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di-zhang-fdu

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AOPS

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[ "Suppose that Sam's average speed during the last $30$ minutes was $x$ mph.\nRecall that a half hour is equal to $30$ minutes. Therefore, Sam drove $60\\cdot0.5=30$ miles during the first half hour, $65\\cdot0.5=32.5$ miles during the second half hour, and $x\\cdot0.5$ miles during the last half hour. We have \\begin{align*} 30+32.5+x\\cdot0.5&=96 \\\\ x\\cdot0.5&=33.5 \\\\ x&=\\boxed{67} ~Haha0201 ~MRENTHUSIASM", "Suppose that Sam's average speed during the last $30$ minutes was $x$ mph.\nNote that Sam's average speed during the entire trip was $\\frac{96}{3/2}=64$ mph. Since Sam drove at $60$ mph, $65$ mph, and $x$ mph for the same duration ( $30$ minutes each), his average speed during the entire trip was the average of $60$ mph, $65$ mph, and $x$ mph. We have \\begin{align*} \\frac{60+65+x}{3}&=64 \\\\ 60+65+x&=192 \\\\ x&=\\boxed{67} ~coolmath_2018 ~MRENTHUSIASM" ]

[ "Suppose that Sam's average speed during the last $30$ minutes was $x$ mph.\nRecall that a half hour is equal to $30$ minutes. Therefore, Sam drove $60\\cdot0.5=30$ miles during the first half hour, $65\\cdot0.5=32.5$ miles during the second half hour, and $x\\cdot0.5$ miles during the last half hour. We have \\begin{align*} 30+32.5+x\\cdot0.5&=96 \\\\ x\\cdot0.5&=33.5 \\\\ x&=\\boxed{67} ~Haha0201 ~MRENTHUSIASM", "Suppose that Sam's average speed during the last $30$ minutes was $x$ mph.\nNote that Sam's average speed during the entire trip was $\\frac{96}{3/2}=64$ mph. Since Sam drove at $60$ mph, $65$ mph, and $x$ mph for the same duration ( $30$ minutes each), his average speed during the entire trip was the average of $60$ mph, $65$ mph, and $x$ mph. We have \\begin{align*} \\frac{60+65+x}{3}&=64 \\\\ 60+65+x&=192 \\\\ x&=\\boxed{67} ~coolmath_2018 ~MRENTHUSIASM" ]

[ "Let's call the distance that Samia had to travel in total as $2x$ , so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both $\\frac{2x}{2}$ , or $x$ \\[\\] She bikes at a rate of $17$ kph, so she travels the distance she bikes in $\\frac{x}{17}$ hours. She walks at a rate of $5$ kph, so she travels the distance she walks in $\\frac{x}{5}$ hours. \\[\\] The total time is $\\frac{x}{17}+\\frac{x}{5} = \\frac{22x}{85}$ . This is equal to $\\frac{44}{60} = \\frac{11}{15}$ of an hour. Solving for $x$ , we have: \\[\\] \\[\\frac{22x}{85} = \\frac{11}{15}\\] \\[\\frac{2x}{85} = \\frac{1}{15}\\] \\[30x = 85\\] \\[6x = 17\\] \\[x = \\frac{17}{6}\\] \\[\\] Since $x$ is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about $\\boxed{2.8}$", "Notice that Samia walks $\\frac {17}{5}$ times slower than she bikes, and that she walks and bikes the same distance. Thus, the fraction of the total time that Samia will be walking is \\[\\frac{\\frac{17}{5}}{\\frac{17}{5}+\\frac{5}{5}} = \\frac{17}{22}\\] Then, multiply this by the time \\[\\frac{17}{22} \\cdot 44 \\text{minutes} = 34 \\text{minutes}\\] 34 minutes is a little greater than $\\frac {1}{2}$ of an hour so Samia traveled \\[\\sim \\frac {1}{2} \\cdot 5 = 2.5 \\text{kilometers}\\] The answer choice a little greater than 2.5 is $\\boxed{2.8}$ by $5$ and gotten the exact answer as well)", "Since the distance between biking and walking is equal, we can use the given rates' harmonic mean to find the average speed.\n\\[\\frac{2ab}{a+b}\\implies\\frac{2 \\cdot 17 \\cdot 5}{17+5}=\\frac{85}{11}\\text{kph}\\]\nThen, multiplying by $\\frac{11}{15}$ hours gives the overall distance $\\frac{17}{3}$ kilometers. Samia only walks half of that, so $\\frac{17}{6}\\approx \\boxed{2.8}$ kilometers." ]

[ "Since the distance traveled by bicycle and foot are equal, we can substitute the time traveled by bike, or $b$ , as $\\frac{5}{17}$ of the time traveled by foot, or $f$\nAccordingly, we have $\\frac{22}{17}f=44$ This comes out to $f=34$ This means that Samia traveled $34$ minutes on foot, and hence, $44-34=10$ minutes on bicycle\nBecause $10$ minutes on bike yields $\\frac{17}{6}$ kilometers, and distance on bike = distance on foot, we have the final answer of $\\frac{17}{6} \\approx \\boxed{2.8}$" ]

[ "Let's call the distance that Samia had to travel in total as $2x$ , so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they are both $\\frac{2x}{2}$ , or $x$ \\[\\] She bikes at a rate of $17$ kph, so she travels the distance she bikes in $\\frac{x}{17}$ hours. She walks at a rate of $5$ kph, so she travels the distance she walks in $\\frac{x}{5}$ hours. \\[\\] The total time is $\\frac{x}{17}+\\frac{x}{5} = \\frac{22x}{85}$ . This is equal to $\\frac{44}{60} = \\frac{11}{15}$ of an hour. Solving for $x$ , we have: \\[\\] \\[\\frac{22x}{85} = \\frac{11}{15}\\] \\[\\frac{2x}{85} = \\frac{1}{15}\\] \\[30x = 85\\] \\[6x = 17\\] \\[x = \\frac{17}{6}\\] \\[\\] Since $x$ is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about $\\boxed{2.8}$", "Notice that Samia walks $\\frac {17}{5}$ times slower than she bikes, and that she walks and bikes the same distance. Thus, the fraction of the total time that Samia will be walking is \\[\\frac{\\frac{17}{5}}{\\frac{17}{5}+\\frac{5}{5}} = \\frac{17}{22}\\] Then, multiply this by the time \\[\\frac{17}{22} \\cdot 44 \\text{minutes} = 34 \\text{minutes}\\] 34 minutes is a little greater than $\\frac {1}{2}$ of an hour so Samia traveled \\[\\sim \\frac {1}{2} \\cdot 5 = 2.5 \\text{kilometers}\\] The answer choice a little greater than 2.5 is $\\boxed{2.8}$ by $5$ and gotten the exact answer as well)", "Since the distance between biking and walking is equal, we can use the given rates' harmonic mean to find the average speed.\n\\[\\frac{2ab}{a+b}\\implies\\frac{2 \\cdot 17 \\cdot 5}{17+5}=\\frac{85}{11}\\text{kph}\\]\nThen, multiplying by $\\frac{11}{15}$ hours gives the overall distance $\\frac{17}{3}$ kilometers. Samia only walks half of that, so $\\frac{17}{6}\\approx \\boxed{2.8}$ kilometers." ]

[ "Since the distance traveled by bicycle and foot are equal, we can substitute the time traveled by bike, or $b$ , as $\\frac{5}{17}$ of the time traveled by foot, or $f$\nAccordingly, we have $\\frac{22}{17}f=44$ This comes out to $f=34$ This means that Samia traveled $34$ minutes on foot, and hence, $44-34=10$ minutes on bicycle\nBecause $10$ minutes on bike yields $\\frac{17}{6}$ kilometers, and distance on bike = distance on foot, we have the final answer of $\\frac{17}{6} \\approx \\boxed{2.8}$" ]

[ "The $5$ sandwiches cost $5\\cdot 3=15$ dollars. The $8$ sodas cost $8\\cdot 2=16$ dollars. In total, the purchase costs $15+16=\\boxed{31}$ dollars." ]

[ "The $5$ sandwiches cost $5\\cdot 3=15$ dollars. The $8$ sodas cost $8\\cdot 2=16$ dollars. In total, the purchase costs $15+16=\\boxed{31}$ dollars." ]

[ "If $65\\%$ of the votes were likes, then $35\\%$ of the votes were dislikes. $65\\%-35\\%=30\\%$ , so $90$ votes is $30\\%$ of the total number of votes. Doing quick arithmetic shows that the answer is $\\boxed{300}$", "Let's consider that Sangho has received 100 votes. This means he has received 65 upvotes and 35 downvotes. Part of these upvotes and downvotes cancel out, so Sangho is now left with a total of 30 upvotes, or a score increase of 30. In order for his score to be 90, Sangho must receive three sets of 100 votes. Therefore, the answer is $\\boxed{300}$", "Let $x$ be the amount of votes cast, $35\\%$ of $x$ would be dislikes and $65\\%$ of $x$ would be likes. Since a like earns the video 1 point and a dislike takes 1 point away from the video, the amount of points Sangho's video will have in terms of $x$ is $65\\%-35\\%=30\\%$ of $x$ , or $\\frac{3}{10}$ of $x$ . Because Songho's video had a score of 90 points, $\\frac{3x}{10} = 90$ . After we solve for $x$ , we get $x = \\boxed{300}$", "Let x be the amount of like votes and y be the amount of dislike votes. Then $x-y=90$ . Since $65\\%$ of the total votes are like votes, $0.65(x+y)=x$ . then $.65y = .35x$ or $13/7$ y = $x$ . Plugging that in to the original equation, $6/7 y$ $90$ and $y =105$ Then $x=195$ and the total number of votes is $105+195 = 300$ . The answer is $\\boxed{300}$", "Obviously, $90/(65\\%-35\\%) = 90/30\\% = 300$ . The answer is $\\boxed{300}$" ]

[ "A staircase with $n$ steps contains $4 + 6 + 8 + ... + 2n + 2$ toothpicks. This can be rewritten as $(n+1)(n+2) -2$\nSo, $(n+1)(n+2) - 2 = 180$\nSo, $(n+1)(n+2) = 182.$\nInspection could tell us that $13 \\cdot 14 = 182$ , so the answer is $13 - 1 = \\boxed{12}$", "Layer $1$ $4$ steps\nLayer $1,2$ $10$ steps\nLayer $1,2,3$ $18$ steps\nLayer $1,2,3,4$ $28$ steps\nFrom inspection, we can see that with each increase in layer the difference in toothpicks between the current layer and the previous increases by $2$ . Using this pattern:\n$4, 10, 18, 28, 40, 54, 70, 88, 108, 130, 154, 180$\nFrom this we see that the solution is $\\boxed{12}$", "We can find a function that gives us the number of toothpicks for every layer. Using finite difference, we know that the degree must be $2$ and the leading coefficient is $1$ . The function is $f(n)=n^2+3n$ where $n$ is the layer and $f(n)$ is the number of toothpicks.\nWe have to solve for $n$ when $n^2+3n=180\\Rightarrow n^2+3n-180=0$ . Factor to get $(n-12)(n+15)$ . The roots are $12$ and $-15$ . Clearly $-15$ is impossible so the answer is $\\boxed{12}$", "Notice that the number of toothpicks can be found by adding all the horizontal and all the vertical toothpicks. We can see that for the case of 3 steps, there are $2(3+3+2+1)=18$ toothpicks. Thus, the equation is $2S + 2(1+2+3...+S)=180$ with $S$ being the number of steps. Solving, we get $S = 12$ , or $\\boxed{12}$ .\n-liu4505", "There are $\\frac{n(n+1)}{2}$ squares. Each has $4$ toothpick sides. To remove overlap, note that there are $4n$ perimeter toothpicks. $\\frac{\\frac{n(n+1)}{2}\\cdot 4-4n}{2}$ is the number of overlapped toothpicks\nAdd $4n$ to get the perimeter (non-overlapping). Formula is $\\text{number of toothpicks} = n^2+3n$ Then you can \"guess\" or factor (also guessing) to get the answer $\\boxed{12}$ .\n~bjc" ]

[ "Let $x$ be the two-digit number, $y$ be the three-digit number. Putting together the given, we have $1000x+y=9xy \\Longrightarrow 9xy-1000x-y=0$ . Using SFFT , this factorizes to $(9x-1)\\left(y-\\dfrac{1000}{9}\\right)=\\dfrac{1000}{9}$ , and $(9x-1)(9y-1000)=1000$\nSince $89 < 9x-1 < 890$ , we can use trial and error on factors of 1000. If $9x - 1 = 100$ , we get a non-integer. If $9x - 1 = 125$ , we get $x=14$ and $y=112$ , which satisifies the conditions. Hence the answer is $112 + 14 = \\boxed{126}$", "As shown above, we have $1000x+y=9xy$ , so $1000/y=9-1/x$ $1000/y$ must be just a little bit smaller than 9, so we find $y=112$ $x=14$ , and the solution is $\\boxed{126}$", "To begin, we rewrite $(10a+b)*(100x+10y+z)*9 = 10000a + 1000b + 100x + 10y + z$\nas\n$(90a+9b-1)(100x+10y+z) = 10000a + 1000b$\nand\n$(90a+9b-1)(100x+10y+z) = 1000(10a + b)$\nThis is the most important part: Notice $(90a+9b-1)$ is $-1 \\pmod{10a+b}$ and $1000(10a + b)$ is $0\\pmod{10a+b}$ . That means $(100x+10y+z)$ is also $0\\pmod{10+b}$ . Rewrite $(100x+10y+z)$ as $n\\times(10a+b)$\n$(90a+9b-1)\\times n(10a+b)= 1000(10a + b)$\n$(90a+9b-1)\\times n= 1000$\nNow we have to find a number that divides 1000 using prime factors 2 or 5 and is $8\\pmod9$ . It is quick to find there is only one: 125. That gives 14 as $10a+b$ and 112 as $100x+10y+z$ . Therefore the answer is $112 + 14 = \\boxed{126}$" ]

[ "We will simulate the process in steps.\nIn the beginning, we have:\nIn the first step we pour $4/2=2$ ounces of coffee from cup $1$ to cup $2$ , getting:\nIn the second step we pour $2/2=1$ ounce of coffee and $4/2=2$ ounces of cream from cup $2$ to cup $1$ , getting:\nHence at the end we have $3+2=5$ ounces of liquid in cup $1$ , and out of these $2$ ounces is cream. Thus the answer is $\\boxed{25}$", "Let's consider this in steps. \nWe have 4 ounces of coffee in the first cup. \nWe hace 4 ounces of cream in the second cup. \nWe take half of the coffee in the first cup(2 ounces), and add it to the second cup, yielding 6 ounces in total in the second cup(in a 1:2 ratio between coffee and cream, respecitvely).\nWe then take half of the second cup and pour it into the first cup. $6/2=3$ , so there is now 5 ounces in the first cup, 2 coffee and 3 the mixture.\nRemember that the mixture is in a 1:2 ratio between coffee and cream. So, coffee has one ounce and cream has 2 ounces. \nIn total, there is 2 ounces in the 5 ounce first cup. Putting 2 over 5, we get the answer. Therefore, the answer is $\\boxed{25}$" ]

[ "We will simulate the process in steps.\nIn the beginning, we have:\nIn the first step we pour $4/2=2$ ounces of coffee from cup $1$ to cup $2$ , getting:\nIn the second step we pour $2/2=1$ ounce of coffee and $4/2=2$ ounces of cream from cup $2$ to cup $1$ , getting:\nHence at the end we have $3+2=5$ ounces of liquid in cup $1$ , and out of these $2$ ounces is cream. Thus the answer is $\\boxed{25}$", "Let's consider this in steps. \nWe have 4 ounces of coffee in the first cup. \nWe hace 4 ounces of cream in the second cup. \nWe take half of the coffee in the first cup(2 ounces), and add it to the second cup, yielding 6 ounces in total in the second cup(in a 1:2 ratio between coffee and cream, respecitvely).\nWe then take half of the second cup and pour it into the first cup. $6/2=3$ , so there is now 5 ounces in the first cup, 2 coffee and 3 the mixture.\nRemember that the mixture is in a 1:2 ratio between coffee and cream. So, coffee has one ounce and cream has 2 ounces. \nIn total, there is 2 ounces in the 5 ounce first cup. Putting 2 over 5, we get the answer. Therefore, the answer is $\\boxed{25}$" ]

[ "Note that since $AB$ is a diameter, $\\angle AEB = 90^{\\circ}$ , which means $AB$ is an altitude of equilateral triangle $ABC$ . It follows that $\\triangle ABE$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, and so $AE = AB \\cdot \\frac{\\sqrt{3}}{2} = (2 \\cdot 1) (\\frac{\\sqrt{3}}{2}) = \\boxed{3}$" ]

[ "$\\triangle ABC$ is isosceles, hence $\\angle ACB = \\angle CAB$\nThe sum of internal angles of $\\triangle ABC$ can now be expressed as $\\angle B + \\frac 52 \\angle B + \\frac 52 \\angle B = 6\\angle B$ , hence $\\angle B = 30^\\circ$ , and each of the other two angles is $75^\\circ$\nNow we know that $\\angle DCE = \\angle ACB = 75^\\circ$\nFinally, $\\triangle CDE$ is isosceles, hence each of the two remaining angles ( $\\angle D$ and $\\angle E$ ) is equal to $\\frac{180^\\circ - 75^\\circ}2 = \\frac{105^\\circ}2 = \\boxed{52.5}$" ]

[ "First scale down the whole cube by $12$ . Let point $P$ have coordinates $(x, y, z)$ , point $A$ have coordinates $(0, 0, 0)$ , and $s$ be the side length. Then we have the equations \\begin{align*} (s-x)^2+y^2+z^2&=\\left(5\\sqrt{10}\\right)^2, \\\\ x^2+(s-y)^2+z^2&=\\left(5\\sqrt{5}\\right)^2, \\\\ x^2+y^2+(s-z)^2&=\\left(10\\sqrt{2}\\right)^2, \\\\ (s-x)^2+(s-y)^2+(s-z)^2&=\\left(3\\sqrt{7}\\right)^2. \\end{align*} These simplify into \\begin{align*} s^2+x^2+y^2+z^2-2sx&=250, \\\\ s^2+x^2+y^2+z^2-2sy&=125, \\\\ s^2+x^2+y^2+z^2-2sz&=200, \\\\ 3s^2-2s(x+y+z)+x^2+y^2+z^2&=63. \\end{align*} Adding the first three equations together, we get $3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575$ .\nSubtracting this from the fourth equation, we get $2(x^2+y^2+z^2)=512$ , so $x^2+y^2+z^2=256$ . This means $PA=16$ . However, we scaled down everything by $12$ so our answer is $16*12=\\boxed{192}$", "Once the equations for the distance between point P and the vertices of the cube have been written, we can add the first, second, and third to receive, \\[2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.\\] Subtracting the fourth equation gives \\begin{align*} 2(x^2 + y^2 + z^2) &= 575 - 63 \\\\ x^2 + y^2 + z^2 &= 256 \\\\ \\sqrt{x^2 + y^2 + z^2} &= 16. \\end{align*} Since point $A = (0,0,0), PA = 16$ , and since we scaled the answer is $16 \\cdot 12 = \\boxed{192}$", "Let $E$ be the vertex of the cube such that $ABED$ is a square.\nUsing the British Flag Theorem , we can easily show that \\[PA^2 + PE^2 = PB^2 + PD^2\\] and \\[PA^2 + PG^2 = PC^2 + PE^2\\] Hence, by adding the two equations together, we get $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$ . Substituting in the values we know, we get $2PA^2 + 7\\cdot 36^2 =10\\cdot60^2 + 5\\cdot 60^2 + 2\\cdot 120^2$\nThus, we can solve for $PA$ , which ends up being $\\boxed{192}$", "For all points $X$ in space, define the function $f:\\mathbb{R}^{3}\\rightarrow\\mathbb{R}$ by $f(X)=PX^{2}-GX^{2}$ . Then $f$ is linear; let $O=\\frac{2A+G}{3}$ be the center of $\\triangle BCD$ . Then since $f$ is linear, \\begin{align*} 3f(O)=f(B)+f(C)+f(D)&=2f(A)+f(G) \\\\ \\left(PB^{2}-GB^{2}\\right)+\\left(PC^{2}-GC^{2}\\right)+\\left(PD^{2}-GD^{2}\\right)&=2\\left(PA^{2}-GA^{2}\\right)+PG^{2} \\\\ \\left(60\\sqrt{10}\\right)^{2}-2x^{2}+\\left(60\\sqrt{5}\\right)^{2}-2x^{2}+\\left(120\\sqrt{2}\\right)^{2}-2x^{2}&=2PA^{2}-2\\cdot 3x^{2}+\\left(36\\sqrt{7}\\right)^{2}, \\end{align*} where $x$ denotes the side length of the cube. Thus \\begin{align*} 36\\text000+18\\text000+28\\text800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\\\ 82\\text800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\\\ 73\\text728&=2PA^{2} \\\\ 36\\text864&=PA^{2} \\\\ PA&=\\boxed{192}" ]

[ "Let $O=\\Gamma$ be the center of the semicircle and $X=\\Omega$ be the center of the circle.\nApplying the Extended Law of Sines to $\\triangle PQR,$ we find the radius of $\\odot X:$ \\[XP=\\frac{QR}{2\\cdot\\sin \\angle QPR}=\\frac{3\\sqrt3}{2\\cdot\\frac{\\sqrt3}{2}}=3.\\] Alternatively, by the Inscribed Angle Theorem, $\\triangle QRX$ is a $30^\\circ\\text{-}30^\\circ\\text{-}120^\\circ$ triangle with base $QR=3\\sqrt3.$ Dividing $\\triangle QRX$ into two congruent $30^\\circ\\text{-}60^\\circ\\text{-}90^\\circ$ triangles, we get that the radius of $\\odot X$ is $XQ=XR=3$ by the side-length ratios.\nLet $M$ be the midpoint of $\\overline{QR}.$ By the Perpendicular Chord Bisector Converse, we have $\\overline{OM}\\perp\\overline{QR}$ and $\\overline{XM}\\perp\\overline{QR}.$ Together, points $O, X,$ and $M$ must be collinear.\nBy the SAS Congruence, we have $\\triangle QXM\\cong\\triangle RXM,$ both of which are $30^\\circ\\text{-}60^\\circ\\text{-}90^\\circ$ triangles. By the side-length ratios, we obtain $RM=\\frac{3\\sqrt3}{2}, RX=3,$ and $XM=\\frac{3}{2}.$ By the Pythagorean Theorem on right $\\triangle ORM,$ we get $OM=\\frac{13}{2}$ and $OX=OM-XM=5.$ By the Pythagorean Theorem on right $\\triangle OXP,$ we get $OP=4.$\nLet $C$ be the foot of the perpendicular from $P$ to $\\overline{QR},$ and $D$ be the foot of the perpendicular from $X$ to $\\overline{PC},$ as shown below: Clearly, quadrilateral $XDCM$ is a rectangle. Since $\\angle XPD=\\angle OXP$ by alternate interior angles, we have $\\triangle XPD\\sim\\triangle OXP$ by the AA Similarity, with the ratio of similitude $\\frac{XP}{OX}=\\frac 35.$ Therefore, we get $PD=\\frac 95$ and $PC=PD+DC=PD+XM=\\frac 95 + \\frac 32 = \\frac{33}{10}.$\nThe area of $\\triangle PQR$ is \\[\\frac12\\cdot QR\\cdot PC=\\frac12\\cdot3\\sqrt3\\cdot\\frac{33}{10}=\\frac{99\\sqrt3}{20},\\] from which the answer is $99+3+20=\\boxed{122}.$", "\nSuppose we label the points as shown in the diagram above, where $C$ is the center of the semicircle and $O$ is the center of the circle tangent to $\\overline{AB}$ . Since $\\angle QPR = 60^{\\circ}$ , we have $\\angle QOR = 2\\cdot 60^{\\circ}=120^{\\circ}$ and $\\triangle QOR$ is a $30-30-120$ triangle, which can be split into two $30-60-90$ triangles by the altitude from $O$ . Since $QR=3\\sqrt{3},$ we know $OQ=OR=\\tfrac{3\\sqrt{3}}{\\sqrt{3}}=3$ by $30-60-90$ triangles. The area of this part of $\\triangle PQR$ is $\\frac{1}{2}bh=\\tfrac{3\\sqrt{3}}{2}\\cdot\\tfrac{3}{2}=\\tfrac{9\\sqrt{3}}{4}$ . We would like to add this value to the sum of the areas of the other two parts of $\\triangle PQR$\nTo find the areas of the other two parts of $\\triangle PQR$ using the $\\sin$ area formula, we need the sides and included angles. Here we know the sides but what we don't know are the angles. So it seems like we will have to use an angle from another triangle and combine them with the angles we already know to find these angles easily. We know that $\\angle QOR = 120^{\\circ}$ and triangles $\\triangle COQ$ and $\\triangle COR$ are congruent as they share a side, $CQ=CR,$ and $OQ=OR$ . Therefore $\\angle COQ = \\angle COR = 120^{\\circ}$ . Suppose $CO=x$ . Then $3^{2}+x^{2}-6x\\cos{120^{\\circ}}=7^{2}$ , and since $\\cos{120^{\\circ}}=-\\tfrac{1}{2}$ , this simplifies to $x^{2}+3x=7^{2}-3^{2}\\rightarrow x^{2}+3x-40=0$ . This factors nicely as $(x-5)(x+8)=0$ , so $x=5$ as $x$ can't be $-8$ . Since $CO=5, OP=3$ and $\\angle OPC=90^{\\circ}$ , we now know that $\\triangle OPC$ is a $3-4-5$ right triangle. This may be useful info for later as we might use an angle in this triangle to find the areas of the other two parts of $\\triangle PQR$\nLet $\\angle POC = \\alpha$ . Then $\\sin\\alpha = \\tfrac{4}{5}, \\cos\\alpha = \\tfrac{3}{5}, \\angle QOP = 120+\\alpha,$ and $\\angle POR = 120-\\alpha$ . The sum of the areas of $\\triangle QOP$ and $\\triangle POR$ is $3\\cdot 3\\cdot\\tfrac{1}{2}\\cdot\\left[\\sin(120-\\alpha)+\\sin(120+\\alpha)\\right]=\\tfrac{9}{2}\\left[\\sin(120-\\alpha)+\\sin(120+\\alpha)\\right],$ which we will add to $\\tfrac{9\\sqrt{3}}{4}$ to get the area of $\\triangle PQR$ . Observe that \\[\\sin(120-\\alpha) = \\sin 120\\cos\\alpha-\\sin\\alpha\\cos 120 = \\tfrac{\\sqrt{3}}{2}\\cdot\\tfrac{3}{5}-\\tfrac{4}{5}\\cdot\\tfrac{-1}{2}=\\tfrac{3\\sqrt{3}}{10}+\\tfrac{4}{10}=\\tfrac{3\\sqrt{3}+4}{10}\\] and similarly $\\sin(120+\\alpha)=\\tfrac{3\\sqrt{3}-4}{10}$ . Adding these two gives $\\tfrac{3\\sqrt{3}}{5}$ and multiplying that by $\\tfrac{9}{2}$ gets us $\\tfrac{27\\sqrt{3}}{10},$ which we add to $\\tfrac{9\\sqrt{3}}{4}$ to get $\\tfrac{54\\sqrt{3}+45\\sqrt{3}}{20}=\\tfrac{99\\sqrt{3}}{20}$ . The answer is $99+3+20=102+20=\\boxed{122}.$", " Define points as shown above, where $N=\\overleftrightarrow{PA}\\cap\\overleftrightarrow{QR}$ . The area of $\\triangle PQR$ is simply \\[\\dfrac{1}{2}PX\\cdot QR=\\dfrac{3\\sqrt{3}}{2}PX;\\] it remains to compute the value of $PX$ . Note that $PX$ is simply a weighted average of $BT$ and $CS;$ it is $\\dfrac{CP}{BP}$ times closer to $BT$ than it is to $CS$ . Observe that \\[CS=\\sqrt{CQ^{2}-\\left(\\dfrac{1}{2}QR\\right)^{2}}=\\sqrt{7^{2}-\\left(\\dfrac{3\\sqrt{3}}{2}\\right)^{2}}=6.5\\] since the radius of $\\Gamma$ is $7$ as its diameter is $14$ . Note also by the Extended Law of Sines the radius of $\\Omega$ is $\\dfrac{3\\sqrt{3}}{2\\sin 60^{\\circ}}=3,$ so $OS=3\\cos 60^{\\circ}=1.5$ . Since $C, O,$ and $S$ are collinear by symmetry we have $CO=CS-OS=5,$ so $CP=\\sqrt{5^{2}-3^{2}}=4$ and $BP=7-4=3$ . Therefore, $\\triangle OPC$ is a $3\\text{-}4\\text{-}5$ right triangle; $\\triangle OPC\\sim\\triangle NSC$ since $\\angle OPC=\\angle CSN=90^{\\circ}$ and $\\angle OCP=\\angle NCS=\\sin^{-1}\\left(\\dfrac{3}{5}\\right)$ . Therefore $\\dfrac{CN}{CS}=\\cfrac{CO}{CP}=\\dfrac{5}{4}$ so $CN=\\dfrac{5}{4}CS=\\dfrac{65}{8}$ . Since $\\triangle BTN\\sim\\triangle CSN,$ we have $\\dfrac{BT}{BN}=\\dfrac{CS}{CN}=\\dfrac{4}{5}$ . Therefore \\[BT=\\dfrac{4}{5}BN=\\dfrac{4}{5}\\left(CN-7\\right)=\\dfrac{4}{5}\\cdot\\dfrac{9}{8}=\\dfrac{36}{40}=0.9;\\] so $PX$ is $\\dfrac{4}{3}$ times as close to $0.9$ as to $6.5;$ we can compute $PX=\\dfrac{4}{7}BT+\\dfrac{3}{7}CS=\\dfrac{4}{7}\\cdot0.9+\\dfrac{3}{7}\\cdot6.5=3.3$ . The area of $\\triangle PQR$ is \\[\\dfrac{3\\sqrt{3}}{2}\\cdot 3.3=\\dfrac{99\\sqrt{3}}{20}\\] and $99+3+20=\\boxed{122}$", "Let $O_{1}$ be the center of $\\odot\\Gamma, O_2$ be the center of $\\odot\\Omega,$ and $M$ be the midpoint of $\\overline{QR}.$ We have $O_{1}M=\\sqrt{7^2-\\left(\\frac{3\\sqrt3}{2}\\right)^2}=\\frac{13}{2}$ and by Extended Law of Sines, the radius of $\\odot\\Omega$ is $\\frac{3\\sqrt3}{2\\sin 60^\\circ}=3$ so $O_{2}M=3\\cos 60^\\circ=\\frac{3}{2}.$ Therefore $O_{1}O_{2}=O_{1}M-O_{2}M=5$ and $O_{1}P=\\sqrt{5^2 - 3^2}=4.$\nLet $X=\\overline{AB}\\cap\\overline{QR}.$ Obviously \\[\\angle O_{1}PO_{2}=\\angle O_{1}MX=90^\\circ~\\text{and}~\\angle PO_{1}O_{2}=\\angle MO_{1}X=\\arcsin\\left(\\frac{3}{5}\\right)\\] so $\\triangle PO_{1}O_{2}\\sim\\triangle MO_{1}X$ with ratio $\\frac{PO_{1}}{MO_{1}}=\\frac{4}{\\tfrac{13}{2}}=\\frac{8}{13}.$ Therefore $O_1X=\\frac{13}{8}\\cdot O_{1}O_{2}=\\frac{13}{8}\\cdot5=\\frac{65}{8}$ and $MX=\\frac{13}{8}\\cdot PO_{2}=\\frac{13}{8}\\cdot3=\\frac{39}{8}.$\nLet $H$ denote the foot of the altitude from $P$ to $\\overline{QR}.$ Because $\\overline{PH}\\parallel\\overline{O_{1}M},$ it follows that $\\triangle PHX\\sim\\triangle O_{1}MX.$ This similarity has ratio \\[\\frac{PX}{O_{1}X}=1-\\frac{O_{1}P}{O_{1}X}=1-\\frac{4}{\\tfrac{65}{8}}=1-\\frac{32}{65}=\\frac{33}{65}.\\] We therefore have $PH=\\frac{33}{65}\\cdot O_{1}M=\\frac{33}{65}\\cdot\\frac{13}{2}=\\frac{33}{10}.$\nFinally, the area of $\\triangle PQR$ is \\[\\frac{1}{2}\\cdot QR\\cdot PH=\\frac{1}{2}\\cdot3\\sqrt3\\cdot\\frac{33}{10}=\\frac{99\\sqrt3}{20},\\] so the answer is $99+3+20=\\boxed{122}.$", "By the Law of Sine in $\\triangle PQR$ and its circumcircle $\\odot \\Omega$ $2r_{\\Omega} = \\frac{QR}{ \\sin 60^{\\circ} } = \\frac{ 3\\sqrt{3} }{ \\frac{ \\sqrt{3} }{2} } = 6$ $r_{\\Omega} = 3$\n\\[\\Gamma \\Omega = \\sqrt{r_{\\Gamma}^2 - \\left( \\frac{ PQ }{2}\\right)^2} - \\sqrt{r_{\\Omega} - \\left( \\frac{PQ}{2}\\right)^2} = \\sqrt{7^2 - \\left( \\frac{ 3 \\sqrt{3} }{2}\\right)^2} - \\sqrt{3^2 - \\left( \\frac{ 3 \\sqrt{3} }{2}\\right)^2} = \\frac{13}{2} - \\frac32 = 5, \\quad \\Gamma P = \\sqrt{5^2 - 3^2} = 4\\]\nBy Power of a Point in $\\odot \\Gamma$ $PQ \\cdot PS = PA \\cdot PB = (7+4)(7-4) = 33$\nBy the Law of Sine in $\\triangle PRS$ $\\frac{PR}{PS} = \\frac{ \\sin \\angle PSR }{ \\sin \\angle PRS }$\nBy the Law of Sine in $\\triangle QRS$ and its circumcircle $\\odot \\Gamma$ $\\frac{QR}{ \\sin \\angle PSR } = 14$ $\\frac{ 3\\sqrt{3} }{ \\sin \\angle PSR } = 14$ $\\sin \\angle PSR = \\frac{ 3\\sqrt{3} }{14}$ $\\cos \\angle PSR = \\frac{ 13 }{14}$\n\\[\\sin \\angle PRS = \\sin ( 60^{\\circ} - \\angle PSR ) = \\sin 60^{\\circ} \\cos \\angle PSR - \\sin \\angle PSR \\cos 60^{\\circ} = \\frac{ \\sqrt{3} }{2} \\cdot \\frac{ 13 }{14} - \\frac{ 3\\sqrt{3} }{14} \\cdot \\frac12 = \\frac{ 5\\sqrt{3} }{14}\\]\n\\[\\frac{PR}{PS} = \\frac{ \\frac{ 3\\sqrt{3} }{14} }{ \\frac{ 5\\sqrt{3} }{14} } = \\frac35, \\quad PQ \\cdot PR = PQ \\cdot PS \\cdot \\frac{PR}{PS} = 33 \\cdot \\frac35 = \\frac{99}{5}\\]\n\\[[PQR] = \\frac12 \\cdot \\sin 60^{\\circ} \\cdot PQ \\cdot PR = \\frac12 \\cdot \\frac12 \\cdot \\frac{99}{5} = \\frac{ 99\\sqrt{3} }{20}, \\quad 99 + 3 + 20 = \\boxed{122}\\]", "Following Solution 4, We have $O_{1}$ (0,0) , $O_{2}$ (4,3). \nWe can write the equation of the two circles as: \\[\\odot\\Gamma  : x^{2}+y^{2}=7^{2}...(1)\\] \\[\\odot\\Omega : (x-4)^{2}+(y-3)^{2}=3^{2}...(2)\\] By substituting (1) into (2), we get \\[8x+6y-65=0...(3)\\] Notice (3) is the relationship between $x$ value and $y$ value, in other words, (3) is the linear equation that go through $R$ and $Q$ . \nLet the height drops from $P$ to $QR$ at $H$ . Therefore, we have \\[Area \\triangle PQR =\\frac{1}{2}\\cdot{QR}\\cdot{PH}\\] So \\[QR=3\\sqrt{3}\\] And by distance formula, $PH$ is the distance from $P$ (4,0) to $\\overline{QR}$ \\[PH={\\frac{|8\\cdot4+6\\cdot0-65|}{\\sqrt{8^{2}+6^{2}}}=\\frac{33}{10}}\\] Thus, We get \\[Area \\triangle PQR =\\frac{1}{2}\\cdot3\\sqrt{3}\\cdot\\frac{33}{10}=\\frac{99\\sqrt{3}}{20}\\] So the answer is $99+3+20=\\boxed{122}.$" ]

[ "Note that since set $A$ has $m$ consecutive integers that sum to $2m$ , the middle integer (i.e., the median) must be $2$ . Therefore, the largest element in $A$ is $2 + \\frac{m-1}{2}$\nFurther, we see that the median of set $B$ is $0.5$ , which means that the \"middle two\" integers of set $B$ are $0$ and $1$ . Therefore, the largest element in $B$ is $1 + \\frac{2m-2}{2} = m$ $2 + \\frac{m-1}{2} > m$ if $m < 3$ , which is clearly not possible, thus $2 + \\frac{m-1}{2} < m$\nSolving, we get \\begin{align*} m - 2 - \\frac{m-1}{2} &= 99\\\\ m-\\frac{m}{2}+\\frac{1}{2}&=101\\\\ \\frac{m}{2}&=100\\frac{1}{2}.\\\\ m &= \\boxed{201}", "Let us give the elements of our sets names: $A = \\{x, x + 1, x + 2, \\ldots, x + m - 1\\}$ and $B = \\{y, y + 1, \\ldots, y + 2m - 1\\}$ . So we are given that \\[2m = x + (x + 1) + \\ldots + (x + m - 1) = mx + (1 + 2 + \\ldots + (m - 1)) = mx + \\frac{m(m -1)}2,\\] so $2 = x + \\frac{m - 1}2$ and $x + (m - 1) = \\frac{m + 3}2$ (this is because $x = 2 - \\frac{m-1}{2}$ so plugging this into $x+(m-1)$ yields $\\frac{m+3}{2}$ ). Also, \\[m = y + (y + 1) + \\ldots + (y + 2m - 1) = 2my + \\frac{2m(2m - 1)}2,\\] so $1 = 2y + (2m - 1)$ so $2m = 2(y + 2m - 1)$ and $m = y + 2m - 1$\nThen by the given, $99 = |(x + m - 1) - (y + 2m - 1)| = \\left|\\frac{m + 3}2 - m\\right| = \\left|\\frac{m - 3}2\\right|$ $m$ is a positive integer so we must have $99 = \\frac{m - 3}2$ and so $m = \\boxed{201}$", "The thing about this problem is, you have some \"choices\" that you can make freely when you get to a certain point, and these choices won't affect the accuracy of the solution, but will make things a lot easier for us.\nFirst, we note that for set $A$\n\\[\\frac{m(f + l)}{2} = 2m\\]\nWhere $f$ and $l$ represent the first and last terms of $A$ . This comes from the sum of an arithmetic sequence.\nSolving for $f+l$ , we find the sum of the two terms is $4$\nDoing the same for set B, and setting up the equation with $b$ and $e$ being the first and last terms of set $B$\n\\[m(b+e) = m\\]\nand so $b+e = 1$\nNow we know, assume that both sequences are increasing sequences, for the sake of simplicity. Based on the fact that set $A$ has half the number of elements as set $B$ , and the difference between the greatest terms of the two two sequences is $99$ (forget about absolute value, it's insignificant here since we can just assume both sets end with positive last terms), you can set up an equation where $x$ is the last term of set A:\n\\[2(x-(-x+4)+1) = 1+(x+99)-(-x-99+1)\\]\nNote how i basically just counted the number of terms in each sequence here. It's made a lot simpler because we just assumed that the first term is negative and last is positive for each set, it has absolutely no effect on the end result! This is a great strategy that can help significantly simplify problems. Also note how exactly i used the fact that the first and last terms of each sequence sum to $4$ and $1$ respectively (add $x$ and $(-x+4)$ to see what i mean).\nSolving this equation we find $x = 102$ . We know the first and last terms have to sum to $4$ so we find the first term of the sequence is $-98$ . Now, the solution is in clear sight, we just find the number of integers between $-98$ and $102$ , inclusive, and it is $m = \\boxed{201}$", "First, calculate the average of set $A$ and set $B$ . It's obvious that they are $2$ and $1/2$ respectively. \nLet's look at both sets. Obviously, there is an odd number of integers in the set with $2$ being in the middle, which means that $m$ is an odd number and that the number of consecutive integers on each side of $2$ are equal. In set $B$ , it is clear that it contains an even number of integers, but since the number in the middle is $1/2$ , we know that the range of the consecutive numbers on both sides will be $(x$ to $0)$ and $(1$ to $-y)$\nNothing seems useful right now, but let's try plugging an odd number, $3$ , for $m$ in set $B$ . We see that there are $6$ consecutive integers and $3$ on both sides of $1/2$ . After plugging this into set $A$ , we find that the set equals \\[{1,2,3}\\] . From there, we find the absolute value of the difference of both of the greatest values, and get 0.\nLet's try plugging in another odd number, $55$ . We see that the resulting set of numbers is $(-54$ to $0)$ , and $(1$ to $55)$ . We then plug this into set $A$ , and find that the set of numbers is $(-25$ to $-29)$ which indeed results in the average being $2$ . We then find the difference of the greatest values to be 26.\nFrom here, we see a pattern that can be proven by more trial and error. When we make $m$ equal to $3$ , then the difference is $0$ whearas when we make it $55$ , then the difference is $26$ $55-3$ equals to $52$ and $26-0$ is just $0$ . We then see that $m$ increases twice as fast as the difference. So when the difference is $99$ , it increased $99$ from when it was $0$ , which means that $m$ increased by $99*2$ which is $198$ . We then add this to our initial $m$ of $3$ , and get $\\boxed{201}$ as our answer.", "Let the first term of $A$ be $a$ and the first term of $B$ be $b$ . There are $m$ elements in $A$ so $A$ is $a, a+1, a+2,...,a+m-1$ . Adding these up, we get $\\frac{2a+m-1}{2}\\cdot m = 2m \\implies 2a+m=5$ . Set $B$ contains the numbers $b, b+1, b+2,...,b+2m-1$ . Summing these up, we get $\\frac{2b+2m-1}{2}\\cdot 2m =m \\implies 2b+2m=2$ . The problem gives us that the absolute value of the difference of the largest terms in $A$ and $B$ is $99$ . The largest term in $A$ is $a+m-1$ and the largest term in $B$ is $b+2m-1$ so $|b-a+m|=99$ . From the first two equations we get, we can get that $2(b-a)+m=-3$ . Now, we make a guess and assume that $b-a+m=99$ (if we get a negative value for $m$ , we can try $b-a+m=-99$ ). From here we get that $b-a=-102$ . Solving for $m$ , we get that the answer is $\\boxed{201}$" ]

[ "$A \\cup B$ will be smallest if $B$ is completely contained in $A$ , in which case all the elements in $B$ would be counted for in $A$ . So the total would be the number of elements in $A$ , which is $\\boxed{20}$", "Assume WLOG that $A={1, 2, 3, 4, \\cdots , 20}$ , and $B={6, 7, 8, 9, 10, \\cdots , 20}$ . Then, all the integers $6$ through $20$ would be redundant in $A \\cup B$ , so $A \\cup B = 1, 2, 3, 4, \\cdots, 20 \\implies \\boxed{20}$" ]

[ "Note that terms of the sequence $(u_k)$ lie in the interval $\\left(0,\\frac12\\right),$ strictly increasing.\nSince the sequence $(u_k)$ tends to the limit $L,$ we set $u_{k+1}=u_k=L>0.$\nThe given equation becomes \\[L=2L-2L^2,\\] from which $L=\\frac12.$\nThe given inequality becomes \\[\\frac12-\\frac{1}{2^{1000}} \\leq u_k \\leq \\frac12+\\frac{1}{2^{1000}},\\] and we only need to consider $\\frac12-\\frac{1}{2^{1000}} \\leq u_k.$\nWe have \\begin{alignat*}{8} u_0 &= \\phantom{1}\\frac14 &&= \\frac{2^1-1}{2^2}, \\\\ u_1 &= \\phantom{1}\\frac38 &&= \\frac{2^2-1}{2^3}, \\\\ u_2 &= \\ \\frac{15}{32} &&= \\frac{2^4-1}{2^5}, \\\\ u_3 &= \\frac{255}{512} &&= \\frac{2^8-1}{2^9}, \\\\ & \\phantom{1111} \\vdots \\end{alignat*} By induction, it can be proven that \\[u_k=\\frac{2^{2^k}-1}{2^{2^k+1}}=\\frac12-\\frac{1}{2^{2^k+1}}.\\] We substitute this into the inequality, then solve for $k:$ \\begin{align*} \\frac12-\\frac{1}{2^{1000}} &\\leq \\frac12-\\frac{1}{2^{2^k+1}} \\\\ -\\frac{1}{2^{1000}} &\\leq -\\frac{1}{2^{2^k+1}} \\\\ 2^{1000} &\\leq 2^{2^k+1} \\\\ 1000 &\\leq 2^k+1. \\end{align*} Therefore, the least such value of $k$ is $\\boxed{10}.$", "If we list out the first few values of $k$ , we get the series $\\frac{1}{4}, \\frac{3}{8}, \\frac{15}{32}, \\frac{255}{512}$ , which always seems to be a negative power of $2$ away from $\\frac{1}{2}$ . We can test this out by setting $u_k=\\frac{1}{2}-\\frac{1}{2^{n_k}}$ , where $n_0=2$\nNow, we get \\begin{align*} u_{k+1} &= 2\\cdot\\left(\\frac{1}{2}-\\frac{1}{2^{n_{k}}}\\right)-2\\cdot\\left(\\frac{1}{2}-\\frac{1}{2^{n_{k}}}\\right)^2 \\\\ &= 1-\\frac{1}{2^{n_k - 1}}-2\\cdot\\left(\\frac{1}{4}-\\frac{1}{2^{n_k}}+\\frac{1}{2^{2 \\cdot n_k}}\\right)\\\\ &= 1-\\frac{1}{2^{n_k - 1}}-\\frac{1}{2}+\\frac{1}{2^{n_k-1}}-\\frac{1}{2^{2 \\cdot n_k-1}} \\\\ &= \\frac{1}{2}-\\frac{1}{2^{2 \\cdot n_k-1}}. \\end{align*} This means that this series approaches $\\frac{1}{2}$ , as the second term is decreasing. In addition, we find that $n_{k+1}=2 \\cdot n_k-1$\nWe claim that $n_k = 2^k+1$ , which can be proven by induction:\nBase Case\nWe have $n_0=2=2^0+1$\nInduction Step\nAssuming that the claim is true, we have $n_{k+1}=2 \\cdot (2^k+1)-1=2^{k+1}+1$\nIt follows that $n_{10}=2^{10}+1>1000$ and $n_9=2^9+1<1000$ . Therefore, the least value of $k$ would be $\\boxed{10}$", "We are given $u_{k+1} = 2u_k - 2{u_k}^2$ . Multiply this equation by $2$ and subtract $1$ from both sides. The equations can then be written nicely as $2u_{k+1} - 1 = -(2u_k-1)^2$ . Let $v_k = 2u_k - 1$ so that $v_{k+1} = -(v_k)^2$\nClearly, $v_0 = 2u_0 - 1 = -\\frac{1}{2}$ . Since the magnitude of $v_0$ is less than $1$ and because our recursive relation for $v_k$ squares the previous term (and negates it), we see that as $k \\rightarrow \\infty, 2u_k - 1 = v_k \\rightarrow 0$ . This means $u_k \\rightarrow \\frac{1}{2}$ , so $L = \\frac{1}{2}$\nIsolating $u_k$ in our relation $2u_k - 1 = v_k$ gives us $u_k = \\frac{v_k + 1}{2}$ .\nSubstituting into the inequality, we have $\\left|\\frac{v_k + 1}{2}-\\frac{1}{2}\\right| \\le \\frac{1}{2^{1000}}$ . Rewriting this, we get $|v_k| \\le \\frac{1}{2^{999}}$\nThe sequence $\\{v_k\\}$ is much easier to handle because of its simple recursive relation. Writing out a few terms shows that $|v_k| = \\frac{1}{2^{2^k}}$ . Now it just comes down to having $2^k > 999$ , so $k = \\boxed{10}$" ]

[ "Let $x$ be the number of elements in $A$ and $B$ which is equal.\nThen we could form equation $2x-1001 = 2007$\n$2x = 3008$\n$x = 1504$\nThe answer is $\\boxed{1504}$", "Let $x$ be the number of elements in $A$ not including the intersection. $2007-1001=1006$ total elements excluding the intersection. Since we know that $A=B$ , we can find that $x=\\frac{1006}2=503$ . Now we need to add the intersection. $503+1001=\\boxed{1504}$" ]

[ "The big cookie has radius $3$ , since the center of the center cookie is the same as that of the large cookie. The difference in areas of the big cookie and the seven small ones is $3^2\\pi-7\\pi=9\\pi-7\\pi=2 \\pi$ . The scrap cookie has this area, so its radius must be $\\boxed{2}$" ]

[ "The volume of each cube follows the pattern of $n^3$ , for $n$ is between $1$ and $7$\nWe see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the $7\\times 7\\times 7$ cube (which is just $7 \\times 7 = 49$ ). The sides areas can be measured as the sum $4\\sum_{n=1}^{7} n^2$ , giving us $560$ . Structurally, if we examine the tower from the top, we see that it really just forms a $7\\times 7$ square of area $49$ . Therefore, we can say that the total surface area is $560 + 49 + 49 = \\boxed{658}$ .\nAlternatively, for the area of the tops, we could have found the sum $\\sum_{n=2}^{7}((n)^{2}-(n-1)^{2})$ , giving us $49$ as well.", "It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7 inclusive.\nFirst, we will calculate the total surface area of the cubes, ignoring overlap. This value is $6 ( 1^2 + 2^2 + \\cdots + 7^2 ) = 6\\sum_{n=1}^{7} n^2 = 6 \\left( \\frac{7(7 + 1)(2 \\cdot 7 + 1)}{6} \\right) = 7 \\cdot 8 \\cdot 15 = 840$ . Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to $2\\sum_{n=1}^{6} n^2 = 182$ . Subtracting the overlapped surface area from the total surface area, we get $840 - 182 = \\boxed{658}$ . ~ emerald_block", "It can be seen that the side lengths of the cubes using cube roots are all integers from $1$ to $7$ , inclusive.\nOnly the cubes with side length $1$ and $7$ have $5$ faces in the surface area and the rest have $4$ . Also, since the cubes are stacked, we have to find the difference between each $n^2$ and $(n-1)^2$ side length as $n$ ranges from $7$ to $2$\nWe then come up with this: $5(49)+13+4(36)+11+4(25)+9+4(16)+7+4(9)+5+4(4)+3+5(1)$\nWe then add all of this and get $\\boxed{658}$", "Notice that the surface area of the top cube is $6s^2$ and the others are $4s^2$ . Then we can directly compute. The edge length for the first cube is $7$ and has a surface area of $294$ . The surface area of the next cube is $144$ . The surface area of the next cube $100$ . The surface area of the next cube is $64$ . The surface area of the next cube is $36$ . The surface area of the next cube is $16$ . The surface area of the next cube is $4$ . We then sum up $294+144+100+64+36+16+4$ to get $\\boxed{658}$ .\n~smartatmath", "First of all, compute the area of the sides, excluding the top and bottoms, of the cubes. The side lengths (cube root the volumes) are 1, 2, 3, 4, 5, 6, 7. \nEach cube's area of the sides can be calculated with $4($ area of one side $)$ $4(l^2)$ so in total that is $4(1+4+16+...+49)$ so $4(140)=560$ the area of all the sides of the cubes is $560$ . \nThen, calculate the bottom face of the largest cube, $7*7=49$ . \nNow, notice that if you stack the cubes up on top of each other, and look directly down on them, the tops of the cubes showing add up to the area of the bottom cube, the 7x7. Therefore, the sum of the area of the tops of the cubes is $7*7=49$\nNow add them all up: $49+49+560=658.$ Therefore, the answer is $\\boxed{658}$" ]

[ "The volume of each cube follows the pattern of $n^3$ , for $n$ is between $1$ and $7$\nWe see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the $7\\times 7\\times 7$ cube (which is just $7 \\times 7 = 49$ ). The sides areas can be measured as the sum $4\\sum_{n=1}^{7} n^2$ , giving us $560$ . Structurally, if we examine the tower from the top, we see that it really just forms a $7\\times 7$ square of area $49$ . Therefore, we can say that the total surface area is $560 + 49 + 49 = \\boxed{658}$ .\nAlternatively, for the area of the tops, we could have found the sum $\\sum_{n=2}^{7}((n)^{2}-(n-1)^{2})$ , giving us $49$ as well.", "It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7 inclusive.\nFirst, we will calculate the total surface area of the cubes, ignoring overlap. This value is $6 ( 1^2 + 2^2 + \\cdots + 7^2 ) = 6\\sum_{n=1}^{7} n^2 = 6 \\left( \\frac{7(7 + 1)(2 \\cdot 7 + 1)}{6} \\right) = 7 \\cdot 8 \\cdot 15 = 840$ . Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to $2\\sum_{n=1}^{6} n^2 = 182$ . Subtracting the overlapped surface area from the total surface area, we get $840 - 182 = \\boxed{658}$ . ~ emerald_block", "It can be seen that the side lengths of the cubes using cube roots are all integers from $1$ to $7$ , inclusive.\nOnly the cubes with side length $1$ and $7$ have $5$ faces in the surface area and the rest have $4$ . Also, since the cubes are stacked, we have to find the difference between each $n^2$ and $(n-1)^2$ side length as $n$ ranges from $7$ to $2$\nWe then come up with this: $5(49)+13+4(36)+11+4(25)+9+4(16)+7+4(9)+5+4(4)+3+5(1)$\nWe then add all of this and get $\\boxed{658}$", "Notice that the surface area of the top cube is $6s^2$ and the others are $4s^2$ . Then we can directly compute. The edge length for the first cube is $7$ and has a surface area of $294$ . The surface area of the next cube is $144$ . The surface area of the next cube $100$ . The surface area of the next cube is $64$ . The surface area of the next cube is $36$ . The surface area of the next cube is $16$ . The surface area of the next cube is $4$ . We then sum up $294+144+100+64+36+16+4$ to get $\\boxed{658}$ .\n~smartatmath", "First of all, compute the area of the sides, excluding the top and bottoms, of the cubes. The side lengths (cube root the volumes) are 1, 2, 3, 4, 5, 6, 7. \nEach cube's area of the sides can be calculated with $4($ area of one side $)$ $4(l^2)$ so in total that is $4(1+4+16+...+49)$ so $4(140)=560$ the area of all the sides of the cubes is $560$ . \nThen, calculate the bottom face of the largest cube, $7*7=49$ . \nNow, notice that if you stack the cubes up on top of each other, and look directly down on them, the tops of the cubes showing add up to the area of the bottom cube, the 7x7. Therefore, the sum of the area of the tops of the cubes is $7*7=49$\nNow add them all up: $49+49+560=658.$ Therefore, the answer is $\\boxed{658}$" ]

[ "We can count the total number of ways to distribute the candies (ignoring the restrictions), and then subtract the overcount to get the answer.\nEach candy has three choices; it can go in any of the three bags.\nSince there are seven candies, that makes the total distribution $3^7=2187$\nTo find the overcount, we calculate the number of invalid distributions: the red or blue bag is empty.\nThe number of distributions such that the red bag is empty is equal to $2^7$ , since it's equivalent to distributing the $7$ candies into $2$ bags.\nWe know that the number of distributions with the blue bag is empty will be the same number because of the symmetry, so it's also $2^7$\nThe case where both the red and the blue bags are empty (all $7$ candies are in the white bag) are included in both of the above calculations, and this case has only $1$ distribution.\nThe total overcount is $2^7+2^7-1=2^8-1$\nThe final answer will be $\\text{total}-\\text{overcount}=2187-(2^8-1) = 2187-256+1=1931+1=\\boxed{1932}$", "We can use to our advantage the answer choices $\\text{AMC}$ has given us, and eliminate the obvious wrong answer choices.\nWe can first figure out how many ways there are to take two candies from seven distinct candies to place them into the red/blue bags: $7\\cdot 6=42$\nNow we can look at the answer choices to find out which ones are divisible by $42$ , since the total number of combinations must be $42$ multiplied by some other number.\nSince answers A, B, D, and E are not divisible by 3 (divisor of 42), the answer must be $\\boxed{1932}$", "Let $r$ be the number of red bag candies. For ${1 \\leq r \\leq 6}$\nSo the number of candies left for the blue bag and the red bag is $7-r$ . Based on the problem, 1 candy must be fixed for the blue bag, which can be done $7-r$ ways. Now, before we continue, we need to realize that fixing a candy can lead to some over counting in cases where none in the white bag overlap. So we should try and find an alternative because we'll be over counting more than twice, and that will become extremely difficult to account for each case's over counting.\nSo, without fixing candies, we can put everything into options and work on the white bag, because once we figure out two bags, the remaining one is decided.\nThe options for the candies in the white bag are two: In the white bag or out of the white bag(by default, in the blue bag).\nNow, for choosing the red bag, we have ${7 \\choose r}$\nThen, we have for the white bag: $2^{7-r}$ . Because we aren't fixing one for blue, the power is $7-r$ instead of $7-r-1$\nWe have: $({7 \\choose r}) \\cdot (2^{7-r})$ Adding them up for ${1 \\leq r \\leq 6}$ , we get 2058.\nNow, for the invalid cases. Because we didn't fix candy for the blue bag, we need to subtract the cases where the blue bag remains empty. We can accomplish this pretty easily.\nWhen $r = 6$ , how many ways can the remaining 1 candy $not$ be placed in the blue bag. This can be done ${7 \\choose 1} = 7$ ways.\nWhen $r = 5$ , how many ways can the remaining 2 candies $not$ be placed in the blue bag. This can be done ${7 \\choose 2} = 21$ ways.\nWhen $r = 4$ , how many ways can the remaining 3 candies $not$ be placed in the blue bag. This can be done ${7 \\choose 3} = 35$ ways.\nAnd because ${7 \\choose 4} = {7 \\choose 3}, {7 \\choose 5} = {7 \\choose 2}, and {7 \\choose 6} = {7 \\choose 1}$ , we just multiply by two.\nFinally, we have: $2058 - 2 \\cdot (7+21+35) = 2058 - 126 = \\boxed{1932}$" ]

[ "First look at the numbers Alice says. $1, 3, 4, 6, 7, 9 \\cdots$ skipping every number that is congruent to $2 \\pmod 3$ . Thus, Barbara says those numbers EXCEPT every second - being $2 + 3^1 \\equiv 5 \\pmod{3^2=9}$ . So Barbara skips every number congruent to $5 \\pmod 9$ . We continue and see:\nAlice skips $2 \\pmod 3$ , Barbara skips $5 \\pmod 9$ , Candice skips $14 \\pmod {27}$ , Debbie skips $41 \\pmod {81}$ , Eliza skips $122 \\pmod {243}$ , and Fatima skips $365 \\pmod {729}$\nSince the only number congruent to $365 \\pmod {729}$ and less than $1,000$ is $365$ , the correct answer is $\\boxed{365}$", "After Alice says all her numbers, the numbers not mentioned yet are \\[\\text{Alice: } 2,5,8,11,14,17,\\cdots,998.\\] After Barbara says all her numbers, the numbers that haven't been said yet are \\[\\text{Barbara: } 5,14,23,32,41,50,\\cdots,995.\\] After Candice, the list is \\[\\text{Candice: } 14,41,68,\\cdots,986.\\] Notice how each list is an arithmetic sequence where the common differene is thrice the common ratio of the previous list and the first term is the second term of the previous list. Now that the pattern is clear, we construct the rest of the lists: \\[\\text{Debbie: } 41,122,203,\\cdots,959\\] \\[\\text{Eliza: } 122,365,608,878\\] \\[\\text{Fatima: } 365\\]\nThus, George says $\\boxed{365}$", "Notice that Alice has skipped the numbers $3n-1$ for $n=1,2,3,...,333$ .\nNamely, \\[3\\cdot1-1,3\\cdot2-1,3\\cdot3-1,...,3\\cdot333-1\\] Thus the numbers that Barbara skips are \\[3\\cdot2-1,3\\cdot5-1,3\\cdot8-1,...\\] or in a more general expression, $3(3n-1)-1$ for $n=1,2,3,...$ .\nNamely, \\[3\\cdot(3\\cdot1-1)-1,3\\cdot(3\\cdot2-1)-1,3\\cdot(3\\cdot3-1)-1,...\\] Repeating the pattern until George, we have the first number he says, \\[3(3(3(3(3(3\\cdot1-1)-1)-1)-1)-1)-1= \\boxed{365}\\] In addition, note that the second number George says exceeds $1000$", "Similar to Solution 1 we find that the only number that George can say must leave a remainder of $2$ when divided by $3$ , and that it must also leave a remainder of $5$ when divided by $9$ . Since we as human beings are usually lazy, and that MAA provides answer choices, we check all the possible numbers and find that our answer is $\\boxed{365}$", "Every integer from 1 to 1000 can be written in the form $t_n+1$ in base 10, where $t_n$ is a trinary integer with no more than 7 significant digits. The important insight is that person $1\\le{k}\\le6$ will not say $n$ if and only if the $k$ th digit from the right of $t_n$ is 1. Therefore, the last 6 digits of $t_n$ must be 1, as the first 6 people never said the number. The only options for $t_n$ are thus $2111111$ $1111111$ , and $0111111$ . But, since George only said one number, the first two must have been too big for it to be $\\le1000$ . Our answer is therefore $111111_3+1=\\frac{3^6-1}{3-1}+1=364+1=\\boxed{365}$" ]

[ "The results of the five remaining games are independent of the first game, so by symmetry, the probability that $A$ scores higher than $B$ in these five games is equal to the probability that $B$ scores higher than $A$ . We let this probability be $p$ ; then the probability that $A$ and $B$ end with the same score in these five games is $1-2p$\nOf these three cases ( $|A| > |B|, |A| < |B|, |A|=|B|$ ), the last is the easiest to calculate (see solution 2 for a way to directly calculate the other cases).\nThere are ${5\\choose k}$ ways to $A$ to have $k$ victories, and ${5\\choose k}$ ways for $B$ to have $k$ victories. Summing for all values of $k$\nThus $p = \\frac 12 \\left(1-\\frac{126}{512}\\right) = \\frac{193}{512}$ . The desired probability is the sum of the cases when $|A| \\ge |B|$ , so the answer is $\\frac{126}{512} + \\frac{193}{512} = \\frac{319}{512}$ , and $m+n = \\boxed{831}$", "You can break this into cases based on how many rounds $A$ wins out of the remaining $5$ games.\nSumming these 6 cases, we get $\\frac{638}{1024}$ , which simplifies to $\\frac{319}{512}$ , so our answer is $319 + 512 = \\boxed{831}$", "We can apply the concept of generating functions here.\nThe generating function for $B$ is $(1 + 0x^{1})$ for the first game where $x^{n}$ is winning n games. Since $B$ lost the first game, the coefficient for $x^{1}$ is 0. The generating function for the next 5 games is $(1 + x)^{5}$ . Thus, the total generating function for number of games he wins is\nThe generating function for $A$ is the same except that it is multiplied by $x$ instead of $(1+0x)$ . \nThus, the generating function for $A$ is\n$1x + 5x^{2} + 10x^{3} + 10x^{4} + 5x^{5} + x^{6}$\nThe probability that $B$ wins 0 games is $\\frac{1}{32}$ . Since the coefficients for all $x^{n}$ where\n$n \\geq 1$ sums to 32, the probability that $A$ wins more games is $\\frac{32}{32}$\nThus, the probability that $A$ has more wins than $B$ is $\\frac{1}{32} \\times \\frac{32}{32} + \\frac{5}{32} \\times \\frac{31}{32} + \\frac{10}{32} \\times \\frac{26}{32} + \\frac{10}{32} \\times \\frac{16}{32} + \\frac{5}{32} \\times \\frac{6}{32} +\\frac{1}{32} \\times \\frac{1}{32} = \\frac{638}{1024} = \\frac{319}{512}$\nThus, $319 + 512 = \\boxed{831}$", "After the first game, there are $10$ games we care about-- those involving $A$ or $B$ . There are $3$ cases of these $10$ games: $A$ wins more than $B$ $B$ wins more than $A$ , or $A$ and $B$ win the same number of games. Also, there are $2^{10} = 1024$ total outcomes. By symmetry, the first and second cases are equally likely, and the third case occurs $\\binom{5}{0}^2+\\binom{5}{1}^2+\\binom{5}{2}^2+\\binom{5}{3}^2+\\binom{5}{4}^2+\\binom{5}{5}^2 = \\binom{10}{5} = 252$ times, by a special case of Vandermonde's Identity . There are therefore $\\frac{1024-252}{2} = 386$ possibilities for each of the other two cases.\nIf $B$ has more wins than $A$ in its $5$ remaining games, then $A$ cannot beat $B$ overall. However, if $A$ has more wins or if $A$ and $B$ are tied, $A$ will beat $B$ overall. Therefore, out of the $1024$ possibilites, $386+252 = 638$ ways where $A$ wins, so the desired probability is $\\frac{638}{1024} = \\frac{319}{512}$ , and $m+n=\\boxed{831}$" ]

[ "Place the first triangle. Now, we can place the second triangle either adjacent to the first, or with one side between them, for a total of $\\boxed{2}$", "Take ${5 \\choose 2}$ to realize there are 10 ways to choose 2 different triangles. Then divide by 5 for each vertex of a pentagon to get $\\boxed{2}$" ]

[ "Neither of the digits $4$ $6$ , and $8$ can be a units digit of a prime. Therefore the sum of the set is at least $40 + 60 + 80 + 1 + 2 + 3 + 5 + 7 + 9 = 207$\nWe can indeed create a set of primes with this sum, for example the following sets work: $\\{ 41, 67, 89, 2, 3, 5 \\}$ or $\\{ 43, 61, 89, 2, 5, 7 \\}$\nThus the answer is $207\\implies \\boxed{207}$" ]

[ "There are two ways for a student to get $11$ $5+5+1$ and $5+3+3$ . Clearly if someone gets one of these combinations someone else could get the other, so we are not guaranteed the most points with $11$\nThere is only one way to get $13$ points: $5+5+3$ . In this case, the largest score another person could get is $5+3+3=11$ , so having $13$ points guarantees having more points than any other person $\\rightarrow \\boxed{13}$", "If someone gets $11$ points, the possible combinations are $5,5,1$ and $5,3,3$ If he gets $5,3,3$ then someone else can be $5,3,3$ which would not guarantee victory.\nIf we have 13 points, the only way to make this is $5,5,3$ . There is no way to get any number of points higher than this, so the answer is $\\boxed{13}$ .---stjwyl" ]

[ "If this is the case, then if there were only $99$ pieces of candy, the bag would have gone around the table a whole number of times. Thus, the number of students is a divisor of $99$ . The only choice that satisfies this is choice $\\boxed{11}$" ]

[ "Right now, she scored $76, 94,$ and $87$ points, for a total of $257$ points. She wants her average to be $81$ for her $5$ tests, so she needs to score $405$ points in total. This means she needs to score a total of $405-257= 148$ points in her next $2$ tests. Since the maximum score she can get on one of her $2$ tests is $100$ , the least possible score she can get is $\\boxed{48}$", "We know that she scored $76, 94,$ and $87$ points on her first $3$ tests for a total of $257$ points and that she wants her average to be $81$ for her $5$ total tests. Therefore, she needs to score a total of $405$ points. In addition, one of the final $2$ tests needs to be the maximum of $100$ points, to make the final test score—the one that we are looking for—the lowest score possible for her to earn. We can now see here that the sum of $76+94+100$ has a units digit of $0$ and that the final test score must have a units digit ending with a $5$ . Now, $87$ needs to be added to a number that makes the sum divisible by $5$ . Among the answer choices of $48, 52, 66, 70,$ and $74$ , only $\\boxed{48}$ has a units digit that works. ( $48+87=135$ , giving us a units digit of $5$ .)" ]

[ "We should notice that we can turn the information we are given into a linear equation and just solve for our set variables. I'll use the variables $x$ and $y$ for the scores on the last two tests. \\[\\frac{76+94+87+x+y}{5} = 81,\\] \\[\\frac{257+x+y}{5} = 81.\\] We can now cross multiply to get rid of the denominator. \\[257+x+y = 405,\\] \\[x+y = 148.\\] Now that we have this equation, we will assign $y$ as the lowest score of the two other tests, and so: \\[x = 100,\\] \\[y=48.\\] Now we know that the lowest score on the two other tests is $\\boxed{48}$", "We can compare each of the scores with the average of $81$ $76$ $\\rightarrow$ $-5$ $94$ $\\rightarrow$ $+13$ $87$ $\\rightarrow$ $+6$ $100$ $\\rightarrow$ $+19$\nSo the last one has to be $-33$ (since all the differences have to sum to $0$ ), which corresponds to $81-33 = \\boxed{48}$" ]

[ "Let $AC=x.$ By Angle Bisector Theorem, we have $\\frac{AB}{AC}=\\frac{BD}{CD},$ from which $BD=CD\\cdot\\frac{AB}{AC}=\\frac{30}{x}.$\nRecall that $x>0.$ We apply the Triangle Inequality to $\\triangle ABC:$\nTaking the intersection of the solutions gives \\[(m,n)=(0,\\infty)\\cap(0,15)\\cap(3,\\infty)=(3,15),\\] so the answer is $m+n=\\boxed{18}.$" ]

[ "Connect C and A, and we have a 3-4-5 right triangle and 5-12-13 right triangle. The area of both is $\\frac{3\\cdot4}{2}+\\frac{5\\cdot12}{2}=36\\Rightarrow\\boxed{36}$" ]

[ "We know that $\\sin(C)=-\\cos(B)=\\frac{3}{5}$ . Since $B$ and $C$ are obtuse, we have $\\sin(180-C)=\\cos(180-B)=\\frac{3}{5}$ . It is known that $\\sin(x)=\\cos(90-x)$ , so $180-C=90-(180-C)=180-B$ . We simplify this as follows:\n\\[-90+C=180-B\\]\n\\[B+C=270^{\\circ}\\]\nSince $B+C=270^{\\circ}$ , we know that $A+D=360-(B+C)=90^{\\circ}$ . Now extend $AB$ and $CD$ as follows:\n\nLet $AB$ and $CD$ intersect at $E$ . We know that $\\angle AED=90^{\\circ}$ because $\\angle E = 180 - (A+D)=180-90 = 90^{\\circ}$\nSince $\\sin BCD = \\frac{3}{5}$ , we get $\\sin ECB=\\sin(180-BCD)=\\sin BCD = \\frac{3}{5}$ . Thus, $EB=3$ and $EC=4$ from simple sin application.\n$AD$ is the hypotenuse of right $\\triangle AED$ , with leg lengths $AB+BE=7$ and $EC+CD=24$ . Thus, $AD=\\boxed{25}$" ]

[ "Let George's bags contain a total of $x$ marbles, so Jane's bag contains $2x$ marbles. This means the total number of non-chipped marbles is $3x \\equiv 0 \\pmod{3}$ , while the total number of marbles is $18+19+21+23+25+34 = 140 \\equiv 2 \\pmod{3}$ , so the number of chipped marbles must also be congruent to $2-0 \\equiv 2 \\pmod{3}$\nThe answer choices are congruent modulo 3 to $0$ $1$ $0$ $2$ , and $1$ respectively, so the only possible number of chipped marbles among these is $23$ . Indeed, if Jane takes the bags containing $19$ $25$ , and $34$ marbles and George takes the remaining bags containing $18$ and $21$ marbles, then Jane will have a total of $19+25+34 = 78$ marbles, which is twice as many as George's $18+21 = 39$ marbles, as desired. Thus the answer is precisely $\\boxed{23}$" ]

[ "Realize that any sequence that works (ascending) can be reversed for descending, so we can just take the amount of sequences that satisfy the ascending condition and multiply by two.\nIf we choose any of the numbers $1$ through $6$ , there are five other spots to put them, so we get $6 \\cdot 5 = 30$ . However, we overcount some cases. Take the example of $132456$ . We overcount this case because we can remove the $3$ or the $2$ . Therefore, any cases with two adjacent numbers swapped is overcounted, so we subtract $5$ cases (namely, $213456, 132456, 124356, 123546, 123465$ ,) to get $30-5=25$ , but we have to add back one more for the original case, $123456$ . Therefore, there are $26$ cases. Multiplying by $2$ gives the desired answer, $\\boxed{052}$", "Similar to above, a $1-1$ correspondence between ascending and descending is established by subtracting each number from $7$\nWe note that the given condition is equivalent to \"cycling\" $123456$ for a contiguous subset of it. For example,\n$12(345)6 \\rightarrow 125346, 124536$\nIt's not hard to see that no overcount is possible, and that the cycle is either $1$ \"right\" or $1$ \"left.\" Therefore, we consider how many elements we flip by. If we flip $1$ or $2$ such elements, then there is one way to cycle them. Otherwise, we have $2$ ways. Therefore, the total number of ascending is $1 + 5 + 2(4 + 3 + 2 + 1) = 26$ , and multiplying by two gives $\\boxed{052}.$", "Similarly to above, we find the number of ascending arrangements and multiply by 2.\nWe can choose $5$ cards to be the ascending cards, therefore leaving $6$ places to place the remaining card. There are $\\binom{6}{5}\\cdot 6=36$ to do this. However, since the problem is asking for the number of arrangements, we overcount cases such as $123456$ . Notice that the only arrangements that overcount are $123456$ (case 1) or if two adjacent numbers of $123456$ are switched (case 2).\n$\\text{Case 1: }$ This arrangement is counted $6$ times. Each time it is counted for any of the $5$ numbers selected. Therefore we need to subtract $5$ cases of overcounting.\n$\\text{Case 2: }$ Each time $2$ adjacent numbers of switched, there is one overcount. For example, if we have $213456$ , both $1$ or $2$ could be removed. Since there are $5$ possible switches, we need to subtract $5$ cases of overcounting.\nTherefore, we have $36-5-5=26$ total arrangements of ascending numbers. We multiply by two (for descending) to get the answer of $\\boxed{052}.$", "Like in previous solutions, we will count the number of ascending arrangements and multiply by 2.\nFirst, consider the arrangement 1-2-3-4-5-6. That gives us 1 arrangement which works.\nNext, we can switch two adjacent cards. There are 5 ways to pick two adjacent cards, so this gives us 5 arrangements.\nNow, we can \"cycle\" 3 adjacent cards. For example, 1-2-3 becomes 2-3-1 which becomes 3-1-2. There are 4 ways to pick a set of 3 adjacent cards, so this gives us 4x2=8 arrangements.\nCycling 4 adjacent cards, we get the new arrangements 2-3-4-1 (which works), 3-4-1-2 (which doesn't work), and 4-1-2-3 (which does work). We get 6 arrangements.\nSimilarly, when cycling 5 cards, we find 2x2=4 arrangements, and when cycling 6 cards, we find 2x1=2 arrangements.\nAdding, we figure out that there are 1+5+8+6+4+2=26 ascending arrangements. Multiplying by 2, we get the answer $\\boxed{052}.$", "First count the number of permutations of the cards such that if one card is removed, the remaining cards will be in ascending order. There is $1$ such permutation where all the cards appear in order: $123456.$ There are $5$ such permutations where two adjacent cards are interchanged, as in $124356.$ The other such permutations arise from removing one card from $123456$ and placing it in a position at least two away from its starting location. There are $4$ such positions to place each of the cards numbered $1$ and $6,$ and $3$ such positions for each of the cards numbered $2, 3, 4,$ and $5.$ This accounts for $2\\cdot4 + 4\\cdot3 =20$ permutations. Thus there are $1 + 5 + 20 = 26$ permutations where one card can be removed so that the remaining cards are in ascending order. There is an equal number of permutations that result in the cards' being in descending order. This gives the total $26 + 26 = \\boxed{52}$", "More generally, suppose there are $n \\geq 4$ cards numbered $1, 2, 3, \\dots, n$ arranged in ascending order. If any one of the $n$ cards is removed and placed in one of the $n$ positions in the arrangement, the resulting permutation will have the property that one card can be removed so that the remaining cards are in ascending order. This accounts for $n\\cdot n = n^2$ permutations. However, the original ascending order has been counted $n$ times, and each order that arises by switching two neighboring cards has been counted twice. Hence the number of arrangements where one card can be removed resulting in the remaining cards' being in ascending order is $n^2-(n-1)-(n-1)=(n-1)^2+1.$ When $n = 6$ , this is $(6-1)^2+1 = 26$ , and the final answer is $2\\cdot26 = \\boxed{52}$", "For ascending, if the $1$ goes in anything but the first two slots, the rest of the numbers have to go in ascending from $2$ , which are $4$ cases if there are $6$ cards. If $1$ goes in the second spot, then you can put any of the rest in the first slot but then the rest are determined, so in the case of $6$ cards, that gives $5$ more. If $1$ goes in the first slot, that means that you are doing the same problem with $n-1$ cards. So the recursion is $a_n=(n-2)+(n-1)+a_{n-1}$ . There's $a_1=1$ and $a_2=2$ , so you get $a_3=2+3=5$ $a_4=5+5=10$ $a_5=7+10=17$ , and $a_6=9+17=26$ . Or you can see that $a_n=(n-1)^2+1$ . We double to account for descending and get $\\boxed{052}$", "First, we know that ascending order and descending order are symmetrical to each other (namely, if we get 132456 where after we take out 3, it will be one scenario; and if we flip it and write 654231, it will be another scenario)\nThus, we only need to consider either descending or ascending and then times 2.\nWLOG let us consider ascending order\nCase 1: after we take out 1, the rest will be in ascending order:\nNotice that 1 can be tucked in any one of the 6 spaces, thus there are 6 scenarios.\nCase 2: after we take out 2, the rest will be in ascending order:\nNotice that if we put 2 next to 1 (to the right or to the left of 1), it will be an overcount, so there are only 4 cases for 2.\nIt is easy to see that this is the same for 3, 4, 5, and 6.\nThus, in total, we have \\[(6+4\\times5)\\times2=\\boxed{052}\\] ~Adali", "We start with five cards in ascending order, then insert the sixth card to obtain a valid arrangement.\nBased on the card to be inserted, we have six cases. As shown below, the red squares indicate the possible positions to insert the sixth card. Note that all arrangements of the six cards are distinct.\nThere are $26$ arrangements in which removing one card will leave the remaining five cards in ascending order. By symmetry, there are $26$ arrangements in which removing one card will leave the remaining five cards in descending order. So, the answer is $26\\cdot2=\\boxed{052}.$", "We first realize that as long as we have an ascending sequence of $5$ numbers, we can just plug in a $6$ th to make a sequence that works. For example, if we have $12345$ , we can plug in a $6$ in any of $6$ spaces, before $1$ , between $1$ and $2$ , and so on to after $5$ . We can also realize that this is completely symmetrical if the sequence is in descending order. For example, we could have $54321$ , and we could plug in a 6 in 6 of the spaces. For the total number of combinations, we have 6 ascending cases multiplied by the $6$ places that can hold whatever number is missing, and multiply by 2 because there is descending and ascending number cases.\nBut wait, what if we have $12345$ and plug in a $6$ at the end and $12346$ and plug in a $5$ at the $5$ th spot? We have overcounted, so we need to subtract off the identical pairs. Assume that one of them is the \"right\" combination. That means that there are $1$ \"right\" combination because all the rest will have $2$ of their combinations taken by the previous one. For example, if I have $12356$ and $12346$ , I would count those $123456$ as $12356$ 's instead of $12346$ 's. Therefore, we have $6+4+4+4+4+4 = 26$ combinations of ascending order, and since we need to count those of descending order as well, we have $26\\cdot2=\\boxed{052}.$", "Like the previous solutions, we calculate ascending possibilities then multiply by two. Then, the configuration can look like +++++ or +-+++ (the minus can go anywhere), where \"+\" indicates an increase, and \"-\" indicates a decrease. Obviously, for the first case, the only possibility is 123456. For the second case, we can start with 123456, and then take either a card after the \"-\" and put it before the \"-\", or we could take a card before the \"-\" and put it after the \"-\". This means that every card other than the one at the position of the \"-\" can be used in one way. There are five ways to have 4 plus's, and one minus, and for each way, there are 5 ways to rearrange 123456 to achieve that. So, our answer is $2(5\\cdot5+1)=\\boxed{052}$", "First, we select $5$ of the $6$ cards to put in ascending or descending order. Then, we must add the other card to the group to arrange the six cards such that one can be removed so that the remaining cards are in ascending or descending order. Then, it would seem that the answer is $(6C5)*2$ (ascending or descending)* $6$ (the number of ways to place the last card). However, this overcounts some cases. For example, 123456 and 654321 can be made with 12345 and 6, 12346 and 5, etc. Furthermore, 654312 can be made with 65432 and 1 or 65431 and 2. Therefore, we first find that $(6C5)*2*4$ $48$ . Then, we add the special cases ( $123456$ $654321$ $654312$ $123465$ ) to obtain $\\boxed{052}$" ]

[ "Consider shifting every person over three seats left after each person has gotten up and sat back down again. Now, instead of each person being seated not in the same chair and not in an adjacent chair, each person will be seated either in the same chair or an adjacent chair. The problem now becomes the number of ways in which six people can sit down in a chair that is either the same chair or an adjacent chair in a circle.\nConsider the similar problem of $n$ people sitting in a chair that is either the same chair or an adjacent chair in a row. Call the number of possibilities for this $F_n$ . Then if the leftmost person stays put, the problem is reduced to a row of $n-1$ chairs, and if the leftmost person shifts one seat to the right, the new person sitting in the leftmost seat must be the person originally second from the left, reducing the problem to a row of $n-2$ chairs. Thus, $F_n = F_{n-1} + F_{n-2}$ for $n \\geq 3$ . Clearly $F_1 = 1$ and $F_2 = 2$ , so $F_3 = 3$ $F_4 = 5$ , and $F_5 = 8$\nNow consider the six people in a circle and focus on one person. If that person stays put, the problem is reduced to a row of five chairs, for which there are $F_5 = 8$ possibilities. If that person moves one seat to the left, then the person who replaces him in his original seat will either be the person originally to the right of him, which will force everyone to simply shift over one seat to the left, or the person originally to the left of him, which reduces the problem to a row of four chairs, for which there are $F_4 = 5$ possibilities, giving $1 + 5 = 6$ possibilities in all. By symmetry, if that person moves one seat to the right, there are another $6$ possibilities, so we have a total of $8+6+6 = \\boxed{20}$ possibilities.", "Label the people sitting at the table $A, B, C, D, E, F,$ and assume that they are initially seated in the order $ABCDEF$ . The possible new positions for $A, B, C, D, E,$ and $F$ are respectively (a dash indicates a non-allowed position):\n\\[\\begin{tabular}{| c | c | c | c | c | c |} \\hline - & - & A & A & A & - \\\\ \\hline - & - & - & B & B & B \\\\ \\hline C & - & - & - & C & C \\\\ \\hline D & D & - & - & - & D \\\\ \\hline E & E & E & - & - & - \\\\ \\hline - & F & F & F & - & - \\\\ \\hline \\end{tabular}\\]\nThe permutations we are looking for should use one letter from each column, and there should not be any repeated letters:\n$\\begin{tabular}{c} CDEFAB \\\\ CEAFBD \\\\ CEFABD \\\\ CEFBAD \\\\ CFEABD \\\\ CFEBAD \\\\ DEAFBC \\\\ DEAFCB \\\\ DEFABC \\\\ DEFACB \\\\ DEFBAC \\\\ DFEABC \\\\ DFEACB \\\\ DFEBAC \\\\ EDAFBC \\\\ EDAFCB \\\\ EDFABC \\\\ EDFACB \\\\ EDFBAC \\\\ EFABCD \\end{tabular}$\nThere are $\\boxed{20}$ such permutations.", "We can represent each rearrangement as a permutation of the six elements $\\{1,2,3,4,5,6\\}$ in cycle notation. Note that any such permutation cannot have a 1-cycle, so the only possible types of permutations are 2,2,2-cycles, 4,2-cycles, 3,3-cycles, and 6-cycles. We deal with each case separately.\nFor 2,2,2-cycles, suppose that one of the 2-cycles switches the people across from each other, i.e. $(14)$ $(25)$ , or $(36)$ . WLOG, we may assume it to be $(14)$ . Then we could either have both of the other 2-cycles be across from each other, giving the permutation $(14)(25)(36)$ or else neither of the other 2-cycles is across from each other, in which case the only possible permutation is $(14)(26)(35)$ . This can happen for $(25)$ and $(36)$ as well. So since the first permutation is not counted twice, we find a total of $1+3=4$ permutations that are 2,2,2-cycles where at least one of the 2-cycles switches people diametrically opposite from each other. Otherwise, since the elements in a 2-cycle cannot differ by 1, 3, or 5 mod 6, they must differ by 2 or 4 mod 6, i.e. they must be of the same parity. But since we have three odd and three even elements, this is impossible. Hence there are exactly 4 such permutations that are 2,2,2-cycles.\nFor 4,2-cycles, we assume for the moment that 1 is part of the 2-cycle. Then the 2-cycle can be $(13)$ $(15)$ , or $(14)$ . The first two are essentially the same by symmetry, and we must arrange the elements 2, 4, 5, 6 into a 4-cycle. However, 5 must have two neighbors that are not next to it, which is impossible, hence the first two cases yield no permutations. If the 2-cycle is $(14)$ , then we must arrange the elements 2, 3, 5, 6 into a 4-cycle. Then 2 must have the neighbors 5 and 6. We find that the 4-cycles $(2536)$ and $(2635)$ satisfy the desired properties, yielding the permutations $(14)(2536)$ and $(14)(2635)$ . This can be done for the 2-cycles $(25)$ and $(36)$ as well, so we find a total of 6 such permutations that are 4,2-cycles.\nFor 3,3-cycles, note that if 1 neighbor 4, then the third element in the cycle will neighbor one of 1 and 4, so this is impossible. Therefore, the 3-cycle containing 1 must consist of elements 1, 3, and 5. Therefore, we obtain the four 3,3-cycles $(135)(246)$ $(153)(246)$ $(135)(264)$ , and $(153)(264)$\nFor 6-cycles, note that the neighbors of 1 can be 3 and 4, 3 and 5, or 4 and 5. In the first case, we may assume that it looks like $(314\\dots)$ -- the form $(413\\dots)$ is also possible but equivalent to this case. Then we must place the elements 2, 5, and 6. Note that 5 and 6 cannot go together, so 2 must go in between them. Also, 5 cannot neighbor 4, so we are left with one possibility, namely $(314625)$ , which has an analogous possibility $(413526)$ . In the second case, we assume that it looks like $(315\\dots)$ . The 2 must go next to the 5, and the 6 must go last (to neighbor the 3), so the only possibility here is $(315246)$ , with the analogous possibility $(513642)$ . In the final case, we may assume that it looks like $(415\\dots)$ . Then the 2 and 3 cannot go together, so the 6 must go in between them. Therefore, the only possibility is $(415362)$ , with the analogous possibility $(514263)$ . We have covered all possibilities for 6-cycles, and we have found 6 of them.\nTherefore, there are $4+6+4+6 = \\boxed{20}$ such permutations." ]

[ "Define the radii of the six congruent circles as $r$ . If we draw all of the radii to the points of external tangency, we get a regular hexagon . If we connect the vertices of the hexagon to the center of the circle $C$ , we form several equilateral triangles . The length of each side of the triangle is $2r$ . Notice that the radius of circle $C$ is equal to the length of the side of the triangle plus $r$ . Thus, the radius of $C$ has a length of $3r = 30$ , and so $r = 10$ $K = 30^2\\pi - 6(10^2\\pi) = 300\\pi$ , so $\\lfloor 300\\pi \\rfloor = \\boxed{942}$" ]

[ "Count the number of sides that are not exposed, where a cube is connected to another cube and subtract it from the total number of faces. There are $5$ places with two adjacent cubes, covering $10$ sides, and $(6)(6)=36$ faces. The exposed surface area is $36-10 = \\boxed{26}$", "We can count the number of showing faces from each side. One thing that we notice is that the front face has the same number of squares as the back face, the side faces have the same surface area, etc. Therefore, we are looking for $2($ front surface area $+$ side surface area $+$ top surface area $)$ . We find that this is $2(5 + 4 + 4) = 2 * 13 = \\boxed{26}$" ]

[ "Looking at the vertical column, the three numbers sum to $23$ . If we make the numbers on either end $9$ and $8$ in some order, the middle number will be $6$ . This is the minimum for the intersection.\nLooking at the horizontal row, the four numbers sum to $12$ . If we minimize the three numbers on the right to $123$ , the first number has a maximum value of $6$ . This is the maximum for the intersection\nThus, the minimum of the intersection is $6$ , and the maximum of the intersection is $6$ . This means the intersection must be $6$ , and the other numbers must be $9$ and $8$ in the column, and $123$ in the row. The sum of all the numbers is $12 + 23 - 6 = 29$ , and the answer is $\\boxed{29}$" ]

[ "For two numbers to have a difference that is a multiple of $5$ , the numbers must be congruent $\\bmod{5}$ (their remainders after division by $5$ must be the same).\n$0, 1, 2, 3, 4$ are the possible values of numbers in $\\bmod{5}$ . Since there are only $5$ possible values in $\\bmod{5}$ and we are picking $6$ numbers, by the Pigeonhole Principle , two of the numbers must be congruent $\\bmod{5}$\nTherefore the probability that some pair of the $6$ integers has a difference that is a multiple of $5$ is $\\boxed{1}$" ]

[ "Let $n$ be the number of women present, and let _ be some positive number of women between groups of men. Since the problem states that every man stands next to another man, there cannot be isolated men. Thus, there are five cases to consider, where $(k)$ refers to a consecutive group of $k$ men:\nFor the first case, we can place the three groups of men in between women. We can think of the groups of men as dividers splitting up the $n$ women. Since there are $n+1$ possible places to insert the dividers, and we need to choose any three of these locations, we have $\\dbinom{n+1}{3}$ ways.\nThe second, third, and fourth cases are like the first, only that we need to insert two dividers among the $n+1$ possible locations. Each gives us $\\dbinom{n+1}{2}$ ways, for a total of $3\\dbinom{n+1}{2}$ ways.\nThe last case gives us $\\dbinom{n+1}{1}=n+1$ ways.\nTherefore, the total number of possible ways where there are no isolated men is\n\\[\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1).\\]\nThe total number of ways where there is a group of at least four men together is the sum of the third, fourth, and fifth case, or\n\\[2\\dbinom{n+1}{2}+(n+1).\\]\nThus, we want to find the minimum possible value of $n$ where $n$ is a positive integer such that\n\\[\\dfrac{2\\dbinom{n+1}{2}+(n+1)}{\\dbinom{n+1}{3}+3\\dbinom{n+1}{2}+(n+1)}\\le\\dfrac{1}{100}.\\]\nAfter simplification, we arrive at \\[\\dfrac{6(n+1)}{n^2+8n+6}\\le\\dfrac{1}{100}.\\]\nSimplifying again, we see that we seek the smallest positive integer value of $n$ such that $n(n-592)\\ge594$ . Clearly $n>592$ , or the left side will not even be positive; we quickly see that $n=593$ is too small but $n=\\boxed{594}$ satisfies the inequality." ]

[ "First, put the six numbers we have in order, since we are concerned with the median: $3, 5, 5, 7, 8, 9$\nWe have three more numbers to insert into the list, and the median will be the $5^{th}$ highest (and $5^{th}$ lowest) number on the list. If we top-load the list by making all three of the numbers greater than $9$ , the median will be the highest it can possibly be. Thus, the maximum median is the fifth piece of data in the list, which is $8$ , giving an answer of $\\boxed{8}$" ]

[ "The pepperoni circles' diameter is $2$ , since $\\dfrac{12}{6} = 2$ . From that we see that the area of the $24$ circles of pepperoni is $\\left ( \\frac{2}{2} \\right )^2 (24\\pi) = 24\\pi$ . The large pizza's area is $6^2\\pi$ . Therefore, the ratio is $\\frac{24\\pi}{36\\pi} = \\boxed{23}$" ]

[ "The sum of the areas is equal to $2\\cdot1+2\\cdot4+2\\cdot9+2\\cdot16+2\\cdot25+2\\cdot36$ . This is equal to $2(1+4+9+16+25+36)$ , which is equal to $2\\cdot91$ . This is equal to our final answer of $\\boxed{182}$" ]

[ "It can be seen that the diameter of the eighth sphere is equal to the radius of the seventh sphere by drawing out a diagram of the insides of the seventh sphere. The radius of the seventh sphere is $2+1=3$ , the radius of the eight sphere is $\\boxed{32}$" ]

[ "The first line divides the plane into two regions. The second line intersects one line, creating two regions. The third line intersects two lines, creating three regions. Similarly, the fourth line intersects three lines and creates four regions, the fifth line intersects four lines and creates five regions, and the sixth line intersects five lines and creates six regions.\nTotaling the regions created results in $2 + 2 + 3 + 4 + 5 + 6 = 22$ regions, which is answer choice $\\boxed{22}$", "1963 AHSME Problem 27.png\nWith careful drawing, one can draw all six lines and count the regions. There are $22$ regions in total, which is answer choice $\\boxed{22}$" ]

[ "We can use the fact that the number of regions that $n$ lines divide a plane is given by the equation $L_n = \\frac{n^2 + n +2}{2}$ , and in this problems, $n=6$ , from which the answer is $\\boxed{22}$" ]

[ "There are $3$ spaces between the 1st and 4th trees, so each of these spaces has $\\frac{60}{3}=20$ feet. Between the first and last trees there are $5$ spaces, so the distance between them is $20\\times5=100$ feet, $\\boxed{100}$" ]

[ "We know the repeating section is made of $2$ red lights and $3$ green lights. The 3rd red light would appear in the 2nd section of this pattern, and the 21st red light would appear in the 11th section. There would then be a total of $44$ lights in between the 3rd and 21st red light, translating to $45$ $6$ -inch gaps. Since the question asks for the answer in feet, the answer is $\\frac{45*6}{12} \\rightarrow \\boxed{22.5}$" ]

[ "Start by buying the largest packs first. After three $24$ -packs, $90-3(24)=18$ cans are left. After one $12$ -pack, $18-12=6$ cans are left. Then buy one more $6$ -pack. The total number of packs is $3+1+1=\\boxed{5}$" ]

[ "If we let $p$ be the number of people initially in the group, then $0.4p$ is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still $p$ , but the number of girls is $0.4p-2$ . Since only $30\\%$ of the group are girls, \\begin{align*} \\frac{0.4p-2}{p}&=\\frac{3}{10}\\\\ 4p-20&=3p\\\\ p&=20\\end{align*} The number of girls initially in the group is $0.4p=0.4(20)=\\boxed{8}$", "Let $x$ be the number of people initially in the group and $g$ the number of girls. $\\frac{2}{5}x = g$ , so $x = \\frac{5}{2}g$ . Also, the problem states $\\frac{3}{10}x = g-2$ . Substituting $x$ in terms of $g$ into the second equation yields that $g = \\boxed{8}$" ]

[ "If we let $p$ be the number of people initially in the group, then $0.4p$ is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still $p$ , but the number of girls is $0.4p-2$ . Since only $30\\%$ of the group are girls, \\begin{align*} \\frac{0.4p-2}{p}&=\\frac{3}{10}\\\\ 4p-20&=3p\\\\ p&=20\\end{align*} The number of girls initially in the group is $0.4p=0.4(20)=\\boxed{8}$", "Let $x$ be the number of people initially in the group and $g$ the number of girls. $\\frac{2}{5}x = g$ , so $x = \\frac{5}{2}g$ . Also, the problem states $\\frac{3}{10}x = g-2$ . Substituting $x$ in terms of $g$ into the second equation yields that $g = \\boxed{8}$" ]

[ "Let $r$ and $b$ be the number of red and blue marbles originally in the bag respectively. After $1$ red marble is removed, there are $r+b-1$ marbles left in the bag and $r-1$ red marbles left. So \\[\\frac{r-1}{r+b-1}=\\frac{1}{7}.\\] When $2$ blue marbles are removed, there are $r$ red marbles and $r+b-2$ total marbles left in the bag. So \\[\\frac{r}{r+b-2}=\\frac{1}{5}.\\] Cross multiplying for each yields \\begin{align*}7r-7=r+b-1&\\implies 7r-6=r+b\\\\ 5r=r+b-2&\\implies 5r+2=r+b.\\end{align*} We can equate each of these expressions to yields \\[7r-6=5r+2\\implies 2r=8\\implies r=4\\implies b=18.\\] Therefore, the total number of marbles is \\[r+b=4+18=\\boxed{22.}\\]" ]

[ "An even number comes from multiplying an even and even, even and odd, or odd and even. Since an odd number only comes from multiplying an odd and odd, there are less cases and it would be easier to find the probability of spinning two odd numbers from $1$ . Multiply the independent probabilities of each spinner getting an odd number together and subtract it from $1$\n\\[1-\\frac24 \\cdot \\frac23 = 1- \\frac13 = \\boxed{23}\\]", "We can make a chart and the we see that the 12 possibilities: 1, 2, 3, 2, 4, 6, 3, 6, 9, 4, 8, and 12. Out of these only 8 work; thus the probability is \\[\\boxed{23}\\]", "We do a little bit of casework. In order to get a product that's even, we need at least one even number. First, we consider the probability of getting an even number on the first spinner, and then multiply by 1 because the second spinner can be anything. \\[\\frac12 \\cdot 1 = \\frac12\\] Next, we look at the chance that we don't get an even on the first spinner, but get an even on the second spinner (we don't do the probability of even on the second spinner multiplied by one because we would be double counting both spinners are even). \\[\\frac12 \\cdot \\frac13 = \\frac16\\] Add these two together to get the total probability, and we get \\[\\frac12 + \\frac16 = \\frac46 = \\boxed{23}\\]" ]

[ "Let $G$ be the foot of the perpendicular from $O$ to $AB$ . Denote $x = EG$ and $y = FG$ , and $x > y$ (since $AE < BF$ and $AG = BG$ ). Then $\\tan \\angle EOG = \\frac{x}{450}$ , and $\\tan \\angle FOG = \\frac{y}{450}$\nBy the tangent addition rule $\\left( \\tan (a + b) = \\frac{\\tan a + \\tan b}{1 - \\tan a \\tan b} \\right)$ , we see that \\[\\tan 45 = \\tan (EOG + FOG) = \\frac{\\frac{x}{450} + \\frac{y}{450}}{1 - \\frac{x}{450} \\cdot \\frac{y}{450}}.\\] Since $\\tan 45 = 1$ , this simplifies to $1 - \\frac{xy}{450^2} = \\frac{x + y}{450}$ . We know that $x + y = 400$ , so we can substitute this to find that $1 - \\frac{xy}{450^2} = \\frac 89 \\Longrightarrow xy = 150^2$\nSubstituting $x = 400 - y$ again, we know have $xy = (400 - y)y = 150^2$ . This is a quadratic with roots $200 \\pm 50\\sqrt{7}$ . Since $y < x$ , use the smaller root, $200 - 50\\sqrt{7}$\nNow, $BF = BG - FG = 450 - (200 - 50\\sqrt{7}) = 250 + 50\\sqrt{7}$ . The answer is $250 + 50 + 7 = \\boxed{307}$", " Let the midpoint of $\\overline{AB}$ be $M$ and let $FB = x$ , so then $MF = 450 - x$ and $AF = 900 - x$ . Drawing $\\overline{AO}$ , we have $\\triangle OEF\\sim\\triangle AOF$ , so \\[\\frac{OF}{EF} = \\frac{AF}{OF}\\Rightarrow (OF)^2 = 400(900 - x).\\] By the Pythagorean Theorem on $\\triangle OMF$ \\[(OF)^2 = 450^2 + (450 - x)^2.\\] Setting these two expressions for $(OF)^2$ equal and solving for $x$ (it is helpful to scale the problem down by a factor of 50 first), we get $x = 250\\pm 50\\sqrt{7}$ . Since $BF > AE$ , we want the value $x = 250 + 50\\sqrt{7}$ , and the answer is $250 + 50 + 7 = \\boxed{307}$", "Let lower case letters be the complex numbers correspond to their respective upper case points in the complex plane, with $o = 0, a = -450 + 450i, b = 450 + 450i$ , and $f = x + 450i$ . Since $EF$ = 400, $e = (x-400) + 450i$ . From $\\angle{EOF} = 45^{\\circ}$ , we can deduce that the rotation of point $F$ 45 degrees counterclockwise, $E$ , and the origin are collinear. In other words, \\[\\dfrac{e^{i \\frac{\\pi}{4}} \\cdot (x + 450i)}{(x - 400) + 450i}\\] is a real number. Simplyfying using the fact that $e^{i \\frac{\\pi}{4}} = \\dfrac{\\sqrt{2}}{2} + i \\dfrac{\\sqrt{2}}{2}$ , clearing the denominator, and setting the imaginary part equal to $0$ , we eventually get the quadratic \\[x^2 - 400x + 22500 = 0\\] which has solutions $x = 200 \\pm 50\\sqrt{7}$ . It is given that $AE < BF$ , so $x = 200 - 50\\sqrt{7}$ and \\[BF = 450 - (200 - 50\\sqrt{7}) = 250 + 50\\sqrt{7} \\Rightarrow \\boxed{307}.\\]", " Let G be a point such that it lies on AB, and GOE is 90 degrees. Let H be foot of the altitude from O to AB.\nSince $\\triangle GOE \\sim \\triangle OHE$ $\\frac{GO}{OE} = \\frac{450}{x}$ , and by Angle Bisector Theorem $\\frac{GF}{FE} = \\frac{450}{x}$ . Thus, $GF = \\frac{450 \\cdot 400}{x}$ $AF = AH-FH = 50+x$ , and $KA = EB$ (90 degree rotation), and now we can bash on 2 similar triangles $\\triangle GAK \\sim \\triangle GHO$\n\\[\\frac{GA}{AK} = \\frac{GH}{OH}\\] \\[\\frac{\\frac{450 \\cdot 400}{x}-50-x}{450-x} = \\frac{\\frac{450 \\cdot 400}{x}+400-x}{450}\\] I hope you like expanding \\[x^2 - 850x + \\frac{81000000}{x} = -450x - 22500 + \\frac{81000000}{x}\\] \\[x^2 - 400x + 22500 = 0\\] Quadratic formula gives us \\[x = 200 \\pm 50 \\sqrt{7}\\] Since AE < BF \\[x = 200 - 50 \\sqrt{7}\\] Thus, \\[BF = 250 + 50 \\sqrt{7}\\] So, our answer is $\\boxed{307}$", "We know that G is on the perpendicular bisector of $EF$ , which means that $EJ=JF=200$ $EG=GF=200\\sqrt{2}$ and $GH=250$ . Now, let $HO$ be equal to $x$ . We can set up an equation with the Pythagorean Theorem:\n\\begin{align*} \\sqrt{x^2+250^2}&=(200\\sqrt{2})^2 \\\\ x^2+62500&=80000 \\\\ x^2&=17500 \\\\ x&=50\\sqrt{7} \\end{align*}\nNow, since $IO=450$\n\\begin{align*} HI&=450-x \\\\ &=450-50\\sqrt{7} \\\\ \\end{align*} \\\\\nSince $HI=AJ$ , we now have:\n\\begin{align*} BF&=AB-AJ-JF \\\\ &=900-(450-50\\sqrt{7})-200 \\\\ &=250+50\\sqrt{7} \\\\ \\end{align*}\nThis means that our answer would be $250+50+7=\\boxed{307}$", "Construct $BO, AO.$ Let $\\angle{FOB} = \\alpha.$ Also let $FB = x$ then $AE = 500-x.$ We then have from simple angle-chasing: \\begin{align*} \\angle{BFO} = 135 - \\alpha \\\\ \\angle{OFE} = 45 + \\alpha \\\\ \\angle{EOA} = 45 - \\alpha \\\\ \\angle{AEO} = 90 + \\alpha \\\\ \\angle{OEF} = 90 - \\alpha. \\end{align*} From AA similarity we have \\[\\triangle{EOB} \\sim \\triangle{EFO}.\\] This gives the ratios, \\[\\dfrac{400 + x}{EO} = \\dfrac{450\\sqrt{2}}{FO}.\\] Similarly from AA similarity \\[\\triangle{FOA} \\sim \\triangle{FEO}.\\] So we get the ratios \\[\\dfrac{EO}{450\\sqrt{2}} = \\dfrac{FO}{900-x}.\\] We can multiply to get \\[\\dfrac{400 + x}{450\\sqrt{2}} = \\dfrac{450\\sqrt{2}}{900 - x}.\\] Cross-multiplying reveals \\[360000 + 500x - x^2 = 405000.\\] Bringing everything to one side we have \\[x^2 - 500x + 45000 = 0.\\] By the quadratic formula we get \\[x = \\dfrac{500 + \\sqrt{500^2 - 4\\cdot45000}}{2} = \\dfrac{500 + \\sqrt{70000}}{2} = \\dfrac{500 + 100\\sqrt{7}}{2} = 250 + 50\\sqrt{7}.\\] Therefore \\[p+q+r = 250 + 50 + 7 = \\boxed{307}.\\] ~aa1024", "Draw AO, OB, and extend OB to D. Let $\\angle{FOB} = \\alpha.$ Then, after angle chasing, we find that \\[\\angle{AEB} = 90 + \\alpha\\] . \nUsing this, we draw a line perpendicular to $AB$ at $E$ to meet $BD$ at $M$ . Since $\\angle{MEO} = \\alpha$ and $\\angle{EMO} = 45$ , we have that \\[\\triangle{EMO} \\sim \\triangle{OBF}\\] Let $FB = x$ . Then $EM = 400+x$ . Since $FB/BO = \\frac{x}{450\\sqrt{2}}$ , and $MO/EM = FB/OB$ , we have \\[MO = \\frac{(400+x)x}{450\\sqrt{2}}\\] Since $\\triangle{EBM}$ is a $45-45-90$ triangle, \\[(400+x)\\sqrt{2} = 450 \\sqrt{2} + \\frac{(400+x)x}{450\\sqrt{2}}\\] Solving for $x$ , we get that $x=250 +- 50s\\sqrt{7}$ , but since $FB>AE$ $FB = 250+50\\sqrt{7}$ , thus \\[p+q+r=\\boxed{307}\\] -dchang0524" ]

[ "Drawing $EF$ , it clearly passes through the center of $ABCD$ . Letting this point be $P$ , we note that $AEBP$ and $CFDP$ are congruent cyclic quadrilaterals, and that $AP=BP=CP=DP=\\frac{13}{\\sqrt{2}}.$ Now, from Ptolemy's, $13\\cdot EP=\\frac{13}{\\sqrt{2}}(12+5)\\implies EP=\\frac{17\\sqrt{2}}{2}$ . Since $EF=EP+FP=2\\cdot EP$ , the answer is $(17\\sqrt{2})^2=\\boxed{578}.$", "We first see that the whole figure is symmetrical and reflections across the center that we will denote as $O$ bring each half of the figure to the other half. Thus we consider a single part of the figure, namely $EO.$\nFirst note that $\\angle BAO = 45^{\\circ}$ since $O$ is the center of square $ABCD.$ Also note that $\\angle EAB = \\arccos{\\left(\\frac{12}{13}\\right)}$ or $\\arcsin{\\left(\\frac{5}{13}\\right)}.$ Finally, we know that $AO =\\frac{13\\sqrt{2}}{2}.$ Now we apply laws of cosines on $\\bigtriangleup AEO.$\nWe have $EO^2 = 12^2 + (\\frac{13\\sqrt{2}}{2})^2 - 2 \\cdot 12 \\cdot \\left(\\frac{13\\sqrt{2}}{2}\\right) \\cdot \\cos{\\angle EAO}.$ We know that $\\angle EAO = 45^{\\circ} + \\arccos{\\left(\\frac{12}{13}\\right)}.$ Thus we have $\\cos{\\angle EAO} = \\cos\\left(45^{\\circ} + \\arccos{\\left(\\frac{12}{13}\\right)}\\right)$ which applying the cosine sum identity yields $\\cos{45^{\\circ}}\\cos{\\arccos\\frac{12}{13}} - \\sin{45^{\\circ}}\\sin\\arcsin{\\frac{5}{12}} =\\frac{12\\sqrt{2}}{26} -\\frac{5\\sqrt{2}}{26} =\\frac{7\\sqrt{2}}{26}.$\nNote that we are looking for $4EO^2$ so we multiply $EO^2 = 12^2 + \\left(\\frac{13\\sqrt{2}}{2}\\right)^2 - 2 \\cdot 12 \\cdot \\left(\\frac{13\\sqrt{2}}{2}\\right) \\cdot \\cos{\\angle EAO}$ by $4$ obtaining $4EO^2 = 576 + 338 - 8 \\cdot\\left(\\frac{13\\sqrt{2}}{2}\\right) \\cdot 12 \\cdot\\frac{7\\sqrt{2}}{26} = 576 + 338 - 4 \\cdot 12 \\cdot 7 = \\boxed{578}.$" ]

[ "As shown below, let $M_1,M_2,M_3,M_4$ be the midpoints of $\\overline{AB},\\overline{BC},\\overline{CD},\\overline{DA},$ respectively, and $G_1,G_2,G_3,G_4$ be the centroids of $\\triangle{ABP},\\triangle{BCP},\\triangle{CDP},\\triangle{DAP},$ respectively. By SAS, we conclude that $\\triangle G_1G_2P\\sim\\triangle M_1M_2P, \\triangle G_2G_3P\\sim\\triangle M_2M_3P, \\triangle G_3G_4P\\sim\\triangle M_3M_4P,$ and $\\triangle G_4G_1P\\sim\\triangle M_4M_1P.$ By the properties of centroids, the ratio of similitude for each pair of triangles is $\\frac23.$\nNote that quadrilateral $M_1M_2M_3M_4$ is a square of side-length $15\\sqrt2.$ It follows that:\nTogether, quadrilateral $G_1G_2G_3G_4$ is a square of side-length $10\\sqrt2,$ so its area is $\\left(10\\sqrt2\\right)^2=\\boxed{200}.$", "This solution refers to the diagram in Solution 1.\nBy SAS, we conclude that $\\triangle G_1G_3P\\sim\\triangle M_1M_3P$ and $\\triangle G_2G_4P\\sim\\triangle M_2M_4P.$ By the properties of centroids, the ratio of similitude for each pair of triangles is $\\frac23.$\nNote that quadrilateral $M_1M_2M_3M_4$ is a square of diagonal-length $30,$ so $\\overline{M_1M_3}\\perp\\overline{M_2M_4}.$ Since $\\overline{G_1G_3}\\parallel\\overline{M_1M_3}$ and $\\overline{G_2G_4}\\parallel\\overline{M_2M_4}$ by the Converse of the Corresponding Angles Postulate, we have $\\overline{G_1G_3}\\perp\\overline{G_2G_4}.$\nTherefore, the area of quadrilateral $G_1G_2G_3G_4$ is \\[\\frac12\\cdot G_1G_3\\cdot G_2G_4 = \\frac12\\cdot\\left(\\frac23\\cdot M_1M_3\\right)\\cdot\\left(\\frac23\\cdot M_2M_4\\right)=\\boxed{200}.\\] ~Funnybunny5246 ~MRENTHUSIASM", "This solution refers to the diagram in Solution 1.\nWe place the diagram in the coordinate plane: Let $A=(0,30),B=(0,0),C=(30,0),D=(30,30),$ and $P=(3x,3y).$\nRecall that for any triangle in the coordinate plane, the coordinates of its centroid are the averages of the coordinates of its vertices. It follows that $G_1=(x,y+10),G_2=(x+10,y),G_3=(x+20,y+10),$ and $G_4=(x+10,y+20).$\nNote that $G_1G_3=G_2G_4=20$ and $\\overline{G_1G_3}\\perp\\overline{G_2G_4}.$ Therefore, the area of quadrilateral $G_1G_2G_3G_4$ is \\[\\frac12\\cdot G_1G_3\\cdot G_2G_4=\\boxed{200}.\\]", "Let $X,Y,Z,W$ be the midpoints of sides $AB,BC,CD,DE$ , respectively.\nNotice that a homothety centered at P with ratio $\\frac{2}{3}$ will send $XYZW$ to $G_{1}G_{2}G_{3}G_{4}$ , so $G_{1}G_{2}G_{3}G_{4}$ is a square with area $\\left(\\frac{2}{3}\\right)^2 [XYZW]$ , but $[XYZW]=\\frac{1}{2}[ABCD]$ so our desired area is \\[\\frac{4}{9}\\cdot\\frac{1}{2}\\cdot900=\\boxed{200}\\]" ]

[ " Call the vertices of the new square A', B', C', and D', in relation to the vertices of $ABCD$ , and define $s$ to be one of the sides of that square. Since the sides are parallel , by corresponding angles and AA~ we know that triangles $AA'D'$ and $D'C'E$ are similar. Thus, the sides are proportional: $\\frac{AA'}{A'D'} = \\frac{D'C'}{C'E} \\Longrightarrow \\frac{1 - s}{s} = \\frac{s}{1 - s - CE}$ . Simplifying, we get that $s^2 = (1 - s)(1 - s - CE)$\n$\\angle EAF$ is $60$ degrees, so $\\angle BAE = \\frac{90 - 60}{2} = 15$ . Thus, $\\cos 15 = \\cos (45 - 30) = \\frac{\\sqrt{6} + \\sqrt{2}}{4} = \\frac{1}{AE}$ , so $AE = \\frac{4}{\\sqrt{6} + \\sqrt{2}} \\cdot \\frac{\\sqrt{6} - \\sqrt{2}}{\\sqrt{6} - \\sqrt{2}} = \\sqrt{6} - \\sqrt{2}$ . Since $\\triangle AEF$ is equilateral $EF = AE = \\sqrt{6} - \\sqrt{2}$ $\\triangle CEF$ is a $45-45-90 \\triangle$ , so $CE = \\frac{AE}{\\sqrt{2}} = \\sqrt{3} - 1$ . Substituting back into the equation from the beginning, we get $s^2 = (1 - s)(2 - \\sqrt{3} - s)$ , so $(3 - \\sqrt{3})s = 2 - \\sqrt{3}$ . Therefore, $s = \\frac{2 - \\sqrt{3}}{3 - \\sqrt{3}} \\cdot \\frac{3 + \\sqrt{3}}{3 + \\sqrt{3}} = \\frac{3 - \\sqrt{3}}{6}$ , and $a + b + c = 3 + 3 + 6 = \\boxed{12}$", "Suppose $\\overline{AB} = \\overline{AD} = x.$ Note that $\\angle EAF = 60$ since the triangle is equilateral, and by symmetry, $\\angle BAE = \\angle DAF = 15.$ Note that if $\\overline{AD} = x$ and $\\angle BAE = 15$ , then $\\overline{AA'}=\\frac{x}{\\tan(15)}.$ Also note that \\[AB = 1 = \\overline{AA'} + \\overline{A'B} = \\frac{x}{\\tan(15)} + x\\] Using the fact $\\tan(15) = 2-\\sqrt{3}$ , this yields \\[x = \\frac{1}{3+\\sqrt{3}} = \\frac{3-\\sqrt{3}}{6} \\rightarrow 3 + 3 + 6 = \\boxed{12}\\]", "Why not solve in terms of the side $x$ only (single-variable beauty)? By similar triangles we obtain that $BE=\\frac{x}{1-x}$ , therefore $CE=\\frac{1-2x}{1-x}$ . Then $AE=\\sqrt{2}*\\frac{1-2x}{1-x}$ . Using Pythagorean Theorem on $\\triangle{ABE}$ yields $\\frac{x^2}{(1-x)^2} + 1 = 2 * \\frac{(1-2x)^2}{(1-x)^2}$ . This means $6x^2-6x+1=0$ , and it's clear we take the smaller root: $x=\\frac{3-\\sqrt{3}}{6}$ . Answer: $\\boxed{12}$", "Since $AEF$ is equilateral, $AE=EF$ . Let $BE=x$ . By the Pythagorean theorem $1+x^2=2(1-x)^2$ . Simplifying, we get $x^2-4x+1=0$ . By the quadratic formula, the roots are $2 \\pm \\sqrt{3}$ . Since $x<1$ , we discard the root with the \"+\", giving $x=2-\\sqrt{3}$ Let the side length of the square be s. Since $MEK$ is similar to $ABE$ $s=\\frac{2-\\sqrt{3}-s}{2-\\sqrt{3}}$ . Solving, we get $s=\\frac{3-\\sqrt{3}}{6}$ and the final answer is $\\boxed{012}$" ]

[ "Without loss of generality, let $(0,0)$ $(2,0)$ $(0,2)$ , and $(2,2)$ be the vertices of the square. Suppose the endpoints of the segment lie on the two sides of the square determined by the vertex $(0,0)$ . Let the two endpoints of the segment have coordinates $(x,0)$ and $(0,y)$ . Because the segment has length 2, $x^2+y^2=4$ . Using the midpoint formula, we find that the midpoint of the segment has coordinates $\\left(\\frac{x}{2},\\frac{y}{2}\\right)$ . Let $d$ be the distance from $(0,0)$ to $\\left(\\frac{x}{2},\\frac{y}{2}\\right)$ . Using the distance formula we see that $d=\\sqrt{\\left(\\frac{x}{2}\\right)^2+\\left(\\frac{y}{2}\\right)^2}= \\sqrt{\\frac{1}{4}\\left(x^2+y^2\\right)}=\\sqrt{\\frac{1}{4}(4)}=1$ . Thus the midpoints lying on the sides determined by vertex $(0,0)$ form a quarter- circle with radius 1.\nThe set of all midpoints forms a quarter circle at each corner of the square. The area enclosed by all of the midpoints is $4-4\\cdot \\left(\\frac{\\pi}{4}\\right)=4-\\pi \\approx .86$ to the nearest hundredth. Thus $100\\cdot k=\\boxed{86}$", "To imagine the area, think of a ladder with a length of $2$ sliding down a wall. It is known that as a ladder slides down a wall, its midpoint traces a quarter circle (if you don't believe me, try it with your pencil). There are $4$ quarter circles, so their area is one circle or $\\pi$ . Thus, they enclose the area of the square minus the area of the quarter circles, which is $4-\\pi \\approx 0.86$ , so $100k = \\boxed{086}$ . ~Extremelysupercooldude" ]

[ "Since the square has side length $3$ , the area of the entire square is $9$\nThe segments divide the square into 3 equal parts, so the area of each part is $9 \\div 3 = 3$\nSince $\\triangle CBM$ has area $3$ and base $CB = 3$ , using the area formula for a triangle:\n$A_{tri} = \\frac{1}{2}bh$\n$3 = \\frac{1}{2}3h$\n$h = 2$\nThus, height $BM = 2$\nSince $\\triangle CBM$ is a right triangle, $CM = \\sqrt{BM^2 + BC^2} = \\sqrt{2^2 + 3^2} = \\boxed{13}$", "Connect $AC$ $S_\\triangle AMC=S_\\triangle ANC$ . To satisfied the three area is equal, we have $2S_\\triangle AMC=S_\\triangle BMC$ $2S_\\triangle ANC=S_\\triangle DNC$ . Thus, $AM=AN=\\frac{1}{2}BM=\\frac{1}{2}AB=1$ $BM=2,BC=3,MC=\\boxed{13}$" ]

[ "Let $O$ be the center of the circle, and $2a$ be the side length of $ABCD$ $2b$ be the side length of $EFGH$ . By the Pythagorean Theorem , the radius of $\\odot O = OC = a\\sqrt{2}$\nNow consider right triangle $OGI$ , where $I$ is the midpoint of $\\overline{GH}$ . Then, by the Pythagorean Theorem,\n\\begin{align*} OG^2 = 2a^2 &= OI^2 + GI^2 = (a+2b)^2 + b^2 \\\\ 0 &= a^2 - 4ab - 5b^2 = (a - 5b)(a + b) \\end{align*}\nThus $a = 5b$ (since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so $\\frac{[EFGH]}{[ABCD]} = \\left(\\frac 15\\right)^2 = \\frac{1}{25}$ , and the answer is $10n + m = \\boxed{251}$", "Let point $A$ be the top-left corner of square $ABCD$ and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let $D$ have coordinates $(0,0)$ and the side length of square $ABCD$ be $a$ . Let $DF$ $b$ and diameter $HI$ go through $J$ the midpoint of $EF$ . Since a diameter always bisects a chord perpendicular to it, $DJ$ $JC$ and since $F$ and $E$ must be symmetric around the diameter, $FJ = JE$ and it follows that $DF = EC = b.$ Hence $FE$ the side of square $EFGH$ has length $a - 2b$ $F$ has coordinates $(b,0)$ and $G$ has coordinates $(b, 2b - a).$ We know that point $G$ must be on the circle $O$ - hence it must satisfy the circle equation. Since the center of the circle is at the center of the square $(a/2, a/2)$ and has radius $a *$ $\\sqrt{2} / 2$ , half the diagonal of the square, $(x - a/2)^2 + (y - a/2)^2 = 1/2a^2$ follows as the circle equation. Then substituting coordinates of $G$ into the equation, $(b - a/2)^2 + (2b - a - a/2)^2 = a^2/2$ . Simplifying and factoring, we get $2a^2-7ab+5b^2 = (2a-5b)(a-b) = 0.$ Since $a = b$ would imply $m = n$ , and $m < n$ in the problem, we must use the other factor. We get $b = 2/5a$ , meaning the ratio of areas $((a-2b)/a)^2$ $(1/5)^2$ $1/25$ $m/n.$ Then $10n + m = 25 * 10 + 1 = \\boxed{251}$" ]

[ "First, let's call the center of both squares $I$ . Then, $\\angle{AIE} = 20$ , and since $\\overline{EI} = \\overline{AI}$ $\\angle{AEI} = \\angle{EAI} = 80$ . Then, we know that $AI$ bisects angle $\\angle{DAB}$ , so $\\angle{BAI} = \\angle{DAI} = 45$ . Subtracting $45$ from $80$ , we get $\\boxed{35}$", "First, label the point between $A$ and $H$ point $O$ and the point between $A$ and $H$ point $P$ . We know that $\\angle{AOP} = 20$ and that $\\angle{A} = 90$ . Subtracting $20$ and $90$ from $180$ , we get that $\\angle{APO}$ is $70$ . Subtracting $70$ from $180$ , we get that $\\angle{OPB} = 110$ . From this, we derive that $\\angle{APE} = 110$ . Since triangle $APE$ is an isosceles triangle, we get that $\\angle{EAP} = (180 - 110)/2 = 35$ . Therefore, $\\angle{EAB} = 35$ . The answer is $\\boxed{35}$", "Call the center of both squares point $O$ , and draw circle $O$ such that it circumscribes the squares. $\\angle{EOF} = 90$ and $\\angle{BOF} = 20$ , so $\\angle{EOB} = 70$ . Since $\\angle{EAB}$ is inscribed in arc $\\overset \\frown {EB}$ $\\angle{EAB} = 70/2 = \\boxed{35}$", "Draw $EA$ : we want to find $\\angle EAB$ . Call $P$ the point at which $AB$ and $EH$ intersect. Reflecting $\\triangle APE$ over $EA$ , we have a parallelogram. Since $\\angle EPB = 70^{\\circ}$ , angle subtraction tells us that two of the angles of the parallelogram are $110^{\\circ}$ . The other two are equal to $2\\angle EAB$ (by properties of reflection).\nSince angles on the transversal of a parallelogram sum to $180^{\\circ}$ , we have $2\\angle EAB + 110 = 180$ , yielding $\\angle EAB = \\boxed{35}$", "We call the point where $AB$ and $EH$ intersect I. We can make an educated guess that triangle AEI is isosceles so $AI=EI$ $\\angle AIE = 110^{\\circ}$ $\\angle AIH = 20^{\\circ}$ , and $\\angle EIB = 70^{\\circ}$ . So, we get $\\angle EAI$ is $(180^{\\circ} - 110^{\\circ})/2 = \\boxed{35}$" ]

[ "Note that if the altitude of the triangle is at most $10$ , then the maximum area of the intersection of the triangle and the square is $5\\cdot10=50$ .\nThis implies that vertex G must be located outside of square $AIME$\nLet $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$ . Clearly, $XY=6$ since the area of trapezoid $XYME$ is $80$ . Also, $\\triangle GXY \\sim \\triangle GEM$\nLet the height of $GXY$ be $h$ . By the similarity, $\\dfrac{h}{6} = \\dfrac{h + 10}{10}$ , we get $h = 15$ . Thus, the height of $GEM$ is $h + 10 = \\boxed{025}$" ]

[ "The sum of the areas of the squares if they were not interconnected is a geometric sequence\nThen subtract the areas of the intersections, which is $\\left(\\frac{1}{4}\\right)^2 + \\ldots + \\left(\\frac{1}{32}\\right)^2$\nThe majority of the terms cancel, leaving $1 + \\frac{1}{4} - \\frac{1}{1024}$ , which simplifies down to $\\frac{1024 + \\left(256 - 1\\right)}{1024}$ . Thus, $m-n = \\boxed{255}$" ]

[ "If the sides of the open box are folded down so that a flat sheet with four corners cut out remains, then the revealed surface would have the same area as the interior of the box. This is equal to the area of the four corners subtracted from the area of the original sheet, which is $((20)(30)-4(5)(5)) = 600-100 = \\boxed{500}$" ]

[ "The area of the entire region in the plane is the area of the figure. However, we cannot simply add the two areas of the squares. We find the area of $\\triangle ABG$ and subtract this from $200$ , the total area of the two squares.\nSince $G$ is the center of $ABCD$ $BG$ is half of the diagonal of the square. The diagonal of $ABCD$ is $10\\sqrt{2}$ so $BG=5\\sqrt{2}$ . Since $EFGH$ is a square, $\\angle G=90^\\circ$ . So $\\triangle ABG$ is an isosceles right triangle. Its area is $\\frac{(5\\sqrt{2})^2}{2}=\\frac{50}{2}=25$ . Therefore, the area of the region is $200-25=\\boxed{175.}$" ]

[ "Letting $AI=a$ and $IB=b$ , we have \\[IJ^{2}=a^{2}+b^{2} \\geq 1008\\] by AM-GM inequality . Also, since $EFGH||ABCD$ , the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since \\[2016=12^{2} \\cdot 14\\] we have the maximum area is \\[2016 \\cdot \\dfrac{11}{12} = 1848\\] (the areas of the squares from largest to smallest are $12^{2} \\cdot 14, 11 \\cdot 12 \\cdot 14, 11^{2} \\cdot 14$ forming a geometric progression).\nThe minimum area is $1008$ (every square is half the area of the square whose sides its vertices touch), so the desired answer is \\[1848-1008=\\boxed{840}\\]" ]

[ "1987 AIME-15a.png\nBecause all the triangles in the figure are similar to triangle $ABC$ , it's a good idea to use area ratios . In the diagram above, $\\frac {T_1}{T_3} = \\frac {T_2}{T_4} = \\frac {441}{440}.$ Hence, $T_3 = \\frac {440}{441}T_1$ and $T_4 = \\frac {440}{441}T_2$ . Additionally, the area of triangle $ABC$ is equal to both $T_1 + T_2 + 441$ and $T_3 + T_4 + T_5 + 440.$\nSetting the equations equal and solving for $T_5$ $T_5 = 1 + T_1 - T_3 + T_2 - T_4 = 1 + \\frac {T_1}{441} + \\frac {T_2}{441}$ . Therefore, $441T_5 = 441 + T_1 + T_2$ . However, $441 + T_1 + T_2$ is equal to the area of triangle $ABC$ ! This means that the ratio between the areas $T_5$ and $ABC$ is $441$ , and the ratio between the sides is $\\sqrt {441} = 21$ . As a result, $AB = 21\\sqrt {440} = \\sqrt {AC^2 + BC^2}$ . We now need $(AC)(BC)$ to find the value of $AC + BC$ , because $AC^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2$\nLet $h$ denote the height to the hypotenuse of triangle $ABC$ . Notice that $h - \\frac {1}{21}h = \\sqrt {440}$ . (The height of $ABC$ decreased by the corresponding height of $T_5$ ) Thus, $(AB)(h) = (AC)(BC) = 22\\cdot 21^2$ . Because $AC^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2 = 21^2\\cdot22^2$ $AC + BC = (21)(22) = \\boxed{462}$", "Let $\\tan\\angle ABC = x$ . Now using the 1st square, $AC=21(1+x)$ and $CB=21(1+x^{-1})$ . Using the second square, $AB=\\sqrt{440}(1+x+x^{-1})$ . We have $AC^2+CB^2=AB^2$ , or \\[441(x^2+x^{-2}+2x+2x^{-1}+2)=440(x^2+x^{-2}+2x+2x^{-1}+3).\\] Rearranging and letting $u=x+x^{-1} \\Rightarrow u^2 - 2 = x^2 + x^{-2}$ gives us $u^2+2u-440=0.$ We take the positive root, so $u=20$ , which means $AC+CB=21(2+x+x^{-1})=21(2+u)=\\boxed{462}$", "Let $\\theta$ be the smaller angle in the triangle. Then the sum of shorter and longer leg is $\\sqrt{441}(2+\\tan{\\theta}+\\cot{\\theta})$ . We observe that the short leg has length $\\sqrt{441}(1+\\tan{\\theta}) = \\sqrt{440}(\\sec{\\theta}+\\sin{\\theta})$ . Grouping and squaring, we get $\\sqrt{\\frac{440}{441}} = \\frac{\\sin{\\theta}+\\cos{\\theta}}{1+\\sin{\\theta}\\cos{\\theta}}$ . Squaring and using the double angle identity for sine, we get, $110(\\sin{2\\theta})^2 + \\sin{2\\theta} - 1 = 0$ . Solving, we get $\\sin{2\\theta} = \\frac{1}{10}$ . Now to find $\\tan{\\theta}$ , we find $\\cos{2\\theta}$ using the Pythagorean\nIdentity, and then use the tangent double angle identity. Thus, $\\tan{\\theta} = 10-3\\sqrt{11}$ . Substituting into the original sum,\nwe get $\\boxed{462}$", "\n\nLabel points as above. Let $x=AC$ $y=BC$ $s_1 = 21$ be the side length of $S_1$ , and $s_2 = \\sqrt{440}$ be the side length of $S_2$\nSince $\\triangle ABC\\sim\\triangle AED$ , we have $\\frac{x}{y} = \\frac{x-s_1}{s_1}$\n$\\implies xs_1=xy-ys_1$\n$\\implies xy=s_1(x+y)$\n$\\implies xy=21(x+y) \\qquad \\qquad (*)$\nSince $\\triangle ABC\\sim\\triangle AWX\\sim\\triangle ZBY$ , we have $s_2 + \\frac{s_2x}{y} + \\frac{s_2y}{x}=\\sqrt{x^2+y^2}$\n$\\implies s_2(x^2+xy+y^2)=xy\\sqrt{x^2+y^2}$\n$\\implies s_2^2(x^2+xy+y^2)^2 = x^2y^2(x^2+y^2)$\n$\\implies 440(x^2+xy+y^2)^2 = x^2y^2(x^2+y^2)$\nLet $t=x+y$ . Repeatedly applying $(*)$ , we get \\[440(t^2-21t)^2 = 441t^2(t^2 - 42t)\\] \\[440(t-21)^2 = 441(t^2-42t)\\] \\[440t^2 - 42\\cdot 440t + 440\\cdot 441 = 441t^2 - 441\\cdot 42t\\] \\[t^2-42t-440\\cdot 441=0\\] \\[(t-21)^2 = 441^2\\] \\[t-21=441\\] \\[t=\\boxed{462}\\]" ]

[ "For every tens digit 2, we subtract 10, and for every units digit 2, we subtract 1. Because 2 appears 10 times as a tens digit and 2 appears 3 times as a units digit, the answer is $10\\cdot 10+1\\cdot 3=\\boxed{103.}$" ]

[ "It takes an even number of steps for the object to reach $(2,2)$ , so the number of steps the object may have taken is either $4$ or $6$\nIf the object took $4$ steps, then it must have gone two steps and two steps , in some permutation. There are $\\frac{4!}{2!2!} = 6$ ways for these four steps of occuring, and the probability is $\\frac{6}{4^{4}}$\nIf the object took $6$ steps, then it must have gone two steps and two steps , and an additional pair of moves that would cancel out, either N/S or W/E . The sequences N,N,N,E,E,S can be permuted in $\\frac{6!}{3!2!1!} = 60$ ways. However, if the first four steps of the sequence are N,N,E,E in some permutation, it would have already reached the point $(2,2)$ in four moves. There are $\\frac{4!}{2!2!}$ ways to order those four steps and $2!$ ways to determine the order of the remaining two steps, for a total of $12$ sequences that we have to exclude. This gives $60-12=48$ sequences of steps. There are the same number of sequences for the steps N,N,E,E,E,W , so the probability here is $\\frac{2 \\times 48}{4^6}$\nThe total probability is $\\frac{6}{4^4} + \\frac{96}{4^6} = \\frac{3}{64}$ , and $m+n= \\boxed{067}$", "Let's let the object wander for 6 steps so we get a constant denominator of $4^{6}$\nIn the first case, we count how many ways the object can end at (2,2), at the end of 6 steps. We will also count it even if we go to (2,2), and go back to (2,2). So, there are 2 different paths for the object to end at (2,2):\n1.To go a permutation of R,R,R,U,U,L or\n2.To go a permutation of R,R,U,U,U,D.\nThere are 60 ways to permute for each case, giving a total of 120 ways for the object to succeed and end at (2,2). In these 120 ways the object could reach (2,2) first and then come back to (2,2). This will be a factor in our second case.\nIn the second case, the object can get to (2,2) first in 4 moves, then move away from (2,2) with the remaing 2 moves. So, there are 6 ways to get to (2,2) in 4 moves, then there are 16 ways the object can \"move around\", but 4 of the ways will return the object back to (2,2). Those 4 ways were already counted in the first case, so we should only count 12 of the 16 ways to prevent over-counting. Thus, there are $12*6 =$ 72 ways in the second case.\nSo, in all, there are 120+72 ways for the object to achieve it's goal of moving to (2,2). Put that over our denominator, we get $\\frac{192}{4^{6}} = \\frac{3}{4^{3}} = \\frac{3}{64}$ , in which adding the numerator and denominator get us an answer of $\\boxed{067}$" ]

[ "We can represent the amount of gold with $g$ and the amount of chests with $c$ . We can use the problem to make the following equations: \\[9c-18 = g\\] \\[6c+3 = g\\]\nWe do this because for 9 chests there are 2 empty and if 9 were in each 9 multiplied by 2 is 18 left.\nTherefore, $6c+3 = 9c-18.$ This implies that $c = 7.$ We therefore have $g = 45.$ So, our answer is $\\boxed{45}$ .\n~CHECKMATE2021", "With $9$ coins, there are $\\frac{9}{9}+2=1+2=3$ chests, by the first condition. These don't fit in with the second condition, so we move onto $27$ coins. By the same first condition, there are $5$ chests( $\\frac{27}{9}+2$ ). This also doesn't fit with the second condition. So, onto $45$ coins. The first condition implies that there are $\\frac{45}{9}+2=7$ chests, which DOES fit with the second condition, since $6\\cdot7+3=42+3=45$ . Thus, the desired value is $\\boxed{45}$" ]

[ "Let $x$ be the number of shots that Candace made in the first half, and let $y$ be the number of shots Candace made in the second half. Since Candace and Steph took the same number of attempts, with an equal percentage of baskets scored, we have $x+y=10+15=25.$ In addition, we have the following inequalities: \\[\\frac{x}{12}<\\frac{15}{20} \\implies x<9,\\] and \\[\\frac{y}{18}<\\frac{10}{10} \\implies y<18.\\] Pairing this up with $x+y=25$ we see the only possible solution is $(x,y)=(8,17),$ for an answer of $17-8 = \\boxed{9}.$", "Clearly, Steph made $15 + 10 = 25$ shots in total. Also, due to parity reasons, the difference between the amount of shots Candace made in the first and second half must be odd. Thus, we can just test 7, 9, and 11, and after doing so we find that the answer is $\\boxed{9}.$", "Steph made 75 percent of his shots in the first half. He makes all of his shots in the second half. The most baskets Candace could have made in the first half is 8 baskets. The most she could have made in the second half is 17 baskets. Steph makes 25 and misses 5 baskets and the only way for Candace to make 25 shots is to make 8 in the first half and 17 in the second. Thus, $17 - 8 = \\boxed{9}.$" ]

[ "We call the three roots (some may be equal to one another) $x_1$ $x_2$ , and $x_3$ . Using Vieta's formulas, we get $x_1+x_2+x_3 = a$ $x_1 \\cdot x_2+x_1 \\cdot x_3+x_2 \\cdot x_3 = \\frac{a^2-81}{2}$ , and $x_1 \\cdot x_2 \\cdot x_3 = \\frac{c}{2}$\nSquaring our first equation we get $x_1^2+x_2^2+x_3^2+2 \\cdot x_1 \\cdot x_2+2 \\cdot x_1 \\cdot x_3+2 \\cdot x_2 \\cdot x_3 = a^2$\nWe can then subtract twice our second equation to get $x_1^2+x_2^2+x_3^2 = a^2-2 \\cdot \\frac{a^2-81}{2}$\nSimplifying the right side:\n\\begin{align*} a^2-2 \\cdot \\frac{a^2-81}{2} &= a^2-a^2+81\\\\ &= 81.\\\\ \\end{align*}\nSo, we know $x_1^2+x_2^2+x_3^2 = 81$\nWe can then list out all the triples of positive integers whose squares sum to $81$\nWe get $(1, 4, 8)$ $(3, 6, 6)$ , and $(4, 4, 7)$\nThese triples give $a$ values of $13$ $15$ , and $15$ , respectively, and $c$ values of $64$ $216$ , and $224$ , respectively.\nWe know that Jon still found two possible values of $c$ when Steve told him the $a$ value, so the $a$ value must be $15$ . Thus, the two $c$ values are $216$ and $224$ , which sum to $\\boxed{440}$", "First things first. Vietas gives us the following:\n\\begin{align} x_1+x_2+x_3 = a\\\\ x_1 \\cdot x_2+x_1 \\cdot x_3+x_2 \\cdot x_3 = \\frac{a^2-81}{2}\\\\ x_1 \\cdot x_2 \\cdot x_3 = \\frac{c}{2} \\end{align}\nFrom $(2)$ $a$ must have odd parity, meaning $a^2-81$ must be a multiple of $4$ , which implies that both sides of $(2)$ are even. Then, from $(1)$ , we see that an odd number of $x_1$ $x_2$ , and $x_3$ must be odd, because we have already deduced that $a$ is odd. In order for both sides of $(2)$ to be even, there must only be one odd number and two even numbers.\nNow, the theoretical maximum value of the left side of $(2)$ is $3 \\cdot \\biggl(\\frac{a}{3}\\biggr)^2=\\frac{a^2}{3}$ . That means that the maximum bound of $a$ is where \\[\\frac{a^2}{3} > \\frac{a^2-81}{2},\\] which simplifies to $\\sqrt{243} > a$ , meaning \\[16 > a.\\] So now we have that $9<a$ from $(2)$ $a<16$ , and $a$ is odd from $(2)$ . This means that $a$ could equal $11$ $13$ , or $15$ . At this point, we have simplified the problem to the point where we can casework+ brute force. As said above, we arrive at our solutions of $(1, 4, 8)$ $(3, 6, 6)$ , and $(4, 4, 7)$ , of which the last two return equal $a$ values. Then, $2(3 \\cdot 6 \\cdot 6+4 \\cdot 4 \\cdot 7)=\\boxed{440}$ AWD.", "Since each of the roots is positive, the local maximum of the function must occur at a positive value of $x$ . Taking $\\frac{d}{dx}$ of the polynomial yields $6x^2-4ax+a^2-81$ , which is equal to $0$ at the local maximum. Since this is a quadratic in $a$ , we can find an expression for $a$ in terms of $x$ . The quadratic formula gives $a=\\frac{4x\\pm\\sqrt{324-8x^2}}{2}$ , which simplifies to $a=2x\\pm\\sqrt{81-2x^2}$ . We know that $a$ is a positive integer, and testing small positive integer values of $x$ yields $a=15$ or $a=1$ when $x=4$ , and $a=15$ or $a=9$ when $x=6$ . Because the value of $a$ alone does not determine the polynomial, $a$ $a$ must equal $15$\nNow our polynomial equals $2x^3-30x^2+144x-c$ . Because one root is less than (or equal to) the $x$ value at the local maximum (picture the graph of a cubic equation), it suffices to synthetically divide by small integer values of $x$ to see if the resulting quadratic also has positive integer roots. Dividing by $x=3$ leaves a quotient of $2x^2-24x+72=2(x-6)^2$ , and dividing by $x=4$ leaves a quotient of $2x^2-22x+56=2(x-4)(x-7)$ . Thus, $c=2\\cdot 3\\cdot 6\\cdot 6=216$ , or $c=2\\cdot 4\\cdot 4\\cdot 7=224$ . Our answer is $216+224=\\boxed{440}$ ~bad_at_mathcounts" ]

[ "Note that cycles exist initially and after each round of erasing.\nLet the parentheses denote cycles. It follows that:\nSince $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: \\[(12\\underline{425}3415251).\\] Therefore, the answer is $4+2+5=\\boxed{11}.$", "After erasing every third digit, the list becomes $1245235134\\ldots$ repeated. After erasing every fourth digit from this list, the list becomes $124235341452513\\ldots$ repeated. Finally, after erasing every fifth digit from this list, the list becomes $124253415251\\ldots$ repeated. Since this list repeats every $12$ digits and $2019,2020,2021$ are $3,4,5$ respectively \nin $\\pmod{12},$ we have that the $2019$ th, $2020$ th, and $2021$ st digits are the $3$ rd, $4$ th, and $5$ th digits respectively. It follows that the answer is $4+2+5= \\boxed{11}.$", "Note that cycles exist initially and after each round of erasing.\nWe will consider one cycle after all three rounds of erasing. Suppose this cycle has length $L$ before any round of erasing. It follows that:\nThe least such positive integer $L$ is $\\operatorname{lcm}(5,3,2,5)=30.$ So, there is a repeating pattern for every $30$ digits on the original list. As shown below, the digits erased in the first, second, and third rounds are colored in red, yellow, and green, respectively: As indicated by the white squares, each group of $30$ digits on the original list has $\\frac25\\cdot30=12$ digits remaining on the final list.\nSince $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively: Therefore, the answer is $4+2+5=\\boxed{11}.$", "As the LCM of $3$ $4$ , and $5$ is $60$ , let us look at a $60$ -digit block of original numbers (many will be erased by Steve). After he erases every third number $\\left(\\dfrac{1}{3}\\right)$ , then every fourth number of what remains $\\left(\\dfrac{1}{4}\\right)$ , then every fifth number of what remains $\\left(\\dfrac{1}{5}\\right)$ , we are left with $\\dfrac{2}{3} \\cdot \\dfrac{3}{4} \\cdot \\dfrac{4}{5} \\cdot 60=24$ digits from this $60$ -digit block. $2019 \\equiv 3 \\pmod {24}, 2020 \\equiv 4 \\pmod {24}, 2021 \\equiv 5 \\pmod {24}$ . Writing out the first few digits of this sequence, we have $\\underbrace{1}_{\\#1}, \\underbrace{2}_{\\#2}, \\cancel{3}, \\underbrace{4}_{\\#3}, \\cancel{5}, \\cancel{1}, \\underbrace{2}_{\\#4}, \\cancel{3}, \\cancel{4}, \\underbrace{5}_{\\#5}, \\dots$ . Therefore, our answer is $4+2+5=\\boxed{11}$", "Lemma: Given a sequence $a_1, a_2, a_3, \\cdots$ , and an positive integer $k>2$ . If we erase every $k$ th item in this sequence, and we name $b_1, b_2, b_3, \\cdots$ as the remaining sequence.\nThen we have \\[b_{(k-1)m+1}=a_{km+1}, b_{(k-1)m+2}=a_{km+2}, \\cdots, b_{(k-1)m+k-1}=a_{km+k-1}.\\]\nProof: For $a_{km+j}$ with some $j, 1\\le j\\le k-1$ , we will have $m$ items removed before this item, so it becomes $b_{(k-1)m+j}$ in the new sequence. Hence, we have $b_{(k-1)m+j}=a_{km+j}$\nIf we start with $a_1, a_2, a_3, \\cdots,$ and let $b_1, b_2, \\cdots$ be the sequence after removing all the indexes that are multiples of $3$ . Then, we have $b_{2n+1}=a_{3n+1},b_{2n+2}=a_{3n+2}$\nSimilarly, if we start with $b_1, b_2, b_3, \\cdots,$ and remove all multiples of $4$ , and get $c_1, c_2, \\cdots$ We have $c_{3n+1}=b_{4n+1},c_{3n+2}=b_{4n+2}, c_{3n+3}=b_{4n+3}$\nIf we start with $c_1, c_2, \\cdots$ and $d_1, d_2, \\cdots$ are remove all multiples of $5$ , we get \\[d_{4n+1}=c_{5n+1}, d_{4n+2}=c_{5n+2}, d_{4n+3}=c_{5n+3}, d_{4n+4}=c_{5n+4}.\\] Therefore, \\begin{align*} d_{2019}&=d_{4\\cdot504+3}=c_{5\\cdot 504+3}=c_{2523}=c_{3\\cdot 840+3}=b_{4\\cdot 840+3}=b_{3363}=b_{2\\cdot 1681+1}=a_{3\\cdot 1681+1}=a_{5044}=a_4=4, \\\\ d_{2020}&=d_{4\\cdot504+4}=c_{5\\cdot 504+4}=c_{2524}=c_{3\\cdot841+1}=b_{4\\cdot841+1}=b_{3365}=b_{2\\cdot1682+1}=a_{3\\cdot 1682+1}=a_{5047}=a_2=2, \\\\ d_{2021}&=d_{4\\cdot505+1}=c_{5\\cdot505+1}=c_{2526}=c_{3\\cdot841+3}=b_{4\\cdot841+3}=b_{3367}=b_{2\\cdot1683+1}=a_{3\\cdot1683+1}=a_{5050}=a_5=5, \\end{align*} and the answer is $4+2+5=\\boxed{11}$" ]

[ "Each of the four hoses hose fills $24,000/4 = 6,000$ gallons of water. At the rate it goes at it will take $6,000/2.5 = 2400$ minutes, or $\\boxed{40}$ hours.", "If all four hoses fill $2.5$ gallons a minute, every minute $10$ gallons would be added. Since every hour has $60$ minutes, $600$ gallons of water would be added every hour. $24000/600=\\boxed{40}$ hours." ]

[ "Altogether, the summer project totaled $(7)(3)+(4)(5)+(5)(9)=21+20+45=86$ days of work for a single student. This equals $744/86=9$ dollars per day per student. The students from Balboa school earned $9(4)(5)=\\boxed{180.00}$" ]

[ "If at least half the guesses are too low, then Norb's age must be greater than $36.$\nIf two of the guesses are off by one, then his age is in between two guesses whose difference is $2.$ It could be $31,37,$ or $48,$ but because his age is greater than $36$ it can only be $37$ or $48.$\nLastly, Norb's age is a prime number so the answer must be $\\boxed{37}$", "Since two guesses are off by one, we know that both $x+1$ and $x-1$ are in the list where $x$ is the age of Norb. Now, we know that $x+1$ and $x-1$ are $28$ and $30$ $30$ and $32$ $36$ and $38$ and $47$ and $49$ . From these values, we know that $x$ must be $29$ $31$ , and $37$ . Since half of the guesses are too low, $24, 28, 30, 32,$ and $36$ are all too low so we can eliminate all numbers in our list lesser than or equal to $36$ . Therefore, our list has only $37$ left so the answer is $\\boxed{37}$" ]

[ "Without loss of generality, assume something costs $100$ dollars. Then with each successive discount, it would cost $90$ dollars, then $72$ dollars. This amounts to a total of $28$ dollars off, so the single discount would be $\\boxed{28}$" ]

[ "Since $20\\% = \\frac{1}{5}$ , multiplying the given condition by $5$ shows that $y$ is $15 \\cdot 5 = \\boxed{75}$ percent of $x$", "Letting $x=100$ (without loss of generality), the condition becomes $0.15\\cdot 100 = 0.2\\cdot y \\Rightarrow 15 = \\frac{y}{5} \\Rightarrow y=75$ . Clearly, it follows that $y$ is $75\\%$ of $x$ , so the answer is $\\boxed{75}$", "We have $15\\%=\\frac{3}{20}$ and $20\\%=\\frac{1}{5}$ , so $\\frac{3}{20}x=\\frac{1}{5}y$ . Solving for $y$ , we multiply by $5$ to give $y = \\frac{15}{20}x = \\frac{3}{4}x$ , so the answer is $\\boxed{75}$", "We are given $0.15x = 0.20y$ , so we may assume without loss of generality that $x=20$ and $y=15$ . This means $\\frac{y}{x}=\\frac{15}{20}=\\frac{75}{100}$ , and thus the answer is $\\boxed{75}$", "$15\\%$ of $x$ is $0.15x$ , and $20\\%$ of $y$ is $0.20y$ . We put $0.15x$ and $0.20y$ into an equation, creating $0.15x = 0.20y$ because $0.15x$ equals $0.20y$ . Solving for $y$ , dividing $0.2$ to both sides, we get $y = \\frac{15}{20}x = \\frac{3}{4}x$ , so the answer is $\\boxed{75}$", "$15\\%$ of $x$ can be written as $\\frac{15}{100}x$ , or $\\frac{15x}{100}$ $20\\%$ of $y$ can similarly be written as $\\frac{20}{100}y$ , or $\\frac{20y}{100}$ . So now, $\\frac{15x}{100} = \\frac{20y}{100}$ . Using cross-multiplication, we can simplify the equation as: $1500x = 2000y$ . Dividing both sides by $500$ , we get: $3x = 4y$ $\\frac{3}{4}$ is the same thing as $75\\%$ , so the answer is $\\boxed{75}$" ]

[ "We have that A is $x\\%$ greater than B, so $A=\\frac{100+x}{100}(B)$ . We solve for $x$ . We get\n$\\frac{A}{B}=\\frac{100+x}{100}$\n$100\\frac{A}{B}=100+x$\n$100\\left(\\frac{A}{B}-1\\right)=x$\n$100\\left(\\frac{A-B}{B}\\right)=x$ $\\boxed{100}$", "The question is basically asking the percentage increase from $B$ to $A$ . We know the formula for percentage increase is $\\frac{\\text{New-Original}}{\\text{Original}}$ . We know the new is $A$ and the original is $B$ . We also must multiple by $100$ to get $x$ out of it's fractional/decimal form. Therefore, the answer is $100\\left(\\frac{A-B}{B}\\right)$ or $\\boxed{100}$", "Without loss of generality, let $A = 125$ and $B = 100,$ forcing $x$ to be $25$ . Plugging our values for $A$ and $B$ into these answer choices, we find that only $\\boxed{100}$ returns $25$" ]

[ "Each can of soda costs $\\frac QS$ quarters, or $\\frac{Q}{4S}$ dollars. Therefore, $D$ dollars can purchase $\\frac{D}{\\left(\\tfrac{Q}{4S}\\right)}=\\boxed{4}$ cans of soda.", "Note that $S$ is in the unit of $\\text{can}.$ On the other hand, $Q$ and $D$ are both in the unit of $\\text{cost}.$\nThe units of $\\textbf{(A)},\\textbf{(B)},\\textbf{(C)},\\textbf{(D)},$ and $\\textbf{(E)}$ are $\\frac{\\text{cost}^2}{\\text{can}},\\text{can},\\frac{1}{\\text{can}},\\frac{\\text{cost}^2}{\\text{can}},$ and $\\text{can},$ respectively. Since the answer is in the unit of $\\text{can},$ we eliminate $\\textbf{(A)},\\textbf{(C)},$ and $\\textbf{(D)}.$ Moreover, it is clear that $D$ dollars can purchase more than $S=\\frac{DS}{4Q}$ cans of soda, so we eliminate $\\textbf{(E)}.$\nFinally, the answer is $\\boxed{4}.$" ]

[ "Because $a + b + c$ is odd, $a$ $b$ $c$ are either one odd and two evens or three odds.\n$\\textbf{Case 1}$ $a$ $b$ $c$ have one odd and two evens.\nWithout loss of generality, we assume $a$ is odd and $b$ and $c$ are even.\nHence, ${\\rm gcd} \\left( a , b \\right)$ and ${\\rm gcd} \\left( a , c \\right)$ are odd, and ${\\rm gcd} \\left( b , c \\right)$ is even.\nHence, ${\\rm gcd} \\left( a , b \\right) + {\\rm gcd} \\left( b , c \\right) + {\\rm gcd} \\left( c , a \\right)$ is even. This violates the condition given in the problem.\nTherefore, there is no solution in this case.\n$\\textbf{Case 2}$ $a$ $b$ $c$ are all odd.\nIn this case, ${\\rm gcd} \\left( a , b \\right)$ ${\\rm gcd} \\left( a , c \\right)$ ${\\rm gcd} \\left( b , c \\right)$ are all odd.\nWithout loss of generality, we assume \\[ {\\rm gcd} \\left( a , b \\right) \\leq {\\rm gcd} \\left( b , c \\right) \\leq {\\rm gcd} \\left( c , a \\right) . \\] $\\textbf{Case 2.1}$ ${\\rm gcd} \\left( a , b \\right) = 1$ ${\\rm gcd} \\left( b , c \\right) = 1$ ${\\rm gcd} \\left( c , a \\right) = 7$\nThe only solution is $(a, b, c) = (7, 9, 7)$\nHence, $a^2 + b^2 + c^2 = 179$\n$\\textbf{Case 2.2}$ ${\\rm gcd} \\left( a , b \\right) = 1$ ${\\rm gcd} \\left( b , c \\right) = 3$ ${\\rm gcd} \\left( c , a \\right) = 5$\nThe only solution is $(a, b, c) = (5, 3, 15)$\nHence, $a^2 + b^2 + c^2 = 259$\n$\\textbf{Case 2.3}$ ${\\rm gcd} \\left( a , b \\right) = 3$ ${\\rm gcd} \\left( b , c \\right) = 3$ ${\\rm gcd} \\left( c , a \\right) = 3$\nThere is no solution in this case.\nTherefore, putting all cases together, the answer is $179 + 259 = \\boxed{438}$", "Let $\\gcd(a,b)=x$ $\\gcd(b,c)=y$ $\\gcd(c,a)=z$ . Without the loss of generality, let $x \\le y \\le z$ . We can split this off into cases:\n$x=1,y=1,z=7$ : let $a=7A, c=7C,$ we can try all possibilities of $A$ and $C$ to find that $a=7, b=9, c=7$ is the only solution.\n$x=1,y=2,z=6$ : No solutions. By $y$ and $z$ , we know that $a$ $b$ , and $c$ have to all be divisible by $2$ . Therefore, $x$ cannot be equal to $1$\n$x=1,y=3,z=5$ : Note that $c$ has to be both a multiple of $3$ and $5$ . Therefore, $c$ has to be a multiple of $15$ . The only solution for this is $a=5, b=3, c=15$\n$x=1,y=4,z=4$ : No solutions. By $y$ and $z$ , we know that $a$ $b$ , and $c$ have to all be divisible by $4$ . Therefore, $x$ cannot be equal to $1$\n$x=2,y=2,z=5$ : No solutions. By $x$ and $y$ , we know that $a$ $b$ , and $c$ have to all be divisible by $2$ . Therefore, $z$ cannot be equal to $5$\n$x=2,y=3,z=4$ : No solutions. By $x$ and $z$ , we know that $a$ $b$ , and $c$ have to all be divisible by $2$ . Therefore, $y$ cannot be equal to $3$\n$x=3,y=3,z=3$ : No solutions. As $a$ $b$ , and $c$ have to all be divisible by $3$ $a+b+c$ has to be divisible by $3$ . This contradicts the sum $a+b+c=23$\nPutting these solutions together, we have $(7^2+9^2+7^2)+(5^2+3^2+15^2)=179+259=\\boxed{438}$", "Since $a+b+c=23$ $\\gcd(a,b,c)=23$ or $\\gcd(a,b,c)=1$\nAs $\\gcd(a,b)+\\gcd(b,c)+\\gcd(c,a)=9$ , it is impossible for $\\gcd(a,b,c)=23$ , so $\\gcd(a,b,c)=1$\nThis means that $\\gcd(a,b)$ $\\gcd(b,c)$ , and $\\gcd(c,a)$ must all be coprime. The only possible ways for this to be true are $1+1+7=9$ and $1+3+5=9$\nWithout loss of generality, let $a\\le b\\le c$ . Since $a+b+c=23$ , then $a=7, b=7, c=9$ or $a=3, b=5, c=15$\n$(7^2+7^2+9^2)+(3^2+5^2+15^2)=179+259=\\boxed{438}$" ]

[ "There are $2$ cases to consider:\nCase $1$ $2$ of $a$ $b$ , and $c$ are positive and the other is negative. Without loss of generality (WLOG), we can assume that $a$ and $b$ are positive and $c$ is negative. In this case, we have that \\[\\frac{a}{|a|}+\\frac{b}{|b|}+\\frac{c}{|c|}+\\frac{abc}{|abc|}=1+1-1-1=0.\\]\nCase $2$ $2$ of $a$ $b$ , and $c$ are negative and the other is positive. WLOG, we can assume that $a$ and $b$ are negative and $c$ is positive. In this case, we have that \\[\\frac{a}{|a|}+\\frac{b}{|b|}+\\frac{c}{|c|}+\\frac{abc}{|abc|}=-1-1+1+1=0.\\]\nNote these are the only valid cases, for neither $3$ negatives nor $3$ positives would work as they cannot sum up to $0$ . In both cases, we get that the given expression equals $\\boxed{0}$" ]

[ "The power of $10$ for any factorial is given by the well-known algorithm \\[\\left\\lfloor \\frac n{5}\\right\\rfloor + \\left\\lfloor \\frac n{25}\\right\\rfloor + \\left\\lfloor \\frac n{125}\\right\\rfloor + \\cdots\\] It is rational to guess numbers right before powers of $5$ because we won't have any extra numbers from higher powers of $5$ . As we list out the powers of 5, it is clear that $5^{4}=625$ is less than 2006 and $5^{5}=3125$ is greater. Therefore, set $a$ and $b$ to be 624. Thus, c is $2006-(624\\cdot 2)=758$ . Applying the algorithm, we see that our answer is $152+152+188= \\boxed{492}$", "Clearly, the power of $2$ that divides $n!$ is larger or equal than the power of $5$ which divides\nit. Hence we are trying to minimize the power of $5$ that will divide $a!b!c!$\nConsider $n! = 1\\cdot 2 \\cdot \\dots \\cdot n$ . Each fifth term is divisible by $5$ , each $25$ -th one\nby $25$ , and so on. Hence the total power of $5$ that divides $n$ is $\\left\\lfloor \\frac n{5}\\right\\rfloor + \\left\\lfloor \\frac n{25}\\right\\rfloor + \\cdots$ . (For any $n$ only finitely many terms in the sum\nare\nnon-zero.)\nIn our case we have $a<2006$ , so the largest power of $5$ that will be less than $a$ is at most $5^4 = 625$ . Therefore the power of $5$ that divides $a!$ is equal to $\\left\\lfloor \\frac a{5}\\right\\rfloor + \\left\\lfloor \\frac a{25}\\right\\rfloor + \\left\\lfloor \\frac a{125}\\right\\rfloor + \\left\\lfloor \\frac a{625}\\right\\rfloor$ . The same\nis true for $b$ and $c$\nIntuition may now try to lure us to split $2006$ into $a+b+c$ as evenly as possible, giving $a=b=669$ and $c=668$ . However, this solution is not optimal.\nTo see how we can do better, let's rearrange the terms as follows:\n\\begin{align*} \\text{result} & = \\Big\\lfloor \\frac a{5}\\Big\\rfloor + \\Big\\lfloor \\frac b{5}\\Big\\rfloor + \\Big\\lfloor \\frac c{5}\\Big\\rfloor \\\\ & + \\Big\\lfloor \\frac a{25}\\Big\\rfloor + \\Big\\lfloor \\frac b{25}\\Big\\rfloor + \\Big\\lfloor \\frac c{25}\\Big\\rfloor \\\\ & + \\Big\\lfloor \\frac a{125}\\Big\\rfloor + \\Big\\lfloor \\frac b{125}\\Big\\rfloor + \\Big\\lfloor \\frac c{125}\\Big\\rfloor \\\\ & + \\Big\\lfloor \\frac a{625}\\Big\\rfloor + \\Big\\lfloor \\frac b{625}\\Big\\rfloor + \\Big\\lfloor \\frac c{625}\\Big\\rfloor \\end{align*}\nThe idea is that the rows of the above equation are roughly equal to $\\left\\lfloor \\frac n{5}\\right\\rfloor$ $\\left\\lfloor \\frac n{25}\\right\\rfloor$ , etc.\nMore precisely, we can now notice that for any positive integers $a,b,c,k$ we can write $a,b,c$ in the form $a=a_0k + a_1$ $b=b_0k+b_1$ $c=c_0k + c_1$ , where all $a_i,b_i,c_i$ are integers and $0\\leq a_1,b_1,c_1<k$\nIt follows that \\[\\Big\\lfloor \\frac a{k}\\Big\\rfloor + \\Big\\lfloor \\frac b{k}\\Big\\rfloor + \\Big\\lfloor \\frac c{k}\\Big\\rfloor = a_0+b_0+c_0\\] and \\[\\Big\\lfloor \\frac {a+b+c}k\\Big\\rfloor = a_0 + b_0 + c_0 + \\Big\\lfloor \\frac {a_1+b_1+c_1}k\\Big\\rfloor \\leq a_0 + b_0 + c_0 + 2\\]\nHence we get that for any positive integers $a,b,c,k$ we have \\[\\Big\\lfloor \\frac a{k}\\Big\\rfloor + \\Big\\lfloor \\frac b{k}\\Big\\rfloor + \\Big\\lfloor \\frac c{k}\\Big\\rfloor \\quad \\geq \\quad \\Big\\lfloor \\frac {a+b+c}k\\Big\\rfloor - 2\\]\nTherefore for any $a,b,c$ the result is at least $\\left\\lfloor \\frac n{5}\\right\\rfloor + \\left\\lfloor \\frac n{25}\\right\\rfloor + \\left\\lfloor \\frac n{125}\\right\\rfloor + \\left\\lfloor \\frac n{625}\\right\\rfloor - 8 = 401 + 80 + 16 + 3 - 8 = 500 - 8 = 492$\nIf we now show how to pick $a,b,c$ so that we'll get the result $492$ , we will be done.\nConsider the row with $625$ in the denominator. We need to achieve sum $1$ in this row,\nhence we need to make two of the numbers smaller than $625$ . Choosing $a=b=624$ does this, and it will give us the largest possible remainders for $a$ and $b$ in\nthe other three rows, so this is a pretty good candidate. We can compute $c=2006-a-b=758$ and verify that this triple gives the desired result $\\boxed{492}$" ]

[ "Since $a=1.5b$ , that means $b=\\frac{a}{1.5}$ . We multiply by $3$ to get a $3b$ term, yielding $3b=2a$ , and $2a$ is $\\boxed{200}$ of $a$", "Without loss of generality, let $b=100$ . Then, we have $a=150$ and $3b=300$ . Thus, $\\frac{3b}{a}=\\frac{300}{150}=2$ , so $3b$ is $200\\%$ of $a$ . Hence the answer is $\\boxed{200}$", "As before, $a = 1.5b$ . Multiply by 2 to obtain $2a = 3b$ . Since $2 = 200\\%$ , the answer is $\\boxed{200}$", "Without loss of generality, let $b=2$ . Then, we have $a=3$ and $3b=6$ . This gives $\\frac{3b}{a}=\\frac{6}{3}=2$ , so $3b$ is $200\\%$ of $a$ , so the answer is $\\boxed{200}$" ]

[ "We are given that $a\\cdot(\\sin{x}+\\sin{(2x)})=\\sin{(3x)}$\nUsing the sine double angle formula combine with the fact that $\\sin{(3x)} = \\sin{x}\\cdot(4\\cos^2{x}-1)$ , which can be derived using sine angle addition with $\\sin{(2x + x)}$ , we have \\[a\\cdot(\\sin{x}+2\\sin{x}\\cos{x})=\\sin{x}\\cdot(4\\cos^2{x}-1)\\] Since $\\sin{x} \\ne 0$ as it is on the open interval $(0, \\pi)$ , we can divide out $\\sin{x}$ from both sides, leaving us with \\[a\\cdot(1+2\\cos{x})=4\\cos^2{x}-1\\] Now, distributing $a$ and rearranging, we achieve the equation \\[4\\cos^2{x} - 2a\\cos{x} - (1+a) = 0\\] which is a quadratic in $\\cos{x}$\nApplying the quadratic formula to solve for $\\cos{x}$ , we get \\[\\cos{x} =\\frac{2a\\pm\\sqrt{4a^2+4(4)(1+a)}}{8}\\] and expanding the terms under the radical, we get \\[\\cos{x} =\\frac{2a\\pm\\sqrt{4a^2+16a+16}}{8}\\] Factoring, since $4a^2+16a+16 = (2a+4)^2$ , we can simplify our expression even further to \\[\\cos{x} =\\frac{a\\pm(a+2)}{4}\\]\nNow, solving for our two solutions, $\\cos{x} = -\\frac{1}{2}$ and $\\cos{x} = \\frac{a+1}{2}$\nSince $\\cos{x} = -\\frac{1}{2}$ yields a solution that is valid for all $a$ , that being $x = \\frac{2\\pi}{3}$ , we must now solve for the case where $\\frac{a+1}{2}$ yields a valid value.\nAs $x\\in (0, \\pi)$ $\\cos{x}\\in (-1, 1)$ , and therefore $\\frac{a+1}{2}\\in (-1, 1)$ , and $a\\in(-3,1)$\nThere is one more case we must consider inside this interval though, the case where $\\frac{a+1}{2} = -\\frac{1}{2}$ , as this would lead to a double root for $\\cos{x}$ , yielding only one valid solution for $x$ . Solving for this case, $a \\ne -2$\nTherefore, combining this fact with our solution interval, $a\\in(-3, -2) \\cup (-2, 1)$ , so the answer is $-3-2+1 = \\boxed{4}$", "We can optimize from the step from \\[a\\cdot(1+2\\cos{x})=4\\cos^2{x}-1\\] in solution 1 by writing\n\\[a = \\frac{4\\cos^2{x}-1}{1+2\\cos{x}} = 2\\cos x - 1\\]\nand then get \\[\\cos x = \\frac{a+1}{2}.\\]\nNow, solving for our two solutions, $\\cos{x} = -\\frac{1}{2}$ and $\\cos{x} = \\frac{a+1}{2}$\nSince $\\cos{x} = -\\frac{1}{2}$ yields a solution that is valid for all $a$ , that being $x = \\frac{2\\pi}{3}$ , we must now solve for the case where $\\frac{a+1}{2}$ yields a valid value.\nAs $x\\in (0, \\pi)$ $\\cos{x}\\in (-1, 1)$ , and therefore $\\frac{a+1}{2}\\in (-1, 1)$ , and $a\\in(-3,1)$\nThere is one more case we must consider inside this interval though, the case where $\\frac{a+1}{2} = -\\frac{1}{2}$ , as this would lead to a double root for $\\cos{x}$ , yielding only one valid solution for $x$ . Solving for this case, $a \\ne -2$\nTherefore, combining this fact with our solution interval, $a\\in(-3, -2) \\cup (-2, 1)$ , so the answer is $-3-2+1 = \\boxed{4}$", "Use the sum to product formula to obtain $2a\\cdot\\sin{\\frac{3x}{2}}\\cos{\\frac{x}{2}}=\\sin{3x}$ . Use the double angle formula on the RHS to obtain $a\\cdot\\sin{\\frac{3x}{2}}\\cos{\\frac{x}{2}}=\\sin{\\frac{3x}{2}}\\cos{\\frac{3x}{2}}$ . From here, it is obvious that $x=\\frac{2\\pi}{3}$ is always a solution, and thus we divide by $\\sin{\\frac{3x}{2}}$ to get \\[a\\cdot\\cos{\\frac{x}{2}}=\\cos{\\frac{3x}{2}}\\] We wish to find all $a$ such that there is at least one more solution to this equation distinct from $x=\\frac{2\\pi}{3}$ . Letting $y=\\cos{\\frac{x}{2}}$ , and noting that $\\cos{\\frac{3x}{2}}=4y^3-3y$ , we can rearrange our equation to $4y^3=y(3+a)$ The smallest value $x$ where $y=0$ is $\\pi$ , which is not in our domain so we divide by $y$ to obtain $4y^2=a+3$ . By the trivial inequality, $a+3\\ge{0}$ . Furthermore, $y\\neq{0}$ , so $a+3>0$ . Also, if $a=-2$ , then the solution to this equation would be shared with $x=\\frac{2\\pi}{3}$ , so there would only be one distinct solution. Finally, because $y\\le{1}$ due to the restrictions of a sine wave, and that $y\\neq{1}$ due to the restrictions on $x$ , we have $-3<a<1$ with $a\\neq{-2}$ . Thus, $p=-3,q=-2, r=1$ , so our final answer is $-3+(-2)+1=\\boxed{4}$" ]

[ "We see that $2.0055$ works but $2.0045$ does not. The digit $d$ can be from $5$ through $9$ , which is $\\boxed{5}$ values." ]

[ "Repeating decimals represent rational numbers . To figure out which rational number, we sum an infinite geometric series $0.d25d25d25\\ldots = \\sum_{n = 1}^\\infty \\frac{d25}{1000^n} = \\frac{100d + 25}{999}$ . Thus $\\frac{n}{810} = \\frac{100d + 25}{999}$ so $n = 30\\frac{100d + 25}{37} =750\\frac{4d + 1}{37}$ . Since 750 and 37 are relatively prime $4d + 1$ must be divisible by 37, and the only digit for which this is possible is $d = 9$ . Thus $4d + 1 = 37$ and $n = \\boxed{750}$", "To get rid of repeating decimals, we multiply the equation by 1000. We get $\\frac{1000n}{810} = d25.d25d25...$ We subtract the original equation from the second to get $\\frac{999n}{810}=d25$ We simplify to $\\frac{37n}{30} = d25$ Since $\\frac{37n}{30}$ is an integer, $n=(30)(5)(2k+1)$ because $37$ is relatively prime to $30$ , and d25 is divisible by $5$ but not $10$ . The only odd number that yields a single digit $d$ and 25 at the end of the three digit number is $k=2$ , so the answer is $\\boxed{750}$", "Similar to Solution 2, we start off by writing that $\\frac{1000n}{810} = d25.d25d25 \\dots$ .Then we subtract this from the original equation to get:\n\\[\\frac{999n}{810} =d25 \\Longrightarrow \\frac{37n}{30} = d25 \\Longrightarrow 37n = d25 \\cdot 30\\]\nSince n is an integer, we have that $37 \\mid d25 \\cdot 30$\nSince $37$ is prime, we can apply Euclid's Lemma (which states that if $p$ is a prime and if $a$ and $b$ are integers and if $p \\mid ab$ , then $p \\mid a$ or $p \\mid b$ ) to realize that $37 \\mid d25$ , since $37 \\nmid 30$ . Then we can expand $d25$ as $25 \\cdot (4d +1)$ . Since $37 \\nmid 25$ , by Euclid, we can arrive at $37 \\mid 4d+1 \\Longrightarrow d=9$ . From this we know that $n= 25 \\cdot 30 = \\boxed{750}$ . (This is true because $37n = 925 \\cdot 30 \\rightarrow n= 25 \\cdot 30 = 750$", "Write out these equations:\n$\\frac{n}{810} = \\frac{d25}{999}$\n$\\frac{n}{30} = \\frac{d25}{37}$\n$37n = 30(d25)$\nThus $n$ divides 25 and 30. The only solution for this under 1000 is $\\boxed{750}$" ]

[ "We can just test all of these statements: \\begin{align*} 3^*+6^* &= \\frac{1}{3}+\\frac{1}{6} \\\\ &= \\frac{1}{2} \\neq 9^* \\\\ 6^*-4^* &= \\frac{1}{6}-\\frac{1}{4} \\\\ &= \\frac{-1}{12} \\neq 2^* \\\\ 2^*\\cdot 6^* &= \\frac{1}{2}\\cdot \\frac{1}{6} \\\\ &= \\frac{1}{12} = 12^* \\\\ 10^* \\div 2^* &= \\frac{1}{10}\\div \\frac{1}{2} \\\\ &= \\frac{1}{5} = 5^* \\end{align*}\nThe last two statements are true and the first two aren't, so $\\boxed{2}$" ]

[ "There are $91 - 19 + 1 = 73$ numbers in the sequence . Since the terms of the sequence can be at most $1$ apart, all of the numbers in the sequence can take one of two possible values. Since $\\frac{546}{73} = 7 R 35$ , the values of each of the terms of the sequence must be either $7$ or $8$ . As the remainder is $35$ $8$ must take on $35$ of the values, with $7$ being the value of the remaining $73 - 35 = 38$ numbers. The 39th number is $\\lfloor r+\\frac{19 + 39 - 1}{100}\\rfloor= \\lfloor r+\\frac{57}{100}\\rfloor$ , which is also the first term of this sequence with a value of $8$ , so $8 \\le r + \\frac{57}{100} < 8.01$ . Solving shows that $\\frac{743}{100} \\le r < \\frac{744}{100}$ , so $743\\le 100r < 744$ , and $\\lfloor 100r \\rfloor = \\boxed{743}$", "Recall by Hermite's Identity that $\\lfloor x\\rfloor +\\lfloor x+\\frac{1}{n}\\rfloor +...+\\lfloor x+\\frac{n-1}{n}\\rfloor = \\lfloor nx\\rfloor$ for positive integers $n$ , and real $x$ . Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, $\\lfloor r\\rfloor \\le ...\\le \\lfloor r+\\frac{18}{100}\\rfloor \\le 7$ and $\\lfloor r+\\frac{92}{100}\\rfloor \\ge ...\\ge \\lfloor r+1\\rfloor \\ge 8$ . We can see that $\\lfloor r\\rfloor +1=\\lfloor r+1\\rfloor$ . Because $\\lfloor r\\rfloor$ is at most 7, and $\\lfloor r+1\\rfloor$ is at least 8, we can clearly see their values are $7$ and $8$ respectively. \nSo, $\\lfloor r\\rfloor = ... = \\lfloor r+\\frac{18}{100}\\rfloor = 7$ , and $\\lfloor r+\\frac{92}{100}\\rfloor = ...= \\lfloor r+1\\rfloor = 8$ . Since there are 19 terms in the former equation and 8 terms in the latter, our answer is $\\lfloor nx\\rfloor = 546+19\\cdot 7+8\\cdot 8=\\boxed{743}$" ]

[ "We can rewrite the given expression as \\[\\sqrt{24^3\\sin^3 x}=24\\cos x\\] Square both sides and divide by $24^2$ to get \\[24\\sin ^3 x=\\cos ^2 x\\] Rewrite $\\cos ^2 x$ as $1-\\sin ^2 x$ \\[24\\sin ^3 x=1-\\sin ^2 x\\] \\[24\\sin ^3 x+\\sin ^2 x - 1=0\\] Testing values using the rational root theorem gives $\\sin x=\\frac{1}{3}$ as a root, $\\sin^{-1} \\frac{1}{3}$ does fall in the first quadrant so it satisfies the interval. \nThere are now two ways to finish this problem.\nFirst way: Since $\\sin x=\\frac{1}{3}$ , we have \\[\\sin ^2 x=\\frac{1}{9}\\] Using the Pythagorean Identity gives us $\\cos ^2 x=\\frac{8}{9}$ . Then we use the definition of $\\cot ^2 x$ to compute our final answer. $24\\cot ^2 x=24\\frac{\\cos ^2 x}{\\sin ^2 x}=24\\left(\\frac{\\frac{8}{9}}{\\frac{1}{9}}\\right)=24(8)=\\boxed{192}$", "Like Solution 1, we can rewrite the given expression as \\[24\\sin^3x=\\cos^2x\\] Divide both sides by $\\sin^3x$ \\[24 = \\cot^2x\\csc x\\] Square both sides. \\[576 = \\cot^4x\\csc^2x\\] Substitute the identity $\\csc^2x = \\cot^2x + 1$ \\[576 = \\cot^4x(\\cot^2x + 1)\\] Let $a = \\cot^2x$ . Then \\[576 = a^3 + a^2\\] .\nSince $\\sqrt[3]{576} \\approx 8$ , we can easily see that $a = 8$ is a solution. Thus, the answer is $24\\cot^2x = 24a = 24 \\cdot 8 = \\boxed{192}$" ]

[ "Assume WLOG that $m$ and $n$ are both $1$ . Plugging into each of the choices, we get $4, 2, 6, 16,$ and $3$ . The only odd integer is $\\boxed{3}$" ]

[ "The four squares we already have assemble nicely into four sides of the cube. Let the central one be the bottom, and fold the other three upwards to get the front, right, and back side. Currently, our box is missing its left side and its top side. We have to count the possibilities that would fold to one of these two places.\nIn total, there are $6\\rightarrow\\boxed{6}$ good possibilities.", "Fold the four squares into the four sides of a cube. Then, there are six edges \"open\" (for lack of better term). For each open edge, we can add a square/side, so the answer is $6\\rightarrow\\boxed{6}$" ]

[ "Let's start off with just $(a_1, b_1), (a_2, b_2)$ and suppose that it satisfies the given condition. We could use $(1, 1), (1, 2)$ for example. We should maximize the number of conditions that the third pair satisfies. We find out that the third pair should equal $(a_1+a_2, b_1+b_2)$\nWe know this must be true: \\[|a_1b_2-a_2b_1| = 1\\]\nSo \\[a_1b_2-a_2b_1 = 1\\]\nWe require the maximum conditions for $(a_3, b_3)$ \\[|a_3b_2-a_2b_3| = 1\\] \\[|a_3b_1-a_1b_3| = 1\\]\nThen one case can be: \\[a_3b_2-a_2b_3 = 1\\] \\[a_3b_1-a_1b_3 = -1\\]\nWe try to do some stuff such as solving for $a_3$ with manipulations: \\[a_3b_2a_1-a_2b_3a_1 = a_1\\] \\[a_3b_1a_2-a_1b_3a_2 = -a_2\\] \\[a_3(a_1b_2-a_2b_1) = a_1+a_2\\] \\[a_3 = a_1+a_2\\] \\[a_3b_2b_1-a_2b_3b_1 = b_1\\] \\[a_3b_1b_2-a_1b_3b_2 = -b_2\\] \\[b_3(a_1b_2-a_2b_1) = b_1+b_2\\] \\[b_3 = b_1+b_2\\]\nWe showed that 3 pairs are a complete graph; however, 4 pairs are not a complete graph. We will now show that: \\[a_4 = a_1+2a_2\\] \\[b_4 = b_1+2b_2\\] \\[|a_1b_1+2a_2b_1-a_1b_1-2a_1b_2| = 1\\] \\[2|a_2b_1-a_1b_2| = 1\\]\nThis is clearly impossible because $1$ is not even and also $|a_2b_1-a_1b_2| = 1$ .\nThe answer is as follows: \\[0+1+2+\\ldots+2\\] $a_1$ has $0$ subtractions that follow condition while $a_2$ has $1$ and then the rest has $2$ .\nThere are $n$ terms, so our answer be $2n-3$ and in case of $n=100$ that means \\[\\boxed{197}.\\] ~Lopkiloinm", "We claim the answer is $197$\nStudy the points $(0, 0), (a_i, b_i), (a_j, b_j)$ . If we let these be the vertices of a triangle, applying shoelace theorem gives us an area of $\\frac{1}{2}|0\\times{b_i}+{a_i}\\times{b_j}+{b_i}\\times{0}-0\\times{a_i}-{b_i}\\times{a_j}-{b_j}\\times{0} = \\frac{1}{2}|a_ib_j - a_j b_i| = \\frac{1}{2}$ . Therefore, the triangle formed by the points $(0, 0), (a_i, b_i), (a_j, b_j)$ must have an area of $\\frac{1}{2}$\nTwo cases follow.\nCase 1: Both $(a_i, b_i), (a_j, b_j)$ have exactly one coordinate equal to $0$ . \nHere, one point must be on the $x$ axis and the other on the $y$ axis in order for the triangle to have a positive area. For the area of the triangle to be $\\frac{1}{2}$ , it follows that the points must be $(1, 0), (0, 1)$ in some order.\nCase 2: At least one of $(a_i, b_i), (a_j, b_j)$ does not have exactly one coordinate equal to $0$ . Define $S[l]$ to be a list of lines such that each line in the list has some two lattice points that, with $(0, 0)$ , form a triangle with area $\\frac{1}{2}$ . Note that for any such line that passes through such two lattice points, we may trivially generate infinite lattice points on the line that have nonnegative coordinates.\nNote that lines $y=1$ and $x=1$ are included in $S[l]$ , because the points $(1, 1), (2, 1)$ serve as examples for $y=1$ and $(1, 1), (1, 2)$ serve as examples for $x=1$ . For the optimal construction, include the points $(1, 0)$ and all the points $(0, 1), (0, 2), (0, 3), ... , (0, 99)$ , in that order. In this case, every adjacent pair of points would count ( $98$ ), as well as picking $(0, 1)$ and a nonadjacent point ( $99$ ), so this would be $98+99=197$\nTo prove that this is the maximum, consider the case where some $n$ number of points were neither on $x=1$ nor on $y=1$ . In this case, we would be removing $n$ adjacent pairs and $n$ options to choose from after choosing $(0, 0)$ , resulting in a net loss of $2n$ . By having $n$ points on some other combination of lines in $S[l]$ , we would trivially have a maximum gain of $n-1$ pairs of points on the lines such that there are no lattice points between those pairs. Because these points are not on $x=1$ or $y=1$ , the altitude from a given point to the line formed by $(0, 0)$ and $(0, 1)$ and $(1, 0)$ is not $1$ , and so the area of the triangle cannot be $\\frac{1}{2}$ . Thus, by not having all points on lines $x=1$ and $y=1$ , we cannot exceed the maximum of $197$ . Thus, $\\boxed{197}$ is our answer." ]

[ "Call the pair $(i, j)$ good if $1\\leq i < j \\leq 100$ and $|a_ib_j-a_jb_i|=1$ . Note that we can reorder the pairs $(a_1, b_1), (a_2, b_2), \\ldots, (a_{100}, b_{100})$ without changing the number of good pairs. Thus, we can reorder them so that $a_1\\leq a_2\\leq\\ldots\\leq a_{100}$ . Furthermore, reorder them so that if $a_i=a_j$ for some $i<j$ , then $b_i<b_j$\nNow I claim the maximum value of $N$ is $\\boxed{197}$ . First, we will show $N\\leq 197$" ]

[ "If we plug in $n=4$ , we get\nBy plugging in $n=3$ , we get\nThis is a system of two equations with two unknowns. Multiplying the second equation by 2 and substituting into the first equation gives $128=5u_4+18 \\Longrightarrow u_4=22$ , therefore $u_5=\\frac{128-22}{2}=53 \\longrightarrow \\textbf{\\boxed{53}$ .\n~ Mathkiddie" ]

[ "For $1\\leq k\\leq 12,$ suppose that cards $1, 2, \\ldots, k$ are picked up on the first pass. It follows that cards $k+1,k+2,\\ldots,13$ are picked up on the second pass.\nOnce we pick the spots for the cards on the first pass, there is only one way to arrange all $\\boldsymbol{13}$ cards.\nFor each value of $k,$ there are $\\binom{13}{k}-1$ ways to pick the $k$ spots for the cards on the first pass: We exclude the arrangement in which the first pass consists of all $13$ cards.\nTherefore, the answer is \\[\\sum_{k=1}^{12}\\left[\\binom{13}{k}-1\\right] = \\left[\\sum_{k=1}^{12}\\binom{13}{k}\\right]-12 = \\left[\\sum_{k=0}^{13}\\binom{13}{k}\\right]-14 = 2^{13} - 14 = \\boxed{8178}.\\]", "Since the $13$ cards are picked up in two passes, the first pass must pick up the first $n$ cards and the second pass must pick up the remaining cards $m$ through $13$ . \nAlso note that if $m$ , which is the card that is numbered one more than $n$ , is placed before $n$ , then $m$ will not be picked up on the first pass since cards are picked up in order. Therefore we desire $m$ to be placed before $n$ to create a second pass, and that after the first pass, the numbers $m$ through $13$ are lined up in order from least to greatest.\nTo construct this, $n$ cannot go in the $n$ th position because all cards $1$ to $n-1$ will have to precede it and there will be no room for $m$ . Therefore $n$ must be in slots $n+1$ to $13$ .\nLet's do casework on which slot $n$ goes into to get a general idea for how the problem works.\nCase 1: With $n$ in spot $n+1$ , there are $n$ available slots before $n$ , and there are $n-1$ cards preceding $n$ . Therefore the number of ways to reserve these slots for the $n-1$ cards is $\\binom{n}{n-1}$ . Then there is only $1$ way to order these cards (since we want them in increasing order). Then card $m$ goes into whatever slot is remaining, and the $13-m$ cards are ordered in increasing order after slot $n+1$ , giving only $1$ way. Therefore in this case there are $\\binom{n}{n-1}$ possibilities.\nCase 2: With $n$ in spot $n+2$ , there are $n+1$ available slots before $n$ , and there are $n-1$ cards preceding $n$ . Therefore the number of ways to reserve slots for these cards are $\\binom{n+1}{n-1}$ . Then there is one way to order these cards. Then cards $m$ and $m+1$ must go in the remaining two slots, and there is only one way to order them since they must be in increasing order. Finally, cards $m+2$ to $13$ will be ordered in increasing order after slot $n+1$ , which yields $1$ way. Therefore, this case has $\\binom{n+1}{n-1}$ possibilities.\nI think we can see a general pattern now. With $n$ in slot $x$ , there are $x-1$ slots to distribute to the previous $n-1$ cards, which can be done in $\\binom{x-1}{n-1}$ ways. Then the remaining cards fill in in just $1$ way. Since the cases of $n$ start in slot $n+1$ and end in slot $13$ , this sum amounts to: \\[\\binom{n}{n-1}+\\binom{n+1}{n-1}+\\binom{n+2}{n-1} + \\cdots + \\binom{12}{n-1}\\] for any $n$\nHmmm ... where have we seen this before?\nWe use wishful thinking to add a term of $\\binom{n-1}{n-1}$ \\[\\binom{n-1}{n-1}+\\binom{n}{n-1}+\\binom{n+1}{n-1}+\\binom{n+2}{n-1} + \\cdots + \\binom{12}{n-1}\\]\nThis is just the hockey stick identity! Applying it, this expression is equal to $\\binom{13}{n}$ . However, we added an extra term, so subtracting it off, the total number of ways to order the $13$ cards for any $n$ is \\[\\binom{13}{n}-1\\]\nFinally, to calculate the total for all $n$ , we sum from $n=0$ to $13$ . This yields us:\n\\[\\sum_{n=0}^{13} \\binom{13}{n}-1 \\implies \\sum_{n=0}^{13} \\binom{13}{n} - \\sum_{n=0}^{13} 1\\] \\[\\implies 2^{13} - 14 = 8192 - 14 = 8178 = \\boxed{8178}.\\]", "To solve this problem, we can use recursion on $n$ . Let $A_n$ be the number of arrangements for $n$ numbers. Now, let's look at how these arrangements are formed by case work on the first number $a_1$\nIf $a_1 = 1$ , the remaining $n-1$ numbers from $2$ to $n$ are arranged in the same way just like number 1 to $n-1$ in the case of $n-1$ numbers. So there are $A_{n-1}$ arrangements.\nIf $a_1 = 2$ , then we need to choose 1 position from position 2 to $n-1$ to put 1, and all remaining numbers must be arranged in increasing order, so there are $\\binom{n-1}{1}$ such arrangements.\nIf $a_1 = k$ , then we need to choose $k-1$ positions from position 2 to $n-1$ to put $1, 2,\\cdots k-1$ , and all remaining numbers must be arranged in increasing order, so there are $\\binom{n-1}{k-1}$ such arrangements.\nSo we can write \\[A_n = A_{n-1} + \\binom{n-1}{1} + \\binom{n-1}{2} + \\cdots + \\binom{n-1}{n-1}\\] which can be simplified to \\[A_n = A_{n-1} + 2^{n-1} - 1\\] We can solve this recursive sequence by summing up $n-1$ lines of the recursive formula \\[A_n - A_{n-1} = 2^{n-1} - 1\\] \\[A_{n-1} - A_{n-2} = 2^{n-2} - 1\\] \\[\\cdots\\] \\[A_2 - A_{1} = 2^{1} - 1\\] to get \\[A_n - A_1 = \\sum_{k=1}^{n-1} (2^k - 1) = 2^n - 2 - (n-1) = 2^n - n - 1\\] since $A_1=0$ , we have \\[A_n = 2^n - n - 1\\] and $A_{13} = 2^{13} - 14 = \\boxed{8178}$", "When we have $3$ cards arranged in a row, after listing out all possible arrangements, we see that we have $4$ ones: $(1, 3, 2), (2, 1, 3), (2, 3, 1),$ and $(3, 1, 2)$ . When we have $4$ cards, we find $11$ possible arrangements: $(1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (2, 1, 3, 4), (2, 3, 1, 4), (2, 3, 4, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 4, 1, 2),$ and $(4, 1, 2, 3).$ Hence, we recognize the pattern that for $n$ cards, we have $2^n - n - 1$ valid arrangements, so our answer is $2^{13} - 13 - 1 = \\boxed{8178}.$ ~eibc", "Notice that for each card \"position\", we can choose for it to be picked up on the first or second pass, for a total of $2^{13}$ options. However, if all of the cards selected to be picked up first are before all of the cards to be picked up second, then this means that the list is in consecutive ascending order (and thus all cards will be picked up on the first pass instead). This can happen in 14 ways, so our answer is $2^{13}-14=\\boxed{8178}$" ]