the entire trip took \\[\\left(\\frac{x+90}{x}+\\frac{D-x-90}{3x/4}\\right)\\text{hours}.\\] We are given that the trip that actually happened resulted in being $3.5$ hours late, while the alternative trip would have resulted in being only $3$ hours late. Therefore the first trip took exactly $\\frac{1}{2}$ hour more: \\[1+\\frac{D-x}{3x/4}=\\frac{x+90}{x}+\\frac{D-x-90}{3x/4}+\\frac{1}{2}.\\] After simplification and cancellation, we get the equation \\[0=\\frac{90}{x}-\\frac{90}{3x/4}+\\frac{1}{2}.\\] Therefore $(90/x)+(1/2)=(120/x)$ , so $1/2=30/x$ . We solve for $x$ and have the original speed of the train: $x=60$\nWe must now solve for the length of the actual trip. If the train had gone $60\\frac{\\text{mi}}{\\text{hr}}$ for the entire trip, then it would have taken $\\frac{D}{60}\\text{hr}$ , and the train would have been on time. But the trip actually took $1+\\frac{D-60}{45}$ hours, not including the half-hour detention, and the train was $3.5$ hours late. Therefore \\[\\frac{D}{60}+3.5=1+\\frac{D-60}{45}+0.5\\Rightarrow 3D+630=180+4D-240+90\\Rightarrow D=600.\\] The length of the trip in miles was $\\boxed{600}$" ]

[ "\\begin{eqnarray*}A' = & (\\sqrt {900}, \\sqrt {300})\\\\ B' = & (\\sqrt {1800}, \\sqrt {600})\\\\ C' = & (\\sqrt {600}, \\sqrt {1800})\\\\ D' = & (\\sqrt {300}, \\sqrt {900}) \\end{eqnarray*}\nFirst we see that lines passing through $AB$ and $CD$ have equations $y = \\frac {1}{3}x$ and $y = 3x$ , respectively. Looking at the points above, we see the equations for $A'B'$ and $C'D'$ are $y^2 = \\frac {1}{3}x^2$ and $y^2 = 3x^2$ , or, after manipulation $y = \\frac {x}{\\sqrt {3}}$ and $y = \\sqrt {3}x$ , respectively, which are still linear functions. Basically the square of the image points gives back the original points and we could plug them back into the original equation to get the equation of the image lines.\nNow take a look at $BC$ and $AD$ , which have the equations $y = - x + 2400$ and $y = - x + 1200$ . The image equations hence are $x^2 + y^2 = 2400$ and $x^2 + y^2 = 1200$ , respectively, which are the equations for circles\n1999 AIME-6.png\nTo find the area between the circles (actually, parts of the circles), we need to figure out the angle of the arc . This could be done by $\\arctan \\sqrt {3} - \\arctan \\frac {1}{\\sqrt {3}} = 60^\\circ - 30^\\circ = 30^\\circ$ . So the requested areas are the area of the enclosed part of the smaller circle subtracted from the area enclosed by the part of the larger circle = $\\frac {30^\\circ}{360^\\circ}(R^2\\pi - r^2\\pi) = \\frac {1}{12}(2400\\pi - 1200\\pi) = 100\\pi$ . Hence the answer is $\\boxed{314}$" ]

[ "Name the trapezoid $ABCD$ , where $AB$ is parallel to $CD$ $AB<CD$ , and $AD<BC$ . Draw a line through $B$ parallel to $AD$ , crossing the side $CD$ at $E$ . Then $BE=AD$ $EC=DC-DE=DC-AB$ . One needs to guarantee that $BE+EC>BC$ , so there are only three possible trapezoids:\n\\[AB=3, BC=7, CD=11, DA=5, CE=8\\] \\[AB=5, BC=7, CD=11, DA=3, CE=6\\] \\[AB=7, BC=5, CD=11, DA=3, CE=4\\]\nIn the first case, by Law of Cosines, $\\cos(\\angle BCD) = (8^2+7^2-5^2)/(2\\cdot 7\\cdot 8) = 11/14$ , so $\\sin (\\angle BCD) = \\sqrt{1-121/196} = 5\\sqrt{3}/14$ . Therefore the area of this trapezoid is $\\frac{1}{2} (3+11) \\cdot 7 \\cdot 5\\sqrt{3}/14 = \\frac{35}{2}\\sqrt{3}$\nIn the second case, $\\cos(\\angle BCD) = (6^2+7^2-3^2)/(2\\cdot 6\\cdot 7) = 19/21$ , so $\\sin (\\angle BCD) = \\sqrt{1-361/441} = 4\\sqrt{5}/21$ . Therefore the area of this trapezoid is $\\frac{1}{2} (5+11) \\cdot 7 \\cdot 4\\sqrt{5}/21 =\\frac{32}{3}\\sqrt{5}$\nIn the third case, $\\angle BCD = 90^{\\circ}$ , therefore the area of this trapezoid is $\\frac{1}{2} (7+11) \\cdot 3 = 27$\nSo $r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5$ , which rounds down to $\\boxed{63}$", "Let the area of the trapezoid be $S$ , the area of the triangle be $S_1$ , the area of the parallelogram be $S_2$\nBy Heron's Formula $S_1 = \\sqrt{\\frac{b+c+d-a}{2} \\cdot \\frac{c+d-a-b}{2} \\cdot \\frac{a+b+d-c}{2} \\cdot \\frac{b+c-a-d}{2}}$\n$S_2 = \\frac{S_1 \\cdot 2}{c-a} \\cdot a = \\frac{2aS_1}{c-a}$\n$S = S_1 + S_2 = S_1(1+\\frac{2a}{c-a}) = S_1 \\cdot \\frac{c+a}{c-a} = \\frac14 \\cdot \\frac{c+a}{c-a} \\cdot \\sqrt{(b+c+d-a)(c+d-a-b)(a+b+d-c)(b+c-a-d)}$\nIf $a = 3$ $b = 7$ $c = 11$ $d = 5$\n$S = \\frac14 \\cdot \\frac{14}{8} \\cdot \\sqrt{(7+11+5-3)(11+5-3-7)(3+7+5-11)(7+11-3-5)} = \\frac{35\\sqrt{3}}{2}$\nIf $a = 3$ $b = 11$ $c = 5$ $d = 7$\n$S = \\frac14 \\cdot \\frac{8}{2} \\cdot \\sqrt{(11+5+7-3)(5+7-3-11)(3+5+7-11)(11+5-3-7)}$ , which is impossible as $5+7-3-11 = -2$\nIf $a = 3$ $b = 5$ $c = 7$ $d = 11$\n$S = \\frac14 \\cdot \\frac{10}{4} \\cdot \\sqrt{(5+7+11-3)(7+11-3-5)(3+5+11-7)(5+7-11-3)}$ , which is impossible as $5+7-11-3 = -2$\nIf $a = 5$ $b = 11$ $c = 7$ $d = 3$\n$S = \\frac14 \\cdot \\frac{12}{2} \\cdot \\sqrt{(11+7+3-5)(7+3-5-11)(5+11+3-7)(11+7-5-3)}$ , which is impossible as $7+3-5-11 = -6$\nIf $a = 5$ $b = 3$ $c = 11$ $d = 7$\n$S = \\frac14 \\cdot \\frac{16}{6} \\cdot \\sqrt{(3+11+7-5)(11+7-5-3)(5+3+7-11)(3+11-5-7)} = \\frac{32\\sqrt{5}}{3}$\nIf $a = 7$ $b = 3$ $c = 11$ $d = 5$\n$S = \\frac14 \\cdot \\frac{18}{4} \\cdot \\sqrt{(3+11+5-7)(11+5-7-)(7+5+5-11)(3+11-7-5)} = 27$\nThus the answer is $\\frac{35}{2} + \\frac{32}{3} + 27 + 3 + 5$ , which rounds down to $\\boxed{63}$" ]

[ "$AB$ is the side of length $10$ , and $CD$ is the median of length $9$ . The altitude of $C$ to $AB$ is $6$ because the 0.5(altitude)(base)=Area of the triangle. $\\theta$ is $\\angle CDE$ . To find $\\sin{\\theta}$ , just use opposite over hypotenuse with the right triangle $\\triangle DCE$ . This is equal to $\\frac69=\\boxed{23}$" ]

[ "The triangle is isosceles. The height of the triangle is therefore given by $h = \\sqrt{10^2 - \\left( \\dfrac{12}{2} \\right)^2} = \\sqrt{64} = 8$\nNow, the area of the triangle is $\\dfrac{bh}{2} = \\dfrac{12*8}{2} = \\dfrac{96}{2} = 48$\nWe have that the area of the rectangle is the same as the area of the triangle, namely $48$ . We also have the width of the rectangle: $4$\nThe length of the rectangle therefore is: $l = \\dfrac{48}{4} = 12$\nThe perimeter of the rectangle then becomes: $2l + 2w = 2*12 + 2*4 = 32$\nThe answer is:\n$\\boxed{32}$", "An alternative way to find the area of the triangle is by using Heron's formula, $A=\\sqrt{(s)(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter of the triangle (meaning half the perimeter). Here, the semi-perimeter is $(10+10+12)/2 = 16$ . Thus the area equals $\\sqrt{(16)(16-10)(16-10)(16-12)} = \\sqrt{16*6*6*4} = 48.$ We know that the width of the rectangle is $4$ , so $48/4 = 12$ , which is the length. The perimeter of the rectangle is $2(4+12) =$ $\\boxed{32}$", "Note that a triangle with side lengths $10,10,12$ is essentially $2$ “6,8,10” right triangles stuck together. Hence, the height is $8$ , and our area is $48$\nSo, the length of the rectangle is $\\frac{48}{4}=12$ , and our perimeter $P=2(4+12)=\\boxed{32}$" ]

[ "\nThe set of all points closer to point $B$ than to point $A$ lie to the right of the perpendicular bisector of $AB$ (line $PZ$ in the diagram), and the set of all points closer to point $B$ than to point $C$ lie below the perpendicular bisector of $BC$ (line $PX$ in the diagram). Therefore, the set of points inside the triangle that are closer to $B$ than to either vertex $A$ or vertex $C$ is bounded by quadrilateral $BXPZ$ . Because $X$ is the midpoint of $BC$ and $Z$ is the midpoint of $AB$ $X=(10,5)$ and $Z=(6,0)$ . The coordinates of point $P$ is the solution to the system of equations defined by lines $PX$ and $PZ$ . Using the point-slope form of a linear equation and the fact that the slope of the line perpendicular to a line with slope $m$ is $-\\frac{1}{m}$ , the equation for line $PX$ is $y=\\frac{2}{5}x+1$ and the equation for line $PZ$ is $x=6$ . The solution of this system is $P=\\left(6,\\frac{17}{5}\\right)$ . Using the shoelace formula on quadrilateral $BXPZ$ and triangle $ABC$ , the area of quadrilateral $BXPZ$ is $\\frac{109}{5}$ and the area of triangle $ABC$ is $60$ . Finally, the probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to vertex $A$ or vertex $C$ is the ratio of the area of quadrilateral $BXPZ$ to the area of $ABC$ , which is $\\frac{\\frac{109}{5}}{60}=\\frac{109}{300}$ . The answer is $109+300=\\boxed{409}$", "Since we know the coordinates of all three vertices of the triangle, we can find the side lengths: $AB=12$ $AC=2\\sqrt{41}$ , and $BC=2\\sqrt{29}$ . We notice that the point where the three distances are the same is the circumcenter - so we use one of the triangle area formulas to find the circumradius, since we know what the area is. \\[\\frac{12 \\cdot 2\\sqrt{41} \\cdot 2\\sqrt{29}}{4 \\cdot R}=\\frac{12 \\cdot 10}{2}.\\] We rearrange to get \\[R=\\frac{\\sqrt{41} \\cdot \\sqrt{29}}{5}.\\] We know that $AP=\\frac{\\sqrt{41} \\cdot \\sqrt{29}}{5}$ , and $AS=6$ , so using the Pythagorean Theorem gives $SP=\\frac{17}{5}$ . This means $[ASP]=[BSP]=\\frac{17}{5} \\cdot 6 \\cdot \\frac{1}{2} = \\frac{51}{5}$ .\nSimilarly, we know that $BP=\\frac{\\sqrt{41} \\cdot \\sqrt{29}}{5}$ , and $BR=\\sqrt{29}$ , so we get that $PR=\\frac{4\\sqrt{29}}{5}$ , and so $[BRP]=[CRP]=\\frac{4\\sqrt{29}}{5} \\cdot \\sqrt{29} \\cdot \\frac{1}{2} = \\frac{58}{5}$ .\nLastly, we know that $CP=\\frac{\\sqrt{41} \\cdot \\sqrt{29}}{5}$ , and $CT=\\sqrt{41}$ , so we get that $PT=\\frac{2\\sqrt{41}}{5}$ , and $[ATP]=[CTP]=\\frac{2\\sqrt{41}}{5} \\cdot \\sqrt{41} \\cdot \\frac{1}{2} = \\frac{41}{5}$ .\nTherefore, our answer is $\\frac{51+58}{2(51+58+41)}=\\frac{109}{300} \\rightarrow \\boxed{409}$", "To start the problem, identify the two midpoints that connect $AB$ and $BC$ . This is because the midpoints of such lines is the mark at which the point will sway closer to vertex $A$ $C$ or vertex $B$ . The midpoint of $AB$ is $(6,0)$ , and the midpoint of $BC$ is $(10,5)$ . Then, determine the line at which the distance between vertex $B$ and vertex $C$ are the same. Assuming that $x$ is the real value of $x$ and $y$ is the real value of $y$ , we can create a simple equation:\n$\\sqrt{|x-8|^2 + (10-y)^2}$ $\\sqrt{(12-x)^2 + y^2}$\nwhere the left side of the equation is for the distance to vertex $C$ and the right side of the equation is the distance to vertex $B$\nSquaring both sides and then distributing, we get\n$x^2 - 16x + 64 + y^2 - 20y + 100 = x^2 - 24x + 144 + y^2$\nNotice that $(x-8)^2 = (-x+8)^2$ , and thus there is no need to create another equation.\nSimplifying, we get $8x + 20 = 20y$\nDivide both sides by 20, then simplify, and the line that represents equivalent distance between vertex $B$ and vertex $C$ is $y = \\frac{2x}{5} + 1$\nThis line starts at the midpoint of $BC$ , which is $(10,5)$ , and ends at the line $x=6$ , as $x=6$ represents equivalent distance between vertex $A$ and vertex $B$ . Plug in $x=6$ to the equation $y = \\frac{2x}{5} + 1$ , and we get $y = \\frac{17}{5}$ . Now that we have our four points that are $(6,0), (6,\\frac{17}{5}), (10,5)$ , and $(12,0)$ , we can calculate the area of the quadrilateral in which a point is closer to vertex $B$ as opposed to either vertex $A$ or vertex $C$ . Simply draw a rectangle that has the points $(6,0), (6,5), (12,5)$ and $(12,0)$ , and then subtract the two triangles that appear in between.\nThus, the area of the quadrilateral is $6\\cdot5 - (\\frac{\\frac{8}{5}\\cdot4}{2} + \\frac{5\\cdot2}{2}) \\rightarrow 30 - \\frac{41}{5} = \\frac{109}{5}$ . \nSince the problem asks us for the probability that a point chosen inside the triangle is inside the quadrilateral, and because the area of $\\triangle ABC$ is $\\frac{12\\cdot10}{2} = 60$ , the probability is $\\frac{\\frac{109}{5}}{60}=\\frac{109}{300}$ , which means the final answer is $109+300=\\boxed{409}$", "Calculate the area of the triangle using the Shoelace Theorem on $(0,0), (12,0), (8,10)$ \\[\\frac{1}{2}|(0+120+0)-(0+0+0)|=60\\]\nGet the four points $(6,0), (6,\\frac{17}{5}), (10,5)$ , and $(12,0)$ by any method from the above solutions. Then use the Shoelace Theorem to find the area of the region we want: \\[\\frac{1}{2}|(0+60+34+0)-(0+0+30+\\frac{102}{5})|=\\frac{109}{5}\\]\nTherefore the probability is $\\frac{\\frac{109}{5}}{60}=\\frac{109}{300}$ . Thus giving the final answer of $109+300=\\boxed{409}$", "Draw the circumradii from the circumcenter to the three vertices. Drop perpendicular from the circumcenter to the sides. Note that since the triangle is isosceles, the perpendicular are in fact perpendicular bisectors. Therefore the region containing the points closer to B are in $\\triangle{OBP_{1}} , \\triangle{OBP_{2}}$ where $O$ is the circumcenter and $P_{1}, P_{2}$ points of contact of the perpendiculars and the sides. Therefore our fraction is $\\frac{109}{300}$ . Our answer is then $\\boxed{409}$", "Using the same graph and methods as Solution 1, find the coordinates of $P$ $X$ $B$ , and $Z$ . Also, note that angles $PXB$ and $PZB$ are right angles, so $PXBZ$ is a cyclic quadrilateral. Then, use Brahmagupta's formula to determine the area of the quadrilateral, which is $\\frac{109}{5}$ . Then find the area of triangle $ABC$ , which is $60$ . The probability is $\\frac{109}{300}$ . Finally, we get our answer of $\\boxed{409}$" ]

[ "The area $K$ of the triangle can be expressed in terms of its inradius $r$ and its semiperimeter $s$ as: \\[K = r \\times s = r \\times \\frac{P}{2}\\]\nSo, $\\frac{P}{K} = \\boxed{2}$" ]

[ "Label the points in the figure as shown below, and draw the segment $CF$ . This segment divides the quadrilateral into two triangles, let their areas be $x$ and $y$\n\nSince triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$\nSince triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$\nSince triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$ , their bases must be in the same ratio, hence $EF:FB = 3:7$\nSince triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$ , their areas must be in the same ratio, hence $x:(y+7) = 3:7$ , which gives us $7x = 3(y+7)$\nSubstituting $y=x+3$ into the second equation we get $7x = 3(x+10)$ , which solves to $x=\\frac{15}{2}$ . Then $y=x+3 = \\frac{15}{2}+3 = \\frac{21}{2}$ , and the total area of the quadrilateral is $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{18}$", "Connect points $E$ and $D$ . Triangles $EFA$ and $FAB$ share an altitude and their areas are in the ratio $3:7$ . Their bases, $EF$ and $FB$ , must be in the same $3:7$ ratio.\nTriangles $EFD$ and $FBD$ share an altitude and their bases are in a $3:7$ ratio. Therefore, their areas are in a $3:7$ ratio and the area of triangle $EFD$ is $3$\nTriangle $CED$ and $DEA$ share an altitude. Therefore, the ratio of their areas is equal to the ratio of bases $CE$ and $EA$ . The ratio is $A:(3+3) \\Rightarrow A:6$ where $A$ is the area of triangle $CED$\nTriangles $CEB$ and $EAB$ also share an altitude. The ratio of their areas is also equal to the ratio of bases $CE$ and $EA$ . The ratio is $(A+3+7):(3+7) \\Rightarrow (A+10):10$\nBecause the two ratios are equal, we get the equation $\\frac{A}{6} = \\frac {A+10}{10} \\Rightarrow 10A = 6A+60 \\Rightarrow A = 15$ . We add the area of triangle $EDF$ to get that the total area of the quadrilateral is $\\boxed{18}$", "We use mass points (similar to above). Let the triangle be $ABC$ with cevians (lines to opposite side) from $B$ and $C$ . Let the points opposite $B$ and $C$ be $D$ and $F$ respectively and the intersection as $P$\nAssign masses of 1 at $B$ and $D$ since $[BPC] = [DPC]$ . Then the mass at $P$ is 2. To find masses at $F$ and $C$ , we let the mass at $F$ be x and the mass at $C$ be y. Then $3x = 7y$ and $y = \\frac{3}{7}x$ . Then $\\frac{10}{7}x = 2$ since we add the masses for the fulcrum mass, and $x = \\frac{7}{5}$ and $y = \\frac{3}{5}$\nTo calculate the mass at a, it is merely $\\frac{7}{5} - 1 = \\frac{2}{5}$ which means $\\frac{[BCF]}{[ACF]} = \\frac{2}{5}$ or $[ACF] = 25$ . It is easy to see the answer is $\\boxed{18}$" ]

[ "We are trying to find the value of $N$ such that \\[1+2+3\\cdots+(N-1)+N=\\frac{N(N+1)}{2}=2016.\\] Noticing that $\\frac{63\\cdot 64}{2}=2016,$ we have $N=63,$ so our answer is $\\boxed{9}.$", "Knowing that each row number can stand for the number of coins there are in the row, we can just add until we get $2016$ .\nNotice that $1 + 2 + 3 \\cdots + 10 = 55.$ Knowing this, we can say that $11 + 12 \\cdots + 20 = 155$ and $21 + \\cdots +30 =255$ and so on. This is a quick way to get to the point that N is between 60 and 70. By subtracting from the sum of the number from 1 through 70, we learn that when we subtract $70, 69, 68, 67, 66, 65,$ and $64, N = 63.$ Adding those two digits, we get the answer $\\boxed{9}.$ - CorgiARMY" ]

[ "We are trying to find the value of $N$ such that \\[1+2+3\\cdots+(N-1)+N=\\frac{N(N+1)}{2}=2016.\\] Noticing that $\\frac{63\\cdot 64}{2}=2016,$ we have $N=63,$ so our answer is $\\boxed{9}.$", "Knowing that each row number can stand for the number of coins there are in the row, we can just add until we get $2016$ .\nNotice that $1 + 2 + 3 \\cdots + 10 = 55.$ Knowing this, we can say that $11 + 12 \\cdots + 20 = 155$ and $21 + \\cdots +30 =255$ and so on. This is a quick way to get to the point that N is between 60 and 70. By subtracting from the sum of the number from 1 through 70, we learn that when we subtract $70, 69, 68, 67, 66, 65,$ and $64, N = 63.$ Adding those two digits, we get the answer $\\boxed{9}.$ - CorgiARMY" ]

[ "Since the three segments are all the same length, the triathlete's average speed is the harmonic mean of the three given rates. Therefore, the average speed is \\[\\frac{3}{\\frac{1}{3}+\\frac{1}{20}+\\frac{1}{10}}=\\frac{3}{\\frac{29}{60}}=\\frac{180}{29}\\approx6\\Rightarrow\\boxed{6}\\]" ]

[ "Diagram borrowed from Solution 1\nApply Pythagorean Theorem on $\\bigtriangleup TOB$ yields \\[BO=\\sqrt{TB^2-TO^2}=3\\] Since $\\bigtriangleup ABC$ is equilateral, we have $\\angle MOB=60^{\\circ}$ and \\[BC=2BM=2(OB\\sin MOB)=3\\sqrt{3}\\] Apply Pythagorean Theorem on $\\bigtriangleup TMB$ yields \\[TM=\\sqrt{TB^2-BM^2}=\\sqrt{5^2-(\\frac{3\\sqrt{3}}{2})^2}=\\frac{\\sqrt{73}}{2}\\] Apply Law of Cosines on $\\bigtriangleup TBC$ we have \\[BC^2=TB^2+TC^2-2(TB)(TC)\\cos BTC\\] \\[(3\\sqrt{3})^2=5^2+5^2-2(5)^2\\cos BTC\\] \\[\\cos BTC=\\frac{23}{50}\\] Apply Law of Cosines on $\\bigtriangleup STB$ using the fact that $\\angle STB=\\angle BTC$ we have \\[SB^2=ST^2+BT^2-2(ST)(BT)\\cos STB\\] \\[SB=\\sqrt{4^2+5^2-2(4)(5)\\cos BTC}=\\frac{\\sqrt{565}}{5}\\] Apply Pythagorean Theorem on $\\bigtriangleup BSM$ yields \\[SM=\\sqrt{SB^2-BM^2}=\\frac{\\sqrt{1585}}{10}\\] Let the perpendicular from $T$ hits $SBC$ at $P$ . Let $SP=x$ and $PM=\\frac{\\sqrt{1585}}{10}-x$ . Apply Pythagorean Theorem on $TSP$ and $TMP$ we have \\[TP^2=TS^2-SP^2=TM^2-PM^2\\] \\[4^2-x^2=(\\frac{\\sqrt{73}}{2})^2-(\\frac{\\sqrt{1585}}{10}-x)^2\\] Cancelling out the $x^2$ term and solving gets $x=\\frac{181}{2\\sqrt{1585}}$\nFinally, by Pythagorean Theorem, \\[TP=\\sqrt{TS^2-SP^2}=\\frac{144}{\\sqrt{1585}}\\] so $\\lfloor m+\\sqrt{n}\\rfloor=\\boxed{183}$" ]

[ "Converting feet to yards and minutes to second, we see that the truck travels $\\dfrac{b}{18}$ yards every $t$ seconds for $180$ seconds. We see that he does $\\dfrac{180}{t}$ cycles of $\\dfrac{b}{18}$ yards. Multiplying, we get $\\dfrac{180b}{18t}$ , or $\\dfrac{10b}{t}$ , or $\\boxed{10}$", "We set a proportion by letting the $x$ being the number of feet the truck travels in $3$ minutes.\n$\\frac{\\frac{b}{6}}{t}=\\frac{x}{180}$\n$\\frac{b}{6t}=\\frac{x}{180}$\n$\\frac{180b}{6t}=x$\n$x=\\frac{30b}{t}$\nRemember $x$ is the number of feet the truck travels, so we divide by $3$ to convert to yards.\n$\\frac{x}{3}=\\frac{10b}{t}$ , which corresponds to $\\boxed{10}$" ]

[ "Consider a trapezoid (label it $ABCD$ as follows) cross-section of the truncate cone along a diameter of the bases:\nAbove, $E,F,$ and $G$ are points of tangency . By the Two Tangent Theorem, $BF = BE = 18$ and $CF = CG = 2$ , so $BC = 20$ . We draw $H$ such that it is the foot of the altitude $\\overline{HD}$ to $\\overline{AB}$\nBy the Pythagorean Theorem \\[r = \\frac{DH}2 = \\frac{\\sqrt{20^2 - 16^2}}2 = 12\\] Therefore, the answer is $\\boxed{6}.$", "Create a trapezoid with inscribed circle $O$ exactly like in Solution #1, and extend lines $\\overline{AD}$ and $\\overline{BC}$ from the solution above and label the point at where they meet $H$ . Because $\\frac{\\overline{GC}}{\\overline{BE}}$ $\\frac{1}{9}$ $\\frac{\\overline{HG}}{\\overline{HE}}$ $\\frac{1}{9}$ . Let $\\overline{HG} = x$ and $\\overline{GE} = 8x$\nBecause these are radii, $\\overline{GO} = \\overline{OE} = \\overline{OF} = 4x$ $\\overline{OF} \\perp \\overline{BH}$ so $\\overline{OF}^2 + \\overline{FH}^2 = \\overline{OH}^2$ . Plugging in, we get $4x^2 + \\overline{FH}^2 = 5x^2$ so $\\overline{FH} = 3x$ .Triangles $OFH$ and $BEH$ are similar so $\\frac{\\overline{OF}}{\\overline{BE}} = \\frac{\\overline{FH}}{\\overline{EH}}$ which gives us $\\frac{4x}{18} = \\frac{3x}{9x}$ . Solving for x, we get \\[x = 1.5\\] and \\[4x =6\\] . Thus, the answer is $\\boxed{6}$" ]

[ "Note that the number $\\underline{xy} = 10x + y.$ By the problem statement, \\[10x + y = x + y^2 \\implies 9x = y^2 - y \\implies 9x = y(y-1).\\] From this we see that $y(y-1)$ must be divisible by $9.$ This only happens when $y=9.$ Then, $x=8.$ Thus, there is only $\\boxed{1}$ cuddly number, which is $89.$", "If the tens digit is $a$ and the ones digit is $b$ then the number is $10a+b$ so we have the equation $10a + b = a + b^2$ . We can guess and check after narrowing the possible cuddly numbers down to $13,14,24,25,35,36,46,47,57,68,78,89,$ and $99$ . (We can narrow it down to these by just thinking about how $a$ 's value affects $b$ 's value and then check all the possiblities.) Checking all of these we get that there is only $\\boxed{1}$ 2-digit cuddly number, and it is $89$ . Yay!!!" ]

[ "The two-inch cube has a volume of $8$ cubic inches, and the three-inch cube has a volume of $27$ cubic inches. Thus, the three-inch cube has a weight that is $\\frac{27}{8}$ times that of the two-inch cube. Then its value is $\\frac{27}{8} \\cdot 200 = \\boxed{675}$" ]

[ "Looking from an overhead view, call the center of the circle $O$ , the tether point to the unicorn $A$ and the last point where the rope touches the tower $B$ $\\triangle OAB$ is a right triangle because $OB$ is a radius and $BA$ is a tangent line at point $B$ . We use the Pythagorean Theorem to find the horizontal component of $AB$ has length $4\\sqrt{5}$\nNow look at a side view and \"unroll\" the cylinder to be a flat surface. Let $C$ be the bottom tether of the rope, let $D$ be the point on the ground below $A$ , and let $E$ be the point directly below $B$ Triangles $\\triangle CDA$ and $\\triangle CEB$ are similar right triangles . By the Pythagorean Theorem $CD=8\\cdot\\sqrt{6}$\nLet $x$ be the length of $CB$ \\[\\frac{CA}{CD}=\\frac{CB}{CE}\\implies \\frac{20}{8\\sqrt{6}}=\\frac{x}{8\\sqrt{6}-4\\sqrt{5}}\\implies x=\\frac{60-\\sqrt{750}}{3}\\]\nTherefore $a=60, b=750, c=3, a+b+c=\\boxed{813}$", "Note that by Power of a Point, the point the unicorn is at has power $4 \\cdot 20 = 80$ which implies that the tangent from that point to the tower is of length $\\sqrt{80}=4\\sqrt{5},$ however this is length of the rope projected into 2-D. If we let $\\theta$ be the angle between the horizontal and the rope, we have that $\\cos\\theta=\\frac{1}{5}$ which implies that $\\sin\\theta=\\frac{2\\sqrt{6}}{5}.$ Note that the portion of rope not on the tower is $4\\sqrt{5} \\cdot \\frac{5}{2\\sqrt{6}}= \\frac{5\\sqrt{30}}{3},$ the requested length of rope is $20-\\frac{5\\sqrt{30}}{3}=\\frac{60-\\sqrt{750}}{3}$ thus the requested sum is $\\boxed{813}.$" ]

[ "There are $60 \\cdot 60 \\cdot 24 = 86400$ seconds in a day, which means that Yasin's blood expires in $10! \\div 86400 = 42$ days. Since there are $31$ days in January (consult a calendar), then $42-31+1$ (Jan 1 doesn't count) is $12$ days into February, so $\\boxed{12}$" ]

[ "The cylinder can be \"unwrapped\" into a rectangle, and we see that the stripe is a parallelogram with base $3$ and height $80$ . Thus, we get $3\\times80=240\\Rightarrow\\boxed{240}$" ]

[ "Since there are $n^2$ little faces on each face of the big wooden cube , there are $6n^2$ little faces painted red.\nSince each unit cube has $6$ faces, there are $6n^3$ little faces total.\nSince one-fourth of the little faces are painted red,\n$\\frac{6n^2}{6n^3}=\\frac{1}{4}$\n$\\frac{1}{n}=\\frac{1}{4}$\n$n=\\boxed{4}$" ]

[ " (Figure not to scale) The area of the square shadow base is $48 + 1 = 49$ , and so the sides of the shadow are $7$ . Using the similar triangles in blue, $\\frac {x}{1} = \\frac {1}{6}$ , and $\\left\\lfloor 1000x \\right\\rfloor = \\boxed{166}$" ]

[ "Observe that if we remove the outer layer of unit cubes from the entire cube, what remains is a smaller cube of side length $(n-2)$ , which contains all of the unpainted cubes and no others. This shows that there are exactly $(n-2)^3$ unpainted cubes. Similarly, taking one face of the cube and removing the outer edge leaves a square of side length $(n-2)$ containing all of the cubes on that face with exactly one face painted. Making the same argument for the other $5$ faces as well, we deduce that there are a total of $6(n-2)^2$ cubes with only one face painted.\nAccordingly, we require \\begin{align*}(n-2)^3 = 6(n-2)^2 &\\iff n-2 = 6 \\qquad \\text{(as } n > 2\\text{, so } n-2 \\neq 0\\text{)} \\\\ &\\iff n = \\boxed{8}" ]

[ "First, figure out all pairs of numbers whose product is 6. Then, using the process of elimination, we can find the following:\n$\\textbf{(A)}$ is possible: $2\\times 3$\n$\\textbf{(C)}$ is possible: $1\\times 6$\n$\\textbf{(D)}$ is possible: $2\\times 6$\n$\\textbf{(E)}$ is possible: $3\\times 6$\nThe only integer that cannot be the sum is $\\boxed{6}.$" ]

[ "From the given information, we can see that Abe can paint $\\frac{1}{15}$ of the room in an hour, Bea can paint $\\frac{1}{15}\\times\\frac{3}{2} = \\frac{1}{10}$ of the room in an hour, and Coe can paint the room in $\\frac{1}{15}\\times 2 = \\frac{2}{15}$ of the room in an hour. After $90$ minutes, Abe has painted $\\frac{1}{15}\\times\\frac{3}{2}=\\frac{1}{10}$ of the room. Working together, Abe and Bea can paint $\\frac{1}{15}+\\frac{1}{10}=\\frac{1}{6}$ of the room in an hour, so it takes then $\\frac{2}{5}\\div \\frac{1}{6}= \\frac{12}{5}$ hours to finish the first half of the room. All three working together can paint $\\frac{1}{6}+\\frac{2}{15}=\\frac{3}{10}$ of the room in an hour, and it takes them $\\frac{1}{2}\\div \\frac{3}{10}=\\frac{5}{3}$ hours to finish the room. The total amount of time they take is \\[90+\\frac{12}{5}\\times 60+\\frac{5}{3}\\times 60 = 90+ 144 + 100 = \\boxed{334}\\]" ]

[ "The favorable responses are either they both show a green bean or they both show a red bean. The probability that both show a green bean is $\\frac{1}{2}\\cdot\\frac{1}{4}=\\frac{1}{8}$ . The probability that both show a red bean is $\\frac{1}{2}\\cdot\\frac{2}{4}=\\frac{1}{4}$ . Therefore the probability is $\\frac{1}{4}+\\frac{1}{8}=\\boxed{38}$", "We can list out all the combinations and we get this: $GG, GY, GR_1, GR_2, RG, RY, RR_1, RR_2$ . There is a total of 8 combinations and 3 that are the same. Hence, we yield the answer $\\boxed{38}$" ]