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[ "We note that the total length must be the same as the total height, as it is given in the problem. Calling the width of each small rectangle $2x$ , and the height $5x$ , we can see that the length is composed of $4$ widths and $5$ bars of length $2$ . This is equal to two heights of the small rectangles as well as $3$ bars of $2$ . Thus, $4(2x) + 5(2) = 2(5x) + 3(2)$ . We quickly find that $x = 2$ . The total side length is $4(4) + 5(2) = 2(10) + 3(2) = 26$ , or $\\boxed{26}$" ]

[ "Let the height of the panes equal $5x$ , and let the width of the panes equal $2x$ . Now notice that the total width of the borders equals $10$ , and the total height of the borders is $6$ . We have \\[10 + 4(2x) = 6 + 2(5x)\\] \\[x = 2\\] Now, the total side length of the window equals \\[10+ 4(2x) = 10 + 16 = \\boxed{26}\\]" ]

[ "In order to get the smallest palindrome greater than $15951$ , we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.\nSo we raise $9$ to the next largest value, $10$ , but obviously, that's not how place value works, so we're in the $16000$ s now. To keep this a palindrome, our number is now $16061$\nSo Megan drove $16061-15951=110$ miles. Since this happened over $2$ hours, she drove at $\\frac{110}{2}=\\boxed{55}$ mph. ~quacker88" ]

[ "Let total distance be $x$ . Her speed in miles per minute is $\\tfrac{x}{180}$ . Then, the distance that she drove before hitting the snowstorm is $\\tfrac{x}{3}$ . Her speed in snowstorm is reduced $20$ miles per hour, or $\\tfrac{1}{3}$ miles per minute. Knowing it took her $276$ minutes in total, we create equation: \\[\\text{Time before Storm}\\, + \\, \\text{Time after Storm} = \\text{Total Time} \\Longrightarrow\\] \\[\\frac{\\text{Distance before Storm}}{\\text{Speed before Storm}} + \\frac{\\text{Distance in Storm}}{\\text{Speed in Storm}} = \\text{Total Time} \\Longrightarrow \\frac{\\tfrac{x}{3}}{\\tfrac{x}{180}} + \\frac{\\tfrac{2x}{3}}{\\tfrac{x}{180} - \\tfrac{1}{3}} = 276\\]\nSolving equation, we get $x=135$ $\\Longrightarrow \\boxed{135}$" ]

[ "On the first day, the candidate moves $[4(0) + 1]^2/2\\ \\text{east},\\, [4(0) + 2]^2/2\\ \\text{north},\\, [4(0) + 3]^2/2\\ \\text{west},\\, [4(0) + 4]^2/2\\ \\text{south}$ , and so on. The E/W displacement is thus $1^2 - 3^2 + 5^2 \\ldots +37^2 - 39^2 = \\left|\\sum_{i=0}^9 \\frac{(4i+1)^2}{2} - \\sum_{i=0}^9 \\frac{(4i+3)^2}{2}\\right|$ . Applying difference of squares , we see that \\begin{align*} \\left|\\sum_{i=0}^9 \\frac{(4i+1)^2 - (4i+3)^2}{2}\\right| &= \\left|\\sum_{i=0}^9 \\frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}\\right|\\\\ &= \\left|\\sum_{i=0}^9 -(8i+4) \\right|. \\end{align*} The N/S displacement is \\[\\left|\\sum_{i=0}^9 \\frac{(4i+2)^2}{2} - \\sum_{i=0}^9 \\frac{(4i+4)^2}{2}\\right| = \\left|\\sum_{i=0}^9 -(8i+6) \\right|.\\] Since $\\sum_{i=0}^{9} i = \\frac{9(10)}{2} = 45$ , the two distances evaluate to $8(45) + 10\\cdot 4 = 400$ and $8(45) + 10\\cdot 6 = 420$ . By the Pythagorean Theorem , the answer is $\\sqrt{400^2 + 420^2} = 29 \\cdot 20 = \\boxed{580}$" ]

[ "From the information given, $35-5-1-1=28$ hits were single. Thus, the percentage was \\[\\dfrac{28}{35}=80\\% \\rightarrow \\boxed{80}.\\]" ]

[ "The original cube has $6$ square surfaces that each have an area of $3^2 = 9$ , for a toal surface area of $6\\cdot 9 = 54$\nSince no two corner cubes touch, we can examine the effect of removing each corner cube individually.\nEach corner cube contribues $3$ faces each of surface area $1$ to the big cube, so the surface area is decreased by $3$ when the cube is removed.\nHowever, when the cube is removed, $3$ faces on the 3x3x3 cube will be revealed, increasing the surface area by $3$\nThus, the surface area does not change with the removal of a corner cube, and it remains $54$ , which is answer $\\boxed{54}$" ]

[ "In addition to the original $12$ edges, each original vertex contributes $3$ new edges.\nThere are $8$ original vertices, so there are $12+3\\times 8=36$ edges in the new figure $\\rightarrow \\boxed{36}$" ]

[ "If Walter had done his chores for $10$ days without doing any of them well, he would have earned $3 \\cdot 10 = 30$ dollars. He got $6$ dollars more than this.\nHe gets a $5 - 3 = 2$ dollar bonus every day he does his chores well. Thus, he did his chores exceptionally well $\\frac{6}{2} = 3$ days, and the answer is $\\boxed{3}$", "If Walter had done his chores for $10$ days exceptionally well, he would have earned $5 \\cdot 10 = 50$ dollars. He got $50 - 36 = 14$ dollars less than this.\nHe gets $2$ dollars docked from his pay if he doesn't do his chores well. Therefore, he didn't do his chores well on $\\frac{14}{2} = 7$ days. The other $10 - 7 = 3$ days, he did them exceptionally well. Therefore, the answer is $\\boxed{3}$", "Let $b$ be the number of days Walter does his chores but doesn't do them well, and let $w$ be the number of days he does his chores exceptionally well.\n$b + w = 10$ since there are $10$ days Walter does chores.\n$3b + 5w = 36$ since $3b$ is the amount he earns from doing his chores not well, and $5w$ is the amount he earnes from doing his chores exceptionally well, and those two sum to $36$ dollars.\nMultiply the first equation by $3$ to get:\n$3b + 3w = 30$\n$3b + 5w = 36$\nSubtract the first equation from the second equation to get:\n$5w - 3w = 36 - 30$\n$2w = 6$\n$w = 3$\nThus, he does his chores well $3$ days, and the answer is $\\boxed{3}$" ]

[ "It is well known that $\\text{Distance}=\\text{Speed} \\cdot \\text{Time}$ . In the question, we want distance. From the question, we have that the time is $60$ minutes or $1$ hour. By the equation derived from $\\text{Distance}=\\text{Speed} \\cdot \\text{Time}$ , we have $\\text{Speed}=\\frac{\\text{Distance}}{\\text{Time}}$ , so the speed is $1$ mile per $x$ minutes. Because we want the distance, we multiply the time and speed together yielding $60\\text{ mins}\\cdot \\frac{1\\text{ mile}}{x\\text{ mins}}$ . The minutes cancel out, so now we have $\\dfrac{60}{x}$ as our distance for the first day. The distance for the following days are: \\[\\dfrac{60}{x},\\dfrac{60}{x+5},\\dfrac{60}{x+10},\\dfrac{60}{x+15}.\\] We know that $x,x+5,x+10,x+15$ are all factors of $60$ , therefore, $x=5$ because the factors have to be in an arithmetic sequence with the common difference being $5$ and $x=5$ is the only solution. \\[\\dfrac{60}{5}+\\dfrac{60}{10}+\\dfrac{60}{15}+\\dfrac{60}{20}=12+6+4+3=\\boxed{25}.\\]" ]

[ "We can begin by labeling the number of initial jellybeans $x$ . If she ate $20\\%$ of the jellybeans, then $80\\%$ is remaining. Hence, after day 1, there are: $0.8 * x$\nAfter day 2, there are: $0.8 * 0.8 * x$ or $0.64x$ jellybeans. $0.64x = 32$ , so $x = \\boxed{50}$", "Testing the answers choices out, we see that the answer is $\\boxed{50}$" ]

[ "We can begin by labeling the number of initial jellybeans $x$ . If she ate $20\\%$ of the jellybeans, then $80\\%$ is remaining. Hence, after day 1, there are: $0.8 * x$\nAfter day 2, there are: $0.8 * 0.8 * x$ or $0.64x$ jellybeans. $0.64x = 32$ , so $x = \\boxed{50}$", "Testing the answers choices out, we see that the answer is $\\boxed{50}$" ]

[ "Note that increasing the length of each edge by $50$ % with result in a cube that is similar to the original cube with scale factor $1.5$ . Therefore, the surface area will increase by a factor of $1.5^2$ , or $2.25$ . Converting this back into a percent, the percent increase will be $125$ %. Therefore, the answer is $\\boxed{125}$" ]

[ "Denote $\\alpha = \\tan^{-1} \\frac{\\sqrt{21}}{\\sqrt{31}}$ .\nDenote by $d$ the length of each side of a rhombus.\nNow, we put the solid to the 3-d coordinate space.\nWe put the bottom face on the $x-O-y$ plane.\nFor this bottom face, we put a vertex with an acute angle $2 \\alpha$ at the origin, denoted as $O$ .\nFor two edges that are on the bottom face and meet at $O$ , we put one edge on the positive side of the $x$ -axis. The endpoint is denoted as $A$ . Hence, $A = \\left( d , 0 , 0 \\right)$ .\nWe put the other edge in the first quadrant of the $x-O-y$ plane. The endpoint is denoted as $B$ . Hence, $B = \\left( d \\cos 2 \\alpha , d \\sin 2 \\alpha , 0 \\right)$\nFor the third edge that has one endpoint $O$ , we denote by $C$ its second endpoint.\nWe denote $C = \\left( u , v , w \\right)$ .\nWithout loss of generality, we set $w > 0$ .\nHence, \\[ u^2 + v^2 + w^2 = d^2 . \\hspace{1cm} (1) \\]\nWe have \\begin{align*} \\cos \\angle AOC & = \\frac{\\overrightarrow{OA} \\cdot \\overrightarrow{OC}}{|OA| \\cdot |OC|} \\\\ & = \\frac{u}{d} , \\hspace{1cm} (2) \\end{align*} and \\begin{align*} \\cos \\angle BOC & = \\frac{\\overrightarrow{OB} \\cdot \\overrightarrow{OC}}{|OB| \\cdot |OC|} \\\\ & = \\frac{u \\cos 2 \\alpha + v \\sin 2 \\alpha}{d} . \\hspace{1cm} (3) \\end{align*}\nCase 1: $\\angle AOC = \\angle BOC = 2 \\alpha$ or $2 \\left( 90^\\circ - \\alpha \\right)$\nBy solving (2) and (3), we get \\begin{align*} u & = \\pm d \\cos 2 \\alpha , \\\\ v & = \\pm d \\cos 2 \\alpha \\frac{1 - \\cos 2 \\alpha}{\\sin 2 \\alpha} \\\\ & = \\pm d \\cos 2 \\alpha \\tan \\alpha . \\end{align*}\nPlugging these into (1), we get \\begin{align*} w & = d \\sqrt{1 - \\cos^2 2 \\alpha - \\cos^2 2 \\alpha \\tan^2 \\alpha} \\\\ & = d \\sqrt{\\sin^2 2 \\alpha - \\cos^2 2 \\alpha \\tan^2 \\alpha} . \\hspace{1cm} (4) \\end{align*}\nCase 2: $\\angle AOC = 2 \\alpha$ and $\\angle BOC = 2 \\left( 90^\\circ - \\alpha \\right)$ , or $\\angle BOC = 2 \\alpha$ and $\\angle AOC = 2 \\left( 90^\\circ - \\alpha \\right)$\nBy solving (2) and (3), we get \\begin{align*} u & = \\pm d \\cos 2 \\alpha , \\\\ v & = \\mp d \\cos 2 \\alpha \\frac{1 + \\cos 2 \\alpha}{\\sin 2 \\alpha} \\\\ & = \\mp d \\cos 2 \\alpha \\cot \\alpha . \\end{align*}\nPlugging these into (1), we get \\begin{align*} w & = d \\sqrt{1 - \\cos^2 2 \\alpha - \\cos^2 2 \\alpha \\cot^2 \\alpha} \\\\ & = d \\sqrt{\\sin^2 2 \\alpha - \\cos^2 2 \\alpha \\cot^2 \\alpha} . \\hspace{1cm} (5) \\end{align*}\nWe notice that $(4) > (5)$ . Thus, (4) (resp. (5)) is the parallelepiped with a larger (resp. smaller) height.\nTherefore, the ratio of the volume of the larger parallelepiped to the smaller one is \\begin{align*} \\frac{(4)}{(5)} & = \\frac{\\sqrt{\\sin^2 2 \\alpha - \\cos^2 2 \\alpha \\tan^2 \\alpha}} {\\sqrt{\\sin^2 2 \\alpha - \\cos^2 2 \\alpha \\cot^2 \\alpha}} \\\\ & = \\sqrt{\\frac{\\tan^2 2 \\alpha - \\tan^2 \\alpha}{\\tan^2 2 \\alpha - \\cot^2 \\alpha}} . \\end{align*}\nRecall that $\\tan \\alpha = \\frac{\\sqrt{21}}{\\sqrt{31}}$ .\nThus, $\\tan 2 \\alpha = \\frac{2 \\tan \\alpha}{1 - \\tan^2 \\alpha} = \\frac{\\sqrt{21 \\cdot 31}}{5}$ .\nPlugging this into the equation above, we get \\begin{align*} \\frac{(4)}{(5)} & = \\frac{63}{62}. \\end{align*}\nTherefore, the answer is $63 + 62 = \\boxed{125}$", "Let one of the vertices be at the origin and the three adjacent vertices be $u$ $v$ , and $w$ . For one of the parallelepipeds, the three diagonals involving the origin have length $\\sqrt {21}$ . Hence, $(u+v)\\cdot (u+v)=u\\cdot u+v\\cdot v+2u\\cdot v=21$ and $(u-v)\\cdot (u-v)=u\\cdot u+v\\cdot v-2u\\cdot v=31$ . Since all of $u$ $v$ , and $w$ have equal length, $u\\cdot u=13$ $v\\cdot v=13$ , and $u\\cdot v=-2.5$ . Symmetrically, $w\\cdot w=13$ $u\\cdot w=-2.5$ , and $v\\cdot w=-2.5$ . Hence the volume of the parallelepiped is given by $\\sqrt{\\operatorname{det}\\begin{pmatrix}13&-2.5&-2.5\\\\-2.5&13&-2.5\\\\-2.5&-2.5&13\\end{pmatrix}}=\\sqrt{\\operatorname{det}\\begin{pmatrix}15.5&-15.5&0\\\\-2.5&13&-2.5\\\\0&-15.5&15.5\\end{pmatrix}}=\\sqrt{15.5^2\\operatorname\\det\\begin{pmatrix}1&-1&0\\\\-2.5&13&-2.5\\\\0&-1&1\\end{pmatrix}}=\\sqrt{15.5^2\\cdot 8}$\nFor the other parallelepiped, the three diagonals involving the origin are of length $\\sqrt{31}$ and the volume is $\\sqrt{\\operatorname{det}\\begin{pmatrix}13&2.5&2.5\\\\2.5&13&2.5\\\\2.5&2.5&13\\end{pmatrix}}=\\sqrt{\\operatorname{det}\\begin{pmatrix}10.5&-10.5&0\\\\2.5&13&2.5\\\\0&-10.5&10.5\\end{pmatrix}}=\\sqrt{10.5^2\\operatorname\\det\\begin{pmatrix}1&-1&0\\\\2.5&13&2.5\\\\0&-1&1\\end{pmatrix}}=\\sqrt{10.5^2\\cdot 18}$\nConsequently, the answer is $\\sqrt\\frac{10.5^2\\cdot 18}{15.5^2\\cdot 8}=\\frac{63}{62}$ , giving $\\boxed{125}$", "Observe that both parallelepipeds have two vertices (one on each base) that have three congruent angles meeting at them. Denote the parallelepiped with three acute angles meeting $P$ , and the one with three obtuse angles meeting $P'$\nThe area of a parallelepiped is simply the base area times the height, but because both parallelepipeds have the same base, what we want is just the ratio of the heights.\nDenote the point with three acute angles meeting at it in $P$ as $A$ , and its neighbors $B$ $C$ , and $D$ . Similarly, denote the point with three obtuse angles meeting at it in $P'$ as $A'$ , and its neighbors $B'$ $C'$ , and $D'$\nWe have the following equations:\n\\[\\textrm{Height of }P\\textrm{ from }ACD = \\frac{\\textrm{Vol}(ABCD) \\cdot 3}{[ACD]},\\] \\[\\textrm{Height of }P'\\textrm{ from }A'C'D' = \\frac{\\textrm{Vol}(A'B'C'D') \\cdot 3}{[A'C'D']}.\\]\nHowever, $ACD$ and $A'C'D'$ are both half the area of a rhombus with diagonals $\\sqrt{31}$ and $\\sqrt{21}$ , so our ratio is really\n\\[\\frac{P}{P'} = \\frac{\\textrm{Vol}(ABCD)}{\\textrm{Vol}(A'B'C'D')}.\\]\nBecause the diagonals of all of the faces are $\\sqrt{31}$ and $\\sqrt{21}$ , each edge of the parallelepipeds is $\\sqrt{13}$ by the Pythagorean theorem.\nWe have $AB = AC = AD = \\sqrt{13}$ , and $BC = CD = BD = \\sqrt{21}$ . When we drop a perpendicular to the centroid of $BCD$ from $A$ (let's call this point $O$ ), we have $BO = \\frac{\\sqrt{21}}{\\sqrt{3}} = \\sqrt{7}$ . Thus,\n\\[AB^2 - BO^2 = AO^2\\] \\[13 - 7 = AO^2 = 6\\] \\[AO = \\sqrt{6}.\\]\nThe area of base $BCD$ is $\\frac{21\\sqrt{3}}{4}$ . Hence,\n\\[\\textrm{Vol}(ABCD) = \\frac{\\sqrt{6}\\cdot\\frac{21\\sqrt{3}}{4}}{3}\\] \\[= \\frac{63\\sqrt{2}}{12}.\\]\nWe can apply a similar approach to $A'B'C'D'$\n$A'B' = A'C' = A'D' = \\sqrt{13}$ , and $B'C' = C'D' = B'D' = \\sqrt{31}$ . When we drop a perpendicular to the centroid of $B'C'D'$ from $A'$ (let's call this point $O'$ ), we have $B'O' = \\frac{\\sqrt{31}}{\\sqrt{3}} = \\sqrt{\\frac{31}{3}}$ . Thus,\n\\[A'B'^2 - B'O'^2 = A'O'^2\\] \\[13 - \\frac{31}{3} = A'O'^2\\] \\[A'O' = \\sqrt{8}{3} = \\frac{2\\sqrt{6}}{3}.\\]\nThe area of base $B'C'D'$ is $\\frac{31\\sqrt{3}}{4}$ . Hence,\n\\[\\textrm{Vol}(A'B'C'D') = \\frac{\\frac{2\\sqrt{6}}{3}\\cdot\\frac{31\\sqrt{3}}{4}}{3}\\] \\[= \\frac{186\\sqrt{2}}{36}\\] \\[= \\frac{62\\sqrt{2}}{12}.\\]\nFinally,\n\\[\\frac{P}{P'} = \\frac{\\textrm{Vol}(ABCD)}{\\textrm{Vol}(A'B'C'D')} = \\frac{\\frac{63\\sqrt{2}}{12}}{\\frac{62\\sqrt{2}}{12}} = \\frac{63}{62}.\\]\nOur answer is $63 + 62 = \\boxed{125}$", "Since the two parallelepipeds have the same base, all we need to do is to find their respective heights.\n\nAs illustrated in the above diagram, drop a perpendicular from $D$ onto the base at $P$ . Denote the center of the base by $O$ . By symmetry, $P$ must be on $AO$ . Now we need to find $DP$\nApply Pythagorean theorem to $\\triangle DPA$ we have \\[DP^2 = AD^2 - AP^2.\\]\nApply Pythagorean theorem to $\\triangle DPC$ and then $\\triangle CPO$ we have \\[DP^2 = DC^2 - CP^2 = DC^2 - (CO^2 + OP^2) = DC^2 - (CO^2 + (AO-AP)^2) = DC^2 - CO^2 - (AO-AP)^2.\\]\nCombining the above two, we have \\[AD^2 - AP^2 = DC^2 - CO^2 - (AO-AP)^2.\\]\nSince $AD=\\sqrt{13}$ $DC=\\sqrt{21}$ $CO=\\frac{\\sqrt{21}}{2}$ $AO=\\frac{\\sqrt{31}}{2}$ , plug them into the above equation and solve for the only unknown variable $AP$ , we get $AP=\\frac{5}{\\sqrt{31}}.$\nThus the height \\[DP = \\sqrt{AD^2 - AP^2} = \\sqrt{13 - \\frac{25}{31}} = \\sqrt{\\frac{378}{31}}.\\]\n\nFor the other parallelepiped, using the same approach and drop a perpendicular from $E'$ onto the base at $P'$ . Similarly applying Pythagorean theorem to $\\triangle E'P'C'$ $\\triangle E'P'A'$ and $\\triangle A'P'O'$ we have \\[C'E'^2 - C'P'^2 = A'E'^2 - A'O'^2 - (C'P'-C'O')^2.\\]\nPlugging known values into the above equation and solve for the only unknown variable $C'P'$ , we get $C'P'=\\frac{5}{\\sqrt{21}}.$\nThus the height \\[E'P' = \\sqrt{C'E'^2 - C'P'^2} = \\sqrt{13 - \\frac{25}{21}} = \\sqrt{\\frac{248}{21}}.\\]\nThe ratio between the two is therefore \\[\\frac{DP}{E'P'} = \\frac{\\sqrt{\\frac{378}{31}}}{\\sqrt{\\frac{248}{21}}} = \\sqrt{\\frac{2\\cdot3^3\\cdot7}{31}\\cdot\\frac{3\\cdot7}{2^3\\cdot31}} = \\frac{3^2\\cdot7}{2\\cdot31} = \\frac{63}{62}\\] , giving $\\boxed{125}$" ]

[ "Each half has $3$ red triangles, $5$ blue triangles, and $8$ white triangles. There are also $2$ pairs of red triangles, so $2$ red triangles on each side are used, leaving $1$ red triangle, $5$ blue triangles, and $8$ white triangles remaining on each half. Also, there are $3$ pairs of blue triangles, using $3$ blue triangles on each side, so there is $1$ red triangle, $2$ blue triangles, and $8$ white triangles remaining on each half. Also, we have $2$ red-white pairs. This obviously can't use $2$ red triangles on one side, since there is only $1$ on each side, so we must use $1$ red triangle and $1$ white triangle per side, leaving $2$ blue triangles and $7$ white triangles apiece. The remaining blue triangles cannot be matched with other blue triangles since that would mean there were more than $3$ blue pairs, so the remaining blue triangles must be paired with white triangles, yielding $4$ blue-white pairs, one for each of the remaining blue triangles. This uses $2$ blue triangles and $2$ white triangles on each side, leaving $5$ white triangles apiece, which must be paired with each other, so there are $5$ white-white pairs, $\\boxed{5}$" ]

[ "If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by $25$ cents.\nIf Jane bought $1$ bagel, then she bought $4$ muffins. Her total cost for the week would be $75\\cdot1+50\\cdot4=275$ cents, or $2.75$ dollars. Clearly, she bought one more bagel but one fewer muffin at a total cost of $3.00$ dollars. Therefore, she bought $\\boxed{2}$ bagels." ]

[ "If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by $25$ cents.\nIf Jane bought $1$ bagel, then she bought $4$ muffins. Her total cost for the week would be $75\\cdot1+50\\cdot4=275$ cents, or $2.75$ dollars. Clearly, she bought one more bagel but one fewer muffin at a total cost of $3.00$ dollars. Therefore, she bought $\\boxed{2}$ bagels." ]

[ "The probability of drawing a white marble from box $k$ is $\\frac{k}{k + 1}$ , and the probability of drawing a red marble from box $k$ is $\\frac{1}{k+1}$\nTo stop after drawing $n$ marbles, we must draw a white marble from boxes $1, 2, \\ldots, n-1,$ and draw a red marble from box $n.$ Thus, \\[P(n) = \\left(\\frac{1}{2} \\cdot \\frac{2}{3} \\cdot \\frac{3}{4} \\cdots \\frac {n - 1}{n}\\right) \\cdot \\frac{1}{n +1} = \\frac{1}{n (n + 1)}.\\]\nSo, we must have $\\frac{1}{n(n + 1)} < \\frac{1}{2010}$ or $n(n+1) > 2010.$\nSince $n(n+1)$ increases as $n$ increases, we can simply test values of $n$ ; after some trial and error, we get that the minimum value of $n$ is $\\boxed{45}$ , since $45(46) = 2070$ but $44(45) = 1980.$" ]

[ "The probability of drawing a white marble from box $k$ is $\\frac{k}{k + 1}$ , and the probability of drawing a red marble from box $k$ is $\\frac{1}{k+1}$\nTo stop after drawing $n$ marbles, we must draw a white marble from boxes $1, 2, \\ldots, n-1,$ and draw a red marble from box $n.$ Thus, \\[P(n) = \\left(\\frac{1}{2} \\cdot \\frac{2}{3} \\cdot \\frac{3}{4} \\cdots \\frac {n - 1}{n}\\right) \\cdot \\frac{1}{n +1} = \\frac{1}{n (n + 1)}.\\]\nSo, we must have $\\frac{1}{n(n + 1)} < \\frac{1}{2010}$ or $n(n+1) > 2010.$\nSince $n(n+1)$ increases as $n$ increases, we can simply test values of $n$ ; after some trial and error, we get that the minimum value of $n$ is $\\boxed{45}$ , since $45(46) = 2070$ but $44(45) = 1980.$" ]

[ "We first examine the possible arrangements for parity of number of balls in each box for $2022$ balls.\nIf a $E$ denotes an even number and a $O$ denotes an odd number, then the distribution of balls for $2022$ balls could be $EEE,EOO,OEO,$ or $OOE$ . With the insanely overpowered magic of cheese, we assume that each case is about equally likely.\nFrom $EEE$ , it is not possible to get to all odd by adding one ball; we could either get $OEE,EOE,$ or $EEO$ . For the other $3$ cases, though, if we add a ball to the exact right place, then it'll work.\nFor each of the working cases, we have $1$ possible slot the ball can go into (for $OEO$ , for example, the new ball must go in the center slot to make $OOO$ ) out of the $3$ slots, so there's a $\\dfrac13$ chance. We have a $\\dfrac34$ chance of getting one of these working cases, so our answer is $\\dfrac34\\cdot\\dfrac13=\\boxed{14.}$" ]

[ "We first examine the possible arrangements for parity of number of balls in each box for $2022$ balls.\nIf a $E$ denotes an even number and a $O$ denotes an odd number, then the distribution of balls for $2022$ balls could be $EEE,EOO,OEO,$ or $OOE$ . With the insanely overpowered magic of cheese, we assume that each case is about equally likely.\nFrom $EEE$ , it is not possible to get to all odd by adding one ball; we could either get $OEE,EOE,$ or $EEO$ . For the other $3$ cases, though, if we add a ball to the exact right place, then it'll work.\nFor each of the working cases, we have $1$ possible slot the ball can go into (for $OEO$ , for example, the new ball must go in the center slot to make $OOO$ ) out of the $3$ slots, so there's a $\\dfrac13$ chance. We have a $\\dfrac34$ chance of getting one of these working cases, so our answer is $\\dfrac34\\cdot\\dfrac13=\\boxed{14.}$" ]

[ "Let $a$ be the number of students that can only sing, $b$ can only dance, and $c$ can only act.\nLet $ab$ be the number of students that can sing and dance, $ac$ can sing and act, and $bc$ can dance and act.\nFrom the information given in the problem, $a + ab + b = 29, b + bc + c = 42,$ and $a + ac + c = 65$\nAdding these equations together, we get $2(a + b + c) + ab + bc + ac = 136$\nSince there are a total of $100$ students, $a + b + c + ab + bc + ac = 100$\nSubtracting these equations, we get $a + b + c = 36$\nOur answer is $ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \\boxed{64}$", "An easier way to solve the problem:\nSince $42$ students cannot sing, there are $100-42=58$ students who can.\nSimilarly $65$ students cannot dance, there are $100-65=35$ students who can.\nAnd $29$ students cannot act, there are $100-29=71$ students who can.\nTherefore, there are $58+35+71=164$ students in all ignoring the overlaps between $2$ of $3$ talent categories.\nThere are no students who have all $3$ talents, nor those who have none $(0)$ , so only $1$ or $2$ talents are viable.\nThus, there are $164-100=\\boxed{64}$ students who have $2$ of $3$ talents." ]

[ "For simplicity, we will name this cube $ABCDEFGH$ by vertices, as shown below. Note that for each face of this cube, two edges are labeled $0,$ and two edges are labeled $1.$ For all twelve edges of this cube, we conclude that six edges are labeled $0,$ and six edges are labeled $1.$\nWe apply casework to face $ABCD.$ Recall that there are $\\binom42=6$ ways to label its edges:\nTherefore, we have $4+16=\\boxed{20}$ such labelings in total.", "Since we want the sum of the edges of each face to be $2$ , we need there to be two $1$ s and two $0$ s on each face. Through experimentation, we find that either $2, 4,$ or all of them have $1$ s adjacent to $1$ s and $0$ s adjacent to $0$ on each face. WLOG, let the first face (counterclockwise) be $0,0,1,1$ . In this case we are trying to have all of them be adjacent to each other. First face: $0,0,1,1$ . Second face: $2$ choices: $1,0,0,1$ or $0,0,1,1$ . After that, it is basically forced and everything will fall in to place. Since we assumed WLOG, we need to multiply $2$ by $4$ to get a total of $8$ different arrangements.\nSecondly, $4$ of the faces have all of them adjacent and $2$ of the faces do not: WLOG counting counterclockwise, we have $0,0,1,1$ . Then, we choose the other face next to it. There are two cases, which are $0,1,0,1$ and $1,0,1,0$ . Therefore, this subcase has $4$ different arrangements. Then, we can choose the face at front to be $1,0,1,0$ . This has $4$ cases. The sides can either be $0,1,1,0$ or $1,1,0,0$ . Therefore, we have another $8$ cases.\nSumming these up, we have $8+4+8 = 20$ . Therefore, our answer is $\\boxed{20}$", "We see that if the $3$ edges connecting to $A$ has two $0$ 's, and one $1$ , it would have the same solutions as if it had two $1$ 's, and one $0$ . The solutions would just be inverted. As case 2.1 and case 2.2.2 are inverses, and case 2.2.1 has case 1 as an inverse, there would not be any additional solutions.\nSimilarly, if the $3$ edges connecting to $A$ has three $0$ 's, it would be the same as the inverse of case 1, or case 2.2.1, resulting in no new solutions.\nPutting all the cases together, we have $4+6+4+6=\\boxed{20}$ solutions.", "The problem states the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$ . That is, the sum of the labels on the $4$ edges of a face is equal to $2$ . The labels can only be $0$ or $1$ , meaning $2$ edges are labeled $1$ , the other $2$ are labeled $0$\nThis problem can be approached by Graph Coloring of Graph Theory . Note that each face of the cube connects to $4$ other faces, each with a shared edge. We use the following graph to represent the problem. Each vertex represents a face, each edge represent the cube's edge. Each vertex has $4$ edges connecting to $4$ other vertices. The edges can be colored red or blue, with red as label $0$ , and blue as label $1$ . Each vertex must have $2$ red edges and $2$ blue edges.\n$\\textbf{Case 1}$ $2$ adjacent red edges from vertex A. There are $4$ ways to choose $2$ red edges adjacent to each other and connect to $2$ vertices with an edge between them as shown below.\n$\\textbf{Case 1.1}$ $2$ adjacent red edges from vertex $A$ form a closed loop with a third red edge. There is only $1$ case as shown below.\n$\\textbf{Case 1.2}$ $2$ adjacent red edges from vertex $A$ does not form a closed loop with a third red edge. There are $3$ cases as shown below.\nIn case $1$ , there are total $4 \\cdot (1 + 3) = 16$ ways.\n$\\textbf{Case 2}$ $2$ red edges from vertex $A$ with $1$ blue edge in between. There are $2$ ways to choose $2$ red edges with $1$ blue edge in between.\nThere are only $2$ cases as shown below.\nIn case $2$ , there are total $2 \\cdot 2 = 4$ ways.\nFrom both case $1$ and case $2$ , there are $16 + 4 = \\boxed{20}$ ways in total." ]

[ "For simplicity purposes, we assume that the balls and the bins are both distinguishable.\nRecall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have \\[p=\\frac{5\\cdot4\\cdot\\binom{20}{3,5,4,4,4}}{5^{20}} \\text{ and } q=\\frac{\\binom{20}{4,4,4,4,4}}{5^{20}}.\\] Therefore, the answer is \\[\\frac pq=\\frac{5\\cdot4\\cdot\\binom{20}{3,5,4,4,4}}{\\binom{20}{4,4,4,4,4}}=\\frac{5\\cdot4\\cdot\\frac{20!}{3!\\cdot5!\\cdot4!\\cdot4!\\cdot4!}}{\\frac{20!}{4!\\cdot4!\\cdot4!\\cdot4!\\cdot4!}}=\\frac{5\\cdot4\\cdot(4!\\cdot4!\\cdot4!\\cdot4!\\cdot4!)}{3!\\cdot5!\\cdot4!\\cdot4!\\cdot4!}=\\frac{5\\cdot4\\cdot4}{5}=\\boxed{16}.\\] ~MRENTHUSIASM ~Jesshuang", "For simplicity purposes, we assume that the balls and the bins are both distinguishable.\nLet $q=\\frac{x}{a},$ where $a$ is the total number of combinations and $x$ is the number of cases where every bin ends up with $4$ balls.\nWe can take $1$ ball from one bin and place it in another bin so that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Note that one configuration of $4{-}4{-}4{-}4{-}4$ corresponds to $5\\cdot4\\cdot4=80$ configurations of $3{-}5{-}4{-}4{-}4.$ On the other hand, one configuration of $3{-}5{-}4{-}4{-}4$ corresponds to $5$ configurations of $4{-}4{-}4{-}4{-}4.$\nTherefore, we have \\[p = \\frac{80}{5}\\cdot\\frac{x}{a} = 16\\cdot\\frac{x}{a},\\] from which $\\frac{p}{q} = \\boxed{16}.$", "Since both of the cases will have $3$ bins with $4$ balls in them, we can leave those out. There are $2 \\cdot \\binom {5}{2} = 20$ ways to choose where to place the $3$ and the $5$ . After that, there are $\\binom {8}{3} = 56$ ways to put the $3$ and $5$ balls being put into the bins. For the $4,4,4,4,4$ case, after we canceled the $4,4,4$ out, we have $\\binom {8}{4} = 70$ ways to put the $4$ balls inside the bins. Therefore, we have $\\frac {56\\cdot 20}{70}$ which is equal to $8 \\cdot 2 = \\boxed{16}$", "Construct the set $A$ consisting of all possible $3{-}5{-}4{-}4{-}4$ bin configurations, and construct set $B$ consisting of all possible $4{-}4{-}4{-}4{-}4$ configurations. If we let $N$ be the total number of configurations possible, it's clear we want to solve for $\\frac{p}{q} = \\frac{\\frac{|A|}{N}}{\\frac{|B|}{N}} = \\frac{|A|}{|B|}$\nConsider drawing an edge between an element in $A$ and an element in $B$ if it is possible to reach one configuration from the other by moving a single ball (Note this process is reversible.). Let us consider the total number of edges drawn.\nFor any element in $A$ , we may choose one of the $5$ balls in the $5$ -bin and move it to the $3$ -bin to get a valid element in $B$ . This implies the number of edges is $5|A|$\nOn the other hand, for any element in $B$ , we may choose one of the $20$ balls and move it to one of the other $4$ -bins to get a valid element in $A$ . This implies the number of edges is $80|B|$\nWe equate the expressions to get $5|A| = 80|B|$ , from which $\\frac{|A|}{|B|} = \\frac{80}{5} = \\boxed{16}$" ]

[ "For simplicity purposes, we assume that the balls and the bins are both distinguishable.\nRecall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have \\[p=\\frac{5\\cdot4\\cdot\\binom{20}{3,5,4,4,4}}{5^{20}} \\text{ and } q=\\frac{\\binom{20}{4,4,4,4,4}}{5^{20}}.\\] Therefore, the answer is \\[\\frac pq=\\frac{5\\cdot4\\cdot\\binom{20}{3,5,4,4,4}}{\\binom{20}{4,4,4,4,4}}=\\frac{5\\cdot4\\cdot\\frac{20!}{3!\\cdot5!\\cdot4!\\cdot4!\\cdot4!}}{\\frac{20!}{4!\\cdot4!\\cdot4!\\cdot4!\\cdot4!}}=\\frac{5\\cdot4\\cdot(4!\\cdot4!\\cdot4!\\cdot4!\\cdot4!)}{3!\\cdot5!\\cdot4!\\cdot4!\\cdot4!}=\\frac{5\\cdot4\\cdot4}{5}=\\boxed{16}.\\] ~MRENTHUSIASM ~Jesshuang", "For simplicity purposes, we assume that the balls and the bins are both distinguishable.\nLet $q=\\frac{x}{a},$ where $a$ is the total number of combinations and $x$ is the number of cases where every bin ends up with $4$ balls.\nWe can take $1$ ball from one bin and place it in another bin so that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Note that one configuration of $4{-}4{-}4{-}4{-}4$ corresponds to $5\\cdot4\\cdot4=80$ configurations of $3{-}5{-}4{-}4{-}4.$ On the other hand, one configuration of $3{-}5{-}4{-}4{-}4$ corresponds to $5$ configurations of $4{-}4{-}4{-}4{-}4.$\nTherefore, we have \\[p = \\frac{80}{5}\\cdot\\frac{x}{a} = 16\\cdot\\frac{x}{a},\\] from which $\\frac{p}{q} = \\boxed{16}.$", "Since both of the cases will have $3$ bins with $4$ balls in them, we can leave those out. There are $2 \\cdot \\binom {5}{2} = 20$ ways to choose where to place the $3$ and the $5$ . After that, there are $\\binom {8}{3} = 56$ ways to put the $3$ and $5$ balls being put into the bins. For the $4,4,4,4,4$ case, after we canceled the $4,4,4$ out, we have $\\binom {8}{4} = 70$ ways to put the $4$ balls inside the bins. Therefore, we have $\\frac {56\\cdot 20}{70}$ which is equal to $8 \\cdot 2 = \\boxed{16}$", "Construct the set $A$ consisting of all possible $3{-}5{-}4{-}4{-}4$ bin configurations, and construct set $B$ consisting of all possible $4{-}4{-}4{-}4{-}4$ configurations. If we let $N$ be the total number of configurations possible, it's clear we want to solve for $\\frac{p}{q} = \\frac{\\frac{|A|}{N}}{\\frac{|B|}{N}} = \\frac{|A|}{|B|}$\nConsider drawing an edge between an element in $A$ and an element in $B$ if it is possible to reach one configuration from the other by moving a single ball (Note this process is reversible.). Let us consider the total number of edges drawn.\nFor any element in $A$ , we may choose one of the $5$ balls in the $5$ -bin and move it to the $3$ -bin to get a valid element in $B$ . This implies the number of edges is $5|A|$\nOn the other hand, for any element in $B$ , we may choose one of the $20$ balls and move it to one of the other $4$ -bins to get a valid element in $A$ . This implies the number of edges is $80|B|$\nWe equate the expressions to get $5|A| = 80|B|$ , from which $\\frac{|A|}{|B|} = \\frac{80}{5} = \\boxed{16}$" ]

[ "Let $S$ be the percent of people who study Spanish, $F$ be the number of people who study French, and let $S \\cap F$ be the number of students who study both. Then $\\left\\lceil 80\\% \\cdot 2001 \\right\\rceil = 1601 \\le S \\le \\left\\lfloor 85\\% \\cdot 2001 \\right\\rfloor = 1700$ , and $\\left\\lceil 30\\% \\cdot 2001 \\right\\rceil = 601 \\le F \\le \\left\\lfloor 40\\% \\cdot 2001 \\right\\rfloor = 800$ . By the Principle of Inclusion-Exclusion\n\\[S+F- S \\cap F = S \\cup F = 2001\\]\nFor $m = S \\cap F$ to be smallest, $S$ and $F$ must be minimized.\n\\[1601 + 601 - m = 2001 \\Longrightarrow m = 201\\]\nFor $M = S \\cap F$ to be largest, $S$ and $F$ must be maximized.\n\\[1700 + 800 - M = 2001 \\Longrightarrow M = 499\\]\nTherefore, the answer is $M - m = 499 - 201 = \\boxed{298}$" ]

[ "The union of two sets is equal to the sum of each set minus their intersection. The number of students that have both a dog and a cat is $20+26-39 = \\boxed{7}$", "We create a diagram: \nLet $x$ be the number of students with both a dog and a cat.\nTherefore, we have\n\\[26+20-x = 39\\] \\[46-x = 39\\] \\[x = \\boxed{7}\\]" ]

[ "In order to maximize the sum of the numbers, the numbers must have their digits ordered in decreasing value. There are only two numbers from the answer choices with this property: $76531$ and $87431$ . To determine the answer we will have to use estimation and the first two digits of the numbers.\nFor $76531$ the number that would maximize the sum would start with $98$ . The first two digits of $76531$ (when rounded) are $77$ . Adding $98$ and $77$ , we find that the first three digits of the sum of the two numbers would be $175$\nFor $87431$ the number that would maximize the sum would start with $96$ . The first two digits of $87431$ (when rounded) are $87$ . Adding $96$ and $87$ , we find that the first three digits of the sum of the two numbers would be $183$\nFrom the estimations, we can say that the answer to this problem is $\\boxed{87431}$", "In order to determine the largest number possible, we have to evenly distribute the digits when adding. The two numbers that show an example of this are $97531$ and $86420$ . The digits can be interchangeable between numbers because we only care about the actual digits.\nThe first digit must be either $9$ or $8$ . This immediately knocks out $\\textbf{(A)}\\ 76531$\nThe second digit must be either $7$ or $6$ . This doesn't cancel any choices.\nThe third digit must be either $5$ or $4$ . This knocks out $\\textbf{(B)}\\ 86724$ and $\\textbf{(D)}\\ 96240$\nThe fourth digit must be $3$ or $2$ . This cancels out $\\textbf{(E)}\\ 97403$\nThis leaves us with $\\boxed{87431}$", "If we use intuition, we know that the digits will be IN ORDER, to maximze the number. This eliminates $\\textbf{(B)}$ $\\textbf{(D)}$ ,and $\\textbf{(E)}$ . Additionally, the 2 two numbers must have 9 and 8 for the first digit to maximize the sum, eliminating $\\textbf{(A)}$ . This leaves $\\boxed{87431}$ .\n-written by litttle_master purely, not copied from anywhere" ]

[ "Since the question asks us about the unifying characteristic of all three equations' roots, we have to first determine them.\n$3x^2-2 = 25$ can be rewritten as $3x^2 - 27 = 0$ , which gives the following roots $+3$ and $-3$\n$(2x-1)^2 = (x-1)^2$ can be expanded to $4x^2-4x+1=x^2-2x+1$ , which in turn leads to $3x^2-2x=0$ . The roots here are $0$ and $\\frac{2}{3}$\n$\\sqrt{x^2-7}=\\sqrt{x-1}$ , when squared, also turns into a quadratic equation: $x^2 - x - 6 = 0$ . Binomial factoring gives us the roots $-2$ and $3$\nWe can clearly see that, between all of the equations, there is $\\boxed{3}$" ]

[ "The largest sum occurs when $13$ is placed in the center. This sum is $13 + 10 + 1 = 13 + 7 + 4 = \\boxed{24}$ . Note: Two other common sums, $18$ and $21$ , are also possible.", "Since the horizontal sum equals the vertical sum, twice this sum will be the sum\nof the five numbers plus the number in the center. When the center number is $13$ , the sum is the largest, \\[[10 + 4 + 1 + 7 + 2(13)]=2S\\\\ 48=2S\\\\ S=\\boxed{24}\\] The other\nfour numbers are divided into two pairs with equal sums." ]

[ "Assume that the area of each square is $1$ . Then, the area of the bolded region in the top left square is $\\dfrac{1}{4}$ . The area of the top right bolded region is $\\dfrac{1}{8}$ . The area of the bottom left bolded region is $\\dfrac{3}{8}$ . And the area of the bottom right bolded region is $\\dfrac{1}{4}$ . Add the four fractions: $\\dfrac{1}{4} + \\dfrac{1}{8} + \\dfrac{3}{8} + \\dfrac{1}{4} = 1$ . The four squares together have an area of $4$ , so the percentage bolded is $\\dfrac{1}{4} \\cdot 100 = \\boxed{25}$" ]

[ "There are different ways to approach this problem, and I'll start with the different factor of the numbers of the set $\\{ 1,2,3,4\\}$\n$1$ has factor $1$\n$2$ has factors $1$ and $2$\n$3$ has factors $1$ and $3$\n$4$ has factors $1$ $2$ , and $4$\nFrom here, we note that even though all numbers have the factor $1$ , only $4$ has another factor other than $1$ in the set (ie. $2$\nWe could therefore have one fraction be $\\frac{4}{2}$ and another $\\frac{3}{1}$\nThe sum of the numerators is $4+3=7=\\boxed{7}$" ]

[ "We can form the following expressions for the sum along each line: \\[\\begin{dcases}A+B+C\\\\A+E+F\\\\C+D+E\\\\B+D\\\\B+F\\end{dcases}\\] Adding these together, we must have $2A+3B+2C+2D+2E+2F=47$ , i.e. $2(A+B+C+D+E+F)+B=47$ . Since $A,B,C,D,E,F$ are unique integers between $1$ and $6$ , we obtain $A+B+C+D+E+F=1+2+3+4+5+6=21$ (where the order doesn't matter as addition is commutative), so our equation simplifies to $42 + B = 47$ . This means $B = \\boxed{5}$ .\n~RJ5303707", "Following the first few steps of Solution 1, we have $2(A+C+D+E+F)+3B=47$ . Because an even number ( $2(A+C+D+E+F)$ ) subtracted from an odd number (47) is always odd, we know that $3B$ is odd, showing that $B$ is odd. Now we know that $B$ is either 1, 3, or 5. If we try $B=1$ , we get $43=47$ which is not true. Testing $B=3$ , we get $45=47$ , which is also not true. Therefore, we have $B = \\boxed{5}$" ]

[ "Since the length ratio is $\\frac{1}{\\sqrt{2}}$ , then the area ratio is $\\frac{1}{2}$ (since the area ratio between two similar 2-dimensional objects is equal to the side ratio of those objects squared). This means that $S_2 = 8$ and $S_3 = \\boxed{4}$" ]

[ "The outer circle has radius $1+1+1=3$ , and thus area $9\\pi$ . The little circles have area $\\pi$ each; since there are 7, their total area is $7\\pi$ . Thus, our answer is $9\\pi-7\\pi=\\boxed{2}$" ]

[ "The outer circle has radius $1+1+1=3$ , and thus area $9\\pi$ . The little circles have area $\\pi$ each; since there are 7, their total area is $7\\pi$ . Thus, our answer is $9\\pi-7\\pi=\\boxed{2}$" ]

[ "There are $3$ people who are friends with only each other who won't be invited, plus $1$ person who has no friends, and $2$ people who are friends of friends of friends who won’t be invited. So the answer is $\\boxed{6}$" ]

[ "If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way to calculate $m/n$ . The Principle of Inclusion-Exclusion still requires us to find the individual probability of each box.\nLet $a, b$ represent the number of marbles in each box, and without loss of generality let $a>b$ . Then, $a + b = 25$ , and since the $ab$ may be reduced to form $50$ on the denominator of $\\frac{27}{50}$ $50|ab$ . It follows that $5|a,b$ , so there are 2 pairs of $a$ and $b: (20,5),(15,10)$\nThus, $m + n = \\boxed{026}$", "Let $w_1, w_2, b_1,$ and $b_2$ represent the white and black marbles in boxes 1 and 2.\nSince there are $25$ marbles in the box:\n$w_1 + w_2 + b_1 + b_2 = 25$\nFrom the fact that there is a $\\frac{27}{50}$ chance of drawing one black marble from each box:\n$\\frac{b_1 \\cdot b_2}{(b_1 + w_1)(b_2 + w_2)} = \\frac{27}{50} = \\frac{54}{100} = \\frac{81}{150}$\nThinking of the numerator and denominator separately, if $\\frac{27}{50}$ was not a reduced fraction when calculating out the probability, then $b_1 \\cdot b_2 = 27$ . Since $b_1 < 25$ , this forces the variables to be $3$ and $9$ in some permutation. Without loss of generality, let $b_1 = 3$ and $b_2 = 9$\nThe denominator becomes: $(3 + w_1)(9 + w_2) = 50$\nSince there have been $12$ black marbles used, there must be $13$ white marbles. Substituting that in:\n$(3 + w_1)(9 + (13 - w_1)) = 50$\n$(3 + w_1)(22 - w_1) = 50$\nSince the factors of $50$ that are greater than $3$ are $5, 10, 25,$ and $50$ , the quantity $3 + w_1$ must equal one of those. However, since $w_1 < 13$ , testing $2$ and $7$ for $w_1$ does not give a correct product. Thus, $\\frac{27}{50}$ must be a reduced form of the actual fraction.\nFirst assume that the fraction was reduced from $\\frac{54}{100}$ , yielding the equations $b_1\\cdot b_2 = 54$ and $(b_1 + w_1)(b_2 + w_2) = 100$ .\nFactoring $b_1 \\cdot b_2 = 54$ and saying WLOG that $b_1 < b_2 < 25$ gives $(b_1, b_2) = (3, 18)$ or $(6, 9)$ . Trying the first pair and setting the denominator equal to 100 gives: $(3 + w_1)(18 + w_2) = 100$\nSince $w_1 + w_2 = 4$ , the pairs $(w_1, w_2) = (1, 3), (2,2),$ and $(3,1)$ can be tried, since each box must contain at least one white marble. Plugging in $w_1 = w_2 = 2$ gives the true equation $(3 + 2)(18 + 2) =100$ , so the number of marbles are $(w_1, w_2, b_1, b_2) = (2, 2, 3, 18)$\nThus, the chance of drawing 2 white marbles is $\\frac{w_1 \\cdot w_2 }{(w_1+ b_1)(w_2 + b_2)} = \\frac{4}{100} = \\frac{1}{25}$ in lowest terms, and the answer to the problem is $1 + 25 = \\boxed{026}.$", "We know that $\\frac{27}{50} = \\frac{b_1}{t_1} \\cdot \\frac{b_2}{t_2}$ , where $b_1$ and $b_2$ are the number of black marbles in the first and the second box respectively, and $t_1$ and $t_2$ is the total number of marbles in the first and the second boxes respectively. So, $t_1 + t_2 = 25$ . Then, we can realize that $\\frac{27}{50} = \\frac{9}{10} \\cdot \\frac{3}{5} = \\frac{9}{10} \\cdot \\frac{9}{15}$ , which means that having 9 black marbles out of 10 total in the first box and 9 marbles out of 15 total the second box is valid. Then there is 1 white marble in the first box and 6 in the second box. So, the probability of drawing two white marbles becomes $\\frac{1}{10} \\cdot \\frac{6}{15} = \\frac{1}{25}$ . The answer is $1 + 25 = \\boxed{026}$" ]

[ "If he has enough money to buy $12$ pieces of red candy, $14$ pieces of green candy, and $15$ pieces of blue candy, then the smallest amount of money he could have is $\\text{lcm}{(12,14,15)} = 420$ cents. Since a piece of purple candy costs $20$ cents, the smallest possible value of $n$ is $\\frac{420}{20} = \\boxed{21}$", "We simply need to find a value of $20n$ that is divisible by $12$ $14$ , and $15$ . Observe that $20 \\cdot 18$ is divisible by $12$ and $15$ , but not $14$ $20 \\cdot 21$ is divisible by $12$ $14$ , and $15$ , meaning that we have exact change (in this case, $420$ cents) to buy each type of candy, so the minimum value of $n$ is $\\boxed{21}$", "We can notice that the number of purple candy times $20$ has to be divisible by $7$ , because of the $14$ green candies, and $3$ , because of the $12$ red candies. $7\\cdot3=21$ , so the answer has to be $\\boxed{21}$" ]

[ "If he has enough money to buy $12$ pieces of red candy, $14$ pieces of green candy, and $15$ pieces of blue candy, then the smallest amount of money he could have is $\\text{lcm}{(12,14,15)} = 420$ cents. Since a piece of purple candy costs $20$ cents, the smallest possible value of $n$ is $\\frac{420}{20} = \\boxed{21}$", "We simply need to find a value of $20n$ that is divisible by $12$ $14$ , and $15$ . Observe that $20 \\cdot 18$ is divisible by $12$ and $15$ , but not $14$ $20 \\cdot 21$ is divisible by $12$ $14$ , and $15$ , meaning that we have exact change (in this case, $420$ cents) to buy each type of candy, so the minimum value of $n$ is $\\boxed{21}$", "We can notice that the number of purple candy times $20$ has to be divisible by $7$ , because of the $14$ green candies, and $3$ , because of the $12$ red candies. $7\\cdot3=21$ , so the answer has to be $\\boxed{21}$" ]

[ "If the first year of the $8$ -year period was the final year of a principal's\nterm, then in the next six years two more principals would serve, and the last year of the\nperiod would be the first year of the fourth principal's term. Therefore, the maximum\nnumber of principals who can serve during an $8$ -year period is $4$ , so the answer is $\\boxed{4}$ if the terms are divided $1\\ |\\ 2\\ 3\\ 4\\ |\\ 5\\ 6\\ 7\\ |\\ 8$" ]

[ "There are $22-12+1=11$ rows of $33$ seats, giving $11\\times 33=\\boxed{363}$ seats." ]

[ " Note that $\\triangle ABC$ is an equilateral triangle . From the Pythagorean Theorem (or by using 30-60-90 triangles), $AD = 6\\sqrt{3}$ . That means $DE = 3\\sqrt{3}$ . Using the Pythagorean Theorem again, $BE = \\sqrt{63}$ , which is answer choice $\\boxed{63}$" ]

[ "Let a \"tile\" denote a $1\\times1$ square and \"square\" refer to $2\\times2$\nWe first have $4!=24$ possible ways to fill out the top left square. We then fill out the bottom right tile. In the bottom right square, we already have one corner filled out (from our initial coloring), and we now have $3$ options left to pick from.\nWe then look at the right middle tile. It is part of two squares: the top right and top left. Among these squares, $3$ colors have already been used, so we only have one more option for it. Similarly, every other square only has one more option, so we have a total of $3\\cdot4!=\\boxed{72}$ ways.", " We can split this problem into $2$ cases as shown above. We can swap a set of equal colors for another set of equal colors to create a new square.\nSquare 1:\nThe first square can be rotated to create another square so we have to multiply the number of arrangements by $2$ . We have $4! = 24$ arrangements without rotating and $24\\cdot 2 = 48$ arrangements in total for the first square.\nSquare 2:\nThere are $4! = 24$ ways to arrange the colors.\nIn total, we have $48 + 24 = \\boxed{72}$ arrangements.", "Let’s call the top-right corner color A, the top-middle color B, the top-right color C, and so on, with color D being the middle row, and right corner square, and color G being the bottom-left square’s color. WLOG A=Red, B=White, D=Blue, and E=Green. We will now consider squares C and F’s colors.\nCase 1 : C=Red and F=Blue\nIn this case, we get that G and H have to be Red and White in some order, and the same for H and I. We can color this in 2 ways.\nCase 2 : C=Blue and F=Red\nIn this case, one of G and H needs to be White and Red, and H and I needs to be White and Blue. There is 1 way to color this.\nIn total, we get 24*(2+1)=72 ways to color the grid. $\\boxed{72}$", "We will choose colors step-by-step:\n1. There are $4$ ways to choose a color in the center.\n2. Then we select any corner and there would be $3$ ways to choose a color as we can't use the same color as the one in the center.\n3. Consider the $2\\times 2$ square that contains the center and the corner we have selected. For the other $2$ squares, there are $2$ ways to choose colors.\n4. Now, consider how many configurations it makes sense to construct the $2\\times 2$ square opposite to the corner we have selected using the $2$ other $2\\times 2$ squares, and we get $3$ configurations.\nFinally, the answer is $4 \\cdot 3 \\cdot 2 \\cdot 3 = \\boxed{72}$", "Note that there can be no overlap between colors in each square.\nThen, we can choose $1$ color to be in the center. ${4 \\choose 1}$ = 4\nNow, we have some casework:\nCase 1: 1 color is placed in 4 corners and then others are placed on opposite edges. $232$ $414$ $232$ There's $3!=6$ ways to do this.\nCase 2: 2 colors are placed with 2 in adjacent corners and 1 edge opposite them. The final color is placed in the remaining 2 edges. $232$ $414$ $323$ The orientation of the 2 colors has 2 possibilities, and there are $3!$ color permutations. $2*3!=12$\nThere can't be any more ways to do this, as we have combined all cases such that each color is used once and only once per $2*2$ square.\nWe multiply the start with the sum of the 2 cases: $4(6+12)=\\boxed{72}$", "\nNote that we could fill the 3 by 3 square with numbers of choices, rather than letters or color names. The top-left corner receives a 3 because there are 3 total choices to choose from: R, G and B. The squares bordering them has values of 2 and 1, regardless of order. 2 indicates that the small square can have any color excluding the one existing color, 1 indicates the remaining color of the 2 by 2 square. Finally, the middle square receives 3, since the first 2 by 2 square has RGB already, and it does not matter what color it has. Moving onto the next 2 by 2 square, we see that there are already 2 decided colors, so the other squares must be 2 and 1, again, regardless of the order. The same applys to the third 2 by 2 square, and finally the last square has the value of one, as it is the only square left. Multiplying the numbers, $2\\times2\\times2\\times3\\time3\\times3$ $\\boxed{72}$" ]

[ "There are two cases for the initial configuration:\nTogether, the answer is $2+20=\\boxed{22}.$" ]

[ "There are two cases for the initial configuration:\nTogether, the answer is $2+20=\\boxed{22}.$" ]

[ "In each class, there are $18-2=16$ more students than rabbits. So for all classrooms, the difference between students and rabbits is $16 \\times 4 = \\boxed{64}$" ]

[ "Multiplying $18$ and $2$ by $4$ we get $72$ students and $8$ rabbits. We then subtract: $72 - 8 = \\boxed{64}.$", "In each class, there are $18-2=16$ more students than rabbits. So for all classrooms, the difference between students and rabbits is $16 \\times 4 = \\boxed{64}$" ]

[ "We can use complementary counting , counting all of the colorings that have at least one red $2\\times 2$ square.\nBy the Principle of Inclusion-Exclusion , there are (alternatively subtracting and adding) $128-40+8-1=95$ ways to have at least one red $2 \\times 2$ square.\nThere are $2^9=512$ ways to paint the $3 \\times 3$ square with no restrictions, so there are $512-95=417$ ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a $2 \\times 2$ red square is $\\frac{417}{512}$ , and $417+512=\\boxed{929}$", "We consider how many ways we can have 2*2 grid\n$(1)$ : All the grids are red-- $1$ case\n$(2)$ : One unit square is blue--The blue lies on the center of the bigger square, makes no 2*2 grid $9-1=8$ cases\n$(3)$ : Two unit squares are blue--one of the squares lies in the center of the bigger square, makes no 2*2 grid, $8$ cases. \nOr, two squares lie on second column, first row, second column third row; second row first column, second row third column, 2 extra cases. $\\binom 9 2-8-2=26$ cases\n$(4)$ Three unit squares are blue. We find that if a 2*2 square is formed, there are 5 extra unit squares can be painted. But cases that three squares in the same column or same row is overcomunted. So in this case, there are $4\\cdot (\\binom 5 3)-4=36$\n$(5)$ Four unit squares are blue, no overcomunted case will be considered. there are $4\\cdot \\binom 5 4=20$\n$(6)$ Five unit squares are blue, $4$ cases in all\nSum up those cases, there are $1+8+26+36+20+4=95$ cases that a 2*2 grid can be formed.\nIn all, there are $2^9=512$ possible ways to paint the big square, so the answer is $1-\\frac{95}{512}=\\frac{417}{512}$ leads to $\\boxed{929}$", "\\[\\begin{array}{|c|c|c|} \\hline C_{11} & C_{12} & C_{13}\\\\ \\hline C_{21} & C_{22} & C_{23}\\\\ \\hline C_{31} & C_{32} & C_{33}\\\\ \\hline \\end{array}\\]\nCase 1: The 3-by-3 unit-square grid has exactly $1$ 2-by-2 red square\nAssume the 2-by-2 red square is at $C_{11}, C_{12}, C_{21}, C_{22}$ . To make sure there are no more 2-by-2 red squares, $C_{31} \\text{and} C_{32}$ can't both be red and $C_{13} \\text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{31} \\text{and} C_{32}$ and $C_{13} \\text{and} C_{23}$ $C_{33}$ can be colored with either colors. However, the coloring method where $C_{23}, C_{32}, C_{33}$ are all red needs to be removed. For exactly one 2-by-2 red square at $C_{11}, C_{12}, C_{21}, C_{22}$ , there are $3 \\cdot 3 \\cdot 2 -1=17$ coloring methods. As there are $4$ locations for the 2-by-2 red square on the 3-by-3 unit-square grid, there are $17 \\cdot 4 = 68$ coloring methods.\nCase 2: The 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares\nCase 2.1: $2$ 2-by-2 red squares take up $6$ unit grids\nAssume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$ . To make sure there are no more 2-by-2 red squares, $C_{13} \\text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{13} \\text{and} C_{23}$ $C_{33}$ can be colored with either colors. However, the coloring method where $C_{13}, C_{23}, C_{33}$ are all red needs to be removed. For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$ , there are $3 \\cdot 2 -1=5$ coloring methods. As there are $4$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $5 \\cdot 4 = 20$ coloring methods.\nCase 2.2: $2$ 2-by-2 red squares take up $7$ unit grids\nAssume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ . To make sure there are no more 2-by-2 red squares, $C_{13}$ and $C_{31}$ can't be red. Meaning that there is $1$ coloring method for $C_{13}$ and $C_{31}$ . For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ , there is $1$ coloring method. As there are $2$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \\cdot 2 = 2$ coloring methods.\nHence, if the 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares, there are $20+2 = 22$ coloring methods.\nCase 3: The 3-by-3 unit-square grid has exactly $3$ 2-by-2 red squares\nAssume the $3$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ . To make sure there are no more 2-by-2 red squares, $C_{33}$ can't be red. Meaning that there is $1$ coloring method for $C_{33}$ . For exactly $3$ 2-by-2 red squares at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ , there is $1$ coloring method. As there are $4$ locations for the $3$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \\cdot 4 = 4$ coloring methods.\nCase 4: The 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares\nIf the 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares, all $9$ unit grids are red and there is $1$ coloring method.\nIn total, there are $68+22+4+1=95$ coloring methods with 2-by-2 red squares. \\[\\frac{m}{n}=1-\\frac{95}{2^9}=\\frac{417}{512}\\]\n\\[m+n=417+512=\\boxed{929}\\]" ]

[ "Let the rate of water flowing through valve $A$ be $a$ , the rate of water flowing through valve $B$ be $b$ , and the rate of water flowing through valve $C$ be $c$ WLOG , let the volume of the tank be 1 liter, and let the units for the rates be liters per hour. With this information, we can write three equations. \\[\\frac{1}{a+b+c} = 1\\] \\[\\frac{1}{a+c} = \\frac{3}{2}\\] \\[\\frac{1}{b+c} = 2\\] Manipulate each equation to get \\[1 = a+b+c\\] \\[\\frac{2}{3} = a+c\\] \\[\\frac{1}{2} = b+c\\] Solving for $a$ yields $a = \\tfrac{1}{2}$ , and solving for $b$ yields $b = \\tfrac{1}{3}$ . The number of hours to fill the tub with only valves $A$ and $B$ on is $\\frac{1}{\\frac{1}{2} + \\frac{1}{3}} = \\frac{6}{5} = \\boxed{1.2}$" ]

[ "Note that the sum of the numbers on each face must be 18, because $\\frac{1+2+\\cdots+8}{2}=18$\nSo now consider the opposite edges (two edges which are parallel but not on same face of the cube);\nthey must have the same sum value too.\nNow think about the points $1$ and $8$ . If they are not on the same edge, they must be endpoints of opposite edges, and we should have $1+X=8+Y$ . However, this scenario would yield no solution for $[7,2]$ , which is a contradiction. (Try drawing out the cube if it doesn't make sense to you.)\nThe points $1$ and $8$ are therefore on the same side and all edges parallel must also sum to $9$\nNow we have $4$ parallel sides $1-8, 2-7, 3-6, 4-5$ .\nThinking about $4$ endpoints, we realize they need to sum to $18$ .\nIt is easy to notice only $1-7-6-4$ and $8-2-3-5$ would work.\nSo if we fix one direction $1-8 ($ or $8-1)$ all other $3$ parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$\nNow, the problem is the same as arranging $4$ points in a two-dimensional square, which is $\\frac{4!}{4}=\\boxed{6.}$", "Again, all faces sum to $18.$ If $x,y,z$ are the vertices next to $1$ , then the remaining vertices are $17-x-y, 17-y-z, 17-x-z, x+y+z-16.$ Now it remains to test possibilities. Note that we must have $x+y+z>17.$ Without loss of generality, let $x<y<z.$ \\[3,7,8\\text{: Does not work.}\\] \\[4,6,8\\text{: Works.}\\] \\[4,7,8\\text{: Works.}\\] \\[5,6,7\\text{: Does not work.}\\] \\[5,6,8\\text{: Does not work.}\\] \\[5,7,8\\text{: Does not work.}\\] \\[6,7,8\\text{: Works.}\\]\nKeeping in mind that a) solutions that can be obtained by rotating each other count as one solution and b) a cyclic sequence is always asymmetrical, we can see that there are two solutions (one with $[x, y, z]$ and one with $[z, y, x]$ ) for each combination of $x$ $y$ , and $z$ from above. So, our answer is $3\\cdot 2=\\boxed{6.}$", "We know the sum of each face is $18.$ If we look at an edge of the cube whose numbers sum to $x$ , it must be possible to achieve the sum $18-x$ in two distinct ways, looking at the two faces which contain the edge. If $8$ and $6$ were on the same edge, it is possible to achieve the desired sum only with the numbers $1$ and $3$ since the values must be distinct. Similarly, if $8$ and $7$ were on the same edge, the only way to get the sum is with $1$ and $2$ . This means that $6$ and $7$ are not on the same edge as $8$ , or in other words they are diagonally across from it on the same face, or on the other end of the cube.\nNow we look at three cases, each yielding two solutions which are reflections of each other:\n1) $6$ and $7$ are diagonally opposite $8$ on the same face.\n2) $6$ is diagonally across the cube from $8$ , while $7$ is diagonally across from $8$ on the same face.\n3) $7$ is diagonally across the cube from $8$ , while $6$ is diagonally across from $8$ on the same face.\nThis means the answer is $3\\cdot 2=\\boxed{6.}$" ]

[ "Note that the sum of the numbers on each face must be 18, because $\\frac{1+2+\\cdots+8}{2}=18$\nSo now consider the opposite edges (two edges which are parallel but not on same face of the cube);\nthey must have the same sum value too.\nNow think about the points $1$ and $8$ . If they are not on the same edge, they must be endpoints of opposite edges, and we should have $1+X=8+Y$ . However, this scenario would yield no solution for $[7,2]$ , which is a contradiction. (Try drawing out the cube if it doesn't make sense to you.)\nThe points $1$ and $8$ are therefore on the same side and all edges parallel must also sum to $9$\nNow we have $4$ parallel sides $1-8, 2-7, 3-6, 4-5$ .\nThinking about $4$ endpoints, we realize they need to sum to $18$ .\nIt is easy to notice only $1-7-6-4$ and $8-2-3-5$ would work.\nSo if we fix one direction $1-8 ($ or $8-1)$ all other $3$ parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$\nNow, the problem is the same as arranging $4$ points in a two-dimensional square, which is $\\frac{4!}{4}=\\boxed{6.}$", "Again, all faces sum to $18.$ If $x,y,z$ are the vertices next to $1$ , then the remaining vertices are $17-x-y, 17-y-z, 17-x-z, x+y+z-16.$ Now it remains to test possibilities. Note that we must have $x+y+z>17.$ Without loss of generality, let $x<y<z.$ \\[3,7,8\\text{: Does not work.}\\] \\[4,6,8\\text{: Works.}\\] \\[4,7,8\\text{: Works.}\\] \\[5,6,7\\text{: Does not work.}\\] \\[5,6,8\\text{: Does not work.}\\] \\[5,7,8\\text{: Does not work.}\\] \\[6,7,8\\text{: Works.}\\]\nKeeping in mind that a) solutions that can be obtained by rotating each other count as one solution and b) a cyclic sequence is always asymmetrical, we can see that there are two solutions (one with $[x, y, z]$ and one with $[z, y, x]$ ) for each combination of $x$ $y$ , and $z$ from above. So, our answer is $3\\cdot 2=\\boxed{6.}$", "We know the sum of each face is $18.$ If we look at an edge of the cube whose numbers sum to $x$ , it must be possible to achieve the sum $18-x$ in two distinct ways, looking at the two faces which contain the edge. If $8$ and $6$ were on the same edge, it is possible to achieve the desired sum only with the numbers $1$ and $3$ since the values must be distinct. Similarly, if $8$ and $7$ were on the same edge, the only way to get the sum is with $1$ and $2$ . This means that $6$ and $7$ are not on the same edge as $8$ , or in other words they are diagonally across from it on the same face, or on the other end of the cube.\nNow we look at three cases, each yielding two solutions which are reflections of each other:\n1) $6$ and $7$ are diagonally opposite $8$ on the same face.\n2) $6$ is diagonally across the cube from $8$ , while $7$ is diagonally across from $8$ on the same face.\n3) $7$ is diagonally across the cube from $8$ , while $6$ is diagonally across from $8$ on the same face.\nThis means the answer is $3\\cdot 2=\\boxed{6.}$" ]

[ "\\[\\text{First, we notice that a rectangle is made from two pairs of vertices 1/2 turn away from each other.}\\]\n\\[\\textit{Note: The image is }\\frac{\\textit{280}}{\\textit{841}}\\approx\\frac{\\textit{1}}{\\textit{3}}\\textit{ size.}\\]\n\\[\\text{For there to be no rectangles, there can be at most one same-colored pair for each color and the rest need to have one red and blue.}\\]\n\\[\\textit{\\underline{Case 1: \\textbf{No pairs}}}\\]\n\\[\\text{Each pair has two ways to color: One red or the other red. Thus, there are }2^6=\\boxed{64}\\]" ]

[ "Note that the condition is equivalent to stating that there are no 2 pairs of oppositely spaced vertices with the same color.\nCase 1: There are no pairs. This yields $2$ options for each vertices 1-6, and the remaining vertices 7-12 are set, yielding $2^6=64$ cases.\nCase 2: There is one pair. Again start with 2 options for each vertex in 1-6, but now multiply by 6 since there are 6 possibilities for which pair can have the same color assigned instead of the opposite. Thus, the cases are: $2^6*6=384$\ncase 3: There are two pairs, but oppositely colored. Start with $2^6$ for assigning 1-6, then multiply by 6C2=15 for assigning which have repeated colors. Divide by 2 due to half the cases having the same colored opposites. $2^6*15/2=480$\nIt is apparent that no other cases exist, as more pairs would force there to be 2 pairs of same colored oppositely spaced vertices with the same color. Thus, the answer is: $64+384+480=\\boxed{928}$", "First, we identify the geometric condition for the sentence ``four vertices colored the\nsame color are the four vertices of a rectangle Consider any four vertices on the dodecagon, $A$ $B$ $C$ $D$ .\nDenote by $O$ the center of the dodecagon.\nBecause $OA = OB = OC$ $\\angle OAB = \\angle OBA$ and $\\angle OBC = \\angle OCB$\nThus, \\begin{align*} \\angle ABC & = \\angle OBA + \\angle OBC \\\\ & = \\frac{\\angle OBA + \\angle OAB}{2} + \\frac{\\angle OBC + \\angle OCB}{2} \\\\ & = \\frac{180^\\circ - \\angle AOB}{2} + \\frac{180^\\circ - \\angle COB}{2} \\\\ & = 180^\\circ - \\frac{\\angle AOB + \\angle COB}{2} \\\\ & = 180^\\circ - \\frac{\\angle AOC}{2} . \\end{align*}\nHence, $\\angle ABC = 90^\\circ$ if and only if $\\angle AOC = 180^\\circ$ .\nSimilarly, $\\angle ADC = 90^\\circ$ if and only if $\\angle AOC = 180^\\circ$ , and $\\angle BCD = 90^\\circ$ (or $\\angle DAB = 90^\\circ$ ) if and only if $\\angle BOD = 180^\\circ$\nTherefore, $ABCD$ is a rectangle if and only if two diagonals both pass through $O$\nNow, we categorize 12 vertices into 6 groups. Each group contains 2 diagonal vertices.\nNext, we compute the number of coloring configurations such that the above same-color rectangles do not exist.\nCase 1: Two vertices in each group has distinct colors.\nFor each group, we only need to determine which vertex is red. The other one must be blue.\nTherefore, the number of configurations in this case is $2^6$\nCase 2: There is one group who vertices have the same color. All other groups are with vertices that have distinct colors.\nWe construct such an instance in the following steps.\nStep 1: We determine which group has two vertices that have the same color.\nThe number of ways is 6.\nStep 2: For the selected group, we choose a color for its two vertices.\nThe number of ways is 2.\nStep 3: For each unselected group, we determine which vertex is red.\nThe number of ways is $2^5$\nFollowing from the rule of product, the total number of configurations in this case is $6 \\cdot 2 \\cdot 2^5 = 6 \\cdot 2^6$\nCase 3: One group has two red vertices, one group has two blue vertices, and each of the other four groups has vertices with distinct colors.\nWe construct such an instance in the following steps.\nStep 1: We determine which group has two vertices that have both red color.\nThe number of ways is 6.\nStep 2: We determine which group has two vertices that have both blue color.\nThe number of ways is 5.\nStep 3: For each unselected group, we determine which vertex is red.\nThe number of ways is $2^4$\nFollowing from the rule of product, the total number of configurations in this case is $6 \\cdot 5 \\cdot 2^4 = 30 \\cdot 2^4$\nPutting all cases together, the total number of feasible configurations is \\[ 2^6 + 6 \\cdot 2^6 + 30 \\cdot 2^4 = \\boxed{928}. \\]", "Note that a rectangle is formed if and only if it's diagonals pass through the center of the dodecagon and the diagonal's endpoints are the same color.\nConsider the $6$ diagonals that pass through the center. A rectangle is formed if the endpoints of some pair of them are all the same color. We can now perform casework on the diagonals:\nCase $1:$ The endpoints of all the diagonals are all different (e.g. for all diagonals, the endpoints are red and blue in some order).\nThe amount of ways to do this is $2^6=64,$ since there are $2$ ways to order which vertice is red and which is blue for each diagonal.\nCase $2:$ There exists one diagonal such that the endpoints are the same color.\nThere are $6$ ways to choose this diagonal, $2$ ways to choose which color it is, and $2^5$ ways to color the rest of the diagonals. Therefore, the total for this case is $6 \\cdot 2 \\cdot 2^5 = 384.$\nCase $3:$ There exists two diagonals such that the endpoints are the same color, but they are different colors\nThen, there are $6 \\cdot 5=30$ ways to choose these two diagonals, and $2^4=16$ ways to color the rest, so the total for this case is $30 \\cdot 16 = 480.$\nSumming all of the cases, we have $64+384+480=\\boxed{928}.$" ]

[ "This solution essentially explains other ways of thinking about the cases stated in Solution 2.\nCase 1:\n${6\\choose5} \\cdot 5!$\n5 colors need to be chosen from the group of 6. Those 5 colors have 5! distinct arrangements on the pentagon's vertices.\nCase 2:\n${6\\choose4} \\cdot4\\cdot5\\cdot3!$\n4 colors need to be chosen from the group of 6. Out of those 4 colors, one needs to be chosen to form a pair of 2 identical colors. Then, for arranging this layout onto the pentagon, one of the five sides (same thing as pair of adjacent vertices) of the pentagon needs to be established for the pair. The remaining 3 unique colors can be arranged 3! different ways on the remaining 3 vertices.\nCase 3:\n${6\\choose3} \\cdot {3\\choose2} \\cdot5\\cdot2$\n3 colors need to be chosen from the group of 6. Out of those 3 colors, 2 need to be chosen to be the pairs. Then, for arranging this layout onto the pentagon, start out by thinking about the 1 color that is not part of a pair, as it makes things easier. It can be any one of the 5 vertices of the pentagon. The remaining 2 pairs of colors can only be arranged 2 ways on the remaining 4 vertices.\nSolving each case and adding them up gets you 3120. $\\boxed{3120}$" ]

[ "Let vertex $A$ be any vertex, then vertex $B$ be one of the diagonal vertices to $A$ $C$ be one of the diagonal vertices to $B$ , and so on. We consider cases for this problem.\nIn the case that $C$ has the same color as $A$ $D$ has a different color from $A$ and so $E$ has a different color from $A$ and $D$ . In this case, $A$ has $6$ choices, $B$ has $5$ choices (any color but the color of $A$ ), $C$ has $1$ choice, $D$ has $5$ choices, and $E$ has $4$ choices, resulting in a possible of $6 \\cdot 5 \\cdot 1 \\cdot 5 \\cdot 4 = 600$ combinations.\nIn the case that $C$ has a different color from $A$ and $D$ has a different color from $A$ $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices (since $A$ and $B$ necessarily have different colors), $D$ has $4$ choices, and $E$ has $4$ choices, resulting in a possible of $6 \\cdot 5 \\cdot 4 \\cdot 4 \\cdot 4 = 1920$ combinations.\nIn the case that $C$ has a different color from $A$ and $D$ has the same color as $A$ $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices, $D$ has $1$ choice, and $E$ has $5$ choices, resulting in a possible of $6 \\cdot 5 \\cdot 4 \\cdot 1 \\cdot 5 = 600$ combinations.\nAdding all those combinations up, we get $600+1920+600=\\boxed{3120}$", "First, notice that there can be $3$ cases. One with all vertices painted different colors, one with one pair of adjacent vertices painted the same color and the final one with two pairs of adjacent vertices painted two different colors.\nCase $1$ : There are $6!$ ways of assigning each vertex a different color. $6! = 720$\nCase $2$ : There are $\\frac {6!}{2!} * 5$ ways. After picking four colors, we can rotate our pentagon in $5$ ways to get different outcomes. $\\frac {6!}{2} * 5 = 1800$\nCase $3$ : There are $\\frac {\\frac {6!}{3!} * 10}{2!}$ ways of arranging the final case. We can pick $3$ colors for our pentagon. There are $5$ spots for the first pair of colors. Then, there are $2$ possible ways we can put the final pair in the last $3$ spaces. But because the two pairs are indistinguishable, we divide by $2!$ $\\frac {\\frac {6!}{3!} * 10}{2!} = 600$\nAdding all the possibilities up, we get $720+1800+600=\\boxed{3120}$", "There are $6$ ways to assign a color to $A$ . WLOG, give vertex $A$ a color; we can multiply by $6$ at the end. Since vertices $A$ and $C$ cannot have the same color, there are $5$ ways to assign colors to vertex $C$ . Using this same logic, there are $5$ ways to assign a color to vertices $E$ $B$ , and $D$ , giving a total of $5^4=625$ ways. However, vertex $D$ cannot be the same color as vertex $A$ . To use complementary counting, we need to find the amount of ways for $D$ and $A$ to have the same color. Notice that this is equivalent to the amount of ways to arrange colors among the vertices of a square. Using the same logic as above, there are $5^3=125$ ways, except we must subtract the number of ways for a triangle. Each time, there is $1$ less vertex, so $5$ times less ways to color. This process stops when there are only $2$ vertices left; in this case there are simply $5$ ways to color this figure. \nSo in conclusion, there are $6(5^4-(5^3-(5^2-(5))))=6(5^4-5^3+5^2-5)=\\boxed{3120}$ ways.", "This problem is a direct application of the chromatic polynomial of a graph, represented by $P_k(G)$ , which returns a polynomial representing the number of ways to color a graph $G$ with $k$ colors such that no two adjacent vertices share the same color. In other words, it gives us a polynomial in terms of $k$ where we can plug in $k=6$ to get our answer.\nWell, if we want to find $P_k$ for a graph $G$ , we should probably first draw the graph, right? We can first draw five vertices (without any edges) in the shape of a pentagon. After connecting the diagonals, we get a star. We attempt to shift the vertices around to simplify the graph, which we quickly realize is isomorphic (can be turned into) the five-cycle, or the pentagon.\nFinding $P_k$ on a graph with a cycle of length at most 3 is straightforward -- we first pick a vertex v. It has 0 colored (visited) neighbors, so we can color it in k ways. We then move on to the vertices adjacent to v, etc, and at the end we multiply all these together. For example, the chromatic polynomial of a triangle is $k(k-1)(k-2)$ . However, cycles with length > 3 introduce ambiguity, and thus the above technique fails. Thus, we need to use the recursive formula $P_k(G) = P_k(G-e)-P_k(G*e)$ , where $G-e$ represents removing e and $G*e$ represents contracting e, or collapsing the two endpoints of e into one. When we hit a graph where the longest cycle has length 3, we can use the first technique to quickly find $P_k$\nAfter about four iterations of the algorithm, we get that the chromatic polynomial is $k(k-1)^4 - k(k-1)^3 + k(k-1)(k-2)$ . Plugging in 6 gives us $\\boxed{3120}$", "In pentagon $ABCDE$ , fix any vertex $A$ . Now draw diagonal $AC$ . There are six choices for vertex $A$ and $5$ choices for vertex $C$\nNow draw diagonal $CE$ . Since $E$ cannot be the same color as vertex $C$ , we have $5$ choices for $E$ . Again, we have five choices for vertex $D$ (draw diagonal $AD$ ).\nThus there are $6 \\cdot 5$ choices for vertices $A$ and $C$ and $5 \\cdot 5$ combinations for $D$ and $E$\nTo determine the final count, we consider two cases for the final $25$ combinations of $D$ and $E$ , which uniquely determines $B$ . Then, we multiply by $30$ since the choices of $A$ and $C$ are independent from these two cases.\nCase $1$ $D$ and $E$ are the same color. There are $4$ possible pairs (This is because $D$ and $E$ are not chosen from the same 5 colors. $D$ cannot be $A$ as they are on a diagonal, but $E$ can be. ~ primegn), and thus we have $5$ choices for $B$ . There are $20$ cases here.\nCase $2$ $D$ and $E$ are different. There are $25-4 = 21$ possible combinations and we have $4$ choices for $B$ (not the color of $D$ nor $E$ ). In this case we have $21 \\cdot 4 = 84$ cases.\nOur final count is $30(84+20) = \\boxed{3120}$", "WLOG, draw $ABCDE$ such that point $A$ is at the top, and write the letters in counterclockwise order. WLOG, fill in $A$ first. There are $6$ ways to do so. From here we proceed with casework on the color of $B$\nSumming up the cases, $600+600+1920=3120 \\Longrightarrow \\boxed{3120}$", "Notice that a minimum of, $3$ and a maximum of $5$ colours can be used.\nThis is straight forward and there are $6P5 = 720$ ways.\nIf $x$ and $y$ have same colours, they have to be adjacent. Let $x$ be directly right of $y$ $4$ consecutive points from $x$ clockwise have different colours. They can be coloured in $6P4=360$ ways. And $x$ can be chosen in $5$ ways, bringing the total to $360 \\times 5=1800$\nNote that $3$ points can't have the same colour as at least $1$ of the lines joining them will be a diagonal. Then the arrangement is $2-2-1$ . The point with distinct colour can be chosen in $5$ ways and can be coloured in $6$ ways. The next two points clockwise can be coloured in $5$ ways and the next two in $4$ ways. The total is $5 \\times 6 \\times 5 \\times 4=600$\nFinally, adding all cases, the answer is $600+1800+720=3120 \\Longrightarrow \\boxed{3120}$", "Name the pentagon $ADBEC$ . Then $A,B;$ $B,C;$ $C,D;$ $D,E$ and $E,A$ have to be of different colours.\n$A$ can be coloured in $6$ ways, $B$ in $5$ ways (Without the one used for $A$ ), $C$ in $5$ ways (Without the one used for $B$ ), $D$ in $5$ ways (Without the one used for $C$ ).\nNow we find the expected value of how many ways can $E$ be coloured.\nIf $A$ and $C$ are of the same colour, $A$ and $D$ are always different. This happens $\\frac{1}{5}$ times and leaves $4$ options for $E$\nIf $A$ and $C$ are of different colours $\\left(\\frac{4}{5} \\ \\text{of the time} \\right)$ then there is a $\\frac{1}{5}$ chance that $A$ and $D$ are of same colour and $\\frac{4}{5}$ chance that they are different. The first one leaves $5$ ways to colour $E$ and the second one leaves $4$\nCalculating all these, we find the expected value for the number of ways $E$ can be coloured is $\\frac{1}{5} \\times 4 + \\frac{4}{5} \\left( \\frac{1}{5} \\times 5 + \\frac{4}{5} \\times 4 \\right) = 4.16$\nTherefore total number or ways $= 6 \\times 5 \\times 5 \\times 5 \\times 4.16 = \\boxed{3120}$", "First, we know that there are $6$ ways to assign a color to vertex $A$ . Then, there are $6$ ways to assign a color to vertex $B$ , because it doesn't depend on vertex $A$ . There are also $5$ ways to assign a color to vertex $C$ , because it can't be the same color as vertex $A$ . Notice that both vertices $E$ and $D$ have 5 or 4 choices, depending on whether $A$ $B$ , and $C$ are the same color. This means that the we know that the amount of choices must be bigger than $6 \\times 6 \\times 5 \\times 4 \\times 4 = 2880$ , which eliminates options $A$ and $B$\nThen, notice that the amount of choices must be smaller than $6 \\times 6 \\times 5 \\times 4.5 \\times 4.5 = 3645$ , because the expected amount of choices $E$ and $D$ have are less than 4.5. (There is a greater probability that $A$ $B$ , and $B$ $C$ are both different). This eliminates option $E$ , which leaves us with options $C$ and $D$\nWe know that the amount of choices must be a multiple of $6$ , so $D$ is eliminated, leaving us with option $\\boxed{3120}$" ]

[ "Let the size of Ed's drink equal $x$ ounces, and let the size of Ann's drink equal $\\frac{3}{2}x$ ounces. After both consume $\\frac{3}{4}$ of their drinks, Ed and Ann have $\\frac{x}{4}$ and $\\frac{3x}{8}$ ounces of their drinks remaining. Ann gives away $\\frac{x}{8} + 2$ ounces to Ed.\nIn the end, Ed drank everything in his original lemonade plus what Ann gave him, and Ann drank everything in her original lemonade minus what she gave Ed. Thus we have \\[x + \\frac{x}{8} + 2 = \\frac{3x}{2} - \\frac{x}{8} - 2\\] \\[x = 16\\] The total amount the two of them drank is simply \\[x + \\frac{3}{2}x = 16 + 24 = \\boxed{40}\\]" ]

[ "Let the biking rate be $b$ , swimming rate be $s$ , jogging rate be $j$ , all in km/h.\nWe have $2b + 3j + 4s = 74,2j + 3s + 4b = 91$ . Subtracting the second from twice the first gives $4j + 5s = 57$ . Mod 4, we need $s\\equiv1\\pmod{4}$ . Thus, $(j,s) = (13,1),(8,5),(3,9)$\n$(13,1)$ and $(3,9)$ give non-integral $b$ , but $(8,5)$ gives $b = 15$ . Thus, our answer is $15^{2} + 8^{2} + 5^{2} = \\boxed{314}$", "Let $b$ $j$ , and $s$ be the biking, jogging, and swimming rates of the two people. Hence, $2b + 3j + 4s = 74$ and $4b + 2j + 3s = 91$ . Subtracting gives us that $2b - j - s = 17$ . Adding three times this to the first equation gives that $8b + s = 125\\implies b\\le 15$ . Adding four times the previous equation to the first given one gives us that $10b - j = 142\\implies b > 14\\implies b\\ge 15$ . This gives us that $b = 15$ , and then $j = 8$ and $s = 5$ . Therefore, $b^2 + s^2 + j^2 = 225 + 64 + 25 = \\boxed{314}$", "Creating two systems, we get $2x+3y+4z=74$ , and $2y+3z+4x=91$ . Subtracting the two expressions we get $y+z-2x=-17$ . Note that $-17$ is odd, so one of $x,y,z$ is odd. We see from our second expression that $z$ must be odd, because $91$ is also odd and $2y$ and $4x$ are odd. Thus, with this information, we can test cases quickly:\nWhen readdressing the first equation, we see that if $2x+3y$ will be a multiple of $6$ $4z \\equiv 2 \\pmod{6} = 5$ , we get that $x=15$ and $y=8$ , which works because of integer values. Therefore, $225+64+25=\\boxed{314}$", "Building on top of Solution 3, we can add $j+s-2b=17$ and $2b+3j+4s=74$ (sorry, I used different variables) to get $4j+5s=57$ . Logically speaking, most athletic people swim a lot faster than 1 km/h (0.62 mph), so we test out the next case that works, which is 5 km/h (3.1 mph). This seems much more logical, so we plug it into the equation to get $4j+25=57\\implies j=8$ . This seems reasonable, as people usually jog at around 8 to 16 km/h. Plugging these values into $2b+3j+4s=74$ , we get $b=15$ . 15 km/h is a little slow (most people bike at 20 km/h), but is still reasonable. So, we get $5^2+8^2+15^2=\\boxed{314}$" ]

[ "We are limited by the number of marbles whose right hand neighbor is not the same color as the marble. By surrounding every green marble with red marbles - RGRGRGRGRGR. That's 10 \"not the same colors\" and 0 \"same colors.\" Now, for every red marble we add, we will add one \"same color\" pair and keep all 10 \"not the same color\" pairs. It follows that we can add 10 more red marbles for a total of $m = 16$ . We can place those ten marbles in any of 6 \"boxes\": To the left of the first green marble, to the right of the first but left of the second, etc. up until to the right of the last. This is a stars-and-bars problem, the solution of which can be found as $\\binom{n+k}{k}$ where n is the number of stars and k is the number of bars. There are 10 stars (The unassigned Rs, since each \"box\" must contain at least one, are not counted here) and 5 \"bars,\" the green marbles. So the answer is $\\binom{15}{5} = 3003$ , take the remainder when divided by 1000 to get the answer: $\\boxed{003}$", "Begin with the above solution's reasoning to find that the ideal sequence is RGRGRGRGRGR + 10 Rs. To count the number of arrangements where the Gs are separate, group a G and an R together (GR) which will be rearranged/counted as one. However this also introduces the case of having GRGR.... which results in too few number of marbles whose right-hand neighbor is different, so we fix an R in the beginning of our sequence which also removes it from our counting calculation. Finally we are looking at the arrangement of 5 (GR)s + 10 Rs, all indistinguishable, which is $\\frac{15!}{5!10!}=3003\\Longrightarrow\\boxed{003}$" ]

[ "One such figure would be\n\nThe perimeter of this figure is $\\boxed{18}$" ]

[ "The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through $\\left(1,\\frac 12\\right)$ and $\\left(\\frac 32,2\\right)$ , which can be easily solved to be $6x = 2y + 5$ . Thus, $a^2 + b^2 + c^2 = \\boxed{065}$", "This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slope 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra we get: $6x = 2y + 5$ . The answer is then $6^2 + 2^2 + 5^2 = \\boxed{065}$ . This solution is an alternate explanation to solution 1." ]

[ "Draw an altitude from both end circles of the diagram with the circles of radius one, and call the lengths you get drawing the altitudes of the circles down to $BC$ $a$ and $b$ . Now we have the length of side $BC$ of being $(2)(2022)+1+1+a+b$ . However, the side $BC$ can also be written as $(6)(68)+34+34+34a+34b$ , due to similar triangles from the second diagram. If we set the equations equal, we have $\\frac{1190}{11} = a+b$ . Call the radius of the incircle $r$ , then we have the side BC to be $r(a+b)$ . We find $r$ as $\\frac{4046+\\frac{1190}{11}}{\\frac{1190}{11}}$ , which simplifies to $\\frac{10+((34)(11))}{10}$ ,so we have $\\frac{192}{5}$ , which sums to $\\boxed{197}$", "Assume that $ABC$ is isosceles with $AB=AC$\nIf we let $P_1$ be the intersection of $BC$ and the leftmost of the eight circles of radius $34$ $N_1$ the center of the leftmost circle, and $M_1$ the intersection of the leftmost circle and $AB$ , and we do the same for the $2024$ circles of radius $1$ , naming the points $P_2$ $N_2$ , and $M_2$ , respectively, then we see that $BP_1N_1M_1\\sim BP_2N_2M_2$ . The same goes for vertex $C$ , and the corresponding quadrilaterals are congruent.\nLet $x=BP_2$ . We see that $BP_1=34x$ by similarity ratios (due to the radii). The corresponding figures on vertex $C$ are also these values. If we combine the distances of the figures, we see that $BC=2x+4046$ and $BC=68x+476$ , and solving this system, we find that $x=\\frac{595}{11}$\nIf we consider that the incircle of $\\triangle ABC$ is essentially the case of $1$ circle with $r$ radius (the inradius of $\\triangle ABC$ , we can find that $BC=2rx$ . From $BC=2x+4046$ , we have:\n$r=1+\\frac{2023}{x}$\n$=1+\\frac{11\\cdot2023}{595}$\n$=1+\\frac{187}{5}$\n$=\\frac{192}{5}$\nThus the answer is $192+5=\\boxed{197}$", "Let $x = \\cot{\\frac{B}{2}} + \\cot{\\frac{C}{2}}$ . By representing $BC$ in two ways, we have the following: \\[34x + 7\\cdot 34\\cdot 2 = BC\\] \\[x + 2023 \\cdot 2 = BC\\]\nSolving we find $x = \\frac{1190}{11}$ . \nNow draw the inradius, let it be $r$ . We find that $rx =BC$ , hence \\[xr = x + 4046 \\implies r-1 = \\frac{11}{1190}\\cdot 4046 = \\frac{187}{5}.\\] Thus \\[r = \\frac{192}{5} \\implies \\boxed{197}.\\] ~AtharvNaphade", "First, let the circle tangent to $AB$ and $BC$ be $O$ and the other circle that is tangent to $AC$ and $BC$ be $R$ . Let $x$ be the distance from the tangency point on line segment $BC$ of the circle $O$ to $B$ . Also, let $y$ be the distance of the tangency point of circle $R$ on the line segment $BC$ to point $C$ . Realize that we can let $n$ be the number of circles tangent to line segment $BC$ and $r$ be the corresponding radius of each of the circles. Also, the circles that are tangent to $BC$ are similar. So, we can build the equation $BC = (x+y+2(n-1)) \\times r$ . Looking at the given information, we see that when $n=8$ $r=34$ , and when $n=2024$ $r=1$ , and we also want to find the radius $r$ in the case where $n=1$ . Using these facts, we can write the following equations:\n$BC = (x+y+2(8-1)) \\times 34 = (x+y+2(2024-1)) \\times 1 = (x+y+2(1-1)) \\times r$\nWe can find that $x+y = \\frac{1190}{11}$ . Now, let $(x+y+2(2024-1)) \\times 1 = (x+y+2(1-1)) \\times r$\nSubstituting $x+y = \\frac{1190}{11}$ in, we find that \\[r = \\frac{192}{5} \\implies \\boxed{197}.\\]", "Define $I, x_1, x_8, y_1, y_{2024}$ to be the incenter and centers of the first and last circles of the $8$ and $2024$ tangent circles to $BC,$ and define $r$ to be the inradius of triangle $\\bigtriangleup ABC.$ We calculate $\\overline{x_1x_8} = 34 \\cdot 14$ and $\\overline{y_1y_{2024}} = 1 \\cdot 4046$ because connecting the center of the circles voids two extra radii.\nWe can easily see that $B, x_1, x_8,$ and $I$ are collinear, and the same follows for $C, y_1, y_2024,$ and $I$ (think angle bisectors).\nWe observe that triangles $\\bigtriangleup I x_1 x_8$ and $\\bigtriangleup I y_1 y_{2024}$ are similar, and therefore the ratio of the altitude to the base is the same, so we note\n\\[\\frac{\\text{altitude}}{\\text{base}} = \\frac{r-34}{34\\cdot 14} = \\frac{r-1}{1\\cdot 4046}.\\]\nSolving yields $r = \\frac{192}{5},$ so the answer is $192+5 = \\boxed{197}.$" ]

[ "This problem can be approached by Graph Theory . Note that each face of the octahedron is connected to 3 other faces. We use the above graph to represent the problem. Each vertex represents a face of the octahedron, each edge represent the octahedron's edge.\nNow the problem becomes how many distinguishable ways to color the $8$ vertices such that two colored graphs are distinguishable if neither can be rotated and reflected to become the other.\nNotice that once the outer 4 vertices are colored, no matter how the inner 4 vertices are colored, the resulting graphs are distinguishable graphs.\nThere are $8$ colors and $4$ outer vertices, therefore there are $\\binom{8}{4}$ ways to color outer 4 vertices. Combination is used because the coloring has to be distinguishable when rotated and reflected. There are $4$ colors left, therefore there are $4!$ ways to color inner 4 vertices. Permutation is used because the coloring of the inner vertices have no restrictions. In total that is $\\binom{8}{4} \\cdot 4! = \\boxed{1680}$", "Let the colors be $1$ to $8$ inclusive, then rotate the octahedron such that color $1$ is on top. You have $7$ choices of what color is on the bottom, WLOG $2$ . Then, there's two rings of each $3$ colors on the top and bottom. For the top ring, you can choose any $3$ out of the $6$ remaining colors, and there's two ways to orient them. The octahedron is now fixed in place, so you can have $3!$ ways to put the three remaining colors in three spaces.\nIn total this is $7 \\cdot \\binom{6}{3} \\cdot 2 \\cdot 3!=\\boxed{1680}$" ]

[ "Since Judi's 7 friends had to pay $2.50 extra each to cover the total amount that Judi should have paid, we multiply $2.50\\cdot7=17.50$ is the bill Judi would have paid if she had money. Hence, to calculate the total amount, we multiply $17.50\\cdot8=\\boxed{140}$ to find the total the 8 friends paid.", "Let $m$ be the total bill the 8 friends split up; each person was supposed to pay $$\\frac{m}{8}$ . Since Judi forgot her money, we have to add $$\\frac{m}{8}+2.50$ to get the amount of money each of the $7$ friends have to pay. After multipling by 7, we have $7(\\frac{m}{8}+2.50)=\\frac{7m}{8} + 17.5$ . This is equal to $m$ because both sides represent the total bill. Solving for $m$ \\[\\frac{7m}{8} + 17.5=m\\] \\[\\frac{m}{8}=17.5\\] \\[m=\\boxed{140}.\\]" ]

[ "To choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. We also know that for any six points we pick, there is only $1$ way to connect the points such that a triangle is formed in the circle's interior (this is because we want no two chords to be nonintersecting ~Williamgolly). Therefore, the answer is ${{8}\\choose{6}}$ , which is equivalent to $\\boxed{28}$", "To make a triangle, where the $3$ points are arranged on a circle, you just need to choose $3$ points because no $3$ points are arranged in a straight line on a circle, meaning that to count the number of triangles we get $\\binom{8}{3}=56$ . However, this is incorrect because of the restriction that no three chords intersect in a single point in the circle meaning that the answer is most likely less than $56$ , and the only answer choice that satisfies this condition is $\\boxed{28}$ ~Batmanstark" ]

[ "The two points are one unit apart at $8$ places around the edge of the square. There are $8 \\choose 2$ $= 28$ ways to choose two points. The probability is\n\\[\\frac{8}{28} = \\boxed{27}\\]", "Arbitrarily pick a point in the grid. Clearly, we see two options for the other point to be placed, so the answer is $\\boxed{27}$" ]

[ "The key is to realize the significance that the figures are spheres, not circles . The 2D analogue of the diagram onto the flat surface will not contain 8 circles tangent to a ninth one; instead the circles will overlap since the middle sphere has a larger radius and will sort of “bulge” out.\n1998 AIME-10a.png\nLet us examine the relation between one of the outside 8 spheres and the center one (with radius $r$ ):\n1998 AIME-10b.png\nIf we draw the segment containing the centers and the radii perpendicular to the flat surface, we get a trapezoid ; if we draw the segment parallel to the surface that connects the center of the smaller sphere to the radii of the larger, we get a right triangle. Call that segment $x$ . Then by the Pythagorean Theorem\n\\[x^2 + (r-100)^2 = (r+100)^2 \\Longrightarrow x = 20\\sqrt{r}\\]\n1998 AIME-10c.png\n$x$ is the distance from one of the vertices of the octagon to the center, so the diagonal of the octagon is of length $2x =40\\sqrt{r}$ . We can draw another right triangle as shown above. One leg has a length of $200$ . The other can be found by partitioning the leg into three sections and using $45-45-90 \\triangle$ s to see that the leg is $100\\sqrt{2} + 200 + 100\\sqrt{2} = 200(\\sqrt{2} + 1)$ . Pythagorean Theorem:\n\\begin{eqnarray*} (40\\sqrt{r})^2 &=& 200^2 + [200(\\sqrt{2}+1)]^2\\\\ 1600r &=& 200^2[(1 + \\sqrt{2})^2 + 1] \\\\ r &=& 100 + 50\\sqrt{2} \\end{eqnarray*}\nThus $a + b + c = 100 + 50 + 2 = \\boxed{152}$", "Isolate a triangle, with base length $200$ (a side of the octagon). This triangle is obviously isoceles. Denote the other side length as $x$ . Since the interior angle is $45$ degrees (due to the shape being an octagon), then we can apply Law of Cosines to this triangle. We get: \\begin{eqnarray*} 200^2 &=& 2x^2 - 2x^2*cos(45^\\circ) \\\\ &=& 2x^2 - 2x^2*\\frac{\\sqrt{2}}{2} \\\\ &=& (2-\\sqrt{2})x^2 \\end{eqnarray*}\nAnd thus \\[x = \\frac{200}{\\sqrt{2-\\sqrt{2}}}\\]\nFrom the above, $x = 20\\sqrt{r}$ , so we get\n\\begin{eqnarray*} r &=& (\\frac{200}{20(\\sqrt{2-\\sqrt{2}})})^2 \\\\ &=& (\\frac{10}{\\sqrt{2-\\sqrt{2}})})^2 \\cdot \\frac{2+\\sqrt{2}}{2+\\sqrt{2}} \\\\ &=& \\frac{200 + 100\\sqrt{2}}{2} \\\\ &=& 100 + 50\\sqrt{2} \\end{eqnarray*}\nAnd hence the answer is $100 + 50 + 2 \\Rightarrow \\boxed{152}$" ]

[ "Since all the eleven members paid the same amount, that means that the total must be divisible by $11$ . We can do some trial-and-error to get $A=3$ , so our answer is $\\boxed{3}$ ~SparklyFlowers", "We know that a number is divisible by $11$ if the odd digits added together minus the even digits added together (or vice versa) is a multiple of $11$ . Thus, we have $1+2-A$ = a multiple of $11$ . The only multiple that works here is $0$ , as $11 \\cdot 0 = 0$ . Thus, $A = \\boxed{3}$ ~fn106068" ]

[ "Consider this position chart: \\[\\textbf{1 2 3 4 5 6 7 8 9 10 11 12}\\] Since there has to be an even number of spaces between each pair of the same color, spots $1$ $3$ $5$ $7$ $9$ , and $11$ contain some permutation of all $6$ colored balls. Likewise, so do the even spots, so the number of even configurations is $6! \\cdot 6!$ (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of $\\frac{12!}{(2!)^6}$ possible arrangements, so the probability is: \\[\\frac{6!\\cdot6!}{\\frac{12!}{(2!)^6}} = \\frac{6!\\cdot2^6}{7\\cdot8\\cdot9\\cdot10\\cdot11\\cdot12} = \\frac{2^4}{7\\cdot11\\cdot3} = \\frac{16}{231},\\] which is in simplest form. So, $m + n = 16 + 231 = \\boxed{247}$", "We can simply use constructive counting. First, let us place the red blocks; choose the first slot in $12$ ways, and the second in $6$ ways, because the number is cut in half due to the condition in the problem. This gives $12 \\cdot 6$ ways to place the red blocks. Similarly, there are $10 \\cdot 5$ ways to place the blue blocks, and so on, until there are $2 \\cdot 1$ ways to place the purple blocks. Thus, the probability is \\[\\frac{12 \\cdot 6 \\cdot 10 \\cdot 5 \\cdot 8 \\cdot 4 \\cdot 6 \\cdot 3 \\cdot 4 \\cdot 2 \\cdot 2 \\cdot 1}{12!}=\\frac{16}{231},\\] and the desired answer is $16+231=\\boxed{247}$" ]

[ "There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\\div40=132$ feet.\nEach of Oscar's leaps covers $132\\div12=11$ feet, and each of Elmer's strides covers $132\\div44=3$ feet.\nTherefore, Oscar's leap is $11-3=\\boxed{8}$ feet longer than Elmer's stride.", "There are $41-1=40$ gaps between the $41$ telephone poles, so Elmer takes $44 \\cdot 40 = 1760$ strides in total, and Oscar takes $12 \\cdot 40 = 480$ leaps in total. Therefore, the answer is $(5280 \\div 480) - (5280 \\div 1760) = 11-3=\\boxed{8}$" ]

[ "There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\\div40=132$ feet.\nEach of Oscar's leaps covers $132\\div12=11$ feet, and each of Elmer's strides covers $132\\div44=3$ feet.\nTherefore, Oscar's leap is $11-3=\\boxed{8}$ feet longer than Elmer's stride.", "There are $41-1=40$ gaps between the $41$ telephone poles, so Elmer takes $44 \\cdot 40 = 1760$ strides in total, and Oscar takes $12 \\cdot 40 = 480$ leaps in total. Therefore, the answer is $(5280 \\div 480) - (5280 \\div 1760) = 11-3=\\boxed{8}$" ]

[ "Suppose that his old car runs at $x$ km per liter. Then his new car runs at $\\frac{3}{2}x$ km per liter, or $x$ km per $\\frac{2}{3}$ of a liter. Let the cost of the old car's fuel be $c$ , so the trip in the old car takes $xc$ dollars, while the trip in the new car takes $\\frac{2}{3}\\cdot\\frac{6}{5}xc = \\frac{4}{5}xc$ . He saves $\\frac{\\frac{1}{5}xc}{xc} = \\boxed{20}$", "Because they do not give you a given amount of distance, we'll just make that distance $3x$ miles. Then, we find that the new car will use $2*1.2=2.4x$ . The old car will use $3x$ . Thus the answer is $(3-2.4)/3=.6/3=20/100= \\boxed{20}$", "You can find that the ratio of fuel used by the old car and the new car for the same amount of distance is $3 : 2$ , and the ratio between the fuel price of these two cars is $5 : 6$ . Therefore, by multiplying these two ratios, we get that the costs of using these two cars is \\[15 : 12 = 5 : 4\\] So the percentage of money saved is $1 - \\frac{4}{5} = \\boxed{20}$", "Assume WLOG that Elmer's old car's range is $100$ miles. So, Elmer's new car's range is $100 \\times 1.5 = 150$ miles. Also, assume that the gas Elmer's old car uses is $$10$ , which means that diesel will cost $$12$ . Now we can deduce that Elmer's old car uses $10 \\div 100 = $0.10$ per mile, and Elmer's new car uses $12 \\div 150 = $0.08$ per mile. Therefore, Elmer's new car saves $\\boxed{20}$ more money than his old car." ]

[ "Let $x$ be the length of the ship.\nThen, in the time that Emily walks $210$ steps, the ship moves $210-x$ steps.\nAlso, in the time that Emily walks $42$ steps, the ship moves $x-42$ steps.\nSince the ship and Emily have the same ratio of absolute speeds in either direction, $\\frac{210}{210-x} = \\frac{42}{x-42}$ . Dividing both sides by $42$ and cross multiplying, we get $5(x-42) = 210-x$ , so $6x = 420$ , and $x = \\boxed{70}$", "Let the speed at which Emily walks be $42$ steps per hour. Let the speed at which the ship is moving be $s$ . Walking in the direction of the ship, it takes her $210$ steps, or $\\frac {210}{42} = 5$ hours, to travel. We can create an equation: \\[d = 5(42-s),\\] where $d$ is the length of the ship. Walking in the opposite direction of the ship, it takes her $42$ steps, or $42/42 = 1$ hour. We can create a similar equation: \\[d = 1(42+s).\\] Now we have two variables and two equations. We can equate the expressions for $d$ and solve for $s$ \\begin{align*} 210-5s &= 42 + s \\\\ s &= 28. \\\\ \\end{align*} Therefore, we have $d = 42 + s = \\boxed{70}$", "Suppose that Emily and the ship take steps simultaneously such that Emily's steps cover a greater length than the ship's steps.\nLet $L$ be the length of the ship, $E$ be Emily's step length, and $S$ be the ship's step length. We wish to find $\\frac LE.$\nWhen Emily walks from the back of the ship to the front, she walks a distance of $210E$ and the front of the ship moves a distance of $210S.$ We have $210E=L+210S$ for this scenario, which rearranges to \\[210E-210S=L. \\hspace{15mm}(1)\\] When Emily walks in the opposite direction, she walks a distance of $42E$ and the back of the ship moves a distance of $42S.$ We have $42E=L-42S$ for this scenario, which rearranges to \\[42E+42S=L. \\hspace{19.125mm}(2)\\] We multiply $(2)$ by $5$ and then add $(1)$ to get $420E=6L,$ from which $\\frac LE = \\boxed{70}.$", "Every time Emily takes a step, the boat also \"takes a step\". Call the length of the boats step $s$ . Call the length of the boat $x$\nWhen Emily is walking in the same direction as the boat, every time she takes a step the boat moves an additional distance of $s$ . This means that she travels a total distance of $x + 210 s$ to reach the other end of the boat.\nWhen Emily is walking in the opposite direction of the boat, every time she takes a step the distance till the end of the boat reduces by $s$ (since the boat is coming towards her and moves a distance of $s$ ). This means that she travels a total distance of $x - 42 s$ to reach the other end of the boat.\nTaking Emily's step as a unit of distance, we now have two equations \\begin{align*} 210 &= x + 210 s, \\\\ 42 &= x - 42s. \\end{align*} Solving for $x$ you get $\\boxed{70}$", "Call the speed of the boat $v_s$ and the speed of Emily $v_e$\nConsider the scenario when Emily is walking along with the boat. Relative to an observer on the boat, her speed is $v_e-v_s$\nConsider the scenario when Emily is walking in the opposite direction. Relative to an observer on the boat, her speed is $v_e+v_s$\nSince Emily takes $210$ steps to walk along with the boat and $42$ steps to walk opposite the boat, that means it takes her $5$ times longer to walk the length of a stationary boat at $v_e-v_s$ compared to $v_e+v_s$\nThis means that $5(v_e-v_s)=v_e+v_s$ , so $v_s = \\frac{2v_e}{3}$\nAs Emily takes $210$ steps to walk the length of the boat at a speed of $v_e- \\frac{2v_e}{3}=\\frac{v_e}{3}$ , she must take $\\frac13$ of the time to walk the length of the boat at a speed of $v_e$ , so our answer is $\\frac{210}{3} = \\boxed{70}$" ]

[ "It is clear that $\\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\\triangle ABC$ , we get that $AC^2 = r^2+1^2-2r\\cos{\\frac{2\\pi}{3}} = r^2+r+1$ . Therefore, the area of $\\triangle ACE$ is $\\frac{\\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle.\nIf we extend $BC$ $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$ $BC$ and $DE$ meet at $Y$ , and $DE$ and $FA$ meet at $Z$ , we find that hexagon $ABCDEF$ is formed by taking equilateral triangle $XYZ$ of side length $r+2$ and removing three equilateral triangles, $ABX$ $CDY$ and $EFZ$ , of side length $1$ . The area of $ABCDEF$ is therefore\n$\\frac{\\sqrt{3}}{4}(r+2)^2-\\frac{3\\sqrt{3}}{4} = \\frac{\\sqrt{3}}{4}(r^2+4r+1)$\nBased on the initial conditions,\n\\[\\frac{\\sqrt{3}}{4}(r^2+r+1) = \\frac{7}{10}\\left(\\frac{\\sqrt{3}}{4}\\right)(r^2+4r+1)\\]\nSimplifying this gives us $r^2-6r+1 = 0$ . By Vieta's Formulas we know that the sum of the possible value of $r$ is $\\boxed{6}$" ]

[ "It is clear that $\\triangle ACE$ is an equilateral triangle. From the Law of Cosines on $\\triangle ABC$ , we get that $AC^2 = r^2+1^2-2r\\cos{\\frac{2\\pi}{3}} = r^2+r+1$ . Therefore, the area of $\\triangle ACE$ is $\\frac{\\sqrt{3}}{4}(r^2+r+1)$ by area of an equilateral triangle.\nIf we extend $BC$ $DE$ and $FA$ so that $FA$ and $BC$ meet at $X$ $BC$ and $DE$ meet at $Y$ , and $DE$ and $FA$ meet at $Z$ , we find that hexagon $ABCDEF$ is formed by taking equilateral triangle $XYZ$ of side length $r+2$ and removing three equilateral triangles, $ABX$ $CDY$ and $EFZ$ , of side length $1$ . The area of $ABCDEF$ is therefore\n$\\frac{\\sqrt{3}}{4}(r+2)^2-\\frac{3\\sqrt{3}}{4} = \\frac{\\sqrt{3}}{4}(r^2+4r+1)$\nBased on the initial conditions,\n\\[\\frac{\\sqrt{3}}{4}(r^2+r+1) = \\frac{7}{10}\\left(\\frac{\\sqrt{3}}{4}\\right)(r^2+4r+1)\\]\nSimplifying this gives us $r^2-6r+1 = 0$ . By Vieta's Formulas we know that the sum of the possible value of $r$ is $\\boxed{6}$" ]

[ "The inradius of $\\triangle ABC$ is $100\\sqrt 3$ and the circumradius is $200 \\sqrt 3$ . Now, consider the line perpendicular to plane $ABC$ through the circumcenter of $\\triangle ABC$ . Note that $P,Q,O$ must lie on that line to be equidistant from each of the triangle's vertices. Also, note that since $P, Q, O$ are collinear, and $OP=OQ$ , we must have $O$ is the midpoint of $PQ$ . Now, Let $K$ be the circumcenter of $\\triangle ABC$ , and $L$ be the foot of the altitude from $A$ to $BC$ . We must have $\\tan(\\angle KLP+ \\angle QLK)= \\tan(120^{\\circ})$ . Setting $KP=x$ and $KQ=y$ , assuming WLOG $x>y$ , we must have $\\tan(120^{\\circ})=-\\sqrt{3}=\\dfrac{\\dfrac{x+y}{100 \\sqrt{3}}}{\\dfrac{30000-xy}{30000}}$ . Therefore, we must have $100(x+y)=xy-30000$ . Also, we must have $\\left(\\dfrac{x+y}{2}\\right)^{2}=\\left(\\dfrac{x-y}{2}\\right)^{2}+120000$ by the Pythagorean theorem, so we have $xy=120000$ , so substituting into the other equation we have $90000=100(x+y)$ , or $x+y=900$ . Since we want $\\dfrac{x+y}{2}$ , the desired answer is $\\boxed{450}$", "Draw a good diagram. Draw $CH$ as an altitude of the triangle. Scale everything down by a factor of $100\\sqrt{3}$ , so that $AB=2\\sqrt{3}$ . Finally, call the center of the triangle U. Draw a cross-section of the triangle via line $CH$ , which of course includes $P, Q$ . From there, we can call $OU=h$ . There are two crucial equations we can thus generate. WLOG set $PU<QU$ , then we call $PU=d-h, QU=d+h$ . First equation: using the Pythagorean Theorem on $\\triangle UOB$ $h^2+2^2=d^2$ . Next, using the tangent addition formula on angles $\\angle PHU, \\angle UHQ$ we see that after simplifying $-d^2+h^2=-4, 2d=3\\sqrt{3}$ in the numerator, so $d=\\frac{3\\sqrt{3}}{2}$ . Multiply back the scalar and you get $\\boxed{450}$ . Not that hard, was it?", "To make numbers more feasible, we'll scale everything down by a factor of $100$ so that $\\overline{AB}=\\overline{BC}=\\overline{AC}=6$ . We should also note that $P$ and $Q$ must lie on the line that is perpendicular to the plane of $ABC$ and also passes through the circumcenter of $ABC$ (due to $P$ and $Q$ being equidistant from $A$ $B$ $C$ ), let $D$ be the altitude from $C$ to $AB$ . We can draw a vertical cross-section of the figure then: We let $\\angle PDI=\\alpha$ so $\\angle QDI=120^{\\circ}-\\alpha$ , also note that $\\overline{PO}=\\overline{QO}=\\overline{CO}=d$ . Because $I$ is the centroid of $ABC$ , we know that ratio of $\\overline{CI}$ to $\\overline{DI}$ is $2:1$ . Since we've scaled the figure down, the length of $CD$ is $3\\sqrt{3}$ , from this it's easy to know that $\\overline{CI}=2\\sqrt{3}$ and $\\overline{DI}=\\sqrt{3}$ . The following two equations arise: \\begin{align} \\sqrt{3}\\tan{\\left(\\alpha\\right)}+\\sqrt{3}\\tan{\\left(120^{\\circ}-\\alpha\\right)}&=2d \\\\ \\sqrt{3}\\tan{\\left(\\alpha\\right)} - d &= \\sqrt{d^{2}-12} \\end{align} Using trig identities for the tangent, we find that \\begin{align*} \\sqrt{3}\\tan{\\left(120^{\\circ}-\\alpha\\right)}&=\\sqrt{3}\\left(\\frac{\\tan{\\left(120^{\\circ}\\right)}+\\tan{\\left(\\text{-}\\alpha\\right)}}{1-\\tan{\\left(120^{\\circ}\\right)}\\tan{\\left(\\text{-}\\alpha\\right)}}\\right) \\\\ &= \\sqrt{3}\\left(\\frac{\\text{-}\\sqrt{3}+\\tan{\\left(\\text{-}\\alpha\\right)}}{1+\\sqrt{3}\\tan{\\left(\\text{-}\\alpha\\right)}}\\right) \\\\ &= \\sqrt{3}\\left(\\frac{\\text{-}\\sqrt{3}-\\tan{\\left(\\alpha\\right)}}{1-\\sqrt{3}\\tan{\\left(\\alpha\\right)}}\\right) \\\\ &= \\frac{\\sqrt{3}\\tan{\\left(\\alpha\\right)}+3}{\\sqrt{3}\\tan{\\left(\\alpha\\right)}-1}.\\end{align*} Okay, now we can plug this into $\\left(1\\right)$ to get: \\begin{align}\\sqrt{3}\\tan{\\left(\\alpha\\right)}+\\frac{\\sqrt{3}\\tan{\\left(\\alpha\\right)}+3}{\\sqrt{3}\\tan{\\left(\\alpha\\right)}-1}&=2d \\\\ \\sqrt{3}\\tan{\\left(\\alpha\\right)} - d &= \\sqrt{d^{2}-12} \\end{align} Notice that $\\alpha$ only appears in the above system of equations in the form of $\\sqrt{3}\\tan{\\left(\\alpha\\right)}$ , we can set $\\sqrt{3}\\tan{\\left(\\alpha\\right)}=a$ for convenience since we really only care about $d$ . Now we have \\begin{align}a+\\frac{a+3}{a-1}&=2d \\\\ a - d &= \\sqrt{d^{2}-12} \\end{align} Looking at $\\left(2\\right)$ , it's tempting to square it to get rid of the square-root so now we have: \\begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\\\ a - 2ad &= \\text{-}12 \\end{align*} See the sneaky $2d$ in the above equation? That we means we can substitute it for $a+\\frac{a+3}{a-1}$ \\begin{align*}a^{2}-2ad+d^{2}&=d^{2}-12 \\\\ a^{2} - a\\left(a+\\frac{a+3}{a-1}\\right) &= \\text{-}12 \\\\ a^{2}-a^{2}-\\frac{a^{2}+3a}{a-1} &=\\text{-}12 \\\\ -\\frac{a^{2}+3a}{a-1}&=\\text{-}12 \\\\ \\text{-}a^{2}-3a&=\\text{-}12a+12 \\\\ 0 &= a^{2}-9a+12 \\end{align*} Use the quadratic formula, we find that $a=\\frac{9\\pm\\sqrt{9^{2}-4\\left(1\\right)\\left(12\\right)}}{2\\left(1\\right)}=\\frac{9\\pm\\sqrt{33}}{2}$ - the two solutions were expected because $a$ can be $\\angle PDI$ or $\\angle QDI$ . We can plug this into $\\left(1\\right)$ \\begin{align*}a+\\frac{a+3}{a-1}&=2d \\\\ \\frac{9\\pm\\sqrt{33}}{2}+\\frac{\\frac{9\\pm\\sqrt{33}}{2}+3}{\\frac{9\\pm\\sqrt{33}}{2}-1}=2d \\\\ \\frac{9\\pm\\sqrt{33}}{2}+\\frac{15\\pm\\sqrt{33}}{7\\pm\\sqrt{33}} &= 2d\\end{align*} I'll use $a=\\frac{9+\\sqrt{33}}{2}$ because both values should give the same answer for $d$ \\begin{align*} \\frac{9+\\sqrt{33}}{2}+\\frac{15+\\sqrt{33}}{7+\\sqrt{33}} &= 2d \\\\ \\frac{\\left(9+\\sqrt{33}\\right)\\left(7+\\sqrt{33}\\right)+\\left(2\\right)\\left(15+\\sqrt{33}\\right)}{\\left(2\\right)\\left(7+\\sqrt{33}\\right)} &= 2d \\\\ \\frac{63+33+16\\sqrt{33}+30+2\\sqrt{33}}{14+2\\sqrt{33}} &= 2d \\\\ \\frac{126+18\\sqrt{33}}{14+2\\sqrt{33}} &= 2d \\\\ 9 &= 2d \\\\ \\frac{9}{2} &= d\\end{align*} Wait! Before you get excited, remember that we scaled the entire figure by $100$ ?? That means that the answer is $d=100\\times\\frac{9}{2}=\\boxed{450}$ .\n-fatant", "We use the diagram from solution 3. From basic angle chasing, \\[180=\\angle{QOC}+\\angle{CO}P=2\\angle{OCP}+2\\angle{OCQ}=2\\angle{QCP}\\] so triangle QCP is a right triangle. This means that triangles $CQI$ and $CPI$ are similar. If we let $\\angle{IDQ}=x$ and $\\angle{PDI}=y$ , then we know $x+y=120$ and \\[\\frac{PG}{GC}=\\frac{GC}{GQ}\\Rightarrow\\frac{100\\sqrt{3}\\tan{y}}{200\\sqrt{3}}=\\frac{200\\sqrt{3}}{100\\sqrt{3}\\tan{x}}\\Rightarrow\\tan{x}\\tan{y}=4\\] We also know that \\[PQ=2d=100\\sqrt{3}(\\tan{x}+\\tan{y})\\] \\[d=50\\sqrt{3}(\\tan{x}+\\tan{y})\\] \\[\\frac{d}{1-\\tan{x}\\tan{y}}=50\\sqrt{3}\\cdot\\frac{\\tan{x}+\\tan{y}}{1-\\tan{x}\\tan{y}}\\] \\[\\frac{d}{-3}=50\\sqrt{3}\\tan{(x+y)}\\] \\[d=-150\\sqrt{3}\\tan{120}=-150\\sqrt{3}(-\\sqrt{3})=\\boxed{450}\\]", "We use the diagram from solution 3.\nLet $BP = a$ and $BQ = b$ . Then, by Stewart's on $BPQ$ , we find \\[2x^3 + 2x^3 = a^2x + b^2x \\implies a^2 + b^2 = 4x^2.\\]\nThe altitude from $P$ to $ABC$ is $\\sqrt{a^2 - (200\\sqrt{3})^2}$ so \\[PQ = 2x = \\sqrt{a^2 - (200\\sqrt{3})^2} + \\sqrt{b^2 - (200\\sqrt{3})^2}.\\]\nFurthermore, the altitude from $P$ to $AB$ is $\\sqrt{a^2 - 300^2}$ , so, by LoC and the dihedral condition, \\[a^2 - 300^2 + b^2 - 300^2 + \\sqrt{a^2 - 300^2}\\sqrt{b^2-300^2} = 4x^2.\\]\nSquaring the equation for $PQ$ and substituting $a^2 + b^2 = 4x^2$ yields \\[2\\sqrt{a^2 - (200\\sqrt{3})^2}\\sqrt{b^2 - (200\\sqrt{3})^2} = 6\\cdot 200^2.\\]\nSubstituting $a^2 + b^2 = 4x^2$ into the other equation, \\[\\sqrt{a^2 - 300^2}\\sqrt{b^2-300^2} = 2\\cdot 300^2.\\]\nSquaring both of these gives \\[a^2b^2-3\\cdot 200^2(a^2 + b^2) + 9\\cdot 200^4 = 9\\cdot 200^4\\] \\[a^2b^2 - 300^2(a^2+b^2) + 300^4 = 4\\cdot 300^4.\\]\nSubstituting $a^2 + b^2 = 4x^2$ and solving for $x$ gives $\\boxed{450}$ , as desired.", "Let $AB = a, M$ be midpoint $BC, I$ be the center of equilateral $\\triangle ABC,$ $IM = b = \\frac {a}{2\\sqrt{3}}, O$ be the center of sphere $ABCPQ.$ Then \\[AI = 2b, AO = BO = PO =QO = d.\\] \\[QA=QB=QC,PA=PB=PC \\implies\\] \\[POIQ\\perp ABC, \\angle PMQ = 120^\\circ.\\] (See upper diagram).\nWe construct the circle PQMD, use the formulas for intersecting chords and get \\[DI = 5b, FI = EO = \\frac{3b}{2}\\] \\[\\implies FM = \\frac{5b}{2}.\\] (See lower diagram).\nWe apply the Law of Sine to $\\triangle PMQ$ and get \\[2EM \\sin 120^\\circ =PQ\\] \\[\\implies r \\sqrt{3} = 2d\\] \\[\\implies 3r^2 = 4d^2.\\] We apply the Pythagorean Law on $\\triangle AOI$ and $\\triangle EFM$ and get \\[d^2 = 4b^2 + OI^2, r^2 = \\frac {25b^2}{4} + EF^2 \\implies\\] \\[r = 3b\\implies d = \\frac {3a}{2} = \\boxed{450}.\\] vladimir.shelomovskii@gmail.com, vvsss", "Let $M$ be the midpoint of $\\overline{AB}$ and $X$ the center of $\\triangle ABC$ . Then \\[P, O, Q, M, X, C\\] all lie in the same vertical plane. We can make the following observations:\nTo make calculations easier, we will denote $100\\sqrt{3}=m$ , so that $MX=m$ and $XC=2m$\n\nDenote $PX=p$ and $QX=q$ , where the tangent addition formula on $\\triangle PMQ$ yields \\[\\frac{\\tan\\measuredangle PMX+\\tan\\measuredangle QMX}{1-\\tan\\measuredangle PMX\\tan\\measuredangle QMX}=\\tan(120^{\\circ})=-\\sqrt{3}.\\] Using $\\tan\\measuredangle PMX=\\frac{p}{m}$ and $\\tan\\measuredangle QMX=\\frac{q}{m}$ , we have \\[\\frac{\\frac{p}{m}+\\frac{q}{m}}{1-\\frac{p}{m}\\cdot\\frac{q}{m}}=-\\sqrt{3}.\\] After multiplying both numerator and denominator by $m^{2}$ we have \\[\\frac{(p+q)m}{m^{2}-pq}=-\\sqrt{3}.\\] But note that $pq=(2m)(2m)=4m^{2}$ by power of a point at $X$ , where we deduce by symmetry that $MM^{\\prime}=MX=m$ on the diagram below: \nThus \\begin{align*} \\frac{(p+q)m}{m^{2}-4m^{2}}=-\\sqrt{3} \\\\ \\frac{p+q}{-3m}=-\\sqrt{3} \\\\ p+q=\\left(-\\sqrt{3}\\right)\\left(-3m\\right) \\\\ p+q=3\\sqrt{3}\\cdot m.\\end{align*} Earlier we assigned the variable $m$ to the length $100\\sqrt{3}$ which implies $PQ=\\left(3\\sqrt{3}\\right)\\left(100\\sqrt{3}\\right)=900$ . Thus the distance $d$ is equal to $\\frac{PQ}{2}=\\boxed{450}$", "Let $Z$ be the center of $\\triangle ABC$ . Let $A’$ be the midpoint of $BC$ . Let $ZA’ = c = 100\\sqrt{3}$ and $ZA = 2c = 200\\sqrt{3}$ . Let $PZ = a$ and $QZ = b$ . We will be working in the plane that contains the points: $A$ $P$ $A’$ $Q$ $O$ , and $Z$\nSince $P$ $O$ , and $Q$ are collinear and $PO = QO = AO$ $\\triangle PAQ$ is a right triangle with $\\angle PAQ = 90^{\\circ}$ . Since $AZ \\perp PQ$ $(PZ)(QZ) = (AZ)^2 = ab = (2c)^2 = 120000$\n$PA’ = \\sqrt{a^2 + c^2}$ $QA’ = \\sqrt{b^2 + c^2}$ $PQ = a + b$ , and $\\angle PAQ = 120^{\\circ}$ . By Law of Cosines \\[(a + b)^2 = a^2 + b^2 + 2c^2 + \\sqrt{a^2b^2 + a^2c^2 + b^2c^2 + c^4}\\] . Substituting $4c^2$ for $ab$ and simplifying, we get \\[6c = \\sqrt{17c^2 + a^2 + b^2}\\] . Squaring and simplifying, we get \\[a^2 + b^2 = 19c^2 = 570000\\] . Adding $2ab = 8c^2$ to both sides we get $PQ = a + b = 900$ . Since $O$ is the midpoint of $PQ$ $d = PO = \\boxed{450}$" ]

[ "Screen Shot 2020-02-17 at 2.24.50 PM.png\nWe create a diagram. Each of the red quadrilaterals are actually 60 degree rotations about point A. We easily get that $\\overline{C E_1} = \\sqrt{11}$ , and $\\overline{C E_3} = \\sqrt{11}$ . If we set $\\angle B A D_2 = \\theta$ , we can start angle chasing. In particular, we will like to find $\\angle D E_4 C$ , and $\\angle D E_2 C$ , since then we will be able to set up some Law Of Cosines. $\\angle D E_4 C = \\angle D E_4 A + \\angle A E_4 C = 90 - \\frac{\\theta}{2} + 30 + \\frac{\\theta}{2} = 120^{\\circ}$ That was convenient! We can do it with the other angle as well. $\\angle D E_2 C = \\angle D E_2 A - \\angle C E_2 A = 90 - \\frac{\\theta}{2} - (30 - \\frac{\\theta}{2}) = 60^{\\circ}$ . That means we are able to set up Law of Cosines, on triangles $\\triangle D E_4 C$ and $\\triangle D E_2 C$ , with some really convenient angles.\nLet $CE_2 = x$ , and $CE_4 = y$ \\[333 = 11 + x^2 - \\sqrt{11} x\\] \\[333 = 11 + y^2 + \\sqrt{11} y\\] We subtract and get: \\[0 = (x+y)(x-y-\\sqrt{11})\\] $x+y$ obviously can't be 0, so $x-y = \\sqrt{11}$ We add and get: \\[666 = 22 + x^2 + y^2 + \\sqrt{11} (y-x)\\] $y-x = -\\sqrt{11}$ . Thus, we can fill in and solve. \\[666 = 22 + x^2 + y^2 - 11\\] \\[655 = x^2 + y^2\\] Thus our answer is $C E_1^2 + C E_2^2 + C E_2^2 + C E_4^2 = 11 + 11 + C E_2^2 + C E_4^2 = 11 + 11 + x^2 + y^2 = 11 + 11 + 655 = \\boxed{677}$" ]

[ "Notice that $\\angle{E} = \\angle{BGC} = 120^\\circ$ because $\\angle{A} = 60^\\circ$ . Also, $\\angle{GBC} = \\angle{GAC} = \\angle{FAE}$ because they both correspond to arc ${GC}$ . So $\\Delta{GBC} \\sim \\Delta{EAF}$\n\\[[EAF] = \\frac12 (AE)(EF)\\sin \\angle AEF = \\frac12\\cdot11\\cdot13\\cdot\\sin{120^\\circ} = \\frac {143\\sqrt3}4.\\]\nBecause the ratio of the area of two similar figures is the square of the ratio of the corresponding sides, $[GBC] = \\frac {BC^2}{AF^2}\\cdot[EAF] = \\frac {12}{11^2 + 13^2 - 2\\cdot11\\cdot13\\cdot\\cos120^\\circ}\\cdot\\frac {143\\sqrt3}4 = \\frac {429\\sqrt3}{433}$ . Therefore, the answer is $429+433+3=\\boxed{865}$", "Solution by e_power_pi_times_i/edited by srisainandan6\nLet the center of the circle be $O$ and the origin. Then, $A (0,2)$ $B (-\\sqrt{3}, -1)$ $C (\\sqrt{3}, -1)$ $D$ and $E$ can be calculated easily knowing $AD$ and $AE$ $D (-\\dfrac{13}{2}, \\dfrac{-13\\sqrt{3}+4}{2})$ $E (\\dfrac{11}{2}, \\dfrac{-11\\sqrt{3}+4}{2})$ . As $DF$ and $EF$ are parallel to $AE$ and $AD$ $F (-1, -12\\sqrt{3}+2)$ $G$ and $A$ is the intersection between $AF$ and circle $O$ . Therefore $G (-\\dfrac{48\\sqrt{3}}{433}, -\\dfrac{862}{433})$ . Using the Shoelace Theorem, $[CBG] = \\dfrac{429\\sqrt{3}}{433}$ , so the answer is $\\boxed{865}$ . Note that although the solution may appear short, actually getting all the coordinates take a while as there is plenty of computation.", "Lines $l_1$ and $l_2$ are constructed such that $AEFD$ is a parallelogram, hence $DF = 13$ . Since $BAC$ is equilateral with angle of $60^{\\circ}$ , angle $D$ is $120^{\\circ}$ . Use law of cosines to find $AF = \\sqrt{433}$ . Then use law of sines to find angle $BAG$ and $GAC$ . Next we use Ptolemy's Theorem on $ABGC$ to find that $CG + BG = AG$ . Next we use law of cosine on triangles $BAG$ and $GAC$ , solving for BG and CG respectively. Subtract the two equations and divide out a $BG + CG$ to find the value of $CG - BG$ . Next, $AG = 2\\cdot R \\cos{\\theta}$ , where R is radius of circle $= 2$ and $\\theta =$ angle $BAG$ . We already know sine of the angle so find cosine, hence we have found $AG$ . At this point it is system of equation yielding $CG = \\frac{26\\sqrt{3}}{\\sqrt{433}}$ and $BG = \\frac{22\\sqrt{3}}{\\sqrt{433}}$ . Given $[CBG] = \\frac{BC \\cdot CG \\cdot BG}{4R}$ , and $BC = 2\\sqrt{3}$ by $30-60-90$ triangle, we can evaluate to find $[CBG] = \\frac{429\\sqrt{3}}{433}$ , to give answer = $\\boxed{865}$", "Note that $AB=2\\sqrt3$ $DF=11$ , and $EF=13$ . If we take a homothety of the parallelogram with respect to $A$ , such that $F$ maps to $G$ , we see that $\\frac{[ABG]}{[ACG]}=\\frac{11}{13}$ . Since $\\angle AGB=\\angle AGC=60^{\\circ}$ , from the sine area formula we have $\\frac{BG}{CG}=\\frac{11}{13}$ . Let $BG=11k$ and $CG=13k$\nBy Law of Cosines on $\\triangle BGC$ , we have \\[12=k^2(11^2+11\\cdot13+13^2)=433k^2\\implies k^2=\\frac{12}{433}\\] Thus, $[CBG]=\\frac12 (11k)(13k)\\sin 120^{\\circ} = \\frac{\\sqrt3}{4}\\cdot 143\\cdot \\frac{12}{433}=\\frac{429\\sqrt3}{433}\\implies\\boxed{865}$" ]

[ "By angle chasing, we conclude that $\\triangle AGF$ is a $30^\\circ\\text{-}30^\\circ\\text{-}120^\\circ$ triangle, and $\\triangle BED$ is a $30^\\circ\\text{-}60^\\circ\\text{-}90^\\circ$ triangle.\nLet $AF=x.$ It follows that $FG=x$ and $EB=FC=840-x.$ By the side-length ratios in $\\triangle BED,$ we have $DE=\\frac{840-x}{2}$ and $DB=\\frac{840-x}{2}\\cdot\\sqrt3.$\nLet the brackets denote areas. We have \\[[AFG]=\\frac12\\cdot AF\\cdot FG\\cdot\\sin{\\angle AFG}=\\frac12\\cdot x\\cdot x\\cdot\\sin{120^\\circ}=\\frac12\\cdot x^2\\cdot\\frac{\\sqrt3}{2}\\] and \\[[BED]=\\frac12\\cdot DE\\cdot DB=\\frac12\\cdot\\frac{840-x}{2}\\cdot\\left(\\frac{840-x}{2}\\cdot\\sqrt3\\right).\\]\nWe set up and solve an equation for $x:$ \\begin{align*} \\frac{[AFG]}{[BED]}&=\\frac89 \\\\ \\frac{\\frac12\\cdot x^2\\cdot\\frac{\\sqrt3}{2}}{\\frac12\\cdot\\frac{840-x}{2}\\cdot\\left(\\frac{840-x}{2}\\cdot\\sqrt3\\right)}&=\\frac89 \\\\ \\frac{2x^2}{(840-x)^2}&=\\frac89 \\\\ \\frac{x^2}{(840-x)^2}&=\\frac49. \\end{align*} Since $0<x<840,$ it is clear that $\\frac{x}{840-x}>0.$ Therefore, we take the positive square root for both sides: \\begin{align*} \\frac{x}{840-x}&=\\frac23 \\\\ 3x&=1680-2x \\\\ 5x&=1680 \\\\ x&=\\boxed{336}", "We express the areas of $\\triangle BED$ and $\\triangle AFG$ in terms of $AF$ in order to solve for $AF.$\nWe let $x = AF.$ Because $\\triangle AFG$ is isosceles and $\\triangle AEF$ is equilateral, $AF = FG = EF = AE = x.$\nLet the height of $\\triangle ABC$ be $h$ and the height of $\\triangle AEF$ be $h'.$ Then we have that $h = \\frac{\\sqrt{3}}{2}(840) = 420\\sqrt{3}$ and $h' = \\frac{\\sqrt{3}}{2}(EF) = \\frac{\\sqrt{3}}{2}x.$\nNow we can find $DB$ and $BE$ in terms of $x.$ $DB = h - h' = 420\\sqrt{3} - \\frac{\\sqrt{3}}{2}x,$ $BE = AB - AE = 840 - x.$ Because we are given that $\\angle DBC = 90,$ $\\angle DBE = 30.$ This allows us to use the sin formula for triangle area: the area of $\\triangle BED$ is $\\frac{1}{2}(\\sin 30)\\left(420\\sqrt{3} - \\frac{\\sqrt{3}}{2}x\\right)(840-x).$ Similarly, because $\\angle AFG = 120,$ the area of $\\triangle AFG$ is $\\frac{1}{2}(\\sin 120)(x^2).$\nNow we can make an equation: \\begin{align*} \\frac{\\triangle AFG}{\\triangle BED} &= \\frac{8}{9} \\\\ \\frac{\\frac{1}{2}(\\sin 120)(x^2)}{\\frac{1}{2}(\\sin 30)\\left(420\\sqrt{3} - \\frac{\\sqrt{3}}{2}x\\right)(840-x)} &= \\frac{8}{9} \\\\ \\frac{x^2}{\\left(420 - \\frac{x}{2}\\right)(840-x)} &= \\frac{8}{9}. \\end{align*} To make further calculations easier, we scale everything down by $420$ (while keeping the same variable names, so keep that in mind). \\begin{align*} \\frac{x^2}{\\left(1-\\frac{x}{2}\\right)(2-x)} &= \\frac{8}{9} \\\\ 8\\left(1-\\frac{x}{2}\\right)(2-x) &= 9x^2 \\\\ 16-16x + 4x^2 &= 9x^2 \\\\ 5x^2 + 16x -16 &= 0 \\\\ (5x-4)(x+4) &= 0. \\end{align*} Thus $x = \\frac{4}{5}.$ Because we scaled down everything by $420,$ the actual value of $AF$ is $\\frac{4}{5}(420) = \\boxed{336}.$", "$\\angle AFE = \\angle AEF = \\angle EAF = 60^{0} \\Rightarrow \\angle AFG = 120^{0}$ So, If $\\Delta AFG$ is isosceles, it means that $AF = FG$\nLet $AF = FG = AE = EF = x$\nSo, $[\\Delta AFG] = \\frac{1}{2} \\cdot x^{2} \\textup{sin} 120^{0} = \\frac{\\sqrt{3}}{4}x^{2}$\nIn $\\Delta BED$ $BE = 840 - x$ , Hence $DE = \\frac{840 - x}{2}$ (because $\\textup{sin} 30^{0} = \\frac{1}{2}$\nTherefore, $[\\Delta BED] = \\frac{1}{2} (840 - x) \\left (\\frac{840-x}{2} \\right) \\textup{sin} 60^{0}$\nSo, $[\\Delta BED] = \\frac{\\sqrt{3}}{4} (840 - x) \\left (\\frac{840-x}{2} \\right) = \\frac{\\sqrt{3}}{8} (840 - x)^{2}$\nNow, as we know that the ratio of the areas of $\\Delta AFG$ and $\\Delta BED$ is $8:9$\nSubstituting the values, we get\n$\\frac{\\frac{\\sqrt{3}}{4}x^{2}}{\\frac{\\sqrt{3}}{8} (840 - x)^{2}} = \\frac{8}{9} \\Rightarrow \\left (\\frac{x}{840 - x} \\right)^{2} = \\frac{4}{9}$ Hence, $\\frac{x}{840 - x} = \\frac{2}{3}$ . Solving this, we easily get $x = 336$\nWe have taken $AF = x$ , Hence, $AF = \\boxed{336}$", "Since $\\triangle AFG$ is isosceles, $AF = FG$ , and since $\\triangle AEF$ is equilateral, $AF = EF$ . Thus, $EF = FG$ , and since these triangles share an altitude, they must have the same area.\nDrop perpendiculars from $E$ and $F$ to line $BC$ ; call the meeting points $P$ and $Q$ , respectively. $\\triangle BEP$ is clearly congruent to both $\\triangle BED$ and $\\triangle FQC$ , and thus each of these new triangles has the same area as $\\triangle BED$ . But we can \"slide\" $\\triangle BEP$ over to make it adjacent to $\\triangle FQC$ , thus creating an equilateral triangle whose area has a ratio of $18:8$ when compared to $\\triangle AEF$ (based on our conclusion from the first paragraph). Since these triangles are both equilateral, they are similar, and since the area ratio $18:8$ reduces to $9:4$ , the ratio of their sides must be $3:2$ . So, because $FC$ and $AF$ represent sides of these triangles, and they add to $840$ $AF$ must equal two-fifths of $840$ , or $\\boxed{336}$" ]

[ "Let $X$ be the intersection of the circles with centers $B$ and $E$ , and $Y$ be the intersection of the circles with centers $C$ and $E$ . Since the radius of $B$ is $3$ $AX =4$ . Assume $AE$ $p$ . Then $EX$ and $EY$ are radii of circle $E$ and have length $4+p$ $AC = 8$ , and angle $CAE = 60$ degrees because we are given that triangle $T$ is equilateral. Using the Law of Cosines on triangle $CAE$ , we obtain\n$(6+p)^2 =p^2 + 64 - 2(8)(p) \\cos 60$\nThe $2$ and the $\\cos 60$ terms cancel out:\n$p^2 + 12p +36 = p^2 + 64 - 8p$\n$12p+ 36 = 64 - 8p$\n$p =\\frac {28}{20} = \\frac {7}{5}$ . The radius of circle $E$ is $4 + \\frac {7}{5} = \\frac {27}{5}$ , so the answer is $27 + 5 = \\boxed{032}$" ]

[ "We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\\omega_A$ is the incircle of $\\triangle AB'C'$ Since the diameter of the circle is the height of this triangle, the height of this triangle is $36$ . We can use inradius or equilateral triangle properties to get the inradius of this triangle is $12$ (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is $12$\nLet $O=\\omega$ be the center of the largest circle. We will set up a coordinate system with $O$ as the origin. The center of $\\omega_A$ will be at $(0,-6)$ because it is directly beneath $O$ and is the length of the larger radius minus the smaller radius, or $18-12 = 6$ . By rotating this point $120^{\\circ}$ around $O$ , we get the center of $\\omega_B$ . This means that the magnitude of vector $\\overrightarrow{O\\omega_B}$ is $6$ and is at a $30$ degree angle from the horizontal. Therefore, the coordinates of this point are $(3\\sqrt{3},3)$ and by symmetry the coordinates of the center of $\\omega_C$ is $(-3\\sqrt{3},3)$\nThe upper left and right circles intersect at two points, the lower of which is $X$ . The equations of these two circles are: \\begin{align*} (x+3\\sqrt3)^2 + (y-3)^2 &= 12^2, \\\\ (x-3\\sqrt3)^2 + (y-3)^2 &= 12^2. \\end{align*} We solve this system by subtracting to get $x = 0$ . Plugging back in to the first equation, we have $(3\\sqrt{3})^2 + (y-3)^2 = 144 \\implies (y-3)^2 = 117 \\implies y-3 = \\pm \\sqrt{117} \\implies y = 3 \\pm \\sqrt{117}$ . Since we know $X$ is the lower solution, we take the negative value to get $X = (0,3-\\sqrt{117})$\nWe can solve the problem two ways from here. We can find $Y$ by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find $OX$ as they lie on the same vertical, $\\angle XOY$ is $120$ degrees so we can make use of $30-60-90$ triangles, and $OX = OY$ because $O$ is the center of triangle $XYZ$ . We can draw the diagram as such: Note that $OX = OY = \\sqrt{117} - 3$ . It follows that \\begin{align*} XY &= 2 \\cdot \\frac{OX\\cdot\\sqrt{3}}{2} \\\\ &= OX \\cdot \\sqrt{3} \\\\ &= (\\sqrt{117}-3) \\cdot \\sqrt{3} \\\\ &= \\sqrt{351}-\\sqrt{27}. \\end{align*} Finally, the answer is $351+27 = \\boxed{378}$", " For equilateral triangle with side length $l$ , height $h$ , and circumradius $r$ , there are relationships: $h = \\frac{\\sqrt{3}}{2} l$ $r = \\frac{2}{3} h = \\frac{\\sqrt{3}}{3} l$ , and $l = \\sqrt{3}r$\nThere is a lot of symmetry in the figure. The radius of the big circle $\\odot \\omega$ is $R = 18$ , let the radius of the small circles $\\odot \\omega_A$ $\\odot \\omega_B$ $\\odot \\omega_C$ be $r$\nWe are going to solve this problem in $3$ steps:\n$\\textbf{Step 1:}$\nWe have $\\triangle A \\omega_A D$ is a $30-60-90$ triangle, and $A \\omega_A = 2 \\cdot \\omega_A D$ $A \\omega_A = 2R-r$ $\\odot \\omega$ and $\\odot \\omega_A$ are tangent), and $\\omega_A D = r$ . So, we get $2R-r = 2r$ and $r = \\frac{2}{3} \\cdot R = 12$\nSince $\\odot \\omega$ and $\\odot \\omega_A$ are tangent, we get $\\omega \\omega_A = R - r = \\frac{1}{3} \\cdot R = 6$\nNote that $\\triangle \\omega_A \\omega_B \\omega_C$ is an equilateral triangle, and $\\omega$ is its center, so $\\omega_B \\omega_C = \\sqrt{3} \\cdot \\omega \\omega_A = 6 \\sqrt{3}$\n$\\textbf{Step 2:}$\nNote that $\\triangle \\omega_C E X$ is an isosceles triangle, so \\[EX = 2 \\sqrt{(\\omega_C E)^2 - \\left(\\frac{\\omega_B \\omega_C}{2}\\right)^2} = 2 \\sqrt{r^2 - \\left(\\frac{\\omega_B \\omega_C}{2}\\right)^2} = 2 \\sqrt{12^2 - (3 \\sqrt{3})^2} = 2 \\sqrt{117}.\\]\n$\\textbf{Step 3:}$\nIn $\\odot \\omega_C$ , Power of a Point gives $\\omega X \\cdot \\omega E = r^2 - (\\omega_C \\omega)^2$ and $\\omega E = EX - \\omega X = 2\\sqrt{117} - \\omega X$\nIt follows that $\\omega X \\cdot (2\\sqrt{117} - \\omega X) = 12^2 - 6^2$ . We solve this quadratic equation: $\\omega X = \\sqrt{117} - 3$\nSince $\\omega X$ is the circumradius of equilateral $\\triangle XYZ$ , we have $XY = \\sqrt{3} \\cdot \\omega X = \\sqrt{3} \\cdot (\\sqrt{117} - 3) = \\sqrt{351}-\\sqrt{27}$\nTherefore, the answer is $351+27 = \\boxed{378}$", "Let $O$ be the center, $R = 18$ be the radius, and $CC'$ be the diameter of $\\omega.$ Let $r$ be the radius, $E,D,F$ are the centers of $\\omega_A, \\omega_B,\\omega_C.$ Let $KGH$ be the desired triangle with side $x.$ We find $r$ using \\[CC' = 2R = C'K + KC = r + \\frac{r}{\\sin 30^\\circ} = 3r.\\] \\[r = \\frac{2R}{3} = 12.\\] \\[OE = R – r = 6.\\] Triangles $\\triangle DEF$ and $\\triangle KGH$ – are equilateral triangles with a common center $O,$ therefore in the triangle $OEH$ $OE = 6, \\angle EOH = 120^\\circ, OH = \\frac{x}{\\sqrt3}.$\nWe apply the Law of Cosines to $\\triangle OEH$ and get \\[OE^2 + OH^2 + OE \\cdot OH = EH^2.\\] \\[6^2 + \\frac{x^2}{3} + \\frac{6x}{\\sqrt3} = 12^2.\\] \\[x^2 + 6x \\sqrt{3} = 324\\] \\[x= \\sqrt{351} - \\sqrt{27} \\implies 351 + 27 = \\boxed{378}\\]", "Let $O$ be the center of $\\omega$ $X$ be the intersection of $\\omega_B,\\omega_C$ further from $A$ , and $O_A$ be the center of $\\omega_A$ . Define $Y, Z, O_B, O_C$ similarly. It is well-known that the $A$ -mixtilinear inradius $R_A$ is $\\tfrac{r}{\\cos^2\\left(\\frac{\\angle A}{2}\\right)} = \\tfrac{9}{\\cos^2\\left(30^{\\circ}\\right)} = 12$ , so in particular this means that $OO_B = 18 - R_B = 6 = OO_C$ . Since $\\angle O_BOO_C = \\angle BOC = 120^\\circ$ , it follows by Law of Cosines on $\\triangle OO_BO_C$ that $O_BO_C = 6\\sqrt{3}$ . Then the Pythagorean theorem gives that the altitude of $O_BO_CX$ is $\\sqrt{117}$ , so $OY = OX = \\text{dist}(X, YZ) - \\text{dist}(O, YZ) = \\sqrt{117} - 3$ and $YZ = \\tfrac{O_BO_C\\cdot OY}{OO_B} = \\tfrac{6\\sqrt{3}(\\sqrt{117} - 3)}{6}=\\sqrt{351} - \\sqrt{27}$ so the answer is $351 + 27 = \\boxed{378}$" ]

[ "We want to figure out the number of chunks in $60$ blocks, so we have $60\\cdot 512 \\approx 30000$ . We divide this by $120$ to determine the number of seconds necessary to transmit. $30000/120 \\approx 250$ , which means that it takes approximately $4$ minutes to transmit. Thus, the answer is $\\boxed{4}$", "This solution is if you are running out of time and just want to write down an answer. So, this is quite unreliable. You can logic it out. It doesn't make sense for the first three options to be the answer since that is way too quick. The last option is way too long. That just leaves $\\boxed{4}$" ]

[ "We could triple the population every $25$ years and make a chart:\nYear: 2000\nPopulation: 200\nYear: 2025\nPopulation: 600\nYear: 2050\nPopulation: 1800\nYear: 2075\nPopulation: 5400\nYear: 2100\nPopulation: 16200\nThe closest year is 2075, or $\\boxed{2075}$", "We could find out how many periods of 25 years we need to triple by dividing our total from our present number.\n$\\frac{6000}{200}=30$\nThe power of $3$ that $30$ is closest to is $27=3^3$\nTherefore, after $3$ periods, we will be closest to $6000$\n$\\boxed{2075}$" ]

[ "$401$ is around $400$ and $.205$ is around $.2$ so the fraction is approximately \\[\\frac{400}{.2}=2000\\rightarrow \\boxed{2000}\\]" ]

[ "The convex polyhedron of the problem can be easily visualized; it corresponds to a dodecahedron (a regular solid with $12$ equilateral pentagons) in which the $20$ vertices have all been truncated to form $20$ equilateral triangles with common vertices. The resulting solid has then $p=12$ smaller equilateral pentagons and $t=20$ equilateral triangles yielding a total of $t+p=F=32$ faces. In each vertex, $T=2$ triangles and $P=2$ pentagons are concurrent. Now, the number of edges $E$ can be obtained if we count the number of sides that each triangle and pentagon contributes: $E=\\frac{3t+5p}{2}$ , (the factor $2$ in the denominator is because we are counting twice each edge, since two adjacent faces share one edge). Thus, $E=60$ . Finally, using Euler's formula we have $V=E-30=30$\nIn summary, the solution to the problem is $100P+10T+V=\\boxed{250}$", "As seen above, $E=V+30$ . Every vertex $V$ , there is a triangle for every $T$ and a pentagon for every $P$ by the given. However, there are three times every triangle will be counted and five times every pentagon will be counted because of their numbers of vertices. From this observation, $\\frac{VT}3+\\frac{VP}5=32\\implies V(5T+3P)=480$ . Also, at every vertex $V$ , there are $T+P$ edges coming out from that vertex (one way to see this is to imagine the leftmost segment of each triangle and pentagon that is connected to the given vertex, and note that it includes every one of the edges exactly once), so $\\frac{V(T+P)}2=E\\implies V(T+P)=2E\\implies V(5T+5P)=10E$ , and subtracting the other equation involving the vertices from this gives $2VP=10E-480\\implies VP=5E-240=5(V+30)-240=5V-90$ $\\implies V(5-P)=90$ .\nSince $V|480$ from the first vertex-related observation and $P>0\\implies5-P<5$ , and it quickly follows that $V=30\\implies E=60\\implies P=2\\implies T=2\\implies100P+10T+V=\\boxed{250}$", "Notice that at each vertex, we must have the sum of the angles be less than $360$ degrees or we will not be able to fold the polyhedron. Therefore, we have $60T + 108P < 360.$ Now, let there be $t$ triangles and $p$ pentagons total such that $t+p = 32.$ From the given, we know that $E = V + 30.$ Lastly, we see that $E = \\frac{3t+5p}{2}$ and $V = \\frac{3t}{T}=\\frac{5p}{P}.$\nNow, we do casework on what $P$ is.\nCase 1: $P = 2$ Notice that we must have $t$ and $p$ integral. Trying $T = 1, 2$ yields a solution with $t=2.$ Trying other cases of $P$ and $T$ yields no solutions. Therefore, $T=2, P=2$ and after solving for $t, p$ we get $V=30.$ Finally, we have $100P+10T+V = \\boxed{250}$", "We know that $V-E = -30 \\implies V = E-30$ based off the problem condition. Furthermore, if we draw out a few pentagons as well as triangles on each of side of the pentagons, it's clear that each vertex has 4 edges connected to it, with two triangles and two pentagons for each vertex. However, each edge is used for two vertices and thus counted twice. Therefore, we have that $E = \\frac{4V}{2} = 2V$ . Plugging this in, we have that $V=30$ and so our answer is $200+20+30 = \\boxed{250}$" ]

[ "Plugging in $2$ as $x$ gives $4^4$ , which is merely $\\boxed{256}$" ]

[ "More generally, let $(x,y,z)=\\left(\\sqrt5,\\sqrt6,\\sqrt7\\right)$ so that $\\left(x^2,y^2,z^2\\right)=(5,6,7).$\nWe rewrite the original expression in terms of $x,y,$ and $z,$ then apply the difference of squares repeatedly: \\begin{align*} (x+y+z)(x+y-z)(x-y+z)(-x+y+z) &= \\left[((x+y)+z)((x+y)-z)\\right]\\left[((z+(x-y))(z-(x-y))\\right] \\\\ &= \\left[(x+y)^2-z^2\\right]\\left[z^2 - (x-y)^2\\right] \\\\ &= \\left[x^2+2xy+y^2-z^2\\right]\\left[z^2-x^2+2xy-y^2\\right] \\\\ &= \\left[2xy + \\left(x^2+y^2-z^2\\right)\\right]\\left[2xy - \\left(x^2+y^2-z^2\\right)\\right] \\\\ &= \\left(2xy\\right)^2 - \\left(x^2+y^2-z^2\\right)^2 \\\\ &= \\left(2\\cdot\\sqrt5\\cdot\\sqrt6\\right)^2 - \\left(5+6-7\\right)^2 \\\\ &= \\boxed{104} Remark", "We group the first and last factors as well as the two middle factors, then apply the difference of squares repeatedly: \\begin{align*} \\left(\\left(\\sqrt{6} + \\sqrt{7}\\right)^2 - \\sqrt{5}^2\\right)\\left(\\sqrt{5}^2 - \\left(\\sqrt{6} - \\sqrt{7}\\right)^2\\right) &= \\left(13 + 2\\sqrt{42} - 5\\right)\\left(5 - \\left(13 - 2\\sqrt{42}\\right)\\right) \\\\ &= \\left(2\\sqrt{42} + 8\\right)\\left(2\\sqrt{42} - 8\\right) \\\\ &= \\left(2\\sqrt{42}\\right)^2 - 8^2 \\\\ &= \\boxed{104} ~Azjps (Solution)", "Notice that in a triangle with side-lengths $2\\sqrt5,2\\sqrt6,$ and $2\\sqrt7,$ by Heron's Formula, the area is the square root of the original expression.\nLet $\\theta$ be the measure of the angle opposite the $2\\sqrt7$ side. By the Law of Cosines, \\[\\cos\\theta=\\frac{\\left(2\\sqrt5\\right)^2+\\left(2\\sqrt{6}\\right)^2-\\left(2\\sqrt7\\right)^2}{2\\cdot 2\\sqrt5\\cdot2\\sqrt6}=\\frac{16}{8\\sqrt{30}}=\\sqrt{\\frac{2}{15}},\\] so $\\sin\\theta=\\sqrt{1-\\cos^2\\theta}=\\sqrt{\\frac{13}{15}}.$\nThe area of the triangle is then \\[\\frac{\\sin\\theta}{2}\\cdot 2\\sqrt5\\cdot 2\\sqrt6=\\sqrt{\\frac{26}{30}}\\cdot 2\\sqrt{30}=\\sqrt{104},\\] so our answer is $\\left(\\sqrt{104}\\right)^2=\\boxed{104}.$" ]

[ "Adding the cases up, we get $27 + 54 + 36 = \\boxed{117}$", "Let's say we have picked two cards. We now compare their attributes to decide how we can pick the third card to make a complement set. For each of the three attributes, should the two values be the same we have one option - choose a card with the same value for that attribute. Furthermore, should the two be different there is only one option- choose the only value that is remaining. In this way, every two card pick corresponds to exactly one set, for a total of $\\binom{27}{2} = 27*13 = 351$ possibilities. Note, however, that each set is generated by ${3\\choose 2} = 3$ pairs, so we've overcounted by a multiple of 3 and the answer is $\\frac{351}{3} = \\boxed{117}$", "Treat the sets as ordered. Then for each of the three criterion, there are $3!=6$ choices if the attribute is different and there are $3$ choices is the attribute is the same. Thus all three attributes combine to a total of $(6+3)^3=729$ possibilities. However if all three attributes are the same then the set must be composed of three cards that are the same, which is impossible. This takes out $3^3=27$ possibilities. Notice that we have counted every set $3!=6$ times by treating the set as ordered. The final solution is then $\\frac{729-27}{6}=\\boxed{117}$" ]

[ "There are $x$ schools. This means that there are $3x$ people. Because no one's score was the same as another person's score, that means that there could only have been $1$ median score. This implies that $x$ is an odd number. $x$ cannot be less than $23$ , because there wouldn't be a $64$ th place if x was. $x$ cannot be greater than $23$ either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is $23 \\Rightarrow \\boxed{23}$" ]

[ "Let $a$ be Andrea's place. We know that she was the highest on her team, so $a < 37$\nSince $a$ is the median, there are $a-1$ to the left and right of the median, so the total number of people is $2a-1$ and the number of schools is $(2a-1)/3$ . This implies that $2a-1 \\equiv 0 \\pmod{3} \\implies a \\equiv 2 \\pmod{3}$\nAlso, since $2a-1$ is the rank of the last-place person, and one of Andrea's teammates already got 64th place, $2a-1 > 64 \\implies a \\ge 33$\nPutting it all together: $33 \\le a < 37$ and $a \\equiv 2 \\pmod{3}$ , so clearly $a = 35$ , and the number of schools as we got before is $(2a-1)/3 = 69/3 = \\boxed{23}$" ]

[ "There are $x$ schools. This means that there are $3x$ people. Because no one's score was the same as another person's score, that means that there could only have been $1$ median score. This implies that $x$ is an odd number. $x$ cannot be less than $23$ , because there wouldn't be a $64$ th place if x was. $x$ cannot be greater than $23$ either, because that would tie Andrea and Beth or Andrea's place would be worse than Beth's. Thus, the only possible answer is $23 \\Rightarrow \\boxed{23}$" ]

[ "Let $a$ be Andrea's place. We know that she was the highest on her team, so $a < 37$\nSince $a$ is the median, there are $a-1$ to the left and right of the median, so the total number of people is $2a-1$ and the number of schools is $(2a-1)/3$ . This implies that $2a-1 \\equiv 0 \\pmod{3} \\implies a \\equiv 2 \\pmod{3}$\nAlso, since $2a-1$ is the rank of the last-place person, and one of Andrea's teammates already got 64th place, $2a-1 > 64 \\implies a \\ge 33$\nPutting it all together: $33 \\le a < 37$ and $a \\equiv 2 \\pmod{3}$ , so clearly $a = 35$ , and the number of schools as we got before is $(2a-1)/3 = 69/3 = \\boxed{23}$" ]

[ "Note that $1+\\sum_{k=1}^{n-1} {k\\cdot k!} = 1+\\sum_{k=1}^{n-1} {((k+1)\\cdot k!- k!)} = 1+\\sum_{k=1}^{n-1} {((k+1)!- k!)} = 1 + ((2! - 1!) + (3! - 2!) + \\cdots + (n! - (n-1)!)) = n!$\nThus for all $m\\in\\mathbb{N}$\n$(32m+16)!-(32m)! = \\left(1+\\sum_{k=1}^{32m+15} {k\\cdot k!}\\right)-\\left(1+\\sum_{k=1}^{32m-1} {k\\cdot k!}\\right) = \\sum_{k=32m}^{32m+15}k\\cdot k!.$\nSo now,\n\\begin{align*} 16!-32!+48!-64!+\\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\\cdots+(2000!-1984!)\\\\ &=16! +\\sum_{m=1}^{62}(32m+16)!-(32m)!\\\\ &=16! +\\sum_{m=1}^{62}\\sum_{k=32m}^{32m+15}k\\cdot k! \\end{align*}\nTherefore we have $f_{16} = 1$ $f_k=k$ if $32m\\le k \\le 32m+15$ for some $m=1,2,\\ldots,62$ , and $f_k = 0$ for all other $k$\nTherefore we have:\n\\begin{align*} f_1-f_2+f_3-f_4+\\cdots+(-1)^{j+1}f_j &= (-1)^{17}\\cdot 1 + \\sum_{m=1}^{62}\\sum_{k=32m}^{32m+15}(-1)^{k+1}k\\\\ &= -1 + \\sum_{m=1}^{62}\\left[\\sum_{j=16m}^{16m+7}(-1)^{2j+1}2j+\\sum_{j=16m}^{16m+7}(-1)^{2j+2}(2j+1)\\right]\\\\ &= -1 + \\sum_{m=1}^{62}\\sum_{j=16m}^{16m+7}[(-1)^{2j+1}2j+(-1)^{2j+2}(2j+1)]\\\\ &= -1 + \\sum_{m=1}^{62}\\sum_{j=16m}^{16m+7}[-2j+(2j+1)]\\\\ &= -1 + \\sum_{m=1}^{62}\\sum_{j=16m}^{16m+7}1\\\\ &= -1 + \\sum_{m=1}^{62}8\\\\ &= -1 + 8\\cdot 62\\\\ &= \\boxed{495}", "This is equivalent to Solution 1. I put up this solution merely for learners to see the intuition.\nLet us consider a base $n$ number system. It’s a well known fact that when we take the difference of two integral powers of $n$ , (such as $10000_{10} - 100_{10}$ ) the result will be an integer in base $n$ composed only of the digits $n - 1$ and $0$ (in this example, $9900$ ). More specifically, the difference $(n^k)_n - (n^j)_n$ $j<k$ , is an integer $k$ digits long (note that $(n^k)_n$ has $k + 1$ digits). This integer is made up of $(k-j)$ $(n - 1)$ ’s followed by $j$ $0$ ’s.\nIt should make sense that this fact carries over to the factorial base, albeit with a modification. Whereas in the general base $n$ , the largest digit value is $n - 1$ , in the factorial base, the largest digit value is the argument of the factorial in that place. (for example, $321_!$ is a valid factorial base number, as is $3210_!$ . However, $31_!$ is not, as $3$ is greater than the argument of the second place factorial, $2$ $31_!$ should be represented as $101_!$ , and is $7_{10}$ .) Therefore, for example, $1000000_! - 10000_!$ is not $990000_!$ , but rather is $650000_!$ . Thus, we may add or subtract factorials quite easily by converting each factorial to its factorial base expression, with a $1$ in the argument of the factorial’s place and $0$ ’s everywhere else, and then using a standard carry/borrow system accounting for the place value.\nWith general intuition about the factorial base system out of the way, we may tackle the problem. We use the associative property of addition to regroup the terms as follows: $(2000! - 1984!) + (1968! - 1952!) + \\cdots + (48! - 32!) + 16!$ we now apply our intuition from paragraph 2. $2000!_{10}$ is equivalent to $1$ followed by $1999$ $0$ ’s in the factorial base, and $1984!$ is $1$ followed by $1983$ $0$ ’s, and so on. Therefore, $2000! - 1984! = (1999)(1998)(1997)\\cdots(1984)$ followed by $1983$ $0$ ’s in the factorial base. $1968! - 1952! = (1967)(1966)\\cdots(1952)$ followed by $1951$ $0$ ’s, and so on for the rest of the terms, except $16!$ , which will merely have a $1$ in the $16!$ place followed by $0$ ’s. To add these numbers, no carrying will be necessary, because there is only one non-zero value for each place value in the sum. Therefore, the factorial base place value $f_k$ is $k$ for all $32m \\leq k \\leq 32m+15$ if $1\\leq m \\in\\mathbb{Z} \\leq 62$ $f_{16} = 1$ , and $f_k = 0$ for all other $k$\nTherefore, to answer, we notice that $1999 - 1998 = 1997 - 1996 = 1$ , and this will continue. Therefore, $f_{1999} - f_{1998} + \\cdots - f_{1984} = 8$ . We have 62 sets that sum like this, and each contains $8$ pairs of elements that sum to $1$ , so our answer is almost $8 \\cdot 62$ . However, we must subtract the $1$ in the $f_{16}$ place, and our answer is $8 \\cdot 62 - 1 = \\boxed{495}$", "Let $S = 16!-32!+\\cdots-1984!+2000!$ . Note that since $|S - 2000!| << 2000!$ (or $|S - 2000!| = 1984! + \\cdots$ is significantly smaller than $2000!$ ), it follows that $1999! < S < 2000!$ . Hence $f_{2000} = 0$ . Then $2000! = 2000 \\cdot 1999! = 1999 \\cdot 1999! + 1999!$ , and as $S - 2000! << 1999!$ , it follows that $1999 \\cdot 1999! < S < 2000 \\cdot 1999!$ . Hence $f_{1999} = 1999$ , and we now need to find the factorial base expansion of\n\\[S_2 = S - 1999 \\cdot 1999! = 1999! - 1984! + 1962! - 1946! + \\cdots + 16!\\]\nSince $|S_2 - 1999!| << 1999!$ , we can repeat the above argument recursively to yield $f_{1998} = 1998$ , and so forth down to $f_{1985} = 1985$ . Now $S_{16} = 1985! - 1984! + 1962! + \\cdots = 1984 \\cdot 1984! + 1962! + \\cdots$ , so $f_{1984} = 1984$\nThe remaining sum is now just $1962! - 1946! + \\cdots + 16!$ . We can repeatedly apply the argument from the previous two paragraphs to find that $f_{16} = 1$ , and $f_k=k$ if $32m\\le k \\le 32m+15$ for some $m=1,2,\\ldots,62$ , and $f_k = 0$ for all other $k$\nNow for each $m$ , we have $-f_{32m} + f_{32m+1} - f_{32m+2} + \\cdots + f_{32m + 31}$ $= -32m + (32m + 1) - (32m + 2) + \\cdots - (32m - 30) + (32 m + 31)$ $= 1 + 1 + \\cdots + 1 + 1$ $= 8$ . Thus, our answer is $-f_{16} + 8 \\cdot 62 = \\boxed{495}$" ]

[ "Let $m$ = cost of movie ticket Let $s$ = cost of soda\nWe can create two equations:\n\\[m = \\frac{1}{5}(A - s)\\] \\[s = \\frac{1}{20}(A - m)\\]\nSubstituting we get:\n\\[m = \\frac{1}{5}(A - \\frac{1}{20}(A - m))\\] which yields: \\[m = \\frac{19}{99}A\\]\nNow we can find s and we get:\n\\[s = \\frac{4}{99}A\\]\nSince we want to find what fraction of $A$ did Roger pay for his movie ticket and soda, we add $m$ and $s$ to get:\n\\[\\frac{19}{99}A + \\frac{4}{99}A \\implies \\boxed{23}\\]", "We have two equations from the problem: $5M=A-S$ and $20S=A-M$ If we replace $A$ with $100$ we get a system of equations, and the sum of the values of $M$ and $S$ is the percentage of $A$ .\nSolving, we get $S=\\frac{400}{99}$ and $M=\\frac{1900}{99}$ .\nAdding, we get $\\frac{2300}{99}$ , which is closest to $23$ which is $\\boxed{23}$", "Let $m$ be the price of a movie ticket and $s$ be the price of a soda.\nThen,\n\\[m=\\frac{A-s}{5}\\] and \\[s=\\frac{A-m}{20}\\] Then, we can turn this into \\[5m=A-s\\] \\[20s=A-m\\]\nSubtracting and getting rid of A, we have $20s-5m=-m+s \\rightarrow 19s=4m$ . Assume WLOG that $s=4$ $m=19$ , thus making a solution for this equation. Substituting this into the 1st equation, we get $A=99$ . Hence, $\\frac{m+s}{A} = \\frac{19+4}{99} \\approx \\boxed{23}$" ]

[ "A dynamics programming approach is quick and easy. The number of ways to climb one stair is $1$ . There are $2$ ways to climb two stairs: $1$ $1$ or $2$ . For 3 stairs, there are $4$ ways: \n( $1$ $1$ $1$ )\n( $1$ $2$ )\n( $2$ $1$ )\n( $3$\nFor four stairs, consider what step they came from to land on the fourth stair. They could have hopped straight from the 1st, done a double from #2, or used a single step from #3. The ways to get to each of these steps are $1+2+4=7$ ways to get to step 4. The pattern can then be extended: $4$ steps: $1+2+4=7$ ways. $5$ steps: $2+4+7=13$ ways. $6$ steps: $4+7+13=24$ ways.\nThus, there are $\\boxed{24}$ ways to get to step $6.$", "Complementary counting is also possible. Considering the six steps, Jo has to land on the last step, so there are $2^5=32$ subsets (hit steps) of the other five steps. After that, subtract the number of ways to climb the steps while taking a leap of $4$ $5$ , or $6$ . The eight possible ways for this is ( $4$ $1$ $1$ ), ( $4$ $2$ ), ( $1$ $4$ $1$ ), ( $1$ $1$ $4$ ), ( $2$ $4$ ), ( $1$ $5$ ), ( $5$ $1$ ), and ( $6$\nAltogether this makes for $32-8= \\boxed{24}$ valid ways for Jo to get to step 6.", "We can set up a recursion to solve this problem. Suppose $f(n)$ represents the number of valid ways to get to the $n$ th step. $f(0)=1$ because there is 1 way for Jo to get to the \" $0$ th\" step (i.e. the ground). There is $1$ way to get to the first step (a $1$ -step), so $f(1)=1$ . There are $2$ ways to get to the second step (two $1$ -steps or one $2$ -step). Thus, $f(2) = 2$ . In general, $f(n) = f(n-1) + f(n-2) + f(n-3)$ . This is because from the $n-3$ th step, Jo can take a $3$ -step to get to the $n$ th step, from the $n-2$ th step, Jo can take a $2$ -step to get to the $n$ th step, and from the $n-1$ th step, Jo can take a $1$ -step to get to the $n$ th step. We now iteratively calculate values of $f(n)$\n$f(0) = 1$\n$f(1) = 1$\n$f(2) = 2$\n$f(3) = f(2) + f(1) + f(0) = 2 + 1 + 1 = 4$\n$f(4) = f(3) + f(2) + f(1) = 4 + 2 + 1 = 7$\n$f(5) = f(4) + f(3) + f(2) = 7 + 4 + 2 = 13$\n$f(6) = f(5) + f(4) + f(3) = 13 + 7 + 4 = \\boxed{24}$" ]

[ "Suppose that such a polygon has $n$ sides. Let the three obtuse angle measures, in degrees, be $o_1$ $o_2$ , and $o_3$ and the $(n-3)$ acute angle measures, again in degrees, be $a_1,a_2,a_3, \\dotsc a_{n-3}$\nSince $90 < o_i < 180$ for each $i$ , we have \\[3(90) = 270 < o_1+o_2+o_3 < 3(180) = 540,\\] and similarly, since $0 < a_i < 90$ for each $i$ \\[0 < a_1+a_2+a_3+\\dotsb+a_{n-3} < 90(n-3) = 90n-270.\\] It follows that \\[270+0 < o_1+o_2+o_3+a_1+a_2+a_3+\\dotsb+a_{n-3} < 540+90n-270,\\] and recalling that the sum of the interior angle measures of an $n$ -gon is $180(n-2) = 180n-360$ , this reduces to $270 < 180n-360 < 90n+270$ . Hence \\[\\frac{540}{180} < n < \\frac{270+360}{90} \\iff 3 < n < 7,\\] so an upper bound is $n \\leq 6$ , and it is easy to check that this bound can be attained by e.g. a convex hexagon with a right angle, $2$ acute angles, and $3$ obtuse angles, as shown below:\nHrKCfF2ETNSQSy3uGlkg hexagonsurvey1.gif\nAccordingly, the maximum possible number of sides of such a polygon is $\\boxed{6}$" ]

[ "We can start to write out some of the inequalities now:\nAnd in general,\nIt is apparent that the bounds are slowly closing in on $x$ , so we can just calculate $x$ for some large value of $n$ (randomly, 10, 11):\n$x < \\frac{F_{7}}{F_{8}} \\cdot 1000 = \\frac{13}{21} \\cdot 1000 = 618.\\overline{18}$\n$x > \\frac{F_{8}}{F_{9}} \\cdot 1000 = \\frac{21}{34} \\cdot 1000 \\approx 617.977$\nThus the sequence is maximized when $x = \\boxed{618}.$", "It is well known that $\\lim_{n\\rightarrow\\infty} \\frac{F_{n-1}}{F_n} = \\phi - 1 =\\frac{1 + \\sqrt{5}}{2} - 1 \\approx .61803$ , so $1000 \\cdot \\frac{F_{n-1}}{F_n}$ approaches $x = \\boxed{618}.$" ]

[ "Let $0<x_{}^{}<1$ . Then we may write $A_{k}^{}={N\\choose k}x^{k}=\\frac{N!}{k!(N-k)!}x^{k}=\\frac{(N-k+1)!}{k!}x^{k}$ . Taking logarithms in both sides of this last equation and using the well-known fact $\\log(a_{}^{}b)=\\log a + \\log b$ (valid if $a_{}^{},b_{}^{}>0$ ), we have\n$\\log(A_{k})=\\log\\left[\\frac{(N-k+1)!}{k!}x^{k}\\right]=\\log\\left[\\prod_{j=1}^{k}\\frac{(N-j+1)x}{j}\\right]=\\sum_{j=1}^{k}\\log\\left[\\frac{(N-j+1)x}{j}\\right]\\, .$\nNow, $\\log(A_{k}^{})$ keeps increasing with $k_{}^{}$ as long as the arguments $\\frac{(N-j+1)x}{j}>1$ in each of the $\\log\\big[\\;\\big]$ terms (recall that $\\log y_{}^{} <0$ if $0<y_{}^{}<1$ ). Therefore, the integer $k_{}^{}$ that we are looking for must satisfy $k=\\Big\\lfloor\\frac{(N+1)x}{1+x}\\Big\\rfloor$ , where $\\lfloor z_{}^{}\\rfloor$ denotes the greatest integer less than or equal to $z_{}^{}$\nIn summary, substituting $N_{}^{}=1000$ and $x_{}^{}=0.2$ we finally find that $k=\\boxed{166}$", "We know that once we have found the largest value of $k$ , all values after $A_k$ are less than $A_k$ . Therefore, we are looking for the smallest possible value such that:\n${\\frac{1}{5}}^k\\cdot {{1000} \\choose {k}}>{\\frac{1}{5}}^{k+1}\\cdot {{1000} \\choose {k+1}}$\nDividing by ${\\frac{1}{5}}^k$ gives:\n${1000\\choose k}>{\\frac{1}{5}}\\cdot{1000\\choose {k+1}}$\nWe can express these binomial coefficients as factorials.\n$\\frac{1000!}{(1000-k)!\\cdot(k)!}>{\\frac{1}{5}}\\cdot\\frac{1000!}{(1000-k-1)!\\cdot{(k+1)!}}$\nWe note that the $1000!$ can cancel. Also, $(1000-k)!=(1000-k)(1000-k-1)!$ . Similarly, $(k+1)!=(k+1)k!$\nCanceling these terms yields,\n$\\frac{1}{(1000-k)}>{\\frac{1}{5}}\\cdot\\frac{1}{(k+1)}$\nCross multiplying gives:\n$5k+5>1000-k \\Rightarrow k>165.8$\nTherefore, since this identity holds for all values of $k>165.8$ , the largest possible value of $k$ is $\\boxed{166}$", "We know that $A_k$ will increase as $k$ increases until certain $k=m$ , where $A_0 < A_1 < A_2 < \\dots < A_{m-2} < A_{m-1} < A_m$ and\n$A_m > A_{m+1} > A_{m+2} > \\dots > A_{1000}.$\nNext, to change $A_{k-1}$ to $A_k$ , we multiply $A_{k-1}$ by $\\frac{1000-k+1}{5k}$ . It follows that the numerator must be greater than the\ndenominator or\n\\[1000-k+1>5k\\] \\[k<166.8\\]\nWe conclude that the answer is $\\boxed{166}$", "Notice that the expansion is the largest the moment BEFORE the $nC_p < 5$ (this reasoning can probably be found in the other solutions; basically, if we have a number k and then k+1, the value is the largest when k+1 is larger than k, or in other words $nC_p*\\frac{1}{5} > 1$\nSay we have ${1000 \\choose 5}$ ... this equals $\\frac{1000*999*998*997*996}{5*4*3*2*1}$ , and if we have k+1, then it's just going to be equivalent to multiplying this fraction by $\\frac{995}{6}$ . Notice that this fraction's numerator plus denominator is equal to $1001$ . Calling the numerator x and the denominator y, we get that \\[x+y = 1001\\] \\[\\frac{x}{y} > 5\\] \\[x>5y\\] \\[6y<1001\\] \\[y<166.83333\\] and since the denominator is what we are specifically choosing, or in other words what k is, we conclude k = $\\boxed{166}$", "Notice the relation between $A_m$ and $A_{m+1}$ . We have that: $A_{m+1} = A_m \\cdot \\frac{1}{5} \\cdot \\frac{1000-m}{m+1}$ . This is true because from $A_m$ to $A_{m+1}$ we have to multiply by $\\frac{1}{5}=0.2$ once,and then we must resolve the factorial issue. To do this, we must realize that\n\\[{1000 \\choose m+1} = \\frac{1000!}{(m+1)!(1000-m-1)!} = \\frac{1000 \\cdots (1001-m)(1000-m)}{(m+1)!}\\] and that \\[{1000 \\choose m} = \\frac{1000!}{m!(1000-m)!} = \\frac{1000 \\cdots (1001-m)}{m!}\\]\n\\[\\Longrightarrow {1000 \\choose m} \\cdot \\frac{1000-m}{m+1}\\]\nSo now, we must find out for which $m$ is $1000-m< 5 \\cdot (m+1)$ (when this happens the numerator is less than the denominator for the fraction $\\frac{1000-m}{m+1}$ so then we will have $A_m > A_{m+1}$ ) Then we find that for all $m > 165.833333$ , the above inequality ( $1000-m< 5 \\cdot (m+1)$ ) holds, so then $m=165$ so $k=m+1= \\boxed{166}$" ]

[ "Let $D$ and $E$ be the points of tangency on circles $A$ and $B$ with line $CD$ $AB=8$ . Also, let $BC=x$ . As $\\angle ADC$ and $\\angle BEC$ are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share $\\angle ACD$ $\\triangle ADC \\sim \\triangle BEC$ . From this we can get a proportion.\n$\\frac{BC}{AC}=\\frac{BE}{AD} \\rightarrow \\frac{x}{x+8}=\\frac{3}{5} \\rightarrow 5x=3x+24 \\rightarrow x=\\boxed{12}$" ]

[ "Since the area $BCD=80=\\frac{1}{2}\\cdot10\\cdot16$ , the perpendicular from $D$ to $BC$ has length $16$\nThe perpendicular from $D$ to $ABC$ is $16 \\cdot \\sin 30^\\circ=8$ . Therefore, the volume is $\\frac{8\\cdot120}{3}=\\boxed{320}$" ]