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[ "To show this statement to be false, we need a non-prime value of $n$ such that $n-2$ is prime. Since $13$ and $23$ are prime, they won't prove anything relating to the truth of the statement.\nNow we just check the statement for $n=9,12,16$ . If $n=12$ or $n=16$ , then $n-2$ is $10$ or $14$ , which aren't prime. However, $n=9$ makes $n-2=7$ , which is prime, so $n=9$ proves the statement false.\nTherefore, the answer is $\\boxed{9}$ , 9." ]

[ "We make use of the associative and commutative properties of addition to rearrange the sum as \\begin{align*} (1+49)+(11+39)+(21+29)+(31+19)+(41+9) &= 50+50+50+50+50 \\\\ &= 250 \\Longrightarrow \\boxed{250}", "Or, we can just combine the first, second, third, fourth, and last terms in each parentheses. This gives us:\n\\[(1+9)+(11+19)+(21+29)+(31+39)+(41+49),\\] which gives us \\[10+30+50+70+90 = 250 = \\boxed{250}.\\]" ]

[ "We see that $1901=1800+101$ $1902=1800+102$ , etc. Each term in the first set of numbers is $1800$ more than the corresponding term in the second set; Because there are $93$ terms in the first set, the expression can be paired up as follows and simplified:\n\\[(1901-101) + (1902-102) + (1903-103) + \\cdots + (1993-193)\\\\ =1800 + 1800 + \\cdots + 1800\\\\ =(1800)(93)\\\\ =\\boxed{167,400}\\]" ]

[ "We use the distributive property to get \\[3\\times 4+2\\times 4+2\\times 3 = 26 \\rightarrow \\boxed{26}\\]", "Since $\\frac12+\\frac13+\\frac14 > \\frac12+\\frac14+\\frac14 = 1$ , we have \\[(2\\times 3\\times 4)\\left(\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}\\right) > 2\\times 3\\times 4 \\times 1 = 24\\] The only answer choice greater than $24$ is $\\boxed{26}$", "We can just bash it out, getting $24(\\frac12+\\frac13+\\frac14)= 12 + 8 + 6 = 26 \\Longrightarrow \\boxed{26}$ -fn106068" ]

[ "We use the order of operations here to get\n\\begin{align*} -15+9\\times (6\\div 3) &= -15+9\\times 2 \\\\ &= -15+18 \\\\ &= 3 \\rightarrow \\boxed{3}" ]

[ "$.4+.02+.006 = .400 + .020 + .006 = .426\\rightarrow \\boxed{.426}$" ]

[ "\\begin{align*} 1,000,000,000,000-777,777,777,777 &= 999,999,999,999-777,777,777,777+1 \\\\ &= 222,222,222,222+1 \\\\ &= 222,222,222,223 \\rightarrow \\boxed{222,222,222,223}" ]

[ "Put the numbers in groups of $4$\n$(1-2-3+4)+(5-6-7+8)+(9-10-11+ 12) + \\cdots + (1993-1994-1995+1996)$\nThe first group has a sum of $0$\nThe second group increases the two positive numbers on the end by $1$ , and decreases the two negative numbers in the middle by $1$ . Thus, the second group also has a sum of $0$\nContinuing the pattern, every group has a sum of $0$ , and thus the entire sum is $0$ , giving an answer of $\\boxed{0}$", "Let any term of the series be $t_n$ . Realize that at every $n\\equiv0 \\pmod4$ , the sum of the series is 0. For $t_{1996}$ we know $1996\\equiv0 \\pmod4$ so the solution is $\\boxed{0}$" ]

[ "$Case \\textrm{ } 1:$ The lines are not parallel to the faces\nA line through the point $(a,b,c)$ must contain $(a \\pm 1, b \\pm 1, c \\pm 1)$ on it as well, as otherwise, the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube.\nWe look at the one from $(1,1,1)$ to $(10,10,10)$ . The lower endpoint of the desired lines must contain both a 1 and a 3, so it can be $(1,1,3), (1,2,3), (1,3,3)$ . If $\\textrm{min} > 0$ then the point $(a-1,b-1,c-1)$ will also be on the line for example, 3 applies to the other end.\nAccounting for permutations, there are $12$ ways, so there are $12 \\cdot 4 = 48$ different lines for this case.\n$Case \\textrm{ } 2:$ The lines where the $x$ $y$ , or $z$ is the same for all the points on the line.\nWLOG, let the $x$ value stay the same throughout. Let the line be parallel to the diagonal from $(1,1,1)$ to $(1,10,10)$ . For the line to have 8 points, the $y$ and $z$ must be 1 and 3 in either order, and the $x$ value can be any value from 1 to 10. In addition, this line can be parallel to 6 face diagonals. So we get $2 \\cdot 10 \\cdot 6 = 120$ possible lines for this case.\nThe answer is, therefore, $120 + 48 = \\boxed{168}$", "Look at one pair of opposite faces of the cube. There are $4$ lines say $l_1, l_2, l_3, l_4$ with exactly $8$ collinear points on the top face. For each of these lines, draw a rectangular plane that consists of one of the $l_i$ for $1 \\leq i \\leq 4$ and perpendicular to the top face.\nThere are $16$ lines in total on this plane. $10$ of which are parallel to one of the edges of the rectangular plane and $6$ of which are diagonals. There are $3$ pairs of opposite faces. So $3 \\cdot 4 \\cdot 16=192$ lines.\nBut we are overcounting the lines of the diagonals of those rectangular planes twice. There are $4$ rectangular planes perpendicular to one pair of opposite faces. Thus $4 \\cdot 6=24$ lines are overcounted.\nSo the answer is $192-24=\\boxed{168}$", "Considered the cases $(1, 2, ..., 8), (2, 3, ...,9), (3, 4, ..., 10)$ and reverse. Also, consider the constant subsequences of length 8 $(1, 1, ..., 1), (2, 2, ..., 2), ..., (10, 10, ..., 10)$ . Of all the triplets that work they cannot be extended to form another point on the line in the $10 \\times 10 \\times 10$ grid but we need to divide by 2 because reversing all the subsequences gives the same line. Thus the answer is \\[\\frac{16^3 - 14^3 - 14^3 + 12^3}{2} = \\boxed{168}\\]", "The lines can be defined as starting from $(a, b, c)$ with \"slope\" (vector) $(d, e, f)$ . We impose the condition that at least one of $a - d, b - e, \\textrm{ or } c - f$ is outside the range of $[1, 10]$ in order to ensure that $(a, b, c)$ is the first valid point on this line. Then, the line ranges from $(a, b, c), (a + d, b + e, c + f), \\ldots, (a + 7d, b + 7e, c + 7f)$ , where $1 \\le a + 7d, b + 7e, c + 7f \\le 10$ , in which case at least one of $a + 8d, b + 8e, \\textrm{ or } c + 8f$ is outside of the range $[1, 10]$ to ensure the line does not contain more than 8 points. For $1 \\le a + 7d, b + 7e, c + 7f \\le 10$ to be satisfied, the pairs $(a, d), (b, e), \\textrm{ and } (c, f)$ can only be $(1, 0), (2, 0), \\ldots (10, 0), (1, 1), (2, 1), (3, 1), (10, -1), (9, -1), \\textrm{ and } (8, -1)$ . Notice that there are only two pairs such that $a + 8d, b + 8e, \\textrm{ or } c + 8f \\not \\in [1, 10]$ , namely $(3, 1)$ and $(8, -1)$ . Thus, our line must contain at least one of these two pairs. In addition, only the two pairs $(1, 1)$ and $(10, -1)$ satisfy $a - d, b - e, \\textrm{ or } c - f \\not \\in [1, 10]$ . Thus, we must also include at least one of these two pairs as well. Let us call these 4 pairs \"important\" pairs. Finally, we can include any valid pair for our third pair.\nCase 1: We repeat one of the important pairs\nThere are 2 ways to choose from $(3, 1)$ and $(8, -1)$ , and 2 ways to choose from $(1, 1)$ and $(10, -1)$ . Then, there are 2 ways to choose the repeated pair. Next, we can arrange these pairs $3!/2! = 3$ ways. So, we have $2 \\cdot 2 \\cdot 2 \\cdot 3 = 24.$\nCase 2: We use 3 distinct important pairs\nThere are ${4 \\choose 3}$ ways to choose the pairs (note that by pigeonhole principle, this guarantees we get at least one of each required pair). Then, there are $3! = 6$ ways to arrange it. We obtain $4 \\cdot 6 = 24$\nCase 3: We use 3 unique pairs, one that is not important.\nAgain, there are $2 \\cdot 2 = 4$ ways to choose the important pairs. Then, there are 12 ways to choose the non-important pair. Again, there are $3! = 6$ ways to arrange it. So, we get $4 \\cdot 12 \\cdot 6 = 288$\nSumming all of it up and dividing by two (since we over counted each line and its reversal), we get $\\frac{24 + 24 + 288}{2} = \\boxed{168}.$" ]

[ "Place one corner of the solid at $(0,0,0)$ and let $(a,b,c)$ be the general coordinates of the diagonally opposite corner of the rectangle, where $a, b, c \\in \\mathbb{Z_{+}}$\nWe consider the vector equation of the diagonal segment represented by: $(x, y, z) =m (a, b, c)$ , where $m \\in \\mathbb{R}, 0 < m \\le 1$\nConsider a point on the diagonal with coordinates $(ma, mb, mc)$ . We have 3 key observations as this point moves from $(0,0,0)$ towards $(a,b,c)$\nThe number of cubes the diagonal passes is equal to the number of points on the diagonal that has one or more positive integers as coordinates.\nIf we slice the solid up by the $x$ -planes defined by $x=1,2,3,4, \\ldots, a$ , the diagonal will cut these $x$ -planes exactly $a$ times (plane of $x=0$ is not considered since $m \\ne 0$ ). Similar arguments for slices along $y$ -planes and $z$ -planes give diagonal cuts of $b$ , and $c$ times respectively. The total cuts by the diagonal is therefore $a+b+c$ , if we can ensure that no more than $1$ positive integer is present in the x, y, or z coordinate in all points $(ma,mb,mc)$ on the diagonal. Note that point $(a,b,c)$ is already one such exception.\nBut for each diagonal point $(ma,mb,mc)$ with 2 (or more) positive integers occurring at the same time, $a+b+c$ counts the number of cube passes as $2$ instead of $1$ for each such point. There are $\\gcd(a,b)+\\gcd(b,c)+\\gcd(c,a)$ points in such over-counting. We therefore subtract one time such over-counting from $a+b+c$\nAnd for each diagonal point $(ma,mb,mc)$ with exactly $3$ integers occurring at the same time, $a+b+c$ counts the number of cube passes as $3$ instead of $1$ ; ie $a+b+c$ over-counts each of such points by $2$ . But since $\\gcd(a,b)+\\gcd(b,c)+\\gcd(c,a)$ already subtracted three times for the case of $3$ integers occurring at the same time (since there are $3$ of these gcd terms that represent all combinations of 3 edges of a cube meeting at a vertex), we have the final count for each such point as $1 = 3-3+k \\Rightarrow k=1$ , where $k$ is our correction term. That is, we need to add $k=1$ time $\\gcd(a,b,c)$ back to account for the case of 3 simultaneous integers.\nTherefore, the total diagonal cube passes is: $D = a+b+c-\\left[ \\gcd(a,b)+\\gcd(b,c)+\\gcd(c,a) \\right]+\\gcd(a,b,c)$\nFor $(a,b,c) = (150, 324, 375)$ , we have: $\\gcd(150,324)=6$ $\\gcd(324,375)=3$ $\\gcd(150,375)=75$ $\\gcd(150,324,375)=3$\nTherefore $D = 150+324+375-(6+3+75)+3 = \\boxed{768}$", "Consider a point travelling across the internal diagonal, and let the internal diagonal have a length of $d$ . The point enters a new unit cube in the $x,y,z$ dimensions at multiples of $\\frac{d}{150}, \\frac{d}{324}, \\frac{d}{375}$ respectively. We proceed by using PIE.\nThe point enters a new cube in the $x$ dimension $150$ times, in the $y$ dimension $324$ times and in the $z$ dimension, $375$ times.\nThe point enters a new cube in the $x$ and $y$ dimensions whenever a multiple of $\\frac{d}{150}$ equals a multiple of $\\frac{d}{324}$ . This occurs $\\gcd(150, 324)$ times. Similarly, a point enters a new cube in the $y,z$ dimensions $\\gcd(324, 375)$ times and a point enters a new cube in the $z,x$ dimensions $\\gcd(375, 150)$ times.\nThe point enters a new cube in the $x,y$ and $z$ dimensions whenever some multiples of $\\frac{d}{150}, \\frac{d}{324}, \\frac{d}{375}$ are equal. This occurs $\\gcd(150, 324, 375)$ times.\nThe total number of unit cubes entered is then $150+324+375-[\\gcd(150, 324)+\\gcd(324, 375)+\\gcd(375, 150)] + \\gcd(150, 324, 375) = \\boxed{768}$" ]

[ "In the middle, we have $\\cdots + 1010-1000+990 -\\cdots$\nIf we match up the back with the front, and then do the same for the rest, we get pairs with $2000$ and $-2000$ , so these will cancel out. In the middle, we have $2000-1000$ which doesn't cancel, but gives us $1000 \\rightarrow \\boxed{1000}$", "We can see that there are $199$ terms in total. We can also see that the first $198$ numbers form groups of two that add to $10$ each. Dividing to see how many pairs we have, $198$ $2$ $99$ groups of ten, or $990$ . However, we have to remember to add the 199th term ( $10$ ), so we get $990$ $10$ $1000$ , which gives us $\\boxed{1000}$" ]

[ "By placing the $2 \\times 3$ rectangle adjacent to the $3 \\times 4$ rectangle with the 3 side of the $2 \\times 3$ rectangle next to the 4 side of the $3 \\times 4$ rectangle, we get a figure that can be completely enclosed in a square with a side length of 5. The area of this square is $5^2 = 25$\nSince placing the two rectangles inside a $4 \\times 4$ square must result in overlap, the smallest possible area of the square is $25$\nSo the answer is $\\boxed{25}$" ]

[ "If a green square cannot share its top or right side with a red square, then a red square can not share its bottom or left side with a green square. Let us split this up into several cases.\nCase 1: There are no green squares. This can be done in $1$ way.\nCase 2: There is one green square and three red squares. This can only be done when the green square's top and right edges are against the edge, so there is $1$ way.\nCase 3: There are two green squares and two red squares. This happens when the two green squares are in the two top squares or two right squares, so there are $2$ ways.\nCase 4: There are three green squares and one red square. Similar to case 2, this happens when the red square's left and bottom edges are against the edge, so there is $1$ way.\nCase 5: There are four green squares and zero red squares. $1$ way.\n\\[1+1+2+1+1 = \\boxed{6}\\]" ]

[ "Find that \\[(81+83+85+87+89+91+93+95+97+99) = 5 \\cdot 180\\] Which gives us \\begin{align*} 2(5 \\cdot 180) &= 10 \\cdot 180\\\\ &= 1800 & \\text{ Thus \\boxed{1800}", "$2(81+83+85+87+89+91+93+95+97+99)$ Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, each adding up to $81+99$ $83+97$ $85+95$ $87+93$ $89+91$ $180$ . Since we have $5$ pairs, we multiply $180$ by $5$ to get $900$ . But since we have to multiply by 2 (remember the 2 at the beginning of the parentheses!), we get $1800$ , which is $\\boxed{1800}$" ]

[ "By the Pythagorean triple $(7,24,25)$ , the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \\boxed{8}$", "We can observe that the above setup forms a right angled triangles whose base is 7ft and whose hypotenuse is 25ft taking the height to be x ft.\n\\[x^2 + 7^2 = 25^2\\] \\[x^2 = 625 - 49\\] \\[x^2 = 576\\] \\[x = 24\\]\nSince the top of the ladder slipped by 4 ft the new height is $24 - 4 = 20 ft$ . The base of the ladder has moved so the new base is say $(7+y)$ . The hypotenuse remains the same at 25ft. So,\n\\[20^2 + (7+y)^2 = 25^2\\] \\[400 + 49 + y^2 + 14y = 625\\] \\[y^2 + 14y - 176 = 0\\] \\[y^2 + 22y - 8y - 176\\] \\[x(y+22) - 8(y+22)\\] \\[(y-8)(y+22)\\]\nDisregarding the negative solution to equation the solution to the problem is $\\boxed{8}$" ]

[ "Taking the first product, we have\n$\\left(1-\\frac{1}{2}\\right)=\\frac{1}{2}$\n$\\frac{1}{2}\\times2=1$\nLooking at the second, we get\n$\\left(1-\\frac{1}{3}\\right)=\\frac{2}{3}$\n$\\frac{2}{3}\\times3=2$\nWe seem to be going up by $1$\nJust to check,\n$1-\\frac{1}{n}=\\frac{n-1}{n}$\n$\\frac{n-1}{n}\\times n=n-1$\nNow that we have discovered the pattern, we have to find the last term.\n$1-\\frac{1}{10}=\\frac{9}{10}$\n$\\frac{9}{10}\\times10=9$\nThe sum of all numbers from $1$ to $9$ is\n$\\frac{9\\cdot10}{2}=45=\\boxed{45}$" ]

[ "We can eliminate B, C, and D, because they are not $21$ subtracted by any multiple of $4$ . Finally, we see that there is no way to have A, so the solution is $\\boxed{5}$", "Let $x$ be the number of $1x1$ tiles. There are $21$ squares and each $2x2$ or $1x4$ tile takes up 4 squares, so $x \\equiv 1 \\pmod{4}$ , so it is either $1$ or $5$ . Color the columns, starting with red, then blue, and alternating colors, ending with a red column. There are $12$ red squares and $9$ blue squares, but each $2x2$ and $1x4$ shape takes up an equal number of blue and red squares, so there must be $3$ more $1x1$ tiles on red squares than on blue squares, which is impossible if there is just one, so the answer is $\\boxed{5}$ , which can easily be confirmed to work.", "Suppose there are $a$ different $2\\times 2$ tiles, $b$ different $4\\times 1$ tiles and $c$ different $1\\times 1$ tiles. Since the areas of these tiles must total up to $21$ (area of the whole grid), we have \\[4a + 4b + c = 21.\\] Reducing modulo $4$ gives $c\\equiv 1\\pmod{4}$ , or $c = 1$ or $c = 5$\nIf $c = 1$ , then $a + b = 5$ . After some testing, there is no valid pair $(a, b)$ that works, so the answer must be $\\boxed{5}$ , which can be constructed in many ways." ]

[ "We can make use of the distributive property as follows: \\begin{align*} 4(299)+3(299)+2(299)+298 &= 4(299)+3(299)+2(299)+1(299)-1 \\\\ &= (4+3+2+1)(299)-1 \\\\ &= 10(299)-1 \\\\ &= 2989 \\\\ \\end{align*}\n$\\boxed{2989}$" ]

[ "After reversing the numbers on the second and fourth rows, the block will look like this:\n$\\begin{tabular}[t]{|c|c|c|c|} \\multicolumn{4}{c}{}\\\\\\hline 1&2&3&4\\\\\\hline 11&10&9&8\\\\\\hline 15&16&17&18\\\\\\hline 25&24&23&22\\\\\\hline \\end{tabular}$\nThe positive difference between the two diagonal sums is then $(4+9+16+25)-(1+10+17+22)=3-1-1+3=\\boxed{4}$", "Notice that at baseline the diagonals sum to the same number ( $52$ ). Therefore we need only compute the effect of the swap. The positive difference between $9$ and $10$ is $1$ and the positive difference between $22$ and $25$ is $3$ . Adding gives $1+3=\\boxed{4}$" ]

[ "After reversing the numbers on the second and fourth rows, the block will look like this:\n$\\begin{tabular}[t]{|c|c|c|c|} \\multicolumn{4}{c}{}\\\\\\hline 1&2&3&4\\\\\\hline 11&10&9&8\\\\\\hline 15&16&17&18\\\\\\hline 25&24&23&22\\\\\\hline \\end{tabular}$\nThe positive difference between the two diagonal sums is then $(4+9+16+25)-(1+10+17+22)=3-1-1+3=\\boxed{4}$", "Notice that at baseline the diagonals sum to the same number ( $52$ ). Therefore we need only compute the effect of the swap. The positive difference between $9$ and $10$ is $1$ and the positive difference between $22$ and $25$ is $3$ . Adding gives $1+3=\\boxed{4}$" ]

[ "Each small cube would have dimensions $1\\times 1\\times 1$ making each cube a unit cube.\nIf there are $16$ cubes per face and there are $5$ faces we are counting, we have $16\\times 5= 80$ cubes.\nSome cubes are on account of overlap between different faces.\nWe could reduce this number by subtracting the overlap areas, which could mean subtracting 4 cubes from each side and 12 from the bottom.\n$80-(4\\times 4+12)=80-28=52=\\boxed{52}$", "You can imagine removing the cubes that do not fit the description of the problem, forming a \"square cup\".\nThere are $4$ cubes in the center of the top face that do not fit the description. Remove those.\nOnce you remove those cubes on top, you must go down one level and remove the cubes in the same position on the second layer. Thus, $4$ more cubes are removed.\nFinally, you repeat this on the third layer, for $4$ more cubes.\nOnce you do the top three layers, you will be on the bottom layer, and you don't remove any more cubes.\nThis means you removed $4 + 4 + 4 = 12$ cubes of the $4 \\times 4 \\times 4 = 64$ cubes.\nThus, $64 - 12 = 52$ cubes remain, and the answer is $\\boxed{52}$", "We can use casework to solve this.\nCase $1$ : The cubes on the sides\nThere are four sides of a cube which means $64$ cubes, but there will be an overlap of $16$ cubes ( $4$ sides $\\times$ $4$ cubes on each side).\nSo there are $16 \\times 4-16=64-16=48$ cubes on the sides.\nCase $2$ : The cubes on the bottom\nThere are $16$ cubes on the bottom, but by counting the cubes on the sides, $12$ of the cubes are already accounted for. $16-12=4$ cubes on the bottom that aren't already counted.\n$48+4=\\boxed{52}$ cubes touching at least one side. Since there is no top, we don't need to count the top face cubes." ]

[ "Firstly, we consider how many different ways possible to divide the $7\\times 1$ board.\nWe ignore the cases of 1 or 2 pieces since we need at least one tile of each color.\nSecondly, we use Principle of Inclusion-Exclusion to consider how many ways to color them:\nFinally, we combine them together: $15\\times 6+20\\times 36+15\\times 150+6\\times 540+1\\times 1806= 8106$\nSo the answer is $\\boxed{106}$", "This solution is basically solution 1 with more things done at once. The game plan:\n$\\sum_{i=0}^{7} ($ the amount of ways to divide the board into $i$ pieces $) \\cdot ($ the amount of ways to color the respective divisions)\nThe amount of ways to divide the board is determined using stars and bars. The colorings are found using PIE giving $3^i-3\\cdot2^i+3$ . Plus, we don't have to worry about the cases where $i=1$ or $I=2$ since they both give no solutions. So our equation becomes:\n$\\sum_{i=3}^{7} \\left(\\dbinom{6}{I}\\right)\\cdot\\left(3^i-3\\cdot2^i+3\\right)$\nWriting it all out and keeping the numbers small with mod 1000, we will eventually arrive at the answer of $\\boxed{106}$", "3 colors is a lot. How many ways can we tile an $n \\times 1$ board with one color? It's going to be $2^{n-1}$ because if you draw out the $n$ spaces, you can decide whether each of the borders between the tiles are either there or not there. There are $n-1$ borders so there are $2^{n-1}$ tilings. Define a one-tiling of an mx1 as $f_1(m)$\nNow let's look at two colors. Let's see if we can get a two-tiling of an $(n+1) \\times 1$ based off a $n \\times 1$ . There are 2 cases we should consider:\n1) The $n \\times 1$ was a one-coloring and the block we are going to add consists of the second color. The number of ways we can do this is $2f_1(n)$\n2) The $n \\times 1$ was a two-color tiling so now we've got 3 cases to form the $(n+1) \\times 1$ : We can either add a block of the first color, the second color, or we can adjoin a block to the last block in the $n \\times 1$\nThis gives us $f_2(n+1)=2f_1(n)+3f_2(n)$\nTime to tackle the 3-color tiling. Again, we split into 2 cases:\n1) The $n \\times 1$ was a two-color tiling, and the block we are adding is of the 3rd color. This gives $f_2(n)$ ways but we have to multiply by $3C2 = 3$ because we have to pick 2 different colors for the two-color tiling.\n2) The $n \\times 1$ was a 3-color tiling, and we have to consider what we can do with the block that we are adding. It can be any of the 3 colors, or we can adjoin it to whatever was the last block in the $n \\times 1$ . This gives $4f_3(n)$\nSo in total we have $f_3(n+1)=3f_2(n)+4f_3(n)$\nFinally, we just sorta bash through the computation to get $f_3(7)=8\\boxed{106}$", "Let $n$ be the length of the board and $x$ be the number of colors. We will find the number of ways to tile the $n \\times 1$ board with no color restrictions (some colors may be unused) and then use PIE.\nBy stars and bars, the number of ways to divide the board into $k$ pieces is ${n-1 \\choose k-1}$ . There are $x^k$ ways to color each of these divisions. Therefore, the total number of ways to divide and color the board is \\begin{align*} \\chi(n, x) &:= \\sum_{k=1}^n {n-1 \\choose k-1} x^k \\\\ &= x\\sum_{k=0}^{n-1} {n-1 \\choose k} x^k \\\\ &= x(x+1)^{n-1}. \\end{align*}\nIn the given problem, we have $n=7$ . By PIE, we have \\begin{align*} &\\quad {3 \\choose 3} \\chi(7, 3) - {3 \\choose 2} \\, \\chi(7, 2) + {3 \\choose 1} \\, \\chi(7, 1) \\\\ &= 3 \\cdot 4^6 - 3(2 \\cdot 3^6) + 3(1 \\cdot 2^6) \\\\ &= 8106 \\rightarrow \\boxed{106}" ]

[ "To simplify the problem, we can group 90’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \\cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax].\n[mathjax]90\\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.\nRearranging the numbers so each pair sums up to 10, we have:\n[mathjax display=true](1 + 9)+(2+8)+(3+7)+(4+6)+5[/mathjax]. [mathjax]4\\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\\boxed{945}[/mathjax].", "We can express each of the terms as a difference from $100$ and then add the negatives using $\\frac{n(n+1)}{2}=1+2+3+\\cdots+(n-1)+n$ to get the answer. \\begin{align*} (100-10)+(100-9)+\\cdots+(100-1) &= 100\\cdot10 -(1+2+\\cdots+9+10)\\\\ &= 1000 - 55\\\\ &= \\boxed{945}", "Instead of breaking the sum then rearranging, we can rearrange directly: \\begin{align*} 90+91+92+\\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\\\ &= 189+189+189+189+189 \\\\ &= \\boxed{945}", "The finite arithmetic sequence formula states that the sum in the sequence is equal to $\\frac{n}{2}\\cdot(a_1+a_n)$ where $n$ is the number of terms in the sequence, $a_1$ is the first term and $a_n$ is the last term.\nApplying the formula, we have: \\[\\frac{10}{2}\\cdot(90+99)=\\boxed{945}\\]", "The expression is equal to the sum of integers from $1$ to $99$ minus the sum of integers from $1$ to $89$ , so it is equal to $\\frac{99(100)}{2} - \\frac{89(90)}{2} = 4950 - 4005 = \\boxed{945}$" ]

[ "Let the number of rocks in $A$ be $a$ $B$ be $b$ $C$ be $c$ . The total weight of $A$ be $40a$ $B$ be $50b$ $C$ be $kc$\nWe can write the information given as, $\\frac{40a + 50b}{a+b} = 43$ $\\frac{40a + kc}{a+c} = 44$ $\\frac{50b + kc}{b+c} = ?$\n$40a + 50b = 43 a + 43 b$ $3a = 7b$\n$40a + kc = 44a + 44 c$ $kc = 4a + 44c = \\frac{28}{3}b + 44c$\n$\\frac{50b + kc}{b+c} = \\frac{50 \\cdot \\frac{3}{7}a + kc}{\\frac{3}{7}a+c} = \\frac{150a + 7kc}{3a + 7c} = \\frac{150a + 28a + 308c}{3a+7c} = \\frac{178a + 308c}{3a+7c} = \\frac{44(3a+7c)+46a}{3a+7c}$\n$= 44 + \\frac{46a}{3a+7c} = 44 + \\frac{46}{3 + \\frac{7}{a}} < 44 + \\frac{46}{3} \\approx 59.3$\n$\\frac{50b + kc}{b+c} \\le \\boxed{59}$" ]

[ "Let the total number of rocks in pile $A$ be $A_n$ , and the total number of rocks in pile $B$ be $B_n$ . Then, by restriction 3 (the average of $A$ and $B$ ), we can establish the equation: \\[\\frac{40A_n+50B_n}{A_n+B_n}=43\\] .\nCross-multiplying, we get: \\[40A_n+50B_n=43A_n+43B_n \\implies 3A_n=7B_n\\] .\nLet's say we have $7k$ rocks in $A$ and $3k$ rocks in $B$ . Hence, we have $280k$ and $150k$ as the total weight of piles $A$ and $B$ , respectively. Let the total weight of $C$ be $m$ , and the total number of rocks in $C$ be $n$ . \nUsing the last restriction regarding the average of piles $A$ and $C$ , we have: \\[\\frac{280k+m}{7k+n}=44 \\implies 280k + m=280k + 28k + 44n \\implies m=28k+44n\\] .\nTo find the average of piles $B$ and $C$ , we can establish the expression: \\[\\frac{150k+28k+44n}{3k+n}=\\frac{178k+44n}{3k+n}=\\frac{132k+44n+46k}{3k+n}=\\frac{44(3k+n)+46k}{3k+n}=44+\\frac{46k}{3k+n}.\\] When we let the final expression equal $59$ , we get: \\[44+\\frac{46k}{3k+n}=59\\implies\\frac{46k}{3k+n}=15\\] Cross-multiplying, we get: \\[46k=45k+14n\\implies k=14n\\] $k$ is still positive here, so 59 works. As this is the greatest option, we can circle $\\textbf{(E)}$ immediately.\nTo show why $59$ is the greatest, consider the following:\nWhen we let the final expression equal $60$ , we get: \\[44+\\frac{46k}{3k+n}=60\\implies\\frac{46k}{3k+n}=16\\] Cross-multiplying, we get: \\[46k=48k+14n\\implies k=-7n\\] Since $k$ is positive, the final expression could not equal 60. It further implies that the final expression could not equal any other integer greater than 60. Therefore, we have our final answer $\\boxed{59}$" ]

[ "Let $A$ do $r_A$ of the job per day, $B$ do $r_B$ of the job per day, and $C$ do $r_C$ of the job per day. These three quantities have unit $\\frac{\\text{job}}{\\text{day}}$ . Therefore our three conditions give us the three equations: \\begin{align*} (2\\text{ days})(r_A+r_B)&=1\\text{ job},\\nonumber\\\\ (4\\text{ days})(r_B+r_C)&=1\\text{ job},\\nonumber\\\\ (2.4\\text{ days})(r_C+r_A)&=1\\text{ job}.\\nonumber \\end{align*} We divide the three equations by the required constant so that the coefficients of the variables become 1: \\begin{align*} r_A+r_B&=\\frac{1}{2}\\cdot\\frac{\\text{job}}{\\text{day}},\\nonumber\\\\ r_B+r_C&=\\frac{1}{4}\\cdot\\frac{\\text{job}}{\\text{day}},\\nonumber\\\\ r_C+r_A&=\\frac{5}{12}\\cdot\\frac{\\text{job}}{\\text{day}}.\\nonumber \\end{align*} If we add these three new equations together and divide the result by two, we obtain an equation with left-hand side $r_A+r_B+r_C$ , so if we subtract $r_B+r_C$ (the value of which we know) from both equations, we obtain the value of $r_A$ , which is what we wish to determine anyways. So we add these three equations and divide by two: \\[r_A+r_B+r_C=\\frac{1}{2}\\cdot\\left(\\frac{1}{2}+\\frac{1}{4}+\\frac{5}{12}\\right)\\cdot\\frac{\\text{job}}{\\text{day}}=\\frac{7}{12}\\cdot\\frac{\\text{job}}{\\text{day}}.\\] Hence: \\begin{align*} r_A &= (r_A+r_B+r_C)-(r_B+r_C)\\nonumber\\\\ &=\\frac{7}{12}\\cdot\\frac{\\text{job}}{\\text{day}}-\\frac{1}{4}\\cdot\\frac{\\text{job}}{\\text{day}}\\nonumber\\\\ &=\\frac{1}{3}\\cdot\\frac{\\text{job}}{\\text{day}}.\\nonumber \\end{align*} This shows that $A$ does one third of the job per day. Therefore, if $A$ were to do the entire job himself, he would require $\\boxed{3}$ days." ]

[ "Since $EF$ is the perpendicular bisector of $\\overline{BB'}$ , it follows that $BE = B'E$ (by SAS). By the Pythagorean Theorem , we have $AB' = 15$ . Similarly, from $BF = B'F$ , we have \\begin{align*} BC^2 + CF^2 = B'D^2 + DF^2 &\\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\\\ BC &= \\frac{70}{3} \\end{align*} Thus the perimeter of $ABCD$ is $2\\left(25 + \\frac{70}{3}\\right) = \\frac{290}{3}$ , and our answer is $m+n=\\boxed{293}$", "Let $A = (0,0), B=(0,25)$ , so $E = (0,8)$ and $F = (l,22)$ , and let $l = AD$ be the length of the rectangle. The slope of $EF$ is $\\frac{14}{l}$ and so the equation of $EF$ is $y -8 = \\frac{14}{l}x$ . We know that $EF$ is perpendicular to and bisects $BB'$ . The slope of $BB'$ is thus $\\frac{-l}{14}$ , and so the equation of $BB'$ is $y -25 = \\frac{-l}{14}x$ . Let the point of intersection of $EF, BB'$ be $G$ . Then the y-coordinate of $G$ is $\\frac{25}{2}$ , so \\begin{align*} \\frac{14}{l}x &= y-8 = \\frac{9}{2}\\\\ \\frac{-l}{14}x &= y-25 = -\\frac{25}{2}\\\\ \\end{align*} Dividing the two equations yields\nThe answer is $\\boxed{293}$ as above.", "Firstly, observe that if we are given that $AE=8$ and $BE=17$ , the length of the triangle is given and the height depends solely on the length of $CF$ . Let Point $A = (0,0)$ . Since $AE=8$ , point E is at (8,0). Next, point $B$ is at $(25,0)$ since $BE=17$ and point $B'$ is at $(0,-15)$ since $BE=AE$ by symmetry. Draw line segment $BB'$ . Notice that this is perpendicular to $EF$ by symmetry. Next, find the slope of EB, which is $\\frac{15}{25}=\\frac{3}{5}$ . Then, the slope of $EF$ is - $\\frac{5}{3}$\nLine EF can be written as y= $-\\frac{5}{3}x+b$ . Plug in the point $(8,0)$ , and we get the equation of EF to be y= $_\\frac{5}{3}x+\\frac{40}{3}$ . Since the length of $AB$ =25, a point on line $EF$ lies on $DC$ when $x=25-3=22$ . Plug in $x=22$ into our equation to get $y=-\\frac{70}{3}$ $|y|=BC=\\frac{70}{3}$ . Therefore, our answer is $2(AB+BC)=2\\left(25+\\frac{70}{3}\\right)=2\\left(\\frac{145}{3}\\right)=\\frac{290}{3}= \\boxed{293}$", "Firstly, note that $B'E=BE=17$ , so $AB'=\\sqrt{17^2-8^2}=15$ . Then let $\\angle BEF=\\angle B'EF=\\theta$ , so $\\angle B'EA = \\pi-2\\theta$ . Then $\\tan(\\pi-2\\theta)=\\frac{15}{8}$ , or\n\\[\\frac{2\\tan(\\theta)}{\\tan^2(\\theta)-1}=\\frac{15}{8}\\] using supplementary and double angle identities. Multiplying though and factoring yields\n\\[(3\\tan(\\theta)-5)(5\\tan(\\theta)+3)=0\\]\nIt is clear from the problem setup that $0<\\theta<\\frac\\pi2$ , so the correct value is $\\tan(\\theta)=\\frac53$ . Next, extend rays $\\overrightarrow{BC}$ and $\\overrightarrow{EF}$ to intersect at $C'$ . Then $\\tan(\\theta)=\\frac{BC'}{17}=\\frac53$ , so $BC'=\\frac{85}{3}$ . By similar triangles, $CC'=\\frac{3}{17}BC'=\\frac{15}{3}$ , so $BC=\\frac{70}{3}$ . The perimeter is $\\frac{140}{3}+50=\\frac{290}{3}\\Longrightarrow \\boxed{293}$", "Use the prepared diagram for this solution.\nCall the intersection of DF and B'C' G. AB'E is an 8-15-17 right triangle, and so are B'DG and C'FG. Since C'F is 3, then using the properties of similar triangles GF is 51/8. DF is 22, so DG is 125/8. Finally, DB can to calculated to be 25/3.\nAdd all the sides together to get $\\boxed{293}$", "Call the intersection of $B'C'$ $BC$ , and $EF$ $G$ . Since $FCBE$ and $FC'B'E$ are congruent, we know that the three lines intersect.\nWe already know $AB$ so we just need to find $CB$ , call it $x$ . Drop an altitude from $F$ to $AB$ and call it $H$ $EH=EB-FC=14$ . Using Pythagorean Theorem, we have $EF=\\sqrt{x^2+14^2}$ . Triangles $EFH$ and $EGB$ are similar (AA), so we get \\[\\frac{HF}{BG}=\\frac{EH}{EB}\\] \\[\\frac{x}{x+GC}=\\frac{14}{17}\\] Simplify and we get $GC=\\frac{3x}{14}$\nWe find the area of $FCBE$ by using the fact that it is a trapezoid. $[FCBE]=\\frac{(3+17)x}{2}=10x$\nA different way to find the area: $[FCBE]=\\frac{1}{2} EG\\cdot($ height of $EGB$ with $EG$ as base $)-[FGC]$\nSince $GBE$ and $G'B'E$ are congruent(SAS), their height from $EG$ is the same. $B'B=\\sqrt{AB'^2+AB^2}=5\\sqrt{34}$ $EG=\\sqrt{EB^2+BG^2}=\\sqrt{(\\frac{17x}{14})^2+17^2}=17\\sqrt{\\frac{x^2}{196}+1}$\n\\[[FCBE]=\\frac{1}{2} \\cdot 17 \\cdot \\sqrt{\\frac{x^2}{196}+1} \\cdot \\frac{5\\sqrt{34}}{2}-\\frac{9x}{28}\\] \\[280x+9x=7\\cdot 5 \\cdot \\sqrt{34} \\cdot 17 \\cdot \\sqrt{\\frac{x^2}{196}+1}\\] \\[17^4 x^2=49 \\cdot 25 \\cdot 34 \\cdot 17^2 \\cdot (\\frac{x^2}{196}+1)\\] \\[17x^2=\\frac{25}{2}x^2+2450\\] \\[x=\\frac{70}{3}\\]\nThe perimeter is $\\frac{140}{3}+50=\\frac{290}{3},$ so our answer is $\\boxed{293}$", "Let the endpoint of the intersection of the fold near $F$ be $G$ . Since trapezoid $BCFE$ is folded, it is congruent to trapezoid $B'C'FE$ . Therefore, $BE=B'E=17$ . Since $\\triangle AB'E$ is a right triangle, $AB'=15$ from the pythagorean theorem. From here, we can see that triangles $\\triangle AEB \\sim \\triangle DGB' \\sim \\triangle C'GF$ by AA similarity. From here, we find $BC$ from a lot of similarities. Let $BC=x$\nSince $\\triangle ABE' \\sim \\triangle DGB'$\n\\[\\frac {AE}{AB'} = \\frac{DB}{DG}\\]\n\\[\\frac {8}{15} = \\frac {x-15}{DG}\\]\n\\[DG = \\frac {15(x-15)}{8}\\]\n\\[GF = DC-DG-FC\\]\n\\[GF = \\frac{-15x+401}{8}\\]\nSince $\\triangle ABE' \\sim \\triangle C'GF'$\n\\[\\frac {AE}{B'E} = \\frac {C'F}{GF}\\]\n\\[\\frac {8}{17} = \\frac{3}{\\frac {-15x+401}{8}}\\]\nfrom which we get $x= \\frac {70}{3}$\nFinally, our answer is $2(\\frac {70}{3}) + 2(25)=\\frac {290}{3}$ , which is $290+3=\\boxed{293}$" ]

[ "We first note that diagonal $\\overline{AC}$ is of length $\\sqrt{6} + \\sqrt{2}$ . It must be that $\\overline{AP}$ divides the diagonal into two segments in the ratio $\\sqrt{3}$ to $1$ . It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimensions $2\\sqrt{3}$ by $\\sqrt{3} + 1$ . The area of the overall region (of the initial and final squares) is therefore twice the area of the original square minus the overlap, or $2 (\\sqrt{3} + 1)^2 - 2 (\\sqrt{3} + 1) = 2 (4 + 2 \\sqrt{3}) - 2 \\sqrt{3} - 2 = 6 + 2 \\sqrt{3}$\nThe area also includes $4$ circular segments. Two are quarter-circles centered at $P$ of radii $\\sqrt{2}$ (the segment bounded by $\\overline{PA}$ and $\\overline{PA'}$ ) and $\\sqrt{6}$ (that bounded by $\\overline{PC}$ and $\\overline{PC'}$ ). Assuming $A$ is the bottom-left vertex and $B$ is the bottom-right one, it is clear that the third segment is formed as $B$ swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when $D$ overshoots the final square's left edge. To find these areas, consider the perpendicular from $P$ to $\\overline{BC}$ . Call the point of intersection $E$ . From the previous paragraph, it is clear that $PE = \\sqrt{3}$ and $BE = 1$ . This means $PB = 2$ , and $B$ swings back inside edge $\\overline{BC}$ at a point $1$ unit above $E$ (since it left the edge $1$ unit below). The triangle of the circular sector is therefore an equilateral triangle of side length $2$ , and so the angle of the segment is $60^{\\circ}$ . Imagining the process in reverse, it is clear that the situation is the same with point $D$\nThe area of the segments can be found by subtracting the area of the triangle from that of the sector; it follows that the two quarter-segments have areas $\\frac{1}{4} \\pi (\\sqrt{2})^2 - \\frac{1}{2} \\sqrt{2} \\sqrt{2} = \\frac{\\pi}{2} - 1$ and $\\frac{1}{4} \\pi (\\sqrt{6})^2 - \\frac{1}{2} \\sqrt{6} \\sqrt{6} = \\frac{3 \\pi}{2} - 3$ . The other two segments both have area $\\frac{1}{6} \\pi (2)^2 - \\frac{(2)^2 \\sqrt{3}}{4} = \\frac{2 \\pi}{3} - \\sqrt{3}$\nThe total area is therefore \\[(6 + 2 \\sqrt{3}) + (\\frac{\\pi}{2} - 1) + (\\frac{3 \\pi}{2} - 3) + 2 (\\frac{2 \\pi}{3} - \\sqrt{3})\\] \\[= 2 + 2 \\sqrt{3} + 2 \\pi + \\frac{4 \\pi}{3} - 2 \\sqrt{3}\\] \\[= \\frac{10 \\pi}{3} + 2\\] \\[= \\frac{1}{3} (10 \\pi + 6)\\]\nSince $a = 10$ $b = 6$ , and $c = 3$ , the answer is $a + b + c = 10 + 6 + 3 = \\boxed{19}$" ]

[ "To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways. First, we can divide regular pentagon $ABCDE$ into five congruent triangles. \nIf $s$ is the side length of the regular pentagon, then each of the triangles $AOB$ $BOC$ $COD$ $DOE$ , and $EOA$ has base $s$ and height 1, so the area of regular pentagon $ABCDE$ is $5s/2$\nNext, we divide regular pentagon $ABCDE$ into triangles $ABC$ $ACD$ , and $ADE$\n Triangle $ACD$ has base $s$ and height $AP = AO + 1$ . Triangle $ABC$ has base $s$ and height $AQ$ . Triangle $ADE$ has base $s$ and height $AR$ . Therefore, the area of regular pentagon $ABCDE$ is also \\[\\frac{s}{2} (AO + AQ + AR + 1).\\] Hence, \\[\\frac{s}{2} (AO + AQ + AR + 1) = \\frac{5s}{2},\\] which means $AO + AQ + AR + 1 = 5$ , or $AO + AQ + AR = \\boxed{4}$" ]

[ "Now, we know that angle $D$ has measure $\\frac{180 \\cdot 3}{5} = 108$ . Since \\[\\sin 54 = \\frac{OP}{DO} = \\frac{1}{DO}, DO = \\frac{1}{\\sin 54}\\] \\[\\tan 54 = \\frac{OP}{DP} = \\frac{1}{DP}, DP = \\frac{1}{\\tan 54}\\] Therefore, $AB = 2DP = \\frac{2}{\\tan 54}$ \\[\\sin 72 = \\frac{AQ}{AB} = AQ \\tan 54 \\cdot \\frac{1}{2}, AQ = \\frac{2 \\sin 72}{\\tan 54}\\] Therefore, $AO + AQ + AR = AO + 2AQ = \\frac{1}{\\sin 54}+\\frac{4 \\sin 72}{\\tan 54} = \\frac{1}{\\sin 54} + 8 \\sin 36 \\cos 54 = \\frac{1}{\\cos 36} + 8-8\\cos^2(36)$ . Recalling that $\\cos 36 = \\frac{1 + \\sqrt{5}}{4}$ gives a final answer of $\\boxed{4}$" ]

[ "\\[R=c\\cdot\\frac{S}T\\]\nfor some constant $c$\nYou know that\n\\[\\frac43=c\\cdot\\frac{3/7}{9/14}=c\\cdot\\frac37\\cdot\\frac{14}9=c\\cdot\\frac23\\,,\\]\nso\n\\[c=\\frac{4/3}{2/3}=2\\,.\\]\nWhen $R=\\sqrt{48}$ and $T=\\sqrt{75}$ we have\n\\[\\sqrt{48}=\\frac{2S}{\\sqrt{75}}\\,,\\]\nso\n\\[S=\\frac12\\sqrt{48\\cdot75}=30\\,.\\] $\\boxed{30}$" ]

[ "The expression becomes $(x-y+z)-(x-y-z) = x-y+z-x+y+z = 2z$ , which is $\\boxed{2}$" ]

[ "Using the left triangle, we have:\n$\\angle 1 + 70 + 40 = 180$\n$\\angle 1 = 180 - 110$\n$\\angle 1 = 70$\nUsing the given fact that $\\angle 1 + \\angle 2 = 180$ , we have $\\angle 2 = 180 - 70 = 110$\nFinally, using the right triangle, and the fact that $\\angle 3 = \\angle 4$ , we have:\n$\\angle 2 + \\angle 3 + \\angle 4 = 180$\n$110 + \\angle 3 + \\angle 4 = 180$\n$110 + \\angle 4+ \\angle 4 = 180$\n$2\\angle 4 = 70$\n$\\angle 4 = 35$\nThus, the answer is $\\boxed{35}$" ]

[ "\\begin{align*} \\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9} &= \\dfrac{(10-9)+(8-7)+(6-5)+(4-3)+(2-1)}{1+(-2+3)+(-4+5)+(-6+7)+(-8+9)} \\\\ &= \\dfrac{1+1+1+1+1}{1+1+1+1+1} \\\\ &= 1 \\rightarrow \\boxed{1}" ]

[ "Convert each fraction to a decimal.\n$\\dfrac{1}{10} + \\dfrac{9}{100} + \\dfrac{9}{1000} + \\dfrac{7}{10000}$\n$0.1+0.09+0.009+0.0007=0.1997$\n$\\boxed{0.1997}$" ]

[ "$1+ 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = \\dfrac{(9)(10)}{2} = 45$\n$\\frac{45+55}{10} = \\dfrac{100}{10} = \\boxed{10}$" ]

[ "Since it doesn't take too long, we can just make a table with all the possible values of the sum \\[\\begin{array}{|c|c|c|} \\multicolumn{3}{}{} \\\\ \\hline \\diamondsuit & \\Delta & \\diamondsuit + \\Delta \\\\ \\hline 36 & 1 & 37 \\\\ \\hline 18 & 2 & 20 \\\\ \\hline 12 & 3 & 15 \\\\ \\hline 9 & 4 & 13 \\\\ \\hline 6 & 6 & 12 \\\\ \\hline \\end{array}\\]\nClearly $37\\rightarrow \\boxed{37}$ is the largest." ]

[ "We have $t_n = \\frac{n (n+1)}{2}$ .\nIf $t_n$ is a perfect square, then it can be written as $\\frac{n (n+1)}{2} = k^2$ ,\nwhere $k$ is a positive integer.\nThus, $n (n+1) = 2 k^2$ . Rearranging, we get $(2n+1)^2-2(2k)^2=1$ , a Pell equation. So $\\frac{2n+1}{2k}$ must be a truncation of the continued fraction for $\\sqrt{2}$\n\\begin{eqnarray*} 1+\\frac12&=&\\frac{2\\cdot1+1}{2\\cdot1}\\\\ 1+\\frac1{2+\\frac1{2+\\frac12}}&=&\\frac{2\\cdot8+1}{2\\cdot6}\\\\ 1+\\frac1{2+\\frac1{2+\\frac1{2+\\frac1{2+\\frac12}}}}&=&\\frac{2\\cdot49+1}{2\\cdot35}\\\\ 1+\\frac1{2+\\frac1{2+\\frac1{2+\\frac1{2+\\frac1{2+\\frac1{2+\\frac12}}}}}}&=&\\frac{2\\cdot288+1}{2\\cdot204} \\end{eqnarray*}\nTherefore, $t_{288} = \\frac{288\\cdot289}{2} = 204^2 = 41616$ , so the answer is $4+1+6+1+6=\\boxed{18}$", "As mentioned above, $t_n = \\frac{n (n+1)}{2}$ . If $t_n$ is a perfect square, one of two things must occur when the fraction is split into a product. Either $\\frac{n}{2}$ and $n+1$ must both be squares, or $n$ and $\\frac{n+1}{2}$ must both be squares, and thus the search for the next perfect square triangular number can be narrowed down by testing values of $n$ that are close to or are perfect squares. After some work, we reach $n = 288$ $1$ less than $289$ , and $t_{288} = \\frac{288\\cdot289}{2} = 144 \\cdot 289 = 41616$ . This product is a perfect square, and thus the sum of the digits of the fourth smallest perfect square triangular number is therefore $4+1+6+1+6=\\boxed{18}$", "We want to find integer $n_i$ and $m_i$ such that $t_{n_i} =\\frac{n_i (n_i + 1)}{2}=m_i^2, n_0 = 0.$\nWe use the formula $\\sqrt{n_{i+1}} = \\sqrt{2n_i} + \\sqrt{n_i + 1}$ and get\n\\[n_1 = ( \\sqrt{2n_0} + \\sqrt{n_0 + 1})^2 = (0+1)^2 = 1,\\] \\[n_2= ( \\sqrt{2n_1} + \\sqrt{n_1 + 1})^2 = ( \\sqrt{2} + \\sqrt{1 + 1})^2 = 8,\\] \\[n_3= ( \\sqrt{2n_2} + \\sqrt{n_2 + 1})^2 = ( \\sqrt{16} + \\sqrt{8 + 1})^2 = 49,\\] \\[n_4 = ( \\sqrt{2n_3} + \\sqrt{n_3 + 1})^2 = ( \\sqrt{98} + \\sqrt{49 + 1})^2 = ((7 + 5)\\sqrt{2})^2 = 288,\\] \\[n_5 = ( \\sqrt{2n_4} + \\sqrt{n_4 + 1})^2 = ( \\sqrt{576} + \\sqrt{289})^2 = (24 + 17)^2 = 1681,\\] \\[n_6 = ( \\sqrt{2n_5} + \\sqrt{n_5 + 1})^2 = ( 41\\sqrt{2} + \\sqrt{1682})^2 = ((41 + 29)\\sqrt{2})^2 = 9800,...\\] Therefore, $t_{n_4} = t_{288} = \\frac{288\\cdot289}{2} = 41616 \\implies 4+1+6+1+6=\\boxed{18}$" ]

[ "According to the problem, we want to find integer $p$ such $\\frac{n(n+1)}{2}=p^2$ , after expanding, we have $n^2+n=2p^2, 4n^2+4n=8p^2, (2n+1)^2-8p^2=1$ , we call $2n+1=q$ , the equation becomes $q^2-8p^2=1$ , obviously $(q,p)=(3,1)$ is the elementary solution for this pell equation, thus the forth smallest solution set $q_4+2\\sqrt{2}p_4=(3+2\\sqrt{2})^4=577+408\\sqrt{2}$ , which indicates $p=204, p^2=41616$ leads to $\\boxed{18}$", "If $n \\choose 2$ is a square, then ${(2n-1)^2 \\choose 2}$ is also a square. We can prove this quite simply:\n\\[{(2n-1)^2 \\choose 2}\\] \\[= \\frac{(2n-1)^2 \\cdot ((2n-1)^2 - 1)}{2}\\] \\[= \\frac{(2n-1)^2 \\cdot (2n \\cdot (2n - 2))}{2}\\] \\[= (2n-1)^2 \\cdot 4{n \\choose 2}.\\]\nTherefore, ${(2 \\cdot 9 - 1)^2 \\choose 2}$ is a square. Note that $T_n = {n+1 \\choose 2}$ . We can easily check all smaller possibilities using a bit of casework, and they don't work. Our solution is thus ${289 \\choose 2} = 204^2 = 41616$ , and so the answer is $\\boxed{18}$" ]

[ "The numerator is $(3!)! = 6!$\nThe denominator is $3! = 6$\nUsing the property that $6! = 6 \\cdot 5!$ in the numerator, the sixes cancel, leaving $5! = 120$ , which is answer $\\boxed{120}$" ]

[ "We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this: \\[\\frac{10^7}{5 \\times 10^4} = \\frac{10^3}{5}\\]\nWe know that $10^3 = 10 \\times 10 \\times 10$ , so\n\\begin{align*} \\frac{10^3}{5} &= \\frac{10\\times 10\\times 10}{5} \\\\ &= 2\\times 10\\times 10 \\\\ &= 200 \\\\ \\end{align*}\nSo the answer is $\\boxed{200}$" ]

[ "\\begin{align*} \\frac{16+8}{4-2} &= \\frac{24}{2} \\\\ &= 12\\rightarrow \\boxed{12}" ]

[ "Each of the fractions simplify to $\\frac{1}{10}$ , so this sum is \\begin{align*} \\frac{1}{10}+\\frac{1}{10}+\\frac{1}{10} &= \\frac{3}{10} \\\\ &= .3 \\rightarrow \\boxed{.3}" ]

[ "\\begin{align*} \\frac{2}{10}+\\frac{4}{100}+\\frac{6}{1000} &= \\frac{200}{1000}+\\frac{40}{1000}+\\frac{6}{1000} \\\\ &= \\frac{246}{1000} \\\\ &= .246 \\rightarrow \\boxed{.246}" ]

[ "$\\frac{2}{25}=\\frac{2\\cdot 4}{25\\cdot 4} = \\frac{8}{100} = 0.08\\rightarrow \\boxed{.08}$" ]

[ "First, note that we can pull the exponents out of every factor, since they are all squared. This results in $\\left(\\frac{(x+1)(x^{2}-x+1)}{x^{3}+1}\\right)^{4}\\cdot\\left(\\frac{(x-1)(x^{2}+x+1)}{x^{3}-1}\\right)^{4}$ Now, multiplying the numerators together gives $\\left(\\frac{x^3+1}{x^3+1}\\right)^{4}\\cdot\\left(\\frac{x^3-1}{x^3-1}\\right)^{4}$ ,\nwhich simplifies to $\\boxed{1}$" ]

[ "$\\log p+\\log q=\\log(p+q)\\implies \\log pq=\\log(p+q)\\implies pq=p+q\\implies \\boxed{1}$" ]

[ "$x$ is at most $999$ , so $x+32$ is at most $1031$ . The minimum value of $x+32$ is $1000$ . However, the only palindrome between $1000$ and $1032$ is $1001$ , which means that $x+32$ must be $1001$\nIt follows that $x$ is $969$ , so the sum of the digits is $\\boxed{24}$", "For $x+32$ to be a four-digit number, $x$ is in between $968$ and $999$ . The palindromes in this range are $969$ $979$ $989$ , and $999$ , so the sum of the digits of $x$ can be $24$ $25$ $26$ , or $27$ . Only $\\boxed{24}$ is an option, and upon checking, $x+32=1001$ is indeed a palindrome.", "Since we know $x+32$ to be $1 a a 1$ and the only palindrome that works is $0 = a$ , that means $x+32 = 1001$ , and so $x = 1001 - 32 = 969$ . So, $9$ $6$ $9$ $\\boxed{24}$ .\n~songmath20" ]

[ "$x$ is at most $999$ , so $x+32$ is at most $1031$ . The minimum value of $x+32$ is $1000$ . However, the only palindrome between $1000$ and $1032$ is $1001$ , which means that $x+32$ must be $1001$\nIt follows that $x$ is $969$ , so the sum of the digits is $\\boxed{24}$", "For $x+32$ to be a four-digit number, $x$ is in between $968$ and $999$ . The palindromes in this range are $969$ $979$ $989$ , and $999$ , so the sum of the digits of $x$ can be $24$ $25$ $26$ , or $27$ . Only $\\boxed{24}$ is an option, and upon checking, $x+32=1001$ is indeed a palindrome.", "Since we know $x+32$ to be $1 a a 1$ and the only palindrome that works is $0 = a$ , that means $x+32 = 1001$ , and so $x = 1001 - 32 = 969$ . So, $9$ $6$ $9$ $\\boxed{24}$ .\n~songmath20" ]

[ "Since $\\angle C = 90^{\\circ}$ and $\\angle A = 20^{\\circ}$ , we have $\\angle ABC = 70^{\\circ}$ . Thus $\\angle DBC = 35^{\\circ}$ . It follows that $\\angle BDC = 90^{\\circ} - 35^{\\circ} = 55^{\\circ}$ , which is $\\boxed{55}$" ]

[ "Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end.\nWe take casework:\nCase 1: 3 lines :\nIn this case, the lines would need to be $2$ of one shape and $1$ of another, so there are $\\frac{3!}{2} = 3$ ways to arrange the lines and $2$ ways to pick which shape has only one line. In total, this is $3\\cdot 2 = 6.$\nCase 2: 2 lines :\nIn this case, the lines would be one line of triangles, one line of circles, and the last one can be anything that includes both shapes. There are $3! = 6$ ways to arrange the lines and $2^3-2 = 6$ ways to choose the last line. (We subtract $2$ from the last line because one arrangement of the last line is all triangles and the other arrangement of the last line is all circles, which causes Case 2 to overlap with Case 1 and further complicating the solution.) In total, this is $6\\cdot 6 = 36.$\nFinally, we add and multiply: $2(36+6)=2(42)=\\boxed{84}$", "We will only consider cases where the three identical symbols are the same column, but at the end we shall double our answer as the same holds true for rows. There are $3$ ways to choose a column with all $\\bigcirc$ 's and $2$ ways to choose a column with all $\\triangle$ 's. The third column can be filled in $2^3=8$ ways. Therefore, we have a total of $3\\cdot2\\cdot8=48$ cases. However, we overcounted the cases with $2$ complete columns of with one symbol and $1$ complete column with another symbol. This happens in $2\\cdot3=6$ cases. $48-6=42$ . However, we have to remember to double our answer, giving us $\\boxed{84}$ ways to complete the grid." ]

[ "The best way to solve this problem is to find patterns and to utilize them to our advantage. For example, we can't really do anything without knowing how many squares there are in the 37th row. But who wants to continue the diagram for 37 rows? And what if the problem said 100,000th row? It'll still be possible - but not if your method is to continue the diagram...\nSo hopefully there's a pattern. We find a pattern by noticing what is changing from row 1 to row 2. Basically, for the next row, we are just adding 2 squares (1 on each side) to the number of squares we had in the previous row. So each time we're adding 2. So how can we find $N$ , if $N$ is the number of squares in the $a^\\text{th}$ row of this diagram? We can't just say that $N = 1 + 2a$ , because it doesn't work for the first row. But since 1 is the first term, we have to EXCLUDE the first term, meaning that we must subtract 1 from a. Thus, $N = 1 + 2\\times(a - 1) = 2a - 1$ . So in the 37th row we will have $2 \\times 37 - 1 = 74 - 1 = 73$\nYou may now be thinking - aha, we're finished. But we're only half finished. We still need to find how many black squares there are in these 73 squares. Well let's see - they alternate white-black-white-black... but we can't divide by two - there aren't exactly as many white squares as black squares... there's always 1 more white square... aha! If we subtract 1 from the number of squares (1 white square), we will have exactly 2 times the number of black squares.\nThus, the number of black squares is $\\frac{73 - 1}{2} = \\frac{72}{2} = 36$\n36 is $\\boxed{36}$", "Note that each row adds one black and one white square to the end of the previous one, and then shifts the new row over a little. Since the first row had no black squares, the number of black squares in any row is one less than the row number.\nNow that this has been established, we just have $37-1=36\\rightarrow \\boxed{36}$", "We can easily spot the pattern. In the first row, there are $0$ black squares. In the second row, there is $1$ black square. So the pattern is that in order to find the number of black squares in a row, you need to subtract $1$ from the row number. Therefore, the number of black squares in the $37\\text{th}$ row we subtract 1. $37-1=36\\rightarrow \\boxed{36}$" ]

[ "Clearly we have people moving at speeds of $6,8$ and $10$ feet/second. Notice that out of the three people, Cy is at the largest disadvantage to begin with and since all speeds are close, it is hardest for him to catch up. Furthermore, Bob is clearly the farthest along. Thus it is reasonable to assume that there is some point when Al is halfway between Cy and Bob. At this time $s$ , we have that\n$\\frac{8(s-4)+10(s-2)}{2}=6s$ After solving, $s=\\frac{26}{3}$ . At this time, Al has traveled $6\\cdot\\frac{26}{3}=52$ feet.\nWe could easily check that Al is in the middle by trying all three possible cases. $\\frac{6s + 8(s-4)}{2} = 10(s-2)$ yields that $s = \\frac 43$ , which can be disregarded since both Bob and Cy hadn't started yet. $\\frac{6s + 10(s-2)}{2} = 8(s-4)$ yields that $-10=-32$ , a contradiction. Thus, the answer is $\\boxed{52}$" ]

[ "As stated in the solution, there are $6$ points along the line $y=-x$ that constitute a sort of \"boundary\". Once the ant reaches one of these $6$ points, it is exactly halfway to $(4, 4)$ . Also notice that the ant will only cross one of the $6$ points during any one of its paths. Therefore we can divide the problem into $3$ cases, focusing on $1$ quadrant; then multiplying the sum by $2$ to get the total (because there is symmetry).\nFor the sake of this explanation, we will focus on the fourth quadrant (it really doesn't matter which quadrant because, again the layout is symmetrical) The three cases are when the ant crosses $(4, -4), (3, -3),$ and $(2, -2)$\nFor each of the cases, notice that the path the ant takes can be expressed as a sequence of steps, such as:\nright, right, up, right,..., etc.\nAlso notice that there are always $8$ steps per sequence (if there were more or less steps, the ant would be breaking the conditions given in the problem). This means we can figure out the number of ways to get to a point based on the particular sequence of steps that denote each path. For example, there is only one way for the ant to pass through $(4, -4);$ it MUST keep traveling right for all $8$ steps. This seems fairly obvious; however, notice that this is equivalent to\n$\\binom{8}{0}$\nNow we consider the number of ways to get from $(4, -4)$ to $(4, 4)$ . by symmetry, there is only $1$ such way. So the number of paths containing $(4, -4)$ is $1^2,$ or $1$\nMoving on to the next case, we see that the ant MUST travel right exactly $7$ times and up exactly once. So each sequence of this type will have $7$ \"right\"s and $1$ \"up\". So, the total number of paths that go through $(3, -3)$ is equivalent to the number of ways to arrange $1$ \"up\" into $8$ spots. This is\n$\\binom{8}{1} = 8$\nSimilarly to the first case, we square this value to account for the second half of the journey: $8^2 = 64$\nFinally, for the third case (ant passes through $(2, -2)$ ) the ant must travel right exactly $6$ times and up exactly $2$ times. This is equivalent to the number of ways to arrange $2$ \"ups\" in a sequence of $8$ movements, or\n$\\binom{8}{2} = 28$\nAgain, we square $28$ $28^2 = 784$ . Adding up all of these cases we get\n$1+64+784 = 849$\npaths through the fourth quadrant. Doubling this number to account for the paths through the second quadrant, we have\n$849*2=1698 \\Rightarrow \\boxed{1698}$" ]

[ "Each path must go through either the second or the fourth quadrant.\nEach path that goes through the second quadrant must pass through exactly one of the points $(-4,4)$ $(-3,3)$ , and $(-2,2)$\nThere is $1$ path of the first kind, ${8\\choose 1}^2=64$ paths of the second kind, and ${8\\choose 2}^2=28^2=784$ paths of the third type. \nEach path that goes through the fourth quadrant must pass through exactly one of the points $(4,-4)$ $(3,-3)$ , and $(2,-2)$ .\nAgain, there is $1$ path of the first kind, ${8\\choose 1}^2=64$ paths of the second kind, and ${8\\choose 2}^2=28^2=784$ paths of the third type.\nHence the total number of paths is $2(1+64+784) = \\boxed{1698}$" ]

[ "In a 3x3x3 cube, there are $8$ cubes with three faces showing, $12$ with two faces showing and $6$ with one face showing. The smallest sum with three faces showing is $1+2+3=6$ , with two faces showing is $1+2=3$ , and with one face showing is $1$ . Hence, the smallest possible sum is $8(6)+12(3)+6(1)=48+36+6=90$ . Our answer is thus $\\boxed{90}$" ]

[ "The total number of balls in the bag must be $4\\times 6=24$ , so there are $24-6=18$ green balls $\\rightarrow \\boxed{18}$" ]

[ "Each bounce is $2/3$ times the height of the previous bounce. The first bounce reaches $2$ meters, the second $4/3$ , the third $8/9$ , the fourth $16/27$ , and the fifth $32/81$ . Half of $81$ is $40.5$ , so the ball does not reach the required height on bounce $\\boxed{5}$" ]

[ "The borrower pays $\\textdollar{120}$ in a single year for a loan of $\\textdollar{114}$ . This means that the bank charges an interest of $\\textdollar{6}$ for a loan of $\\textdollar{114}$ over a single year, so that the annual interest rate is $100*\\frac{\\textdollar{6}}{\\textdollar{114}} = \\frac{100}{19} \\approx 5$ percentage points. Therefore, our answer is $\\boxed{5}$ and we are done." ]

[ "What we want to find is the number of hamburgers sold in the winter. Since we don't know what it is, let's call it $x$ . From the graph, we know that in Spring, 4.5 million hamburgers were sold, in the Summer was 5 million and in the Fall was 4 million. We know that the number of hamburgers sold in Fall is exactly $\\frac{1}{4}$ of the total number of hamburgers sold, so we can say that...\n$4 \\times \\text{Fall} = \\text{Spring} + \\text{Winter} + \\text{Fall} + \\text{Summer}$\n$4 \\times 4 = 4.5 + 4 + x + 5$\n$16 = x + 13.5$\n$2.5 = x$\nThe answer is 2.5, or $\\boxed{2.5}$" ]

[ "The walls are $13*5=65$ and $10*5=50$ in area, and the ceiling has an area of $10*13=130$\n$((65+50)2)2+130=590 \\Rightarrow \\boxed{590}$" ]

[ "To be a three digit number in base-10:\n$10^{2} \\leq n \\leq 10^{3}-1$\n$100 \\leq n \\leq 999$\nThus there are $900$ three-digit numbers in base-10\nTo be a three-digit number in base-9:\n$9^{2} \\leq n \\leq 9^{3}-1$\n$81 \\leq n \\leq 728$\nTo be a three-digit number in base-11:\n$11^{2} \\leq n \\leq 11^{3}-1$\n$121 \\leq n \\leq 1330$\nSo, $121 \\leq n \\leq 728$\nThus, there are $608$ base-10 three-digit numbers that are three digit numbers in base-9 and base-11.\nTherefore the desired probability is $\\frac{608}{900}\\approx 0.7 \\Rightarrow\\boxed{0.7}$" ]

[ "On one team they play $3N$ games in their division and $4M$ games in the other. This gives $3N+4M=76$\nSince $M>4$ we start by trying M=5. This doesn't work because $56$ is not divisible by $3$\nNext, $M=6$ does not work because $52$ is not divisible by $3$\nWe try $M=7$ does work by giving $N=16$ $~M=7$ and thus $3\\times 16=\\boxed{48}$ games in their division.", "$76=3N+4M > 10M$ , giving $M \\le 7$ .\nSince $M>4$ , we have $M=5,6,7$ .\nSince $4M$ is $1$ $\\pmod{3}$ , we must have $M$ equal to $1$ $\\pmod{3}$ , so $M=7$\nThis gives $3N=48$ , as desired. The answer is $\\boxed{48}$", "Notice that each team plays $N$ games against each of the three other teams in its division. So that's $3N$\nSince each team plays $M$ games against each of the four other teams in the other division, that's $4M$\nSo $3N+4M=76$ , with $M>4, N>2M$\nLet's start by solving this Diophantine equation. In other words, $N=\\frac{76-4M}{3}$\nSo $76-4M\\equiv0 \\pmod{3}$ (remember: $M$ must be divisible by 3 for $N$ to be an integer!). Therefore, after reducing $76$ to $1$ and $-4M$ to $2M$ (we are doing things in $\\pmod{3}$ ), we find that $M\\equiv1 \\pmod{3}$\nSince $M>4$ , so the minimum possible value of $M$ is $7$ . However, remember that $N>2M$ ! To find the greatest possible value of M, we assume that $N=2M$ and that is the upper limit of $M$ (excluding that value because $N>2M$ ). Plugging $N=2M$ in, $10M=76$ . So $M<7.6$ . Since you can't have $7.6$ games, we know that we can only check $M=7$ since we know that since $M>4, M<7.6, M\\equiv1 \\pmod{3}$ . After checking $M=7$ , we find that it works.\nSo $M=7, N=16$ . So each team plays 16 games against each team in its division. Since they are asking for games in it division, which equals $3n = 48$ . Select $\\boxed{48}$" ]

[ "We graph the $10$ shots on a grid. Suppose that a made shot is represented by a step of $(0,1)$ , and a missed shot is represented by $(1,0)$ . Then the basketball player's shots can be represented by the number of paths from $(0,0)$ to $(6,4)$ that always stay below the line $y=\\frac{2x}{3}$ . We can find the number of such paths using a Pascal's Triangle type method below, computing the number of paths to each point that only move right and up. Therefore, there are $23$ ways to shoot $4$ makes and $6$ misses under the given conditions. The probability of each possible sequence occurring is $(.4)^4(.6)^6$ . Hence the desired probability is \\[\\frac{23\\cdot 2^4\\cdot 3^6}{5^{10}},\\] and the answer is $(23+2+3+5)(4+6+10)=\\boxed{660}$", "The first restriction is that $a_{10} = .4$ , meaning that the player gets exactly 4 out of 10 baskets. The second restriction is $a_n\\le.4$ . This means that the player may never have a shooting average over 40%. Thus, the first and second shots must fail, since $\\frac{1}{1}$ and $\\frac{1}{2}$ are both over $.4$ , but the player may make the third basket, since $\\frac{1}{3} \\le .4$ In other words, the earliest the first basket may be made is attempt 3. Using similar reasoning, the earliest the second basket may be made is attempt 5, the earliest the third basket may be made is attempt 8, and the earliest the fourth basket may be made is attempt 10.\nUsing X to represent a basket and O to represent a failure, this 'earliest' solution may be represented as:\nOOXOXOOXOX\nTo simplify counting, note that the first, second, and tenth shots are predetermined. The first two shots must fail, and the last shot must succeed. Thus, only slots 3-9 need to be counted, and can be abbreviated as follows:\nXOXOOXO\nThe problem may be separated into five cases, since the first shot may be made on attempt 3, 4, 5, 6, or 7. The easiest way to count the problem is to remember that each X may slide to the right, but NOT to the left.\nFirst shot made on attempt 3:\nXOXOOXO\nXOXOOOX\nXOOXOXO\nXOOXOOX\nXOOOXXO\nXOOOXOX\nXOOOOXX\nTotal - 7\nFirst shot made on attempt 4:\nNote that all that needs to be done is change each line in the prior case from starting with \"XO.....\" to \"OX.....\".\nTotal - 7\nFirst shot made on attempt 5:\nOOXXOXO\nOOXXOOX\nOOXOXXO\nOOXOXOX\nOOXOOXX\nTotal - 5\nFirst shot made on attempt 6:\nOOOXXXO\nOOOXXOX\nOOOXOXX\nTotal - 3\nFirst shot made on attempt 7:\nOOOOXXX\nTotal - 1\nThe total number of ways the player may satisfy the requirements is $7+7+5+3+1=23$\nThe chance of hitting any individual combination (say, for example, OOOOOOXXXX) is $\\left(\\frac{3}{5}\\right)^6\\left(\\frac{2}{5}\\right)^4$\nThus, the chance of hitting any of these 23 combinations is $23\\left(\\frac{3}{5}\\right)^6\\left(\\frac{2}{5}\\right)^4 = \\frac{23\\cdot3^6\\cdot2^4}{5^{10}}$\nThus, the final answer is $(23+3+2+5)(6+4+10)=\\boxed{660}$", "Note $a_{10}=.4$ . Therefore the player made 4 shots out of 10. He must make the 10th shot, because if he doesn't, then $a_9=\\frac{4}{9}>.4$ . Since $a_n\\leq .4$ for all $n$ less than 11, we know that $a_1=a_2=0$ . Now we must look at the 3rd through 9th shot.\nNow let's take a look at those un-determined shots. Let's put them into groups: the 3rd, 4th, and 5th shots in group A, and the 6th, 7th, 8th, and 9th shots in group B. The total number of shots made in groups A and B must be 3, since the player makes the 10th shot. We cannot have all three shots made in group A, since $a_5\\leq .4$ . Therefore we can have two shots made, one shot made, or no shots made in group A.\nCase 1: Group A contains no made shots.\nThe number of ways this can happen in group A is 1. Now we must arrange the shots in group B accordingly. There are four ways to arrange them total, and all of them work. There are $\\textbf{4}$ possibilities here.\nCase 2: Group A contains one made shot.\nThe number of ways this could happen in group A is 3. Now we must arrange the shots in group B accordingly. There are six ways to arrange them total, but the arrangement \"hit hit miss miss\" fails, because that would mean $a_7=\\frac{3}{7}>.4$ . All the rest work. Therefore there are $3\\cdot5=\\textbf{15}$ possibilities here.\nCase 3: Group A contains two made shots.\nThe number of ways this could happen in group A is 2 (hit hit miss doesn't work but the rest do). Now we must arrange the shots in group B accordingly. Note hit miss miss miss and miss hit miss miss fail. Therefore there are only 2 ways to do this, and there are $2\\cdot 2=\\textbf{4}$ total possibilities for this case.\nTaking all these cases into account, we find that there are $4+15+4=23$ ways to have $a_{10} = .4$ and $a_n\\leq .4$ . Each of these has a probability of $.4^4\\cdot.6^6=\\frac{2^4\\cdot 3^6}{5^{10}}$ . Therefore the probability that we have $a_{10} = .4$ and $a_n\\leq .4$ is $\\frac{23\\cdot2^4\\cdot3^6}{5^{10}}$ . Now we are asked to find the product of the sum of the primes and the sum of the exponents, which is $(23+2+3+5)(4+6+10)=33\\cdot20=\\boxed{660}$" ]

[ "The number of points could have been 10, 11, 12, 13, 14, or 15. This is because the minimum is 2*5=10 and the maximum is 3*5=15. The numbers between 10 and 15 are possible as well. Thus, the answer is $\\boxed{6}$" ]

[ "Let $x$ be the number of two point shots attempted and $y$ the number of three point shots attempted. Because each two point shot is worth two points and the team made 50% and each three point shot is worth 3 points and the team made 40%, $0.5(2x)+0.4(3y)=54$ or $x+1.2y=54$ . Because the team attempted 50% more two point shots then threes, $x=1.5y$ . Substituting $1.5y$ for $x$ in the first equation gives $1.5y+1.2y=54$ , which equals $2.7y=54$ so $y=$ $\\boxed{20}$" ]

[ "At each point of reflection, we pretend instead that the light continues to travel straight. Note that after $k$ reflections (excluding the first one at $C$ ) the extended line will form an angle $k \\beta$ at point $B$ . For the $k$ th reflection to be just inside or at point $B$ , we must have $k\\beta \\le 180 - 2\\alpha \\Longrightarrow k \\le \\frac{180 - 2\\alpha}{\\beta} = 70.27$ . Thus, our answer is, including the first intersection, $\\left\\lfloor \\frac{180 - 2\\alpha}{\\beta} \\right\\rfloor + 1 = \\boxed{071}$" ]

[ "Let $x=e^{i\\pi/6}$ , a $30^\\circ$ counterclockwise rotation centered at the origin. Notice that $P_k$ on the complex plane is:\n\\[1+2x+3x^2+\\cdots+(k+1)x^k\\]\nWe need to find the magnitude of $P_{2015}$ on the complex plane. This is an arithmetico-geometric series\n\\begin{align*} S &=1+2x+3x^2+\\cdots+2015x^{2014} \\\\ xS &=x+2x^2+3x^3+\\cdots+2015x^{2015} \\\\ (1-x)S &=1+x+x^2+\\cdots+x^{2014}-2015x^{2015} \\\\ S &= \\frac{1-x^{2015}}{(1-x)^2}-\\frac{2015x^{2015}}{1-x} \\end{align*}\nWe want to find $|S|$ . First, note that $x^{2015}=x^{11}=x^{-1}$ because $x^{12}=1$ . Therefore\n\\[S =\\frac{1-\\frac{1}{x}}{(1-x)^2}-\\frac{2015}{x(1-x)}=-\\frac{1}{x(1-x)}-\\frac{2015}{x(1-x)}=-\\frac{2016}{x(1-x)}.\\]\nHence, since $|x|=1$ , we have $|S| = \\frac{2016}{|1-x|}.$\nNow we just have to find $|1-x|$ . This can just be computed directly:\n\\[1-x=1-\\frac{\\sqrt{3}}{2}-\\frac{1}{2}i\\]\n\\[|1-x|^2=\\left(1-\\sqrt{3}+\\frac{3}{4}\\right)+\\frac{1}{4}=2-\\sqrt{3}={\\left(\\frac{\\sqrt{6}-\\sqrt{2}}{2}\\right)}^2\\]\n\\[|1-x|=\\frac{\\sqrt{6}-\\sqrt{2}}{2}\\]\nTherefore $|S|=2016\\cdot\\frac{2}{\\sqrt{6}-\\sqrt{2}}=2016\\left(\\frac{\\sqrt{6}+\\sqrt{2}}{2}\\right)=1008\\sqrt{2}+1008 \\sqrt{6}$\nThus the answer is $1008+2+1008+6=\\boxed{2024}$", "Here is an alternate solution that does not use complex numbers:\nThe distance from $P_{2015}$ to $P_0$ can be evaluated using the Pythagorean theorem . Assuming $P_0$ lies at the origin, we can calculate the distance the bee traveled to $P_{2015}$ by evaluating the distance the bee traveled in the x-direction and the y-direction. Let's start by summing each movement:\n$x=1\\cos{0}+2\\cos{30}+3\\cos{60}+\\cdots+2014\\cos{270}+2015\\cos{300}$\nA movement of $p$ units at $q$ degrees is the same thing as a movement of $-p$ units at $q-180$ degrees, so we can adjust all the cosines with arguments greater than $180$ as follows:\n$x=1\\cos{0}+2\\cos{30}+3\\cos{60}+4\\cos{90}+5\\cos{120}+6\\cos{150}-7\\cos{0}-8\\cos{30}-\\cdots-2015\\cos{120}$\nGrouping terms with like-cosines and factoring out the cosines:\n$x=(1-7+13-\\cdots+2005-2011)\\cos{0}+\\cdots+(6-12+18-\\cdots-2004+2010)\\cos{150}$\nEach sum in the parentheses has $336$ terms (except the very last one, which has $335$ ). By pairing each term, we see there are $\\frac{336}{2}$ pairs of $-6$ . Therefore, each sum equals $168\\cdot-6=-1008$ except the very last sum, which has $167$ pairs of $-6$ plus an extra 2010 and equals $167\\cdot-6+2010=1008$ . Plugging in these values:\n$x=-1008\\cos{0}-1008\\cos{30}-1008\\cos{60}-1008\\cos{90}-1008\\cos{120}+1008\\cos{150}$ $x=1008(-1-\\frac{\\sqrt{3}}{2}-\\frac{1}{2}-0+\\frac{1}{2}-\\frac{\\sqrt{3}}{2})=-1008(1+\\sqrt{3})$\nWe can find how far the bee traveled in the y-direction using the same logic as above, we arrive at the sum:\n$y=-1008\\sin{0}-1008\\sin{30}-1008\\sin{60}-1008\\sin{90}-1008\\sin{120}+1008\\sin{150}$\n$y=1008(0-\\frac{1}{2}-\\frac{\\sqrt{3}}{2}-1-\\frac{\\sqrt{3}}{2}+\\frac{1}{2})=-1008(1+\\sqrt{3})$\nFinally, we use the Pythagorean to find the distance from $P_0$ . This distance is given by:\n$\\sqrt{x^2+y^2}=\\sqrt{(-1008(1+\\sqrt{3}))^2+(-1008(1+\\sqrt{3}))^2}=\\sqrt{2\\cdot1008^2\\cdot(1+\\sqrt{3})^2}=1008(1+\\sqrt{3})\\sqrt{2}=1008\\sqrt{2}+1008\\sqrt{6}$ , so the answer is $1008+2+1008+6=\\boxed{2024}$", "Suppose that the bee makes a move of distance $i$ . After $6$ turns it will be facing the opposite direction and move $i+6$ units. Combining these opposite movements gives a total movement of $-6$ units in the original direction. This means that every $12$ moves, the bee will move $-6$ units in each direction of $0^\\circ, 30^\\circ, 60^\\circ, 90^\\circ, 120^\\circ, 150^\\circ$\nWe want to find the displacement vector for every $12$ moves. Factoring out the $-6$ for now (which flips the direction), we draw a quick diagram of one unit in each direction. Using the 30-60-90 triangles, it is clear that the displacement vector (factoring back in the $-6$ ) is $-6\\left\\langle 1, 2+\\sqrt{3}\\right\\rangle$\nTo compute the distance to $P_{2015}$ , we can compute the position of $P_{2016}$ (a multiple of $12$ moves) and then subtract the vector from $P_{2015}$ to $P_{2016}$\nThe bee reaches $P_{2016}$ after $\\frac{2016}{12} = 168$ sets of $12$ moves, so the total displacement vector to $P_{2016}$ is $168(-6)\\left\\langle 1, 2+\\sqrt{3}\\right\\rangle = \\left\\langle -1008, -2006-1008\\sqrt{3}\\right\\rangle$\nThe bee moves at an angle of $-30^\\circ$ from $P_{2015}$ to $P_{2016}$ , so subtracting it means moving an angle of $150^\\circ$ . Since the vector is $2016$ units long, by a 30-60-90 triangle, it is $\\left\\langle -1008\\sqrt{3}, 1008\\right\\rangle$\nTherefore the total displacement vector to $P_{2015}$ is $\\left\\langle -1008, -2006-1008\\sqrt{3}\\right\\rangle + \\left\\langle -1008\\sqrt{3}, 1008\\right\\rangle = \\left\\langle -1008-1008\\sqrt{3}, -1008-1008\\sqrt{3}\\right\\rangle$ . The displacement is thus $\\sqrt{2}\\left|-1008-1008\\sqrt{3}\\right| = 1008\\sqrt{2}+1008\\sqrt{6} \\implies 1008+2+1008+6 = \\boxed{2024}$", "Let $P_0$ be the origin. East would be the real axis in the positive direction. Then we can assign each $P_n$ a complex value. The displacement would then be the magnitude of the complex number.\nNotice that after the $n$ th move the value of $P_n$ is $P_{n-1}+ne^{\\frac{i(n-1)\\pi}{6}}$ . Also notice that after six moves the bee is facing in the opposite direction. And because we have found a recursion, we can add these up.\nThen we have \\[P_n=e^{0}+2e^{\\frac{i\\pi}{6}}+\\cdots+2015e^{\\frac{2014i\\pi}{6}},\\] and this becomes \\[(1-7+13-\\cdots-2011)e^0+(2-8+\\cdots-2012)e^{\\frac{i\\pi}{6}}+\\cdots+(6-12+\\cdots+2010)e^{\\frac{5i\\pi}{6}}.\\]\nSimplifying, we have \\[-6\\cdot168-6\\cdot168e^{\\frac{i\\pi}{6}}-6\\cdot168e^{\\frac{2i\\pi}{6}}-6\\cdot168e^{\\frac{3i\\pi}{6}}-6\\cdot168e^{\\frac{4i\\pi}{6}}-(2010-6\\cdot167)e^{\\frac{5i\\pi}{6}},\\] which eventually simplifies to \\[-1008-(2+\\sqrt3)1008i,\\] and this is a $15-75-90$ triangle which has ratios of $1:(2+\\sqrt3):(\\sqrt2+\\sqrt6)$ so the magnitude is $1008\\sqrt2+1008\\sqrt6$ and the answer is $1008+2+1008+6=\\boxed{2024}$", "After each 12 moves, the bee will be facing the same direction as it started. Let $P_0$ be the origin and let $P_n$ (with $n$ divisible by $12$ ) be $(x,y)$ . We now notice that each of the move pairs with lengths $n+1$ $n+7$ $n+2$ $n+8$ $n+3$ $n+9$ $n+4$ $n+10$ $n+5$ $n+11$ $n+6$ $n+12$ will move the bee 6 units in the directions corresponding to the moves with lengths $n+7$ $n+8$ $n+9$ $n+10$ $n+11$ , and $n+12$ . This equates to moving the bee from $(x,y)$ to $(x-6,y-12-6\\sqrt{3})$ , a move that repeats every 12 moves. Since $\\frac{2016}{12} = 168$ , we have that $P_(2016) = (-1008, -2016-1008\\sqrt{3})$ . It follows that $P_(2015) = (-1008-1008\\sqrt{3}, -1008-1008\\sqrt{3})$ so the distance to $P_0$ is $1008(\\sqrt{2} + \\sqrt{6})$ so the answer is $1008 + 2 + 1008 + 6 = \\boxed{2024}$" ]

[ "We first notice that if the bee is turning 30 degrees each turn, it will take 12 turns to be looking in the same direction when the bee initially left. This means we simply need to answer the question; how far will the bee be when the bee is facing in the same direction?\nFirst we use the fact that after 3 turns, the bee will be facing in a direction perpendicular to the the initial direction. From here we can draw a perpendicular from $P_2$ to the line $\\overline{P_0P_1}$ intersecting a point $C_0$ . We will also place the point $C_1$ at the intersection of $\\overline{P_0P_1}$ and $\\overline{P_3P_4}$ . In addition, the point $C_2$ is placed at the perpendicular dropped from $P_2$ to the line $\\overline{P_3C_1}$ . We will also set the distance $\\overline{P_0P_1} = n$ and thus $\\overline{P_1P_2} = n+1$ . With this perpendicular we see that the triangle $\\triangle{P_1P_2C_0}$ is a 30-60-90 triangle. This means that the length $\\overline{P_1C_0} = \\frac{(n+1)\\sqrt{3}}{2}$ and the length $\\overline{C_1C_2} = \\frac{n+1}{2}$ . We can also see that the triangle $\\triangle{P_2C_1P_3}$ is a 30-60-90 triangle and thus $\\overline{C_0C_1} = \\frac{n+2}{2}$ and $\\overline{C_2P_3} = \\frac{(n+2)\\sqrt{3}}{2}$ . Now if we continue this across all $P_i$ and set the point $P_0$ to the coordinates $(0, 0)$ . As you can see, we are inherently putting a “box” around the figure. Doing similar calculations for all four “sides” of this spiral we get that the length\n\\[\\overline{P_0C_1} = n + \\frac{(n+1)\\sqrt{3}}{2} + \\frac{n+2}{2}\\] \\[\\overline{C_1C_4} = \\frac{(n+1)}{2} + \\frac{(n+2)\\sqrt{3}}{2} + (n+3) + \\frac{(n+4)\\sqrt{3}}{2} + \\frac{n+5}{2}\\] \\[\\overline{C_4C_7} = \\frac{(n+4)}{2} + \\frac{(n+5)\\sqrt{3}}{2} + (n+6) + \\frac{(n+7)\\sqrt{3}}{2} + \\frac{n+8}{2}\\] \\[\\overline{C_7C_{10}} = \\frac{(n+7)}{2} + \\frac{(n+8)\\sqrt{3}}{2} + (n+9) + \\frac{(n+10)\\sqrt{3}}{2} + \\frac{n+11}{2}\\] , and finally \\[\\overline{C_{10}P_{12}} = \\frac{(n+10)}{2} + \\frac{(n+11)\\sqrt{3}}{2}\\]\nHere the point $C_4$ is defined as the intersection of lines $\\overline{P_3P_4}$ and $\\overline{P_6P_7}$ . The point $C_7$ is defined as the intersection of lines $\\overline{P_6P_7}$ and $\\overline{P_9P_{10}}$ . Finally, the point $C_10$ is defined as the intersection of lines $\\overline{P_{9}P_{10}}$ and $\\overline{P_{12}P_{13}}$ . Note that our spiral stops at $P_{12}$ before the next spiral starts. Calculating the offset from the x and the y direction, we see that the offset, or the new point $P_{12}$ , is $({-6}, {-6}-12 \\sqrt{3})$ . This is an interesting property that the points’ coordinate changes by a constant offset no matter what $n$ is. Since the new point’s subscript changes by 12 each time and we see that 2016 is divisible by 12, the point $P_{2016} = ({-168} \\cdot {6}, {168} \\cdot ({-6} \\sqrt{3} {-12}))$ . Using similar 30-60-90 triangle properties, we see that $P_{2015} = ({-6} \\cdot 168-1008 \\sqrt{3}, 168({-6} \\sqrt{3} - 12) + 1008)$ . Using the distance formula, the numbers cancel out nicely (1008 is divisible by 168, so take 168 when using the distance formula) and we see that the final answer is $(1008)(1+\\sqrt{3})(\\sqrt{2})$ which gives us a final answer of $\\boxed{2024}$" ]

[ "We see that $a \\, \\diamondsuit \\, a = 1$ , and think of division. Testing, we see that the first condition $a \\, \\diamondsuit \\, (b \\, \\diamondsuit \\, c) = (a \\, \\diamondsuit \\, b) \\cdot c$ is satisfied, because $\\frac{a}{\\frac{b}{c}} = \\frac{a}{b} \\cdot c$ . Therefore, division can be the operation $\\diamondsuit$ . Solving the equation, \\[\\frac{2016}{\\frac{6}{x}} = \\frac{2016}{6} \\cdot x = 336x = 100\\implies x=\\frac{100}{336} = \\frac{25}{84},\\] so the answer is $25 + 84 = \\boxed{109}$", "If the given conditions hold for all nonzero numbers $a, b,$ and $c$\nLet $a=b=c.$ From the first two givens, this implies that\n\\[a\\diamondsuit\\, (a\\diamondsuit\\, {a})=(a\\diamondsuit\\, a)\\cdot{a}.\\]\nFrom $a\\diamondsuit\\,{a}=1,$ this equation simply becomes $a\\diamondsuit\\,{1}=a.$\nLet $c=b.$ Substituting this into the first two conditions, we see that\n\\[a\\diamondsuit\\, (b\\diamondsuit\\, {c})=(a\\diamondsuit\\, {b})\\cdot{c} \\implies a\\diamondsuit\\, (b\\diamondsuit\\, {b})=(a\\diamondsuit\\, {b})\\cdot{b}.\\]\nSubstituting $b\\diamondsuit\\, {b} =1$ , the second equation becomes\n\\[a\\diamondsuit\\, {1}=(a\\diamondsuit\\, {b})\\cdot{b} \\implies a=(a\\diamondsuit\\,{b})\\cdot{b}.\\]\nSince $a, b$ and $c$ are nonzero, we can divide by $b$ which yields,\n\\[\\frac{a}{b}=(a\\diamondsuit\\, {b}).\\]\nNow we can find the value of $x$ straightforwardly:\n\\[\\frac{2016}{(\\frac{6}{x})}=100 \\implies 2016=\\frac{600}{x} \\implies x=\\frac{600}{2016} = \\frac{25}{84}.\\]\nTherefore, $a+b=25+84=\\boxed{109}$", "One way to eliminate the $\\diamondsuit$ in this equation is to make $a = b$ so that $a\\,\\diamondsuit\\, (b\\,\\diamondsuit \\,c) = c$ . In this case, we can make $b = 2016$\n\\[2016 \\,\\diamondsuit\\, (6\\,\\diamondsuit\\, x)=100\\implies (2016\\, \\diamondsuit\\, 6) \\cdot x = 100\\]\nBy multiplying both sides by $\\frac{6}{x}$ , we get:\n\\[(2016\\, \\diamondsuit\\, 6) \\cdot 6 = \\frac{600}{x}\\implies 2016 \\, \\diamondsuit\\, (6\\, \\diamondsuit\\, 6) = \\frac{600}{x}\\]\nBecause $6\\, \\diamondsuit\\, 6 = 2016\\, \\diamondsuit\\, 2016 = 1:$\n\\[2016 \\, \\diamondsuit\\, (2016\\, \\diamondsuit\\, 2016) = \\frac{600}{x}\\implies (2016\\, \\diamondsuit\\, 2016) \\cdot 2016 = \\frac{600}{x}\\implies 2016 = \\frac{600}{x}\\]\nTherefore, $x = \\frac{600}{2016} = \\frac{25}{84}$ , so the answer is $25 + 84 = \\boxed{109.}$", "We can manipulate the given identities to arrive at a conclusion about the binary operator $\\diamondsuit$ . Substituting $b = c$ into the first identity yields \\[( a\\ \\diamondsuit\\ b) \\cdot b = a\\ \\diamondsuit\\ (b\\ \\diamondsuit\\ b) = a\\ \\diamondsuit\\ 1 = a\\ \\diamondsuit\\ ( a\\ \\diamondsuit\\ a) = ( a\\ \\diamondsuit\\ a) \\cdot a = a.\\] Hence, $( a\\ \\diamondsuit\\ b) \\cdot b = a,$ or, dividing both sides of the equation by $b,$ $( a\\ \\diamondsuit\\ b) = \\frac{a}{b}.$\nHence, the given equation becomes $\\frac{2016}{\\frac{6}{x}} = 100$ . Solving yields $x=\\frac{100}{336} = \\frac{25}{84},$ so the answer is $25 + 84 = \\boxed{109.}$", "$2016 \\diamondsuit (6 \\diamondsuit x) = (2016 \\diamondsuit 6) \\cdot x = 100$\n$2016 \\diamondsuit (2016 \\diamondsuit 1) = (2016 \\diamondsuit 2016) \\cdot 1 = 1 \\cdot 1 = 1$\n$2016 \\diamondsuit 2016 = 1$ $2016 \\diamondsuit (2016 \\diamondsuit 1) = 1$ , so $2016 \\diamondsuit 1 = 2016$\n$2016 \\diamondsuit 1 = (2016 \\diamondsuit 6) \\cdot 6$ $2016 \\diamondsuit 6 = \\frac{2016 \\diamondsuit 1}{6} = 336$\n$x = \\frac{100}{2016 \\diamondsuit 6} = \\frac{100}{336} = \\frac{25}{84}$ $24 + 85 = \\boxed{109}$", "Notice that $2016 \\diamondsuit (6 \\diamondsuit 6)=(2016 \\diamondsuit 6) \\cdot 6 = 2016$ . Hence, $2016 \\diamondsuit 6 = 336$ . Thus, $2016 \\diamondsuit (6 \\diamondsuit x)=100 \\implies (2016 \\diamondsuit 6) \\cdot x = 100 \\implies 336x=100 \\implies x=\\frac{25}{84}$ . Therefore, the answer is $\\boxed{109}$" ]

[ "We see that $a \\, \\diamondsuit \\, a = 1$ , and think of division. Testing, we see that the first condition $a \\, \\diamondsuit \\, (b \\, \\diamondsuit \\, c) = (a \\, \\diamondsuit \\, b) \\cdot c$ is satisfied, because $\\frac{a}{\\frac{b}{c}} = \\frac{a}{b} \\cdot c$ . Therefore, division can be the operation $\\diamondsuit$ . Solving the equation, \\[\\frac{2016}{\\frac{6}{x}} = \\frac{2016}{6} \\cdot x = 336x = 100\\implies x=\\frac{100}{336} = \\frac{25}{84},\\] so the answer is $25 + 84 = \\boxed{109}$", "If the given conditions hold for all nonzero numbers $a, b,$ and $c$\nLet $a=b=c.$ From the first two givens, this implies that\n\\[a\\diamondsuit\\, (a\\diamondsuit\\, {a})=(a\\diamondsuit\\, a)\\cdot{a}.\\]\nFrom $a\\diamondsuit\\,{a}=1,$ this equation simply becomes $a\\diamondsuit\\,{1}=a.$\nLet $c=b.$ Substituting this into the first two conditions, we see that\n\\[a\\diamondsuit\\, (b\\diamondsuit\\, {c})=(a\\diamondsuit\\, {b})\\cdot{c} \\implies a\\diamondsuit\\, (b\\diamondsuit\\, {b})=(a\\diamondsuit\\, {b})\\cdot{b}.\\]\nSubstituting $b\\diamondsuit\\, {b} =1$ , the second equation becomes\n\\[a\\diamondsuit\\, {1}=(a\\diamondsuit\\, {b})\\cdot{b} \\implies a=(a\\diamondsuit\\,{b})\\cdot{b}.\\]\nSince $a, b$ and $c$ are nonzero, we can divide by $b$ which yields,\n\\[\\frac{a}{b}=(a\\diamondsuit\\, {b}).\\]\nNow we can find the value of $x$ straightforwardly:\n\\[\\frac{2016}{(\\frac{6}{x})}=100 \\implies 2016=\\frac{600}{x} \\implies x=\\frac{600}{2016} = \\frac{25}{84}.\\]\nTherefore, $a+b=25+84=\\boxed{109}$", "One way to eliminate the $\\diamondsuit$ in this equation is to make $a = b$ so that $a\\,\\diamondsuit\\, (b\\,\\diamondsuit \\,c) = c$ . In this case, we can make $b = 2016$\n\\[2016 \\,\\diamondsuit\\, (6\\,\\diamondsuit\\, x)=100\\implies (2016\\, \\diamondsuit\\, 6) \\cdot x = 100\\]\nBy multiplying both sides by $\\frac{6}{x}$ , we get:\n\\[(2016\\, \\diamondsuit\\, 6) \\cdot 6 = \\frac{600}{x}\\implies 2016 \\, \\diamondsuit\\, (6\\, \\diamondsuit\\, 6) = \\frac{600}{x}\\]\nBecause $6\\, \\diamondsuit\\, 6 = 2016\\, \\diamondsuit\\, 2016 = 1:$\n\\[2016 \\, \\diamondsuit\\, (2016\\, \\diamondsuit\\, 2016) = \\frac{600}{x}\\implies (2016\\, \\diamondsuit\\, 2016) \\cdot 2016 = \\frac{600}{x}\\implies 2016 = \\frac{600}{x}\\]\nTherefore, $x = \\frac{600}{2016} = \\frac{25}{84}$ , so the answer is $25 + 84 = \\boxed{109.}$", "We can manipulate the given identities to arrive at a conclusion about the binary operator $\\diamondsuit$ . Substituting $b = c$ into the first identity yields \\[( a\\ \\diamondsuit\\ b) \\cdot b = a\\ \\diamondsuit\\ (b\\ \\diamondsuit\\ b) = a\\ \\diamondsuit\\ 1 = a\\ \\diamondsuit\\ ( a\\ \\diamondsuit\\ a) = ( a\\ \\diamondsuit\\ a) \\cdot a = a.\\] Hence, $( a\\ \\diamondsuit\\ b) \\cdot b = a,$ or, dividing both sides of the equation by $b,$ $( a\\ \\diamondsuit\\ b) = \\frac{a}{b}.$\nHence, the given equation becomes $\\frac{2016}{\\frac{6}{x}} = 100$ . Solving yields $x=\\frac{100}{336} = \\frac{25}{84},$ so the answer is $25 + 84 = \\boxed{109.}$", "$2016 \\diamondsuit (6 \\diamondsuit x) = (2016 \\diamondsuit 6) \\cdot x = 100$\n$2016 \\diamondsuit (2016 \\diamondsuit 1) = (2016 \\diamondsuit 2016) \\cdot 1 = 1 \\cdot 1 = 1$\n$2016 \\diamondsuit 2016 = 1$ $2016 \\diamondsuit (2016 \\diamondsuit 1) = 1$ , so $2016 \\diamondsuit 1 = 2016$\n$2016 \\diamondsuit 1 = (2016 \\diamondsuit 6) \\cdot 6$ $2016 \\diamondsuit 6 = \\frac{2016 \\diamondsuit 1}{6} = 336$\n$x = \\frac{100}{2016 \\diamondsuit 6} = \\frac{100}{336} = \\frac{25}{84}$ $24 + 85 = \\boxed{109}$", "Notice that $2016 \\diamondsuit (6 \\diamondsuit 6)=(2016 \\diamondsuit 6) \\cdot 6 = 2016$ . Hence, $2016 \\diamondsuit 6 = 336$ . Thus, $2016 \\diamondsuit (6 \\diamondsuit x)=100 \\implies (2016 \\diamondsuit 6) \\cdot x = 100 \\implies 336x=100 \\implies x=\\frac{25}{84}$ . Therefore, the answer is $\\boxed{109}$" ]

[ "Of the $70$ fish caught in September, $40\\%$ were not there in May, so $42$ fish were there in May. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, $\\frac{3}{42} = \\frac{60}{x} \\Longrightarrow \\boxed{840}$", "First, we notice that there are 45 tags left, after 25% of the original fish have went away/died. Then, some $x$ percent of fish have been added such that $\\frac{x}{x+75} = 40 \\%$ , or $\\frac{2}{5}$ . Solving for $x$ , we get that $x = 50$ , so the total number of fish in September is $125 \\%$ , or $\\frac{5}{4}$ times the total number of fish in May.\nSince $\\frac{3}{70}$ of the fish in September were tagged, $\\frac{45}{5n/4} = \\frac{3}{70}$ , where $n$ is the number of fish in May. Solving for $n$ , we see that $n = \\boxed{840}$" ]

[ "Let the lengths of the three sides of the rectangular solid after the cutting be $a,b,c$ , so that the desired volume is $abc$ . Note that each cut reduces one of the dimensions by one, so that after ten cuts, $a+b+c = 10 + 13 + 14 - 10 = 27$ . By AM-GM $\\frac{a+b+c}{3} = 9 \\ge \\sqrt[3]{abc} \\Longrightarrow abc \\le \\boxed{729}$ . Equality is achieved when $a=b=c=9$ , which is possible if we make one slice perpendicular to the $10$ cm edge, four slices perpendicular to the $13$ cm edge, and five slices perpendicular to the $14$ cm edge.", "A more intuitive way to solve it is by seeing that to keep the volume of the rectangular cheese the greatest, we must slice the cheese off to decrease the greatest length of the cheese (this is easy to check). Here are the ten slices:\n${10, 13, 14} \\rightarrow {10, 13, 13} \\rightarrow {10, 12, 13} \\rightarrow {10, 12, 12} \\rightarrow {10, 11, 12} \\rightarrow {10, 11, 11} \\rightarrow {10, 10, 11} \\rightarrow {10, 10, 10} \\rightarrow {9, 10, 10} \\rightarrow {9, 9, 10} \\rightarrow {9, 9, 9}.$\nSo the greatest possible volume is thus $9 \\times 9 \\times 9 = \\boxed{729}$" ]

[ "Label the points where the plane intersects the top face of the cylinder as $C$ and $D$ , and the center of the cylinder as $O$ , such that $C,O,$ and $A$ are collinear. Let $N$ be the center of the bottom face, and $M$ the midpoint of $\\overline{AB}$ . Then $ON=4$ $MN=3$ (because of the 120 degree angle), and so $OM=5$\nProject $C$ and $D$ onto the bottom face to get $X$ and $Y$ , respectively. Then the section $ABCD$ (whose area we need to find), is a stretching of the section $ABXY$ on the bottom face. The ratio of stretching is $\\frac{OM}{MN}=\\frac{5}{3}$ , and we do not square this value when finding the area because it is only stretching in one direction. Using 30-60-90 triangles and circular sectors, we find that the area of the section $ABXY$ is $18\\sqrt{3}\\ + 12 \\pi$ . Thus, the area of section $ABCD$ is $20\\pi + 30\\sqrt{3}$ , and so our answer is $20+30+3=\\boxed{053}$", "Label the points same as in the first sentence above. Consider a view of the cylinder such that height is disregarded, i.e. a top view. From this view, note that Cylinder $O$ has become a circle with $\\overarc{AB}$ $\\overarc{CD}$ $120^\\text{o}$ . Using 30-60-90 triangles, we get rectangle $ABCD$ to have a horizontal component of $6$ . Now, consider a side view, such that $A$ and $B$ coincide at the bottom of the diagram. From this view, consider the right triangle composed of hypotenuse $AD$ and a point along the base of the viewpoint, which will be labeled as $E$ . From the top view, $AE = 6$ . Because of the height of the cylinder, $DE$ is equal to $8$ . This makes $AD$ equal to $10$\nNow, the use of simple calculus is required. Conceptualize an infinite number of lines perpendicular to $AE$ intersecting both $AE$ and $AD$ . Consider the area between point $A$ and the first vertical line. Label the point where the line intersects AE as E', and the point where the line intersects AD as D'. The area of the part of the initial unpainted face within these two positions approaches a rectangle with length AD' and width $w$ . The area of the base within these two positions approaches a rectangle with length AE' and width $w$ . The ratio of AD':AE' is 10:6, since the ratio of AD:AE is 10:6. This means that the area of the initial unpainted surface within these two positions to the area of the base within these two positions is equal to 10 $w$ :6 $w$ = 10:6. Through a similar argument, the areas between each set of vertical lines also maintains a ratio of 10:6. Therefore, the ratio of the area we wish to find to the area of the base between AB and CD (from the top perspective) is 10:6. Using 30-60-90 triangles and partial circles, the area of the base between AB and CD is calculated to be $18\\sqrt{3}\\ + 12 \\pi$ . The area of the unpainted surface therefore becomes $20\\pi + 30\\sqrt{3}$ , and so our answer is $\\boxed{053}$", "This problem can be calculus-bashed (for those like me who never noticed the surface was merely a stretch of its projection). Label points as in the first paragraph of Solution 1 ( $A$ and $B$ as given, $M$ the midpoint of $AB$ $O$ the center of the cylinder, $T$ the center of the bottom face of the cylinder). Because of the 120 degrees and right triangle calculations, we find $MT$ = 3, $OT$ = 4, $OM$ = 5). We will be integrating with respect to the y-coordinate which we define as distance downwards from $O$ (in this system, the $y$ -coordinate of the bottom face would be 4).\nWe note that by similar triangles, we have that the length from $O$ to the point on the unpainted surface of coordinate $y$ is $\\ell = \\frac{5}{4} y$ , and therefore $d\\ell = \\frac{5}{4} dy$ . Define the segment $A'B'$ to be the intersection of the painted surface with the circular cross section of the cylinder of coordinate $y$ , with endpoints $A'$ and $B'$ and midpoint $M'$ , with $T'$ the center of this circular cross section. Then, by similar triangles, $T'M' = \\frac{3}{4} y$ and thus \\[A'B' = 2A' M' = 2 \\sqrt{ 6^2 - \\left( \\frac{3}{4}y\\right)^2 } = \\frac{3}{2} \\sqrt{ 64 - y^2 }\\] . We know that $A'B'$ is perpendicular to $OM$\nNow we can set up our integral: we will integrate $y$ from 0 to 4 and multiply by two because the total height is 8. \\[A = 2\\int_0^4 \\left(\\frac{3}{2}\\sqrt{ 64 - y^2 }\\right) \\left( \\frac{5}{4} dy\\right)\\] \\[A = \\frac{15}{4} \\int_0^4 \\sqrt{ 64 - y^2 }dy\\]\nThen we substitute $y = 8\\sin \\theta$ with $dy = 8 \\cos \\theta d \\theta$ , changing the bounds to 0 to $\\frac{\\pi}{6}$ as appropriate. \\[A = \\frac{15}{4} \\int_0^\\frac{\\pi}{6} \\sqrt{ 64 - 64\\sin^2 \\theta }\\cdot 8\\cos\\theta d\\theta\\] \\[A = 240 \\int_0^\\frac{\\pi}{6} \\cos^2 \\theta d\\theta\\] \\[A = 240 \\left[ \\frac{\\theta}{2} + \\frac{\\sin 2\\theta}{4} \\right]_0^\\frac{\\pi}{6} = 240 \\left[ \\frac{\\pi}{12} + \\frac{\\sqrt{3}}{8} \\right] = 20{\\pi} + 30\\sqrt{3}\\]\nTherefore, $a + b + c = 20 + 30 + 3 = \\boxed{053}$" ]

[ "Since the bricks are $1$ foot high, there will be $7$ rows. To minimize the number of blocks used, rows $1, 3, 5,$ and $7$ will look like the bottom row of the picture, which takes $\\frac{100}{2} = 50$ bricks to construct. Rows $2, 4,$ and $6$ will look like the upper row pictured, which has $49$ 2-foot bricks in the middle, and $2$ 1-foot bricks on each end for a total of $51$ bricks.\nFour rows of $50$ bricks and three rows of $51$ bricks totals $4\\cdot 50 + 3\\cdot 51 = 200 + 153 = 353$ bricks, giving the answer $\\boxed{353}.$" ]

[ "Since the arrow must land in one of the three regions, the sum of the probabilities must be 1. Thus the answer is $1-\\frac{1}{2}-\\frac{1}{3}=\\boxed{16}$" ]

[ "Assuming that there are fractions of compact discs, it would take $412/56 ~= 7.357$ CDs to have equal reading time. However, since the number of discs must be a whole number, there are at least 8 CDs, in which case there would be $412/8 = 51.5$ minutes of reading on each of the 8 discs. The answer is $\\boxed{51.5}$", "We look at the options, and see which one is divisible by 412. We try A, and see that you don't get a whole number. Next, we try B. We see that the B is divisible by 412 and find that $\\boxed{51.5}$ is our answer." ]

[ "On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of $4$ , leaving only lockers $2 \\pmod{8}$ and $6 \\pmod{8}$ . Then he goes ahead and opens all lockers $2 \\pmod {8}$ , leaving lockers either $6 \\pmod {16}$ or $14 \\pmod {16}$ . He then goes ahead and opens all lockers $14 \\pmod {16}$ , leaving the lockers either $6 \\pmod {32}$ or $22 \\pmod {32}$ . He then goes ahead and opens all lockers $6 \\pmod {32}$ , leaving $22 \\pmod {64}$ or $54 \\pmod {64}$ . He then opens $54 \\pmod {64}$ , leaving $22 \\pmod {128}$ or $86 \\pmod {128}$ . He then opens $22 \\pmod {128}$ and leaves $86 \\pmod {256}$ and $214 \\pmod {256}$ . He then opens all $214 \\pmod {256}$ , so we have $86 \\pmod {512}$ and $342 \\pmod {512}$ , leaving lockers $86, 342, 598$ , and $854$ , and he is at where he started again. He then opens $86$ and $598$ , and then goes back and opens locker number $854$ , leaving locker number $\\boxed{342}$ untouched. He opens that locker.", "We can also solve this with recursion. Let $L_n$ be the last locker he opens given that he started with $2^n$ lockers. Let there be $2^n$ lockers. After he first reaches the end of the hallway, there are $2^{n-1}$ lockers remaining. There is a correspondence between these unopened lockers and if he began with $2^{n-1}$ lockers. The locker $y$ (if he started with $2^{n-1}$ lockers) corresponds to the locker $2^n+2-2y$ (if he started with $2^n$ lockers). It follows that $L_{n} = 2^{n} +2 -2L_{n-1}$ as they are corresponding lockers. We can compute $L_1=2$ and use the recursion to find $L_{10}=\\boxed{342}$", "List all the numbers from $1$ through $1024$ , then do the process yourself!!! It will take about 25 minutes (if you don't start to see the pattern), but that's okay, eventually, you will get $\\boxed{342}$" ]

[ "We extend line segments $\\ell,m,$ and $n$ to their point of concurrency, as shown below: We claim that lines $\\ell,m,$ and $n$ are concurrent: In the lateral faces of the bowl, we know that lines $\\ell$ and $m$ must intersect, and lines $\\ell$ and $n$ must intersect. In addition, line $\\ell$ intersects the top plane of the bowl at exactly one point. Since lines $m$ and $n$ are both in the top plane of the bowl, we conclude that lines $\\ell,m,$ and $n$ are concurrent.\nIn the lateral faces of the bowl, the dashed red line segments create equilateral triangles. So, the dashed red line segments all have length $1.$ In the top plane of the bowl, we know that $\\overleftrightarrow{m}\\perp\\overleftrightarrow{n}.$ So, the dashed red line segments create an isosceles triangle with leg-length $1.$\nNote that octagon has four pairs of parallel sides, and the successive side-lengths are $1,\\sqrt2,1,\\sqrt2,1,\\sqrt2,1,\\sqrt2,$ as shown below: The area of the octagon is \\[3^2-4\\cdot\\left(\\frac12\\cdot1^2\\right)=\\boxed{7}.\\]", "(This is an alternate way of analyzing the red extension line segments drawn in Solution 1.)\nThe perimeter of the square bottom of the bowl is $4$ .\nHalfway up the bowl, the boundary is still a square, with perimeter $4$ times the hexagon circumradius, aka $4 \\times 2 = 8$ times the hexagon (also square) side length (1), an increase of $4$\nExtending the bottom half of the bowl to twice its height (full height of the bowl) would increase the perimeter by the same amount again, forming a square with perimeter $4 + (8-4)\\times 2= 12$ . Thus the top octagon is cut out of a square of side length $12\\div4=3$ and thus area $9$\nThe difference between the above-constructed square and the octagon is four right triangles, and (by rotational and reflection symmetry), each is isosceles with equal-length perpendicular bases of length $(3-1)/2 = 1$ , and thus having area $\\frac12$ . Therefore the area of the octagon is $9-(4\\times 1/2) = \\boxed{7}$", "Note that the octagon is equiangular by symmetry, but it is not equilateral. $4$ of its sides are shared with the hexagon's sides, so each of those sides have side length $1$ . However, the other $4$ sides are touching the triangles, so we wish to find the length of these sides.\nNotice that when two adjacent hexagons meet at a side, their planes make the same dihedral angle at the bottom-most point of intersection and at the top-most point of intersection by symmetry. Therefore, the triangle that is wedged between the two hexagons has the same angle as the square at the bottom wedged between the hexagons. Thus, the triangle is a $45-45-90$ isosceles triangle.\nThis conclusion can also be reached by cutting the bottom square across a diagonal and noticing that each resulting triangle is congruent to each triangle wedged between the hexagons by symmetry. \nFurthermore, notice that if you take a copy of this bowl and invert it and place it on top of this bowl, you will get a polyhedron with faces of hexagons and squares, a truncated octahedron, and therefore this triangle has a $90^\\circ$ -angle:\nNow that we have come to this conclusion, by simple Pythagorean theorem, we have that the other $4$ sides of the octagon are $\\sqrt{2}$\nWe can draw a square around the octagon so that the area of the octagon is the area of the square minus each corner triangle. The hypotenuse of these corner triangles are $1$ and they are $45-45-90$ triangles because the octagon is equiangular, so each has dimensions $\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2},1$ . \nThe side length of the square is $\\sqrt{2}$ for the larger sides of the octagon, and adding $2$ of $\\frac{\\sqrt{2}}{2}$ for each width of the triangle. Therefore, the area of the square is: \\[\\left(\\sqrt{2} + 2 \\cdot \\frac{\\sqrt{2}}{2}\\right)^2 \\implies \\left(2\\sqrt{2}\\right)^2 = 8\\] The area of each triangle is $\\frac{1}{2} \\cdot \\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{2}}{2} = \\frac{1}{4}$ and there are $4$ of them, so we subtract $1$ from the area of the square. The area of the octagon is thus $\\boxed{7}$", "Denote a square $ABCD, AB = 1,\\pi$ is the plane $ABC,$ regular hexagons $ABFKSE, BCHMLF, CDGPNH, ADGQRE,$ triangles $FKL, ESR, GPQ, HMN.$\nThe main diagonal of each regular hexagon $EF = 2 \\implies EFGH$ is square with side $2$ parallel to $\\pi.$\nThe area of this square $[EFGH] = 4 \\implies [EFGH] - [ABCD] = 3.$\nThe difference $3$ is the area of the projection of $4$ half of hexagons on the plane $\\pi.$\nSo the area of the projections of another $4$ half of hexagons is $3.$\nIt is evident (may be not only for me) that projections of the coincide sides of hexagons are along diagonals of $ABCD$ (for example $A, E, C,$ and $H$ are collinear.)\nSo the projections on $\\pi$ of the coincide sides of hexagons and triangles are perpendicular to this lines $(SE \\perp AE).$\nTherefore in plane $\\pi$ projections of points $S, E,$ and $R$ are collinear and plane of $\\triangle ESR$ is perpendicular to $\\pi.$\nWe get $1 +3 +3 = \\boxed{7}.$", "Through observation, we can reasonably assume that each of the triangles on this shape is a right triangle. Since each side length of the hexagons is $1$ , the hypotenuse of the triangles would be $\\sqrt2$ . Now we know the side lengths of the octagon whose area we are solving for. The octagon can be broken into nine pieces. We have four triangles whose side lengths are 1, and their hypotenuse is a side whose length is $\\sqrt2$ . Next, we have $5$ $1$ by $1$ squares. The triangles each have an area of $\\frac{1}{2}$ , and the squares each have an area of $1$\nThen, we add these up, so we get $\\boxed{7}.$" ]

[ "The first box has volume $2\\times3\\times5=30\\text{ cm}^3$ , and the second has volume $(2\\times2)\\times(3\\times3)\\times(5)=180\\text{ cm}^3$ . The second has a volume that is $6$ times greater, so it holds $6\\times40=\\boxed{240}$ grams." ]

[ "We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting $<15$ of each color by applying the pigeonhole principle and through this we get a perfect guarantee. \nNamely, we can draw up to $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls, for a total of $75$ balls, without drawing $15$ balls of any one color. Drawing one more ball guarantees that we will get $15$ balls of one color — either red, green, or yellow. Thus, the answer is $75 + 1 = \\boxed{76}$" ]

[ "We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting $<15$ of each color by applying the pigeonhole principle and through this we get a perfect guarantee. \nNamely, we can draw up to $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls, for a total of $75$ balls, without drawing $15$ balls of any one color. Drawing one more ball guarantees that we will get $15$ balls of one color — either red, green, or yellow. Thus, the answer is $75 + 1 = \\boxed{76}$" ]

[ "Let $a$ be the amount of triangular tiles and $b$ be the amount of square tiles.\nTriangles have $3$ edges and squares have $4$ edges, so we have a system of equations.\nWe have $a + b$ tiles total, so $a + b = 25$\nWe have $3a + 4b$ edges total, so $3a + 4b = 84$\nMultiplying the first equation by $3$ on both sides gives $3a + 3b = 3(25) = 75$\nSecond equation minus the first equation gives $b = 9$ , so the answer is $\\boxed{9}$", "If all of the tiles were triangles, there would be $75$ edges. This is not enough, so there needs to be some squares. Trading a triangle for a square results in one additional edge each time, so we must trade out $9$ triangles for squares. Answer: $\\boxed{9}$", "Let $x$ be the number of square tiles. A square has $4$ edges, so the total number of edges from the square tiles is $4x$ . There are $25$ total tiles, which means that there are $25-x$ triangle tiles. A triangle has $3$ edges, so the total number of edges from the triangle tiles is $3(25-x)$ . Together, the total number of edges is $4x+3(25-x)=84$ . Solving our equation, we get that $x=9$ which means that our answer is $\\boxed{9}$" ]

[ "Let the number of white be $2x$ . The number of blue is then $x-y$ for some constant $y$ . So we want $2x+x-y=55\\rightarrow 3x-y=55$ . We take mod 3 to find y. $55=1\\pmod{3}$ , so blue is 1 more than half white. The number of whites is then 36, and the number of blues is 19. So $19*3=\\boxed{57}$" ]

[ "The counting numbers that leave a remainder of $4$ when divided by $6$ are $4, 10, 16, 22, 28, 34, \\cdots$ The counting numbers that leave a remainder of $3$ when\ndivided by $5$ are $3,8,13,18,23,28,33, \\cdots$ So $28$ is the smallest possible number\nof coins that meets both conditions. Because 28 is divisible by 7, there are $\\boxed{0}$ coins left\nwhen they are divided among seven people.", "If there were two more coins in the box, the number of coins would be divisible\nby both $6$ and $5$ . The smallest number that is divisible by $6$ and $5$ is $30$ , so the\nsmallest possible number of coins in the box is $28$ and the remainder when divided by $7$ is $\\boxed{0}$", "We can set up a system of modular congruencies: \\[g\\equiv 4 \\pmod{6}\\] \\[g\\equiv 3 \\pmod{5}\\] We can use the division algorithm to say $g=6n+4$ $\\Rightarrow$ $6n\\equiv 4 \\pmod{5}$ $\\Rightarrow$ $n\\equiv 4 \\pmod{5}$ . If we plug the division algorithm in again, we get $n=5q+4$ . This means that $g=30q+28$ , which means that $g\\equiv 28 \\pmod{30}$ . From this, we can see that $28$ is our smallest possible integer satisfying $g\\equiv 28 \\pmod{30}$ $28$ $\\div$ $7=4$ , making our remainder $0$ . This means that there are $\\boxed{0}$ coins left over when equally divided amongst $7$ people." ]

[ "The boy buys $3$ oranges for $10$ cents or $1$ orange for $\\frac{10}{3}$ cents. He sells them at $\\frac{20}{5}=4$ cents each. \nThat means for every orange he sells, he makes a profit of $4-\\frac{10}{3}=\\frac{2}{3}$ cents.\nTo make a profit of $100$ cents, he needs to sell $\\frac{100}{\\frac{2}{3}}=\\boxed{150}$" ]

[ "The boy buys $3$ oranges for $10$ cents. He sells them at $5$ for $20$ cents. So, he buys $15$ for $50$ cents and sells them at $15$ for $60$ cents, so he makes $10$ cents of profit on every $15$ oranges. To make $100$ cents of profit, he needs to sell $15 \\cdot \\frac{100}{10} = \\boxed{150}$ oranges." ]

[ "\nCrawling from $-2$ to $-6$ takes it a distance of $4$ units. Crawling from $-6$ to $5$ takes it a distance of $11$ units. Add $4$ and $11$ to get $\\boxed{15}$" ]

[ "\nCrawling from $-2$ to $-6$ takes it a distance of $4$ units. Crawling from $-6$ to $5$ takes it a distance of $11$ units. Add $4$ and $11$ to get $\\boxed{15}$" ]

[ "Let $P_n$ represent the probability that the bug is at its starting vertex after $n$ moves. If the bug is on its starting vertex after $n$ moves, then it must be not on its starting vertex after $n-1$ moves. At this point it has $\\frac{1}{2}$ chance of reaching the starting vertex in the next move. Thus $P_n=\\frac{1}{2}(1-P_{n-1})$ $P_0=1$ , so now we can build it up:\n$P_1=0$ $P_2=\\frac{1}{2}$ $P_3=\\frac{1}{4}$ $P_4=\\frac{3}{8}$ $P_5=\\frac{5}{16}$ $P_6=\\frac{11}{32}$ $P_7=\\frac{21}{64}$ $P_8=\\frac{43}{128}$ $P_9=\\frac{85}{256}$ $P_{10}=\\frac{171}{512}$\nThus the answer is $171+512=$ $\\boxed{683}$", "Consider there to be a clockwise and a counterclockwise direction around the triangle. Then, in order for the ant to return to the original vertex, the net number of clockwise steps must be a multiple of 3, i.e., $\\#CW - \\#CCW \\equiv 0 \\pmod{3}$ . Since $\\#CW + \\#CCW = 10$ , it is only possible that $(\\#CW,\\, \\#CCW) = (5,5), (8,2), (2,8)$\nIn the first case, we pick $5$ out of the ant's $10$ steps to be clockwise, for a total of ${10 \\choose 5}$ paths. In the second case, we choose $8$ of the steps to be clockwise, and in the third case we choose $2$ to be clockwise. Hence the total number of ways to return to the original vertex is ${10 \\choose 5} + {10 \\choose 8} + {10 \\choose 2} = 252 + 45 + 45 = 342$ . Since the ant has two possible steps at each point, there are $2^{10}$ routes the ant can take, and the probability we seek is $\\frac{342}{2^{10}} = \\frac{171}{512}$ , and the answer is $\\boxed{683}$", "Label the vertices of the triangle $A,B,C$ with the ant starting at $A$ . We will make a table of the number of ways to get to $A,B,C$ in $n$ moves $n\\leq10$ . The values of the table are calculated from the fact that the number of ways from a vertex say $A$ in $n$ steps equals the number of ways to get to $B$ in $n-1$ steps plus the number of ways to get to $C$ in $n-1$ steps.\n\\[\\begin{array}{|l|ccc|} \\multicolumn{4}{c}{\\text{Table}}\\\\\\hline \\text{Step}&A&B&C \\\\\\hline 1 &0 &1 &1 \\\\ 2 &2 &1 &1 \\\\ 3 &2 &3 &3\\\\ \\vdots &\\vdots&\\vdots&\\vdots \\\\ 10 &342 &341 &341 \\\\\\hline \\end{array}\\] Therefore, our answer is $512+171=\\boxed{683}.$", "As a note, do NOT do this on the exam as it will eat up your time, but feel free to experiment around with this if you have a good enough understanding of linear algebra. This writeup will be extremely lengthy because I am assuming just very basic concepts of linear algebra. The concepts here extend to higher levels of mathematics, so feel free to explore more in depth so that you can end up solving almost any variation of this classic problem.\nThere are a possible of 3 states for this problem, so we can model the problem as a stochastic process. The resulting process has a transition matrix of:\n\\[\\hat{T} = \\begin{bmatrix} 0 & 0.5 & 0.5\\\\ 0.5 & 0 & 0.5\\\\ 0.5 & 0.5 & 0 \\end{bmatrix}\\]\nWe aim to diagonalize this transition matrix to make it easier to exponentiate by converting it into what's known as it's Jordan Canonical Form.\nIn order to do this, we must extract the eigenvalues and eigenvectors of the matrix. The eigenpolynomial for this matrix is obtained by calculating this matrix's determinant with $0-\\lambda$ about it's main diagonal like so: \\[\\hat{T}_{\\lambda} = \\begin{bmatrix} 0-\\lambda & 0.5 & 0.5\\\\ 0.5 & 0-\\lambda & 0.5\\\\ 0.5 & 0.5 & 0-\\lambda \\end{bmatrix}\\]\nWe have the matrix's eigenpolynomial to be $\\lambda^3 - \\frac{3\\lambda}{4} + \\frac{1}{4}$ , and extracting eigenvalues by setting the polynomial equal to $0$ , we have 2 eigenvalues: $\\lambda_1 = 1$ of multiplicity 1, and $\\lambda_2 = -\\frac{1}{2}$ of multiplicity 2. To extract the eigenvectors, we must assess the kernel of this matrix (also known as the null space), or the linear subspace of the domain of $\\hat{T}$ where everything gets mapped to the null vector.\nWe first do this for $\\lambda_1$ . Taking $-\\lambda_1$ across the diagonals to get $\\hat{T}_{\\lambda_1} = \\begin{bmatrix} -1 & 0.5 & 0.5\\\\ 0.5 & -1 & 0.5\\\\ 0.5 & 0.5 & -1 \\end{bmatrix}$ , we first reduce it to reduced row echelon form, which is $\\begin{bmatrix} 1 & 0 & -1\\\\ 0 & 1 & -1\\\\ 0 & 0 & 0 \\end{bmatrix}$ . From here, we compute the kernel by setting \\[\\begin{bmatrix} 1 & 0 & -1\\\\ 0 & 1 & -1\\\\ 0 & 0 & 0 \\end{bmatrix} \\begin{bmatrix} x_1\\\\ x_2\\\\ x_3 \\end{bmatrix} = 0\\] . So if we take our free variable $x_3 = 0 = t$ , then that means that in the same fashion, $x_1 - x_1 = x_2 - x_2 = 0 = t$ , so hence, the kernel of $\\hat{T}_{\\lambda} = \\begin{bmatrix} t\\\\ t\\\\ t \\end{bmatrix}$ , or more simply, $t\\begin{bmatrix} 1\\\\ 1\\\\ 1 \\end{bmatrix}$ $\\begin{bmatrix} 1\\\\ 1\\\\ 1 \\end{bmatrix}$ is the eigenvector corresponding to $\\lambda_1$ . We do the same computations for our second unique eigenvalue, but I will save the computation to you. There are actually 2 eigenvectors for $\\lambda_2$ , because the reduced row echelon form for $\\hat{T}_{\\lambda_2}$ has 2 free variables instead of 1, so our eigenvectors for $\\lambda_2$ are $\\begin{bmatrix} -1\\\\ 1\\\\ 0 \\end{bmatrix}, \\begin{bmatrix} -1\\\\ 0\\\\ 1 \\end{bmatrix}$ . We are now ready to begin finding the Jordan canonical form\nIn linear algebra, the JCF (which also goes by the name of Jordan Normal Form) is an upper triangular matrix representing a linear transformation over a finite-dimensional complex vector space. Any square matrix that has a Jordan Canonical Form has its field of coefficients extended into a field containing all it's eigenvalues. You can find more information about them on google, as well as exactly how to find them but for now let's get on with the problem. I will skip the computation in this step, largley because this writeup is already gargantuan for a simple AIME problem, and because there are countless resources explaining how to do so.\nWe aim to decompose $\\hat{T}$ into the form $\\hat{T} = SJS^{-1}$ , where $S$ is a matrix whose columns consist of the eigenvectors of $\\hat{T}$ $J$ is the Jordan matrix, and $S^{-1}$ is, well, the inverse of $S$ . We have 1 eigenvalue of multiplicity 1, and 1 of multiplicity 2, so based on this info, we set our eigenvalues along the diagonals like so.\nWe have: \\[J = \\begin{bmatrix} -\\frac{1}{2} & 0 & 0\\\\ 0 & -\\frac{1}{2} & 0\\\\ 0 & 0 & 1 \\end{bmatrix}, S = \\begin{bmatrix} -1 & -1 & 1\\\\ 0 & 1 & 1\\\\ 1 & 0 & 1 \\end{bmatrix}, S^{-1} = \\frac{1}{3}\\begin{bmatrix} -1 & -1 & 2\\\\ -1 & 2 & -1\\\\ 1 & 1 & 1 \\end{bmatrix}\\] and so: \\[\\hat{T} = \\begin{bmatrix} 0 & 0.5 & 0.5\\\\ 0.5 & 0 & 0.5\\\\ 0.5 & 0.5 & 0 \\end{bmatrix} = SJS^{-1} = \\begin{bmatrix} -1 & -1 & 1\\\\ 0 & 1 & 1\\\\ 1 & 0 & 1 \\end{bmatrix}\\begin{bmatrix} -\\frac{1}{2} & 0 & 0\\\\ 0 & -\\frac{1}{2} & 0\\\\ 0 & 0 & 1 \\end{bmatrix}\\begin{bmatrix} -\\frac{1}{3} & -\\frac{1}{3} & \\frac{2}{3}\\\\ -\\frac{1}{3} & \\frac{2}{3} & -\\frac{1}{3}\\\\ \\frac{1}{3} & \\frac{1}{3} & \\frac{1}{3} \\end{bmatrix}\\] Now that we have converted to Jordan Canonical Form, it is extremely easy to compute $\\hat{T}^n$\nIt is an important fact that for any matrix $K$ with Jordan decomposition $SJ_kS^{-1}$ , we have that $K^n = S(J_k)^nS^{-1}$ Using this fact, we aim to find the general solution for the problem. $J^n = \\begin{bmatrix} \\left(-\\frac{1}{2}\\right)^n & 0 & 0\\\\ 0 & \\left(-\\frac{1}{2}\\right)^n & 0\\\\ 0 & 0 & 1 \\end{bmatrix}$ , and using the laws of matrix multiplication, \\[SJ^n = \\begin{bmatrix} (-1)^{n+1}\\left(\\frac{1}{2}\\right)^n & (-1)^{n+1}\\left(\\frac{1}{2}\\right)^n & 1\\\\ 0 & \\left(-\\frac{1}{2}\\right)^n & 1\\\\ \\left(-\\frac{1}{2}\\right)^n & 0 & 1 \\end{bmatrix}\\] So finally, our final, generalized transition matrix after any number of steps $n$ is: \\[\\hat{T}^n = \\begin{bmatrix} \\frac{1}{3} - \\frac{1}{3}(-1)^{1+n}(2)^{1-n} & \\frac{1}{3} + \\frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)} & \\frac{1}{3} + \\frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)}\\\\ \\frac{1}{3} + \\frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)} & \\frac{1}{3} - \\frac{1}{3}(-1)^{1+n}(2)^{1-n} & \\frac{1}{3} + \\frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)}\\\\ \\frac{1}{3} + \\frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)} & \\frac{1}{3} + \\frac{1}{3}(-1)^{n+1}(2)^{1-(1+n)} & \\frac{1}{3} - \\frac{1}{3}(-1)^{1+n}(2)^{1-n} \\end{bmatrix}\\]\nFor the sake of this problem, we seek the top left element, which is $\\frac{1}{3} - \\frac{1}{3}(-1)^{1+n}(2)^{1-n}$ . Substituting $n=10$ readily gives the probability of the bug reaching it's starting position within 10 moves to be $\\frac{171}{512} \\implies m+n = \\boxed{683}$", "This method does not rigorously get the answer, but it works. As the bug makes more and more moves, the probability of it going back to the origin approaches closer and closer to 1/3. Therefore, after 10 moves, the probability gets close to $341.33/1024$ . We can either round up or down. If we round down, we see $341/1024$ cannot be reduced any further and because the only answers on the AIME are below 1000, this cannot be the right answer. However, if we round up, $342/1024$ can be reduced to $171/512$ where the sum 171+512= $\\boxed{683}$ is an accepted answer.", "Start of with any vertex, say $A$ . Denote $a_n$ the number of paths ending where it started. Then notice that for a path to end in the vertex you started on, you must have for the $(n-1)$ case that of the $2^{n-1}$ total paths, we must take away the paths that end in the $(n-1)$ -st term where it started off. (Because imagine on the $(n-1)$ move that it landed on vertex $A$ . Then if we wanted to find the number of length $n$ paths that end at $A$ , we would be able to find that path, because on the $(n-1)$ -st move it landed at $A$ . You can't go from $A$ to $A$ ). Hence we have the recursion $a_n=2^{n-1}-a_{n-1}$ , with $a_3 = 2$ . Now reiterating gives us $a_{10} = 342$ , so that the probability is $\\frac{342}{2^{10}} = \\frac{171}{512}$ . So we have $171 + 512 = \\boxed{683}$ .\n~th1nq3r" ]

[ "\nThere is $1$ way to get to any of the red arrows. From the first (top) red arrow, there are $2$ ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) red arrow, there are $3$ ways to get to each of the first and the second blue arrows. So there are in total $5$ ways to get to each of the blue arrows.\nFrom each of the first and second blue arrows, there are respectively $4$ ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively $8$ ways to get to each of the first and the second green arrows. Therefore there are in total $5 \\cdot (4+4+8+8) = 120$ ways to get to each of the green arrows.\nFinally, from each of the first and second green arrows, there are respectively $2$ ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are $3$ ways to get to the first orange arrow. Therefore there are $120 \\cdot (2+2+3+3) = 1200$ ways to get to each of the orange arrows, hence $2400$ ways to get to the point $B$ $\\boxed{2400}$", "\nFor every blue arrow, there are $2\\cdot 2=4$ ways to reach it without using the reverse arrow since the bug can choose any of $2$ red arrows to pass through and $2$ black arrows to pass through. If the bug passes through the white arrow, the red arrow that the bug travels through must be the closest to the first black arrow. Otherwise, the bug will have to travel through both red segments, which is impossible because now there is no path to take after the bug emerges from the reverse arrow. Similarly, with the blue segments, the second black arrow taken must be the one that is closest to the blue arrow that was taken. Also, it is trivial that the two black arrows taken must be different. Therefore, if the reverse arrow is taken, the blue arrow taken determines the entire path and there is $1$ path for every arrow. Since the bug cannot return once it takes a blue arrow, the answer must be divisible by $5$ $\\boxed{2400}$ is the only answer that is." ]

[ "We use casework. \nThe main thing to notice is that, if the bug does not go backwards, then every vertical set of arrows can be used, as the bug could travel straight downwards and then use any arrow to continue right.\nNote: The motivation is quite weird so follow my numbers as they start from the left (point A) and go right (point B).\nCase 1: Bug does not go backwards.\nThe number of cases for this is just each vertical set of arrows multiplied to each other (if you don't understand where I'm coming from, try to understand where these numbers come from!)\nAnd so we have $2*2*4*4*4*2*2 = 2^{10}$ cases.\nCase 2: The bug goes backwards once, either at the first arrow or third arrow.\nHere, we have to count the fact that there is a horizontal midline that the bug could not cross, or otherwise it would be stepping on the same edge twice.\nBack on first arrow: $2*1*2*4*4*2*2 = 2^{8}$ cases.\nBack on third arrow: $2*2*4*4*4*1 = 2^{8}$ cases.\nCase 3: The bug goes backwards once, at the second arrow.\nSame thing as above, except since there are 4 arrows in the vertical set (plus one for the backwards arrow), then the calculations are a bit different.\nWe have $2*2*4*1*2*4*2*2 = 2^{9}$ cases.\nNotice that the first and third back arrows decrease the number of cases by a factor of $2^2$ and the second back arrow decreases the number of cases by $2^1$ . Hence,\n1st + 2nd = $2^7$\n2nd + 3rd = $2^7$\n1st + 3rd = $2^6$\n1st + 2nd + 3rd = $2^5$\nAnd so the number of cases in total is $2^{10} + 2^9 + 2^8 + 2^8 + 2^7 + 2^7 + 2^6 + 2^5 \\Rightarrow 1024 + 512 + 256 + 256 + 128 + 128 + 64 + 32 = \\boxed{2400}$" ]

[ "\nThere is $1$ way to get to any of the red arrows. From the first (top) red arrow, there are $2$ ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) red arrow, there are $3$ ways to get to each of the first and the second blue arrows. So there are in total $5$ ways to get to each of the blue arrows.\nFrom each of the first and second blue arrows, there are respectively $4$ ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively $8$ ways to get to each of the first and the second green arrows. Therefore there are in total $5 \\cdot (4+4+8+8) = 120$ ways to get to each of the green arrows.\nFinally, from each of the first and second green arrows, there are respectively $2$ ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are $3$ ways to get to the first orange arrow. Therefore there are $120 \\cdot (2+2+3+3) = 1200$ ways to get to each of the orange arrows, hence $2400$ ways to get to the point $B$ $\\boxed{2400}$", "\nFor every blue arrow, there are $2\\cdot 2=4$ ways to reach it without using the reverse arrow since the bug can choose any of $2$ red arrows to pass through and $2$ black arrows to pass through. If the bug passes through the white arrow, the red arrow that the bug travels through must be the closest to the first black arrow. Otherwise, the bug will have to travel through both red segments, which is impossible because now there is no path to take after the bug emerges from the reverse arrow. Similarly, with the blue segments, the second black arrow taken must be the one that is closest to the blue arrow that was taken. Also, it is trivial that the two black arrows taken must be different. Therefore, if the reverse arrow is taken, the blue arrow taken determines the entire path and there is $1$ path for every arrow. Since the bug cannot return once it takes a blue arrow, the answer must be divisible by $5$ $\\boxed{2400}$ is the only answer that is." ]

[ "We use casework. \nThe main thing to notice is that, if the bug does not go backwards, then every vertical set of arrows can be used, as the bug could travel straight downwards and then use any arrow to continue right.\nNote: The motivation is quite weird so follow my numbers as they start from the left (point A) and go right (point B).\nCase 1: Bug does not go backwards.\nThe number of cases for this is just each vertical set of arrows multiplied to each other (if you don't understand where I'm coming from, try to understand where these numbers come from!)\nAnd so we have $2*2*4*4*4*2*2 = 2^{10}$ cases.\nCase 2: The bug goes backwards once, either at the first arrow or third arrow.\nHere, we have to count the fact that there is a horizontal midline that the bug could not cross, or otherwise it would be stepping on the same edge twice.\nBack on first arrow: $2*1*2*4*4*2*2 = 2^{8}$ cases.\nBack on third arrow: $2*2*4*4*4*1 = 2^{8}$ cases.\nCase 3: The bug goes backwards once, at the second arrow.\nSame thing as above, except since there are 4 arrows in the vertical set (plus one for the backwards arrow), then the calculations are a bit different.\nWe have $2*2*4*1*2*4*2*2 = 2^{9}$ cases.\nNotice that the first and third back arrows decrease the number of cases by a factor of $2^2$ and the second back arrow decreases the number of cases by $2^1$ . Hence,\n1st + 2nd = $2^7$\n2nd + 3rd = $2^7$\n1st + 3rd = $2^6$\n1st + 2nd + 3rd = $2^5$\nAnd so the number of cases in total is $2^{10} + 2^9 + 2^8 + 2^8 + 2^7 + 2^7 + 2^6 + 2^5 \\Rightarrow 1024 + 512 + 256 + 256 + 128 + 128 + 64 + 32 = \\boxed{2400}$" ]

[ "We declare a point $(x, y)$ to make up for the extra steps that the bug has to move. If the point $(x, y)$ satisfies the property that $|x - 3| + |y + 2| + |x + 3| + |y - 2| \\le 20$ , then it is in the desirable range because $|x - 3| + |y + 2|$ is the length of the shortest path from $(x,y)$ to $(3, -2)$ and $|x + 3| + |y - 2|$ is the length of the shortest path from $(x,y)$ to $(-3, 2)$\nIf $-3\\le x \\le 3$ , then $-7\\le y \\le 7$ satisfy the property. there are $15 \\times 7 = 105$ lattice points here.\nelse let $3< x \\le 8$ (and for $-8 \\le x < -3$ because it is symmetrical) We set 8 as the upper bound for x because the shortest distance from $(-3, 2)$ to $(x, y)$ added to the shortest distance from $(x, y)$ to $(3, -2)$ is $|x - 3| + |y + 2| + |x + 3| + |y - 2|$ . Since the minimum value for the difference between the y-coordinates is at $y = 0$ , we get $2x + 4 = 20$ or $-2x + 4 = 20$ . Thus, the upper and lower bounds for $x$ are $8$ and $-8$ , respectively.\nNow we test each value for x satisfying $3< x \\le 8$ and double the result because of symmetry.\nFor $x = 4$ , the possibles values of y are such that $|2y| \\le 12$ for a total of $13$ lattice points,\nfor $x = 5$ , the possibles values of y are such that $|2y| \\le 10$ for a total of $11$ lattice points,\nfor $x = 6$ , the possibles values of y are such that $|2y| \\le 8$ for a total of $9$ lattice points,\nfor $x = 7$ , the possibles values of y are such that $|2y| \\le 6$ for a total of $7$ lattice points,\nfor $x = 8$ , the possibles values of y are such that $|2y| \\le 4$ for a total of $5$ lattice points,\nHence, there are a total of $105 + 2 ( 13 + 11 + 9 + 7 + 5) = \\boxed{195}$ lattice points.", "(Anyone mind making a diagram for this)\nNotice that the bug is basically moving from A to B (length 10) but going on a detour in the middle.\nSpecifically the detour would be of length 5 to some point and then going back by retracing its path (also length 5). Just to simplify things we can observe that the coordinates don't matter here and all we need to remember is that A and B are the diagonally opposite vertices of a 4 by 6 rectangle. The bug can start the \"detour\" from any point on or inside the rectangle. Notice that the bug can go 5 steps in the y direction, 5 steps in the x direction, or anything in between, so the points covered by possible detours from any point would look like a rhombus or square rotated 45 degrees (with centre at a point on or inside the rectangle). Drawing this out we would get an octagon.\nFinding the final answer is then easy, for this solution I will slice the octagon into 4 rectangles (2 of which are squares) and 4 isosceles triangles. There are $(4+1)\\cdot (6+1) = 35$ points on or inside the original triangle, $5\\cdot 7$ points covered by the rectangles above and below the original one and $5\\cdot 5$ points for the squares to the right and left of the original triangle. Lastly each of the four isosceles triangles cover $4+3+2+1 = 10$ points. (Notice that although the length of the detour is 5, the points on the edge of the triangles were already counted).\nAdding these up, we get $3\\cdot 35 + 2\\cdot 25 + 4\\cdot 10 = 195 => \\boxed{195}$" ]

[ "\nWe notice that the moves cycle every 6 moves, so we plot out the first 6 moves on the coordinate grid with point $O$ as the origin. We will keep a tally of the x-coordinate and y-coordinate separately. Then, we will combine them and account for the cycling using the formula for an infinite geometric series.\nFirst move: The ant moves right $5$ . \nSecond move: We use properties of a $30-60-90$ triangle to get $\\frac{5}{4}$ right, $\\frac{5\\sqrt{3}}{4}$ up. \nThird move: $\\frac{5}{8}$ left, $\\frac{5\\sqrt{3}}{8}$ up.\nFourth move: $\\frac{5}{8}$ left.\nFifth move: $\\frac{5}{32}$ left, $\\frac{5\\sqrt{3}}{32}$ down.\nSixth move: $\\frac{5}{64}$ right, $\\frac{5\\sqrt{3}}{64}$ down.\nTotal of x-coordinate: $5 + \\frac{5}{4} - \\frac{5}{8} - \\frac{5}{8} - \\frac{5}{32} + \\frac{5}{64} = \\frac{315}{64}$ .\nTotal of y-coordinate: $0 + \\frac{5\\sqrt{3}}{4} + \\frac{5\\sqrt{3}}{8} + 0 - \\frac{5\\sqrt{3}}{32} - \\frac{5\\sqrt{3}}{64} = \\frac{105\\sqrt{3}}{64}$\nAfter this cycle of six moves, all moves repeat with a factor of $(\\frac{1}{2})^6 = \\frac{1}{64}$ . Using the formula for a geometric series, multiplying each sequence by $\\frac{1}{1-\\frac{1}{64}} = \\frac{64}{63}$ will give us the point $P$\nNow, knowing the initial $x$ and $y,$ we plug this into the geometric series formula ( $\\frac{a}{1-r}$ ), and we get $\\frac{315}{64} \\cdot \\frac{64}{63} = 5$ $\\frac{105\\sqrt{3}}{64} \\cdot \\frac{64}{63} = \\frac{5\\sqrt{3}}{3}$ .\nTherefore, the coordinates of point $P$ are $(5,\\frac{5\\sqrt{3}}{3})$ , so using the Pythagorean Theorem, $OP^2 = \\frac{100}{3}$ , for an answer of $\\boxed{103}$", "We place the ant at the origin of the complex plane with its first move being in the positive real direction. Then the ant's journey can be represented as the infinite series \\[5\\left(1 + \\frac{e^{\\frac{i\\pi}{3}}}{2} + \\left(\\frac{e^{\\frac{i\\pi}{3}}}{2}\\right)^2 + \\cdots\\right)\\] Using the formula for an infinite geometric series, this is equal to \\[\\frac{5}{1 - \\frac12e^{\\frac{i\\pi}{3}}} = \\frac{5}{1 - \\frac{1 + i\\sqrt{3}}{4}} = \\frac{20}{3 - i\\sqrt{3}} = 5 + \\frac{5i\\sqrt{3}}{3}\\] We are looking for the square of the modulus of this value: \\[\\left|\\frac{5 + 5i\\sqrt{3}}{3}\\right|^2 = 25 + \\frac{25}{3} = \\frac{100}{3}\\] so the answer is $100 + 3 = \\boxed{103}$", "The ant goes in the opposite direction every $3$ moves, going $(1/2)^3=1/8$ the distance backwards. Using geometric series, he travels $1-1/8+1/64-1/512...=(7/8)(1+1/64+1/4096...)=(7/8)(64/63)=8/9$ the distance of the first three moves over infinity moves. Now, we use coordinates meaning $(5+5/4-5/8, 0+5\\sqrt3/4+5\\sqrt3/8)$ or $(45/8, 15\\sqrt3/8)$ . Multiplying these by $8/9$ , we get $(5, 5\\sqrt3/3)$ $\\implies$ $\\boxed{103}$", "The bug goes forward and backward in three directions: straight east, northeast, and northwest. It travels $5-\\frac{5}{8}+\\frac{5}{64}-\\cdots=\\frac{40}{9}$ units east. Thus, it goes northeast $\\frac{\\frac{40}{9}}{2}=\\frac{20}{9}$ units northeast and $\\frac{10}{9}$ units northwest. Now, the bug travels a total of $\\frac{40}{9}+\\frac{10}{9}-\\frac{5}{9}=\\frac{45}{9}=5$ units east and a total of $\\frac{10\\sqrt{3}}{9}+\\frac{5\\sqrt{3}}{9}=\\frac{15\\sqrt{3}}{9}=\\frac{5\\sqrt{3}}{3}$ units north because of the 30-60-90 right triangles formed. Now, $OP^2=5^2+\\frac{5^2}{3}=\\frac{100}{3}$ by the Pythagorean Theorem, and the answer is $\\boxed{103}$", "The bug's bearings on each traversal are $0^\\circ, 60^\\circ, 120^\\circ,$ and so on; in general, the $n-$ th traversal has length $5\\cdot (1/2)^{n-1}$ and bearing $60(n-1).$ This means that the $x$ and $y$ displacements for the $n-$ th traversal are\n\\[(\\Delta x_n, \\Delta y_n)=(5\\cdot (1/2)^{n-1}\\cos (60(n-1))^\\circ,5\\cdot (1/2)^{n-1}\\sin (60(n-1))^\\circ).\\]\nSumming this over all the displacements, we get\n\\[x_P=\\sum_{n=1}^{\\infty} 5\\cdot (1/2)^{n-1}\\cos (60(n-1))^\\circ, y_P=\\sum_{n=1}^{\\infty} 5\\cdot (1/2)^{n-1}\\sin (60(n-1))^\\circ.\\]\nWe then have\n\\begin{align} OP^2 &= x_P^2+y_P^2 \\\\ &=\\sum_{n=1}^{\\infty} (5\\cdot (1/2)^{n-1}\\cos ^2(60(n-1))^\\circ) + (5\\cdot (1/2)^{n-1}\\sin ^2(60(n-1))^\\circ) \\\\ &= \\sum_{n=1}^{\\infty} (5\\cdot (1/2)^{n-1})^2(\\cos ^2 (60(n-1))^\\circ+\\sin ^2 (60(n-1))^\\circ) \\\\ &= \\sum_{n=1}^{\\infty} (25\\cdot (1/4)^{n-1}) \\\\ &= \\dfrac{25}{1-1/4} \\\\ &= 100/3. \\end{align}\nThus, the answer is $100+3=\\boxed{103}.$ --MenuThreeOne" ]

[ "There are $30$ grams of filler, so there are $120-30= 90$ grams that aren't filler. $\\frac{90}{120}=\\frac{3}{4}=\\boxed{75}$" ]

[ "Initially, suppose that the bus is at Stop $0$ (starting point) and Zia is at Stop $3.$\nWe construct the following table of $5$ -minute intervals: \\[\\begin{array}{c||c|c} & & \\\\ [-2.5ex] \\textbf{Time} & \\textbf{Bus's Location} & \\textbf{Zia's Location} \\\\ [0.5ex] \\hline & & \\\\ [-2ex] \\boldsymbol{5} \\ \\textbf{Minutes} & \\text{Stop} \\ 2 \\ \\text{(Waiting)} & \\text{Stop} \\ 4 \\\\ \\boldsymbol{10} \\ \\textbf{Minutes} & \\text{Midpoint of Stops} \\ 3 \\ \\text{and} \\ 4 & \\text{Stop} \\ 5 \\\\ \\boldsymbol{15} \\ \\textbf{Minutes} & \\text{Stop} \\ 5 \\ \\text{(Leaving)} & \\text{Stop} \\ 6 \\end{array}\\] Note that Zia will wait for the bus after $15$ minutes, and the bus will arrive $2$ minutes later.\nTherefore, the answer is $15+2=\\boxed{17}.$", "Since Zia will wait for the bus if the bus is at the previous stop, we can create an equation to solve for when the bus is at the previous stop. The bus travels $\\frac{1}{3}$ of a stop per minute, and Zia travels $\\frac{1}{5}$ of a stop per minute. Now we create the equation, $\\frac{1}{3}m = \\frac{1}{5}m + 3 - 1$ (the $-1$ accounts for us wanting to find when the bus reaches the stop before Zia's). Solving, we find that $m=15.$ Now Zia has to wait $2$ minutes for the bus to reach her, so our answer is $15+2=\\boxed{17}.$" ]

[ "A can of soup will feed $5$ children so $15$ children are feed by $3$ cans of soup. Therefore, there are $5-3=2$ cans for adults, so $3 \\times 2 =\\boxed{6}$ adults are fed." ]

[ "The car takes $120 \\text{ miles }\\cdot\\dfrac{1 \\text{ hr }}{30 \\text{ miles }}=4 \\text{ hr}$ to get from $A$ to $B$ . Also, it takes $120 \\text{ miles }\\cdot\\dfrac{1 \\text{ hr }}{40 \\text{ miles }}=3 \\text{ hr}$ to get from $B$ to $A$ . Therefore, the average speed is $\\dfrac{240\\text{ miles }}{7 \\text{ hr}}=34\\dfrac{2}{7}\\text{ mph}$ , which is closest to $\\boxed{34}$" ]