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[ "\\[a^x=b\\] \\[b^y=a\\] \\[{a^x}^y=a\\] \\[xy=1\\] \\[\\log_a b\\log_b a=1\\] As a result, the answer should be $\\boxed{1}$", "Apply the change of base formula to $\\log_a b$ and $\\log_b a$ . For simplicity, let us convert to base-10 log.\nBy change of base, the expression becomes $\\frac{\\log b}{\\log a} * \\frac{\\log a}{\\log b} = \\boxed{1}$" ]

[ "Let $x^2 + px + m = 0$ have roots $a$ and $b$ . Then\n\\[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\\]\nso $p = -(a+b)$ and $m = ab$ . Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$ , so\n\\[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\\]\nand $m = -2(a+b)$ and $n = 4ab$ . Thus $\\frac{n}{p} = \\frac{4ab}{-(a+b)} = \\frac{4m}{\\frac{m}{2}} = \\boxed{8}$", "If the roots of $x^2 + mx + n = 0$ are $2a$ and $2b$ and the roots of $x^2 + px + m = 0$ are $a$ and $b$ , then using Vieta's formulas, \\[2a + 2b = -m\\] \\[a + b = -p\\] \\[2a(2b) = n\\] \\[a(b) = m\\] Therefore, substituting the second equation into the first equation gives \\[m = 2(p)\\] and substituting the fourth equation into the third equation gives \\[n = 4(m)\\] Therefore, $n = 8p$ , so $\\frac{n}{p}= \\boxed{8}$" ]

[ "Let $x^2 + px + m = 0$ have roots $a$ and $b$ . Then\n\\[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\\]\nso $p = -(a+b)$ and $m = ab$ . Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$ , so\n\\[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\\]\nand $m = -2(a+b)$ and $n = 4ab$ . Thus $\\frac{n}{p} = \\frac{4ab}{-(a+b)} = \\frac{4m}{\\frac{m}{2}} = \\boxed{8}$", "If the roots of $x^2 + mx + n = 0$ are $2a$ and $2b$ and the roots of $x^2 + px + m = 0$ are $a$ and $b$ , then using Vieta's formulas, \\[2a + 2b = -m\\] \\[a + b = -p\\] \\[2a(2b) = n\\] \\[a(b) = m\\] Therefore, substituting the second equation into the first equation gives \\[m = 2(p)\\] and substituting the fourth equation into the third equation gives \\[n = 4(m)\\] Therefore, $n = 8p$ , so $\\frac{n}{p}= \\boxed{8}$" ]

[ "We assume that the clicks are heard at the head of the train. Then if the train's speed in miles per hour is $x$ , we can convert it to clicks per minute: \\[\\frac{x\\text{ mile}}{\\text{hr}}\\cdot\\left(\\frac{1\\text{ hr}}{60\\text{ min}}\\right)\\cdot\\left(\\frac{5280\\text{ ft}}{1\\text{ mile}}\\right)\\cdot\\left(\\frac{1\\text{ click}}{30\\text{ ft}}\\right)=\\frac{5280x}{1800}\\cdot\\frac{\\text{click}}{\\text{min}}.\\] Therefore every minute, on average, $5280x/1800$ clicks are heard. The number of clicks heard in $y$ minutes is $y\\cdot 5280x/1800$ , so the number of clicks heard in $1800/5280$ minutes is equal to $x$ . In other words, the speed of the train in miles per hour is equal to the number of clicks heard in $1800/5280$ minutes, which is approximately one-third of a minute, or $\\boxed{20}$" ]

[ "We factor $\\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ as $\\frac{10^{2000}(1+100)}{10^{2001}(1+1)}=\\frac{101}{20}$ . As $\\frac{101}{20}=5.05$ , our answer is $\\boxed{5}$" ]

[ "Let $m$ be Mary's age. Then $\\frac{m}{30}=\\frac{3}{5}$ . Solving for $m$ , we obtain $m=\\boxed{18}.$", "We can see this is a combined ratio of $8$ $(5+3)$ . We can equalize by doing $30\\div5=6$ , and $6\\cdot3=\\boxed{18}$ . With the common ratio of $8$ and difference ratio of $6$ , we see $6\\cdot8=30+18$ . Therefore, we can see our answer is correct." ]

[ "Let $m$ be Mary's age. Then $\\frac{m}{30}=\\frac{3}{5}$ . Solving for $m$ , we obtain $m=\\boxed{18}.$", "We can see this is a combined ratio of $8$ $(5+3)$ . We can equalize by doing $30\\div5=6$ , and $6\\cdot3=\\boxed{18}$ . With the common ratio of $8$ and difference ratio of $6$ , we see $6\\cdot8=30+18$ . Therefore, we can see our answer is correct." ]

[ "Let the number of boys be $2x$ . It follows that the number of girls is $3x$ . These two values add up to $30$ students, so \\[2x+3x=5x=30\\Rightarrow x=6\\]\nThe difference between the number of girls and the number of boys is $3x-2x=x$ , which is $6$ , so the answer is $\\boxed{6}$" ]

[ "The formula for the measure of the interior angle of a regular polygon with $n$ -sides is $180 - \\frac{360}{n}$ . Letting our two polygons have side length $r$ and $k$ , we have that the ratio of the interior angles is $\\frac{180 - \\frac{360}{r}}{180 - \\frac{360}{k}} = \\frac{(r-2) \\cdot k}{(k-2) \\cdot r} = \\frac{3}{2}$ . Cross multiplying both sides, we have $2rk-4k = 3kr-6r \\Rightarrow -rk-4k+6r = 0$ . Using Simon's Favorite Factoring Trick, we have $-k(r+4)+6r+24 = 24 \\Rightarrow (r+4)(6-k)=24$ . Because $k$ and $r$ are both more than $2$ , we know that $6-k < r+4$ . Now, we just set these factors equal to the factors of 24. We can set $6-k$ to $1$ $2$ $3$ , or $4$ and $r+4$ to $24$ $12$ $8$ , or $6$ respectively to get the following pairs for $(k, r)$ $(5, 20)$ $(4, 8)$ $(3, 4)$ , and $(2, 2)$ . However, we have to take out the solution with $k = 2$ , because $k$ and $r$ are both more than $2$ , leaving us with $\\boxed{3}$ as the correct answer." ]

[ "We can set up a system of equations where $x$ and $y$ are the two acute angles. WLOG, assume that $x$ $<$ $y$ in order for the complement of $x$ to be greater than the complement of $y$ . Therefore, $5x$ $=$ $4y$ and $90$ $-$ $x$ $=$ $2$ $(90$ $-$ $y)$ . Solving for $y$ in the first equation and substituting into the second equation yields \\[\\begin{split} 90 - x & = 2 (90 - 1.25x) \\\\ 1.5x & = 90 \\\\ x & = 60 \\end{split}\\]\nSubstituting this $x$ value back into the first equation yields $y$ $=$ $75$ , leaving $x$ $+$ $y$ equal to $\\boxed{135}$", "We let the measures be $5x$ and $4x$ giving us the ratio of $5:4$ . We know $90-4x>90-5x$ since this inequality gives $x>0$ , which is true since the measures of angles are never negative. We also know the bigger complement is twice the smaller, so\n$90-4x=2(90-5x)$\n$90-4x=180-10x$\n$6x=90$\n$x=15$\nTherefore, the angles are $75$ and $60$ , which sum to $\\boxed{135}$" ]

[ "By: dragonfly\nWe set up equations to find each angle. The larger angle will be represented as $x$ and the smaller angle will be represented as $y$ , in degrees. This implies that\n$4x=5y$\nand\n$2\\times(90-x)=90-y$\nsince the larger the original angle, the smaller the complement.\nWe then find that $x=75$ and $y=60$ , and their sum is $\\boxed{135}$", "We can visualize the problem like so: \\[5x+1y = 90^\\circ = 4x+2y\\] Moving like terms to the same side gets $x = y$ , and substituting this back gets $6x = 90^\\circ \\implies x = \\frac{90^\\circ}{6} = 15^\\circ$ , so the sum of the degree measures is $5x + 4x = 9x = 9(15) = \\boxed{135}$ . ~ emerald_block" ]

[ "The ratio means that for every $11$ games won, $4$ are lost, so the team has won $11x$ games, lost $4x$ games, and played $15x$ games for some positive integer $x$ . The percentage of games lost is just $\\dfrac{4x}{15x}\\times100=\\dfrac{4}{15}\\times 100=26.\\overline{6}\\%\\approx\\boxed{27}$" ]

[ "The Won/Lost ratio is 11/4 so, for some number $N$ , the team won $11N$ games and lost $4N$ games. Thus, the team played $15N$ games and the fraction of games lost is $\\dfrac{4N}{15N}=\\dfrac{4}{15}\\approx 0.27=\\boxed{27}.$ -Clara Garza" ]

[ "If the radius of the circle is $r$ , then the perimeter of the first triangle is $3\\left(\\frac{2r}{\\sqrt3}\\right)=2r\\sqrt3$ , and the perimeter of the second is $3r\\sqrt3$ . So the ratio is $\\boxed{23}$" ]

[ "We square $x+\\frac{1}{x}=\\sqrt5$ to get $x^2+2+\\frac{1}{x^2}=5$ . We subtract 2 on both sides for $x^2+\\frac{1}{x^2}=3$ and square again, and see that $x^4+2+\\frac{1}{x^4}=9$ so $x^4+\\frac{1}{x^4}=7$ . We can factor out $x^7$ from our original expression of $x^{11}-7x^7+x^3$ to get that it is equal to $x^7(x^4-7+\\frac{1}{x^4})$ . Therefore because $x^4+\\frac{1}{x^4}$ is 7, it is equal to $x^7(0)=\\boxed{0}$", "Multiplying both sides by $x$ and using the quadratic formula, we get $\\frac{\\sqrt{5} \\pm 1}{2}$ . We can assume that it is $\\frac{\\sqrt{5}+1}{2}$ , and notice that this is also a solution the equation $x^2-x-1=0$ , i.e. we have $x^2=x+1$ . Repeatedly using this on the given (you can also just note Fibonacci numbers), \\begin{align*} (x^{11})-7x^7+x^3 &= (x^{10}+x^9)-7x^7+x^3 \\\\ &=(2x^9+x^8)-7x^7+x^3 \\\\ &=(3x^8+2x^7)-7x^7+x^3 \\\\ &=(3x^8-5x^7)+x^3 \\\\ &=(-2x^7+3x^6)+x^3 \\\\ &=(x^6-2x^5)+x^3 \\\\ &=(-x^5+x^4+x^3) \\\\ &=-x^3(x^2-x-1) = \\boxed{0}", "We can immediately note that the exponents of $x^{11}-7x^7+x^3$ are an arithmetic sequence, so they are symmetric around the middle term. So, $x^{11}-7x^7+x^3 = x^7(x^4-7+\\frac{1}{x^4})$ . We can see that since $x+\\frac{1}{x} = \\sqrt{5}$ $x^2+2+\\frac{1}{x^2} = 5$ and therefore $x^2+\\frac{1}{x^2} = 3$ . Continuing from here, we get $x^4+2+\\frac{1}{x^4} = 9$ , so $x^4-7+\\frac{1}{x^4} = 0$ . We don't even need to find what $x^7$ is! This is since $x^7\\cdot0$ is evidently $\\boxed{0}$ , which is our answer.", "We begin by multiplying $x+\\frac{1}{x} = \\sqrt{5}$ by $x$ , resulting in $x^2+1 = \\sqrt{5}x$ . Now we see this equation: $x^{11}-7x^{7}+x^3$ . The terms all have $x^3$ in common, so we can factor that out, and what we're looking for becomes $x^3(x^8-7x^4+1)$ . Looking back to our original equation, we have $x^2+1 = \\sqrt{5}x$ , which is equal to $x^2 = \\sqrt{5}x-1$ . Using this, we can evaluate $x^4$ to be $5x^2-2\\sqrt{5}x+1$ , and we see that there is another $x^2$ , so we put substitute it in again, resulting in $3\\sqrt{5}x-4$ . Using the same way, we find that $x^8$ is $21\\sqrt{5}x-29$ . We put this into $x^3(x^8-7x^4+1)$ , resulting in $x^3(0)$ , so the answer is $\\boxed{0}$", "The equation we are given is $x+\\tfrac{1}{x}=\\sqrt{5}...$ Yuck. Fractions and radicals! We multiply both sides by $x,$ square, and re-arrange to get \\[x^2+1=\\sqrt{5}x \\implies x^4+2x^2+1=5x^2 \\implies x^4-3x^2+1=0.\\] Now, let us consider the expression we wish to acquire. Factoring out $x^3,$ we have \\[x^3\\left(x^8-7x^4+1\\right) = x^3\\left(x^8+2x^4+1-9x^4\\right).\\] Then, we notice that $x^8+2x^4+1=\\left(x^4+1\\right)^2.$ Furthermore, \\[x^4+1=3x^2 \\implies \\left(x^4+1\\right)^2=x^8+2x^4+1=9x^4.\\] Thus, our answer is \\[x^3\\left(9x^4-9x^4\\right) = x^3 \\cdot 0 = \\boxed{0} ~ 0.\\] ~peace09", "Multiplying by x and solving, we get that $x = \\frac{\\sqrt{5} \\pm 1}{2}.$ Note that whether or not we take $x = \\frac{\\sqrt{5} + 1}{2}$ or we take $\\frac{\\sqrt{5} - 1}{2},$ our answer has to be the same. Thus, we take $x = \\frac{\\sqrt{5} - 1}{2} \\approx 0.62$ . Since this number is small, taking it to high powers like $11$ $7$ , and $3$ will make the number very close to $0$ , so the answer is $\\boxed{0}.$ ~AtharvNaphade" ]

[ "We note that $8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3$ . Therefore, we have that $9x^3 = (x+1)^3$ , so it follows that $x\\sqrt[3]{9} = x+1$ . Solving for $x$ yields $\\frac{1}{\\sqrt[3]{9}-1} = \\frac{\\sqrt[3]{81}+\\sqrt[3]{9}+1}{8}$ , so the answer is $\\boxed{98}$", "Let $r$ be the real root of the given polynomial . Now define the cubic polynomial $Q(x)=-x^3-3x^2-3x+8$ . Note that $1/r$ must be a root of $Q$ . However we can simplify $Q$ as $Q(x)=9-(x+1)^3$ , so we must have that $(\\frac{1}{r}+1)^3=9$ . Thus $\\frac{1}{r}=\\sqrt[3]{9}-1$ , and $r=\\frac{1}{\\sqrt[3]{9}-1}$ . We can then multiply the numerator and denominator of $r$ by $\\sqrt[3]{81}+\\sqrt[3]{9}+1$ to rationalize the denominator, and we therefore have $r=\\frac{\\sqrt[3]{81}+\\sqrt[3]{9}+1}{8}$ , and the answer is $\\boxed{98}$", "It is clear that for the algebraic degree of $x$ to be $3$ that there exists some cubefree integer $p$ and positive integers $m,n$ such that $a = m^3p$ and $b = n^3p^2$ (it is possible that $b = n^3p$ , but then the problem wouldn't ask for both an $a$ and $b$ ). Let $f_1$ be the automorphism over $\\mathbb{Q}[\\sqrt[3]{a}][\\omega]$ which sends $\\sqrt[3]{a} \\to \\omega \\sqrt[3]{a}$ and $f_2$ which sends $\\sqrt[3]{a} \\to \\omega^2 \\sqrt[3]{a}$ (note : $\\omega$ is a cubic root of unity ).\nLetting $r$ be the root, we clearly we have $r + f_1(r) + f_2(r) = \\frac{3}{8}$ by Vieta's formulas. Thus it follows $c=8$ .\nNow, note that $\\sqrt[3]{a} + \\sqrt[3]{b} + 1$ is a root of $x^3 - 3x^2 - 24x - 64 = 0$ . Thus $(x-1)^3 = 27x + 63$ so $(\\sqrt[3]{a} + \\sqrt[3]{b})^3 = 27(\\sqrt[3]{a} + \\sqrt[3]{b}) + 90$ . Checking the non-cubicroot dimension part, we get $a + b = 90$ so it follows that $a + b + c = \\boxed{98}$", "We have $cx-1=\\sqrt[3]{a}+\\sqrt[3]{b}.$ Therefore $(cx-1)^3=(\\sqrt[3]{a}+\\sqrt[3]{b})^3=a+b+3\\sqrt[3]{ab}(\\sqrt[3]{a}+\\sqrt[3]{b})=a+b+3\\sqrt[3]{ab}(cx-1).$ We have \\[c^3x^3-3c^2x^2-(3c\\sqrt[3]{ab}-3c)x-(a+b+1-3\\sqrt[3]{ab})=0.\\] We will find $a,b,c$ so that the equation is equivalent to the original one. Let $\\dfrac{3c^2}{c^3}=\\dfrac{3}{8}, \\dfrac{3c\\sqrt[3]{ab}-3c}{c^3}=\\dfrac{3}{8}, \\dfrac{a+b+1-3\\sqrt[3]{ab}}{c^3}=\\dfrac{1}{8}.$ Easily, $c=8, \\sqrt[3]{ab}=9,$ and $a+b=90.$ So $a + b + c = 90+8=\\boxed{98}$" ]

[ "\nOur triangular pyramid has base $12\\sqrt{3} - 13\\sqrt{3} - 13\\sqrt{3} \\triangle$ . The area of this isosceles triangle is easy to find by $[ACD] = \\frac{1}{2}bh$ , where we can find $h_{ACD}$ to be $\\sqrt{399}$ by the Pythagorean Theorem . Thus $A = \\frac 12(12\\sqrt{3})\\sqrt{399} = 18\\sqrt{133}$\n\nTo find the volume, we want to use the equation $\\frac 13Bh = 6\\sqrt{133}h$ , so we need to find the height of the tetrahedron . By the Pythagorean Theorem, $AP = CP = DP = \\frac{\\sqrt{939}}{2}$ . If we let $P$ be the center of a sphere with radius $\\frac{\\sqrt{939}}{2}$ , then $A,C,D$ lie on the sphere. The cross section of the sphere that contains $A,C,D$ is a circle, and the center of that circle is the foot of the perpendicular from the center of the sphere. Hence the foot of the height we want to find occurs at the circumcenter of $\\triangle ACD$\nFrom here we just need to perform some brutish calculations. Using the formula $A = 18\\sqrt{133} = \\frac{abc}{4R}$ (where $R$ is the circumradius ), we find $R = \\frac{12\\sqrt{3} \\cdot (13\\sqrt{3})^2}{4\\cdot 18\\sqrt{133}} = \\frac{13^2\\sqrt{3}}{2\\sqrt{133}}$ (there are slightly simpler ways to calculate $R$ since we have an isosceles triangle). By the Pythagorean Theorem,\n\\begin{align*}h^2 &= PA^2 - R^2 \\\\ &= \\left(\\frac{\\sqrt{939}}{2}\\right)^2 - \\left(\\frac{13^2\\sqrt{3}}{2\\sqrt{133}}\\right)^2\\\\ &= \\frac{939 \\cdot 133 - 13^4 \\cdot 3}{4 \\cdot 133} = \\frac{13068 \\cdot 3}{4 \\cdot 133} = \\frac{99^2}{133}\\\\ h &= \\frac{99}{\\sqrt{133}} \\end{align*}\nFinally, we substitute $h$ into the volume equation to find $V = 6\\sqrt{133}\\left(\\frac{99}{\\sqrt{133}}\\right) = \\boxed{594}$", "Let $X$ be the apex of the pyramid and $M$ be the midpoint of $\\overline{CD}$ . We find the side lengths of $\\triangle XMP$\n$MP = \\frac{13\\sqrt3}{2}$ $PX$ is half of $AC$ , which is $\\frac{\\sqrt{3\\cdot13^2+3\\cdot12^2}}{2} = \\frac{\\sqrt{939}}{2}$ . To find $MX$ , consider right triangle $XMD$ ; since $XD=13\\sqrt3$ and $MD=6\\sqrt3$ , we have $MX=\\sqrt{3\\cdot13^2-3\\cdot6^2} = \\sqrt{399}$\nLet $\\theta=\\angle XPM$ . For calculating trig, let us double all sides of $\\triangle XMP$ . By Law of Cosines, $\\cos\\theta = \\frac{3\\cdot13^2+939-4\\cdot399}{2\\cdot13\\cdot3\\sqrt{313}}=\\frac{-150}{6\\cdot13\\sqrt{313}}=-\\frac{25}{13\\sqrt{313}}$\nHence, \\[\\sin\\theta = \\sqrt{1-\\cos^2\\theta}\\] \\[=\\sqrt{1-\\frac{625}{13^2\\cdot313}}\\] \\[=\\frac{\\sqrt{13^2\\cdot313-625}}{13\\sqrt{313}}\\] \\[=\\frac{\\sqrt{3\\cdot132^2}}{13\\sqrt{313}}\\] \\[=\\frac{132\\sqrt3}{13\\sqrt{313}}\\] Thus, the height of the pyramid is $PX\\sin\\theta=\\frac{\\sqrt{939}}{2}\\cdot\\frac{132\\sqrt3}{13\\sqrt{313}}=\\frac{66\\cdot3}{13}$ . Since $[CPD]=3\\sqrt{3}\\cdot13\\sqrt{3}=9\\cdot13$ , the volume of the pyramid is $\\frac13 \\cdot 9\\cdot13\\cdot \\frac{66\\cdot3}{13}=\\boxed{594}$" ]

[ "The area of the quadrilateral $ABDF$ is equal to the areas of the two right triangles $\\triangle BCD$ and $\\triangle EFD$ subtracted from the area of the rectangle $ABCD$ . Because $B$ and $F$ are midpoints, we know the dimensions of the two right triangles.\n\n\\[(20)(32)-\\frac{(16)(20)}{2}-\\frac{(10)(32)}{2} = 640-160-160 = \\boxed{320}\\]" ]

[ "In order to solve this problem, we must first visualize what the region looks like. We know that, in a three dimensional space, the region consisting of all points within $3$ units of a point would be a sphere with radius $3$ . However, we need to find the region containing all points within $3$ units of a segment. It can be seen that our region is a cylinder with two hemispheres/endcaps on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal $216 \\pi$ ):\n$\\frac{4 \\pi }{3} \\cdot 3^3+9 \\pi x=216 \\pi$ , where $x$ is equal to the length of our line segment.\nSolving, we find that $x = \\boxed{20}$", "Because this is just a cylinder and $2$ hemispheres (\"half spheres\"), and the radius is $3$ , the volume of the $2$ hemispheres is $\\frac{4(3^3)\\pi}{3} = 36 \\pi$ . Since we also know that the volume of this whole thing is $216 \\pi$ , we do $216-36$ to get $180 \\pi$ as the volume of the cylinder. Thus the height is $180 \\pi$ divided by the area of the base, or $\\frac{180 \\pi}{9\\pi}=20$ , so our answer is $\\boxed{20}.$" ]

[ "First of all, note that $J$ must be $1$ $5$ , or $9$ to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have:\n\nWe also notice that $A+E = B+F = C+G = D+H$\nWLOG, assume that $J = 1$ . Thus the pairs of vertices must be $9$ and $2$ $8$ and $3$ $7$ and $4$ , and $6$ and $5$ . There are $4! = 24$ ways to assign these to the vertices. Furthermore, there are $2^{4} = 16$ ways to switch them (i.e. do $2$ $9$ instead of $9$ $2$ ).\nThus, there are $16(24) = 384$ ways for each possible J value. There are $3$ possible J values that still preserve symmetry: $384(3) = \\boxed{1152}$", "As in solution 1, $J$ must be $1$ $5$ , or $9$ giving us 3 choices. Additionally $A+E = B+F = C+G = D+H$ . This means once we choose $J$ there are $8$ remaining choices. Going clockwise from $A$ we count, $8$ possibilities for $A$ . Choosing $A$ also determines $E$ which leaves $6$ choices for $B$ , once $B$ is chosen it also determines $F$ leaving $4$ choices for $C$ . Once $C$ is chosen it determines $G$ leaving $2$ choices for $D$ . Choosing $D$ determines $H$ , exhausting the numbers. Additionally, there are three possible values for $J$ . To get the answer we multiply $2\\cdot4\\cdot6\\cdot8\\cdot3=\\boxed{1152}$" ]

[ "To determine a remainder when a number is divided by $5$ , you only need to look at the last digit. If the last digit is $0$ or $5$ , the remainder is $0$ . If the last digit is $1$ or $6$ , the remainder is $1$ , and so on.\nTo determine the last digit of $1492\\cdot 1776\\cdot 1812\\cdot 1996$ , you only need to look at the last digit of each number in the product. Thus, we compute $2\\cdot 6\\cdot 2\\cdot 6 = 12^2 = 144$ . The last digit of the number $1492\\cdot 1776\\cdot 1812\\cdot 1996$ is also $4$ , and thus the remainder when the number is divided by $5$ is also $4$ , which gives an answer of $\\boxed{4}$", "$1492\\cdot 1776\\cdot 1812\\cdot 1996 \\mod 5$\n$\\equiv 2\\cdot 1\\cdot 2\\cdot 1 \\mod 5$\n$=4$ , which is option $\\boxed{4}$" ]

[ "Notice repeating decimals can be written as the following:\n$0.\\overline{ab}=\\frac{10a+b}{99}$\n$0.\\overline{abc}=\\frac{100a+10b+c}{999}$\nwhere a,b,c are the digits. Now we plug this back into the original fraction:\n$\\frac{10a+b}{99}+\\frac{100a+10b+c}{999}=\\frac{33}{37}$\nMultiply both sides by $999*99.$ This helps simplify the right side as well because $999=111*9=37*3*9$\n$9990a+999b+9900a+990b+99c=33/37*37*3*9*99=33*3*9*99$\nDividing both sides by $9$ and simplifying gives:\n$2210a+221b+11c=99^2=9801$\nAt this point, seeing the $221$ factor common to both a and b is crucial to simplify. This is because taking $mod 221$ to both sides results in:\n$2210a+221b+11c \\equiv 9801 \\mod 221 \\iff 11c \\equiv 77 \\mod 221$\nNotice that we arrived to the result $9801 \\equiv 77 \\mod 221$ by simply dividing $9801$ by $221$ and seeing $9801=44*221+77.$ Okay, now it's pretty clear to divide both sides by $11$ in the modular equation but we have to worry about $221$ being multiple of $11.$ Well, $220$ is a multiple of $11$ so clearly, $221$ couldn't be. Also, $221=13*17.$ Now finally we simplify and get:\n$c \\equiv 7 \\mod 221$\nBut we know $c$ is between $0$ and $9$ because it is a digit, so $c$ must be $7.$ Now it is straightforward from here to find $a$ and $b$\n$2210a+221b+11(7)=9801 \\iff 221(10a+b)=9724 \\iff 10a+b=44$\nand since a and b are both between $0$ and $9$ , we have $a=b=4$ . Finally we have the $3$ digit integer $\\boxed{447}$", "Note that $\\frac{33}{37}=\\frac{891}{999} = 0.\\overline{891}$ . Also note that the period of $0.abab\\overline{ab}+0.abcabc\\overline{abc}$ is at most $6$ . Therefore, we only need to worry about the sum $0.ababab+ 0.abcabc$ . Adding the two, we get \\[\\begin{array}{ccccccc}&a&b&a&b&a&b\\\\ +&a&b&c&a&b&c\\\\ \\hline &8&9&1&8&9&1\\end{array}\\] From this, we can see that $a=4$ $b=4$ , and $c=7$ , so our desired answer is $\\boxed{447}$", "Noting as above that $0.\\overline{ab} = \\frac{10a + b}{99}$ and $0.\\overline{abc} = \\frac{100a + 10b + c}{999}$ , let $u = 10a + b$ .\nThen \\[\\frac{u}{99} + \\frac{10u + c}{999} = \\frac{33}{37}\\]\n\\[\\frac{u}{11} + \\frac{10u + c}{111} = \\frac{9\\cdot 33}{37}\\]\n\\[\\frac{221u + 11c}{11\\cdot 111} = \\frac{9\\cdot 33}{37}\\]\n\\[221u + 11c = \\frac{9\\cdot 33\\cdot 11\\cdot 111}{37}\\]\n\\[221u + 11c = 9\\cdot 33^2.\\]\nSolving for $c$ gives\n\\[c = 3\\cdot 9\\cdot 33 - \\frac{221u}{11}\\]\n\\[c = 891 - \\frac{221u}{11}\\]\nBecause $c$ must be integer, it follows that $u$ must be a multiple of $11$ (because $221$ clearly is not). Inspecting the equation, one finds that only $u = 44$ yields a digit $c, 7$ . Thus $abc = 10u + c = \\boxed{447}.$", "We note as above that $0.\\overline{ab} = \\frac{10a + b}{99}$ and $0.\\overline{abc} = \\frac{100a + 10b + c}{999},$ so\n\\[\\frac{10a + b}{99} + \\frac{100a + 10b + c}{999} = \\frac{33}{37} = \\frac{891}{999}.\\]\nAs $\\frac{10a + b}{99}$ has a factor of $11$ in the denominator while the other two fractions don't, we need that $11$ to cancel, so $11$ divides $10a + b.$ It follows that $a = b,$ so $\\frac{10a + b}{99} = \\frac{11a}{99} = \\frac{111a}{999},$ so\n\\[\\frac{111a}{999} + \\frac{110a+c}{999} = \\frac{891}{999}.\\]\nThen $111a + 110a + c = 891,$ or $221a + c = 891.$ Thus $a = b = 4$ and $c = 7,$ so the three-digit integer $abc$ is $\\boxed{447}.$" ]

[ "Let $a$ $b$ $c$ be the three roots of the polynomial. The lengthened prism's volume is \\[V = (a+2)(b+2)(c+2) = abc+2ac+2ab+2bc+4a+4b+4c+8 = abc + 2(ab+ac+bc) + 4(a+b+c) + 8.\\] By Vieta's formulas, we know that a cubic polynomial $Ax^3+Bx^2+Cx+D$ with roots $a$ $b$ $c$ satisfies: \\begin{alignat*}{8} a+b+c &= -\\frac{B}{A} &&= \\frac{39}{10}, \\\\ ab+ac+bc &= \\hspace{2mm}\\frac{C}{A} &&= \\frac{29}{10}, \\\\ abc &= -\\frac{D}{A} &&= \\frac{6}{10}. \\end{alignat*} We can substitute these into the expression, obtaining \\[V = \\frac{6}{10} + 2\\left(\\frac{29}{10}\\right) + 4\\left(\\frac{39}{10}\\right) + 8 = \\boxed{30}.\\]", "From the answer choices, we can assume the roots are rational numbers, and therefore this polynomial should be easily factorable.\nThe coefficients of $x$ must multiply to $10$ , so these coefficients must be $5,2,1$ or $10,1,$ in some order. We can try one at a time, and therefore write the factored form as follows: \\[(5x-p)(2x-q)(x-r).\\] Note that $p, q, r$ have to multiply to $6$ , so they must be either $3,2,1$ or $6,1,1$ in some order. Again, we can try one at a time in different positions and see if they multiply correctly.\nWe try $(5x-2)(2x-1)(x-3)$ and multiply the $x-$ terms, and sure enough they add up to $29$ . You can try to add up the $x^2$ terms and they add up to $-39$ . Therefore the roots are $\\frac{2}{5}$ $\\frac{1}{2}$ and $3$ . Now if you add $2$ to each root, you get the volume is $\\frac{12}{5} \\cdot \\frac{5}{2} \\cdot 5 = 6 \\cdot 5 = 30 = \\boxed{30}$", "We can find the roots of the cubic using the Rational Root Theorem, which tells us that the rational roots of the cubic must be in the form $\\frac{p}{q}$ , where $p$ is a factor of the constant $(-6)$ and $q$ is a factor of the leading coefficient $(10)$ . Therefore, $p$ is $\\pm (1, 2, 3, 6)$ and q is $\\pm (1, 2, 5, 10).$\nDoing Synthetic Division, we find that $3$ is a root of the cubic: \\[\\begin{array}{c|rrrr}&10&-39&29&-6\\\\3&&30&-27&6\\\\\\hline\\\\&10&-9&2&0\\\\\\end{array}.\\]\nThen, we have a quadratic $10x^2-9x+2.$ Using the Quadratic Formula, we can find the other two roots: \\[x=\\frac{9 \\pm \\sqrt{(-9)^2-4(10)(2)}}{2 \\cdot 10},\\] which simplifies to $x=\\frac{1}{2}, \\frac{2}{5}.$\nTo find the new volume, we add $2$ to each of the roots we found: \\[(3+2)\\cdot\\left(\\frac{1}{2}+2\\right)\\cdot\\left(\\frac{2}{5}+2\\right).\\] Simplifying, we find that the new volume is $\\boxed{30}.$", "Let $P(x) = 10x^3 - 39x^2 + 29x - 6$ , and let $a, b, c$ be the roots of $P(x)$ . The roots of $P(x-2)$ are then $a + 2, b + 2, c + 2,$ so the product of the roots of $P(x-2)$ is the area of the desired rectangular prism.\n$P(x-2)$ has leading coefficient $10$ and constant term $P(0-2) = P(-2) = 10(-2)^3 - 39(-2)^2 + 29(-2) - 6 = -300$\nThus, by Vieta's Formulas, the product of the roots of $P(x-2)$ is $\\frac{-(-300)}{10} = \\boxed{30}$", "Let $P(x) = 10x^3 - 39x^2 + 29x - 6$ . This can be factored m as $P(x) = 10(x-a)(x-b)(x-c)$ , where $a$ $b$ , and $c$ are the roots of $P(x)$ . We want $V = (a+2)(b+2)(c+2)$\n\"Luckily\" $P(-2) = 10(-2-a)(-2-b)(-2-c) = -10V$ $P(-2) = -300$ , giving $V = \\boxed{30}$", "By Vieta's, we can see that $ABC = \\frac{6}{10}$ .\nUsing this, we can see that if each side $ABC$ is the same length, the length is between $0.8$ $0.512$ ) and $0.9$ $0.729$ ). Adding $2$ to these numbers would give us three numbers that are close to $3$ . Rounding up, we will just assume they are all three. \nIf we multiply all of them, it gives us $27$ .\nThe closest answer choice is $\\boxed{30},$ as all of the other choices are far from this number (the second closest answer choice being $11$ away)." ]

[ "Let $a$ $b$ $c$ be the three roots of the polynomial. The lengthened prism's volume is \\[V = (a+2)(b+2)(c+2) = abc+2ac+2ab+2bc+4a+4b+4c+8 = abc + 2(ab+ac+bc) + 4(a+b+c) + 8.\\] By Vieta's formulas, we know that a cubic polynomial $Ax^3+Bx^2+Cx+D$ with roots $a$ $b$ $c$ satisfies: \\begin{alignat*}{8} a+b+c &= -\\frac{B}{A} &&= \\frac{39}{10}, \\\\ ab+ac+bc &= \\hspace{2mm}\\frac{C}{A} &&= \\frac{29}{10}, \\\\ abc &= -\\frac{D}{A} &&= \\frac{6}{10}. \\end{alignat*} We can substitute these into the expression, obtaining \\[V = \\frac{6}{10} + 2\\left(\\frac{29}{10}\\right) + 4\\left(\\frac{39}{10}\\right) + 8 = \\boxed{30}.\\]", "From the answer choices, we can assume the roots are rational numbers, and therefore this polynomial should be easily factorable.\nThe coefficients of $x$ must multiply to $10$ , so these coefficients must be $5,2,1$ or $10,1,$ in some order. We can try one at a time, and therefore write the factored form as follows: \\[(5x-p)(2x-q)(x-r).\\] Note that $p, q, r$ have to multiply to $6$ , so they must be either $3,2,1$ or $6,1,1$ in some order. Again, we can try one at a time in different positions and see if they multiply correctly.\nWe try $(5x-2)(2x-1)(x-3)$ and multiply the $x-$ terms, and sure enough they add up to $29$ . You can try to add up the $x^2$ terms and they add up to $-39$ . Therefore the roots are $\\frac{2}{5}$ $\\frac{1}{2}$ and $3$ . Now if you add $2$ to each root, you get the volume is $\\frac{12}{5} \\cdot \\frac{5}{2} \\cdot 5 = 6 \\cdot 5 = 30 = \\boxed{30}$", "We can find the roots of the cubic using the Rational Root Theorem, which tells us that the rational roots of the cubic must be in the form $\\frac{p}{q}$ , where $p$ is a factor of the constant $(-6)$ and $q$ is a factor of the leading coefficient $(10)$ . Therefore, $p$ is $\\pm (1, 2, 3, 6)$ and q is $\\pm (1, 2, 5, 10).$\nDoing Synthetic Division, we find that $3$ is a root of the cubic: \\[\\begin{array}{c|rrrr}&10&-39&29&-6\\\\3&&30&-27&6\\\\\\hline\\\\&10&-9&2&0\\\\\\end{array}.\\]\nThen, we have a quadratic $10x^2-9x+2.$ Using the Quadratic Formula, we can find the other two roots: \\[x=\\frac{9 \\pm \\sqrt{(-9)^2-4(10)(2)}}{2 \\cdot 10},\\] which simplifies to $x=\\frac{1}{2}, \\frac{2}{5}.$\nTo find the new volume, we add $2$ to each of the roots we found: \\[(3+2)\\cdot\\left(\\frac{1}{2}+2\\right)\\cdot\\left(\\frac{2}{5}+2\\right).\\] Simplifying, we find that the new volume is $\\boxed{30}.$", "Let $P(x) = 10x^3 - 39x^2 + 29x - 6$ , and let $a, b, c$ be the roots of $P(x)$ . The roots of $P(x-2)$ are then $a + 2, b + 2, c + 2,$ so the product of the roots of $P(x-2)$ is the area of the desired rectangular prism.\n$P(x-2)$ has leading coefficient $10$ and constant term $P(0-2) = P(-2) = 10(-2)^3 - 39(-2)^2 + 29(-2) - 6 = -300$\nThus, by Vieta's Formulas, the product of the roots of $P(x-2)$ is $\\frac{-(-300)}{10} = \\boxed{30}$", "Let $P(x) = 10x^3 - 39x^2 + 29x - 6$ . This can be factored m as $P(x) = 10(x-a)(x-b)(x-c)$ , where $a$ $b$ , and $c$ are the roots of $P(x)$ . We want $V = (a+2)(b+2)(c+2)$\n\"Luckily\" $P(-2) = 10(-2-a)(-2-b)(-2-c) = -10V$ $P(-2) = -300$ , giving $V = \\boxed{30}$", "By Vieta's, we can see that $ABC = \\frac{6}{10}$ .\nUsing this, we can see that if each side $ABC$ is the same length, the length is between $0.8$ $0.512$ ) and $0.9$ $0.729$ ). Adding $2$ to these numbers would give us three numbers that are close to $3$ . Rounding up, we will just assume they are all three. \nIf we multiply all of them, it gives us $27$ .\nThe closest answer choice is $\\boxed{30},$ as all of the other choices are far from this number (the second closest answer choice being $11$ away)." ]

[ "Let the regular price of one tire be $x$ . We have \\begin{align*} 3x+3=240 &\\Rightarrow 3x=237 \\\\ &\\Rightarrow x=79 \\end{align*}\n$\\boxed{79}$ Good Job!" ]

[ "The price Jack rings up is $\\textdollar{(90.00)(1.06)(0.80)}$ . The price Jill rings up is $\\textdollar{(90.00)(0.80)(1.06)}$ . By the commutative property of multiplication, these quantities are the same, and the difference is $\\boxed{0}$" ]

[ "Let $b_i=19\\text{log}_2a_i$ . Then $b_0=0, b_1=1,$ and $b_n=b_{n-1}+2b_{n-2}$ for all $n\\geq 2$ . The characteristic polynomial of this linear recurrence is $x^2-x-2=0$ , which has roots $2$ and $-1$\nTherefore, $b_n=k_12^{n}+k_2(-1)^n$ for constants to be determined $k_1, k_2$ . Using the fact that $b_0=0, b_1=1,$ we can solve a pair of linear equations for $k_1, k_2$\n$k_1+k_2=0$ $2k_1-k_2=1$\nThus $k_1=\\frac{1}{3}$ $k_2=-\\frac{1}{3}$ , and $b_n=\\frac{2^n-(-1)^n}{3}$\nNow, $a_1a_2\\cdots a_k=2^{\\frac{(b_1+b_2+\\cdots+b_k)}{19}}$ , so we are looking for the least value of $k$ so that\n$b_1+b_2+\\cdots+b_k \\equiv 0 \\pmod{19}$\nNote that we can multiply all $b_i$ by three for convenience, as the $b_i$ are always integers, and it does not affect divisibility by $19$\nNow, for all even $k$ the sum (adjusted by a factor of three) is $2^1+2^2+\\cdots+2^k=2^{k+1}-2$ . The smallest $k$ for which this is a multiple of $19$ is $k=18$ by Fermat's Little Theorem, as it is seen with further testing that $2$ is a primitive root $\\pmod{19}$\nNow, assume $k$ is odd. Then the sum (again adjusted by a factor of three) is $2^1+2^2+\\cdots+2^k+1=2^{k+1}-1$ . The smallest $k$ for which this is a multiple of $19$ is $k=17$ , by the same reasons. Thus, the minimal value of $k$ is $\\boxed{17}$", "Since the product $a_1a_2\\cdots a_k$ is an integer, it must be a power of $2$ , so the sum of the base- $2$ logarithms must be an integer. Multiply all of these logarithms by $19$ (to make them integers), so the sum must be a multiple of $19$\nThe logarithms are $b_n = 19\\log_2 a_n$ . Using the recursion $b_0 = 0, b_1 = 1, b_n = b_{n-1}+2b_{n-2}$ (modulo $19$ to save calculation time), we get the sequence \\[0,1,1,3,5,11,2,5,9,0,-1,-1,-3,-5,-11,-2,-5,-9,0,\\dots\\] Listing the numbers out is expedited if you notice $b_{n+1}=2b_n+(-1)^n$\nThe cycle repeats every $9+9=18$ terms. Notice that since $b_n+b_{n+9} \\equiv 0 \\pmod{19}$ , the first $18$ terms sum up to a multiple of $19$ . Since $b_{18}=0$ , we only need at most the first $\\boxed{17}$ terms to sum up to a multiple of $19$ , and this is the lowest answer choice.", "Like in Solution 2 , calculate the first few terms of the sequence, but also keep a running sum $c_n$ of the logarithms (not modulo $19$ here): \\[0,1,2,5,10,21,42,\\dots\\] Notice that $c_n=2c_{n-1}+1$ for odd $n$ and $c_n=2c_{n-1}$ for even $n$ . Since $2$ is relatively prime to $19$ , we can ignore even $n$ and calculate odd $n$ using $c_1 = 1, c_{n} = 4c_{n-2}+1$ (modulo $19$ ): \\[,1,,5,,2,,9,,-1,,-3,,8,,-5,,0,\\dots\\] $c_n$ is first a multiple of $19$ at $n = \\boxed{17}$ . ~ emerald_block" ]

[ "We can now simply start to compute the values $b_i$ by hand:\n\\begin{align*} b_1 & = \\frac 35 \\\\ b_2 & = \\frac 45\\cdot \\frac 35 + \\frac 35 \\sqrt{1 - \\left(\\frac 35\\right)^2} = \\frac{24}{25} \\\\ b_3 & = \\frac 45\\cdot \\frac {24}{25} + \\frac 35 \\sqrt{1 - \\left(\\frac {24}{25}\\right)^2} = \\frac{96}{125} + \\frac 35\\cdot\\frac 7{25} = \\frac{117}{125} \\\\ b_4 & = \\frac 45\\cdot \\frac {117}{125} + \\frac 35 \\sqrt{1 - \\left(\\frac {117}{125}\\right)^2} = \\frac{468}{625} + \\frac 35\\cdot\\frac {44}{125} = \\frac{600}{625} = \\frac{24}{25} \\end{align*}\nWe now discovered that $b_4=b_2$ . And as each $b_{i+1}$ is uniquely determined by $b_i$ , the sequence becomes periodic. In other words, we have $b_3=b_5=b_7=\\cdots=\\frac{117}{125}$ , and $b_2=b_4=\\cdots=b_{10}=\\cdots=\\frac{24}{25}$\nTherefore the answer is\n\\begin{align*} \\lfloor a_{10} \\rfloor & = \\left\\lfloor 2^{10} b_{10} \\right\\rfloor = \\left\\lfloor \\dfrac{1024\\cdot 24}{25} \\right\\rfloor = \\left\\lfloor \\dfrac{1025\\cdot 24}{25} - \\dfrac{24}{25} \\right\\rfloor \\\\ & = \\left\\lfloor 41\\cdot 24 - \\dfrac{24}{25} \\right\\rfloor = 41\\cdot 24 - 1 = \\boxed{983}", "After we do the substitution, we can notice the fact that $\\left( \\frac 35 \\right)^2 + \\left( \\frac 45 \\right)^2 = 1$ , which may suggest that the formula may have something to do with the unit circle. Also, the expression $\\sqrt{1-x^2}$ often appears in trigonometry, for example in the relationship between the sine and the cosine. Both observations suggest that the formula may have a neat geometric interpretation.\nConsider the equation: \\[y = \\frac45 x + \\frac 35\\sqrt{1 - x^2}\\]\nNote that for $t=\\sin^{-1} \\frac 35$ we have $\\sin t=\\frac 35$ and $\\cos t = \\frac 45$ . Now suppose that we have $x=\\sin s$ for some $s$ . Then our equation becomes:\n\\[y=\\cos t \\cdot \\sin s + \\sin t \\cdot |\\cos s|\\]\nDepending on the sign of $\\cos s$ , this is either the angle addition, or the angle subtraction formula for sine. In other words, if $\\cos s \\geq 0$ , then $y=\\sin(s+t)$ , otherwise $y=\\sin(s-t)$\nWe have $b_0=0=\\sin 0$ . Therefore $b_1 = \\sin(0+t) = \\sin t$ $b_2 = \\sin(t+t) = \\sin (2t)$ , and so on. (Remember that $t$ is the constant defined as $t=\\sin^{-1} \\frac 35$ .)\nThis process stops at the first $b_k = \\sin (kt)$ , where $kt$ exceeds $\\frac{\\pi}2$ . Then we'll have $b_{k+1} = \\sin(kt - t) = \\sin ((k-1)t) = b_{k-1}$ and the sequence will start to oscillate.\nNote that $\\sin \\frac{\\pi}6 = \\frac 12 < \\frac 35$ , and $\\sin \\frac{\\pi}4 = \\frac{\\sqrt 2}2 > \\frac 35$ , hence $t$ is strictly between $\\frac{\\pi}6$ and $\\frac{\\pi}4$ . Then $2t\\in\\left(\\frac{\\pi}3,\\frac{\\pi}2 \\right)$ , and $3t\\in\\left( \\frac{\\pi}2, \\frac{3\\pi}4 \\right)$ . Therefore surely $2t < \\frac{\\pi}2 < 3t$\nHence the process stops with $b_3 = \\sin (3t)$ , we then have $b_4 = \\sin (2t) = b_2$ . As in the previous solution, we conclude that $b_{10}=b_2$ , and that the answer is $\\lfloor a_{10} \\rfloor = \\left\\lfloor 2^{10} b_{10} \\right\\rfloor = \\boxed{983}$" ]

[ "The best way to solve this problem is to get the iterated part out of the exponent: \\[5^{(a_{n + 1} - a_n)} = \\frac {1}{n + \\frac {2}{3}} + 1\\] \\[5^{(a_{n + 1} - a_n)} = \\frac {n + \\frac {5}{3}}{n + \\frac {2}{3}}\\] \\[5^{(a_{n + 1} - a_n)} = \\frac {3n + 5}{3n + 2}\\] \\[a_{n + 1} - a_n = \\log_5{\\left(\\frac {3n + 5}{3n + 2}\\right)}\\] \\[a_{n + 1} - a_n = \\log_5{(3n + 5)} - \\log_5{(3n + 2)}\\] Plug in $n = 1, 2, 3, 4$ to see the first few terms of the sequence: \\[\\log_5{5},\\log_5{8}, \\log_5{11}, \\log_5{14}.\\] We notice that the terms $5, 8, 11, 14$ are in arithmetic progression. Since $a_1 = 1 = \\log_5{5} = \\log_5{(3(1) + 2)}$ , we can easily use induction to show that $a_n = \\log_5{(3n + 2)}$ . So now we only need to find the next value of $n$ that makes $\\log_5{(3n + 2)}$ an integer. This means that $3n + 2$ must be a power of $5$ . We test $25$ \\[3n + 2 = 25\\] \\[3n = 23\\] This has no integral solutions, so we try $125$ \\[3n + 2 = 125\\] \\[3n = 123\\] \\[n = \\boxed{041}\\]", "We notice that by multiplying the equation from an arbitrary $a_n$ all the way to $a_1$ , we get: \\[5^{a_n-a_1}=\\dfrac{n+\\tfrac23}{1+\\tfrac23}\\] This simplifies to \\[5^{a_n}=3n+2.\\] We can now test powers of $5$\n$5$ - that gives us $n=1$ , which is useless.\n$25$ - that gives a non-integer $n$\n$125$ - that gives $n=\\boxed{41}$" ]

[ "The sum of the first $1$ numbers is $1$\nThe sum of the next $2$ numbers is $2 + 1$\nThe sum of the next $3$ numbers is $2 + 2 + 1$\nIn general, we can write \"the sum of the next $n$ numbers is $1 + 2(n-1)$ \", where the word \"next\" follows the pattern established above.\nThus, we first want to find what triangular numbers $1234$ is between. By plugging in various values of $n$ into $f(n) = \\frac{n(n+1)}{2}$ , we find:\n$f(50) = 1275$\n$f(49) = 1225$\nThus, we want to add up all those sums from \"next $1$ number\" to the \"next $49$ numbers\", which will give us all the numbers up to and including the $1225^{th}$ number. Then, we can manually tack on the remaining $2$ s to hit $1234$\nWe want to find:\n$\\sum_{n=1}^{49} 1 + 2(n-1)$\n$\\sum_{n=1}^{49} 2n - 1$\n$\\sum_{n=1}^{49} 2n - \\sum_{n=1}^{49} 1$\n$2 \\sum_{n=1}^{49} n - 49$\n$2\\cdot \\frac{49\\cdot 50}{2} - 49$\n$49^2$\n$2401$\nThus, the sum of the first $1225$ terms is $2401$ . We have to add $9$ more $2$ s to get to the $1234^{th}$ term, which gives us $2419$ , or option $\\boxed{2419}$" ]

[ "The $k$ th appearance of 1 is at position $1 + 2 + \\dots + k = \\frac{k(k + 1)}{2}$ . Then there are $k$ 1's and $\\frac{k(k + 1)}{2} - k = \\frac{k(k - 1)}{2}$ 2's among the first $\\frac{k(k + 1)}{2}$ numbers, so the sum of these $\\frac{k(k + 1)}{2}$ terms is $k + k(k - 1) = k^2$\nWhen $k = 49$ $\\frac{k(k + 1)}{2} = 1225$ , and when $k = 50$ $\\frac{k(k + 1)}{2} = 1275$\nThe sum of the first 1225 terms is $49^2 = 2401$ . The numbers in positions 1226 through 1234 are all 2's, so their sum is $(1234 - 1226 + 1) \\cdot 2 = 18$ . Therefore, the sum of the first 1234 terms is $2401 + 18 = \\boxed{2419}$" ]

[ "$S_9 = 110$ $S_7 = 42$\n$S_8 = S_9 - S_ 7 = 110 - 42 = 68$\n$S_6 = S_8 - S_7 = 68 - 42 = 26$\n$S_5 = S_7 - S_6 = 42 - 26 = 16$\n$S_4 = S_6 - S_5 = 26 - 16 = 10$\nTherefore, the answer is $\\boxed{10}$" ]

[ "Rearranging the definitions, we have \\[\\frac{a_n}{a_{n-1}} = \\frac{a_{n-1}}{a_{n-2}} + 1,\\quad \\frac{b_n}{b_{n-1}} = \\frac{b_{n-1}}{b_{n-2}} + 1\\] from which it follows that $\\frac{a_n}{a_{n-1}} = 1+ \\frac{a_{n-1}}{a_{n-2}} = \\cdots = (n-1) + \\frac{a_{1}}{a_0} = n$ and $\\frac{b_n}{b_{n-1}} = (n-1) + \\frac{b_{1}}{b_0} = n+2$ . These recursions $a_{n} = na_{n-1}$ and $b_{n} = (n+2)b_{n-1}$ , respectively, correspond to the explicit functions $a_n = n!$ and $b_n = \\frac{(n+2)!}{2}$ (after applying our initial conditions). It follows that $\\frac{b_{32}}{a_{32}} = \\frac{\\frac{34!}{2}}{32!} = \\frac{34 \\cdot 33}{2} = \\boxed{561}$" ]

[ "We can rewrite the given equation as $2^{a_7-27}=a_7$ . Hence, $a_7$ must be a power of $2$ and larger than $27$ . The first power of 2 that is larger than $27$ , namely $32$ , does satisfy the equation: $2^{32 - 27} = 2^5 = 32$ . In fact, this is the only solution; $2^{a_7-27}$ is exponential whereas $a_7$ is linear, so their graphs will not intersect again.\nNow, let the common difference in the sequence be $d$ . Hence, $a_0 = 32 - 7d$ and $a_2 = 32 - 5d$ . To minimize $a_2$ , we maxmimize $d$ . Since the sequence contains only positive integers, $32 - 7d > 0$ and hence $d \\leq 4$ . When $d = 4$ $a_2 = \\boxed{12}$" ]

[ "\\[\\log_8 a_1+\\log_8 a_2+\\ldots+\\log_8 a_{12}= \\log_8 a+\\log_8 (ar)+\\ldots+\\log_8 (ar^{11}) \\\\ = \\log_8(a\\cdot ar\\cdot ar^2\\cdot \\cdots \\cdot ar^{11}) = \\log_8 (a^{12}r^{66})\\]\nSo our question is equivalent to solving $\\log_8 (a^{12}r^{66})=2006$ for $a, r$ positive integers $a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}$ so $a^{2}r^{11}=2^{1003}$\nThe product of $a^2$ and $r^{11}$ is a power of 2. Since both numbers have to be integers, this means that $a$ and $r$ are themselves powers of 2. Now, let $a=2^x$ and $r=2^y$\n\\begin{eqnarray*}(2^x)^2\\cdot(2^y)^{11}&=&2^{1003}\\\\ 2^{2x}\\cdot 2^{11y}&=&2^{1003}\\\\ 2x+11y&=&1003\\\\ y&=&\\frac{1003-2x}{11} \\end{eqnarray*}\nFor $y$ to be an integer, the numerator must be divisible by $11$ . This occurs when $x=1$ because $1001=91*11$ . Because only even integers are being subtracted from $1003$ , the numerator never equals an even multiple of $11$ . Therefore, the numerator takes on the value of every odd multiple of $11$ from $11$ to $1001$ . Since the odd multiples are separated by a distance of $22$ , the number of ordered pairs that work is $1 + \\frac{1001-11}{22}=1 + \\frac{990}{22}=46$ . (We must add 1 because both endpoints are being included.) So the answer is $\\boxed{046}$", "Using the above method, we can derive that $a^{2}r^{11} = 2^{1003}$ .\nNow, think about what happens when r is an even power of 2. Then $a^{2}$ must be an odd power of 2 in order to satisfy the equation which is clearly not possible. Thus the only restriction r has is that it must be an odd power of 2, so $2^{1}$ $2^{3}$ $2^{5}$ .... all work for r, until r hits $2^{93}$ , when it gets greater than $2^{1003}$ , so the greatest value for r is $2^{91}$ . All that's left is to count the number of odd integers between 1 and 91 (inclusive), which yields $\\boxed{046}$", "Using the method from Solution 1, we get $\\log_8a^{12}r^{66}=2006 \\implies a^{12}r^{66}=8^{2006}=2^{6018}$\nSince $a$ and $r$ both have to be powers of $2$ , we can rewrite this as $12x+66y=6018$\n$6018 \\equiv 66 \\equiv 6\\pmod{12}$ . So, when we subtract $12$ from $6018$ , the result is divisible by $66$ . Evaluating that, we get $(1,91)$ as a valid solution. Since $66 \\cdot 2 = 12 \\cdot 11$ , when we add $11$ to the value of $a$ , we can subtract $2$ from the value of $r$ to keep the equation valid. Using this, we get $(1,91),(12,89),(23,87), \\cdots (541,1)$ . In order to count the number of ordered pairs, we can simply count the number of $y$ values. Every odd number from $1$ to $91$ is included, so we have $\\boxed{046}$ solutions." ]

[ "Let $m$ be the arithmetic mean of $a_1$ and $a_2$ . We can then write $a_1=m-x$ and $a_2=m+x$ for some $x$\nBy definition, $a_3=m$\nNext, $a_4$ is the mean of $m-x$ $m+x$ and $m$ , which is again $m$\nRealizing this, one can easily prove by induction that $\\forall n\\geq 3;~ a_n=m$\nIt follows that $m=a_9=99$ . From $19=a_1=m-x$ we get that $x=80$ . And thus $a_2 = m+x = \\boxed{179}$", "Let $a_1=a$ and $a_2=b$ . Then, $a_3=\\frac{a+b}{2}$ $a_4=\\frac{a+b+\\frac{a+b}{2}}{3}=\\frac{a+b}{2},$ and so on.\nSince $a_3=a_4$ $a_n=a_3$ for all $n\\geq3.$\nHence, $a_9=\\frac{a_1+a_2}{2}=\\frac{a+b}{2}=99, a+b=198.$ We also know that $a_1=a=19.$\nSubtracting $a_1$ from $198,$ we get $b=a_2=\\boxed{179}.$" ]

[ "Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for $b_2$ . When we get to $b_2=9$ and $a_2=91$ , we have $a_4=271$ and $b_4=729$ , which works, therefore, the answer is $b_3+a_3=81+181=\\boxed{262}$", "Using the same reasoning ( $100$ isn't very big), we can guess which terms will work. The first case is $k=3$ , so we assume the second and fourth terms of $c$ are $100$ and $1000$ . We let $r$ be the common ratio of the geometric sequence and write the arithmetic relationships in terms of $r$\nThe common difference is $100-r - 1$ , and so we can equate: $2(99-r)+100-r=1000-r^3$ . Moving all the terms to one side and the constants to the other yields\n$r^3-3r = 702$ , or $r(r^2-3) = 702$ . Simply listing out the factors of $702$ shows that the only factor $3$ less than a square that works is $78$ . Thus $r=9$ and we solve from there to get $\\boxed{262}$", "The reason for bashing in this context can also be justified by the fact 100 isn't very big.\nLet the common difference for the arithmetic sequence be $a$ , and the common ratio for the geometric sequence be $b$ . The sequences are now $1, a+1, 2a+1, \\ldots$ , and $1, b, b^2, \\ldots$ . We can now write the given two equations as the following:\n$1+(k-2)a+b^{k-2} = 100$\n$1+ka+b^k = 1000$\nTake the difference between the two equations to get $2a+(b^2-1)b^{k-2} = 900$ . Since 900 is divisible by 4, we can tell $a$ is even and $b$ is odd. Let $a=2m$ $b=2n+1$ , where $m$ and $n$ are positive integers. Substitute variables and divide by 4 to get:\n$m+(n+1)(n)(2n+1)^{k-2} = 225$\nBecause very small integers for $n$ yield very big results, we can bash through all cases of $n$ . Here, we set an upper bound for $n$ by setting $k$ as 3. After trying values, we find that $n\\leq 4$ , so $b=9, 7, 5, 3$ . Testing out $b=9$ yields the correct answer of $\\boxed{262}$ . Note that even if this answer were associated with another b value like $b=3$ , the value of $k$ can still only be 3 for all of the cases." ]

[ "Consider any square that meets the requirements described in the problem. Then, take the vertices of the square and translate them to the first quadrant (This is the \"mapping\" described in Solution 2). For example, consider a square with vertices $(7, 7), (-7, 7), (-7, -7),$ and $(7, -7)$\n\nAfter following the mapping described in Solution 2, the square looks like this:\n\nThe position of each vertex within their corresponding grid has not changed.For example, the point $(-7, 7)$ is still the top-left point in a grid, albeit a change in quadrant. Trying this out with a couple of other squares, we see that the following property holds:\n\\[\\text{\"For any and all squares with vertices in each of the four quadrants, the 'mapped' version is also a square.\"}\\]\nTherefore the logical inverse is true:\n\\[\\text{\"For any 'mapped' square with all four vertices in the first quadrant,}\\] \\[\\text{there exists at least one corresponding 'unmapped' square with a vertex in each of the four quadrants.\"}\\]\nBut how many \"unmapped\" squares, to be exact?\nThis might seem complicated at first, but with some intuitive thinking, we realize that there are exactly $4$ \"unmapped\" squares that correspond with a \"mapped\" square. This is because given a \"mapped\" square, there are $4$ choices for the vertex that will remain in the first quadrant; but once that point is chosen, there is only $1$ distribution of the other $3$ vertices that will result in a square. So, we want four times the number of squares we can make in the first quadrant grid.\nWe divide our counting method into two cases: squares with side length $0$ after mapping (which means all four vertices are in the same position relative to their own grids) and squares with side length $1-4$ after mapping.\nCase 1: There are $25$ such squares of length $0$ (this is equivalent to counting the number of points on the grid). However, in this scenario, all of the vertices have been mapped onto the same point. So instead of $4$ choices for the first quadrant vertex, there is only one. Subsequently there are only $25$ such squares that correspond to them.\nCase 2: Let a square with sides parallel to the axes be known as $A$ squares. These $A$ squares can have side length $1, 2, 3,$ or $4$ . However, the number of $A$ squares possible depends on the side length. For example, there is only $1$ possible $A$ square of side length $4$ , but $16$ squares of side length $1$ . To be exact, there are $(5-s)^2$ possible $A$ squares of side length $s$ . So, the total number of $A$ squares is\n$\\sum_{s=1}^4 (5-s)^2$\nBut what about \"tilted\" squares? Notice that \"tilted squares\" can always be inscribed (drawn within) another, bigger square. Let a square inscribed within an $A$ square be called a $B$ square. How many $B$ squares are there? Well, this also depends on the side length. We only want squares whose vertices are lattice points (integer value coordinates), so the number of $B$ squares should increase along with side length. We defined $B$ squares to be inscribed within $A$ squares, so we can say that all $B$ squares have their vertices on the side on an $A$ square. Consider an $A$ square with side length $4$ . There are $3$ other lattice points along the side of the $A$ square, not counting the vertices. Therefore, we can say that there are $s-1$ possible $B$ squares for every $A$ square with side length $s$ . We can multiply $(s-1)$ times the number of $A$ squares to get the number of $B$ squares. This is\n$\\sum_{s=1}^4 (s-1)(5-s)^2$\ntotal $B$ squares. But we need to add these two quantities to get the number of squares for Case 2:\n$\\sum_{s=1}^4 (s-1)(5-s)^2 + \\sum_{s=1}^4 (5-s)^2$\nBy distributive property, the expression becomes\n$\\sum_{s=1}^4 s(5-s)^2$\nSolving, we get $50$ \"mapped\" squares, both $A$ and $B$ . Multiplying this by $4$ to get the corresponding number of \"unmapped\" squares, then adding to get the number of squares for Case 1, we get\n$50*4 + 25 = 225 \\Rightarrow \\boxed{225}$" ]

[ "We need to find a reasonably easy way to count the squares.\nFirst, obviously the maximum distance between two points in the same quadrant is $4\\sqrt 2 < 6$ , hence each square has exactly one vertex in each quadrant.\nGiven any square, we can circumscribe another axes-parallel square around it. In the picture below, the original square is red and the circumscribed one is blue.\n\nLet's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it?\nAnswering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations.\n\nThe size of the blue square can range from $6\\times 6$ to $14\\times 14$ , and for the intermediate sizes, there is more than one valid placement. We will now examine the cases one after another. Also, we can use symmetry to reduce the number of cases.\nSumming the last column, we get that the answer is $\\boxed{225}$" ]

[ "The least common multiple of $18$ and $48$ is $144$ , so define $n = e^{2\\pi i/144}$ . We can write the numbers of set $A$ as $\\{n^8, n^{16}, \\ldots n^{144}\\}$ and of set $B$ as $\\{n^3, n^6, \\ldots n^{144}\\}$ $n^x$ can yield at most $144$ different values. All solutions for $zw$ will be in the form of $n^{8k_1 + 3k_2}$\n$8$ and $3$ are relatively prime, and by the Chicken McNugget Theorem , for two relatively prime integers $a,b$ , the largest number that cannot be expressed as the sum of multiples of $a,b$ is $a \\cdot b - a - b$ . For $3,8$ , this is $13$ ; however, we can easily see that the numbers $145$ to $157$ can be written in terms of $3,8$ . Since the exponents are of roots of unities, they reduce $\\mod{144}$ , so all numbers in the range are covered. Thus the answer is $\\boxed{144}$", "The 18 and 48th roots of $1$ can be found by De Moivre's Theorem . They are $\\text{cis}\\,\\left(\\frac{2\\pi k_1}{18}\\right)$ and $\\text{cis}\\,\\left(\\frac{2\\pi k_2}{48}\\right)$ respectively, where $\\text{cis}\\,\\theta = \\cos \\theta + i \\sin \\theta$ and $k_1$ and $k_2$ are integers from $0$ to $17$ and $0$ to $47$ , respectively.\n$zw = \\text{cis}\\,\\left(\\frac{\\pi k_1}{9} + \\frac{\\pi k_2}{24}\\right) = \\text{cis}\\,\\left(\\frac{8\\pi k_1 + 3 \\pi k_2}{72}\\right)$ . Since the trigonometric functions are periodic every $2\\pi$ , there are at most $72 \\cdot 2 = \\boxed{144}$ distinct elements in $C$ . As above, all of these will work.", "The values in polar form will be $(1, 20x)$ and $(1, 7.5x)$ . Multiplying these gives $(1, 27.5x)$ . Then, we get $27.5$ $55$ $82.5$ $110$ $...$ up to $3960$ $(\\text{lcm}(55,360)) \\implies \\frac{3960 \\cdot 2}{55}=\\boxed{144}$" ]

[ "In the above diagram, we focus on the line that appears closest and is parallel to $BC$ . All the blue lines are perpendicular lines to $BC$ and their other points are on $AB$ , the main diagonal. The green lines are projections of the blue lines onto the bottom face; all of the green lines originate in the corner and reach out to $AC$ , and have the same lengths as their corresponding blue lines. So we want to find the shortest distance between $AC$ and that corner, which is $\\frac {wl}{\\sqrt {w^2 + l^2}}$\nSo we have: \\[\\frac {lw}{\\sqrt {l^2 + w^2}} = \\frac {10}{\\sqrt {5}}\\] \\[\\frac {hw}{\\sqrt {h^2 + w^2}} = \\frac {30}{\\sqrt {13}}\\] \\[\\frac {hl}{\\sqrt {h^2 + l^2}} = \\frac {15}{\\sqrt {10}}\\]\nNotice the familiar roots: $\\sqrt {5}$ $\\sqrt {13}$ $\\sqrt {10}$ , which are $\\sqrt {1^2 + 2^2}$ $\\sqrt {2^2 + 3^2}$ $\\sqrt {1^2 + 3^2}$ , respectively. (This would give us the guess that the sides are of the ratio 1:2:3, but let's provide the complete solution.)\n\\[\\frac {l^2w^2}{l^2 + w^2} = \\frac {1}{\\frac {1}{l^2} + \\frac {1}{w^2}} = 20\\] \\[\\frac {h^2w^2}{h^2 + w^2} = \\frac {1}{\\frac {1}{h^2} + \\frac {1}{w^2}} = \\frac {900}{13}\\] \\[\\frac {h^2l^2}{h^2 + l^2} = \\frac {1}{\\frac {1}{h^2} + \\frac {1}{l^2}} = \\frac {45}{2}\\]\nWe invert the above equations to get a system of linear equations in $\\frac {1}{h^2}$ $\\frac {1}{l^2}$ , and $\\frac {1}{w^2}$ \\[\\frac {1}{l^2} + \\frac {1}{w^2} = \\frac {45}{900}\\] \\[\\frac {1}{h^2} + \\frac {1}{w^2} = \\frac {13}{900}\\] \\[\\frac {1}{h^2} + \\frac {1}{l^2} = \\frac {40}{900}\\]\nWe see that $h = 15$ $l = 5$ $w = 10$ . Therefore $V = 5 \\cdot 10 \\cdot 15 = \\boxed{750}$" ]

[ "By the Median Formula $PM = \\frac12\\sqrt{2PQ^2+2PR^2-QR^2}$\nPlugging in the numbers given in the problem, we get \\[\\frac72=\\frac12\\sqrt{2\\cdot4^2+2\\cdot7^2-QR^2}\\]\nSolving, \\[7=\\sqrt{2(16)+2(49)-QR^2}\\] \\[49=32+98-QR^2\\] \\[QR^2=81\\] \\[QR=9=\\boxed{9}\\]" ]

[ "Let the shorter segment be $x$ and the altitude be $y$ . The larger segment is then $80-x$ . By the Pythagorean Theorem \\[30^2-y^2=x^2 \\qquad(1)\\] and \\[(80-x)^2=70^2-y^2 \\qquad(2)\\] Adding $(1)$ and $(2)$ and simplifying gives $x=15$ . Therefore, the answer is $80-15=\\boxed{65}$" ]

[ "By Triangle Inequality $6.5 + s >10$ and therefore $s>3.5$ . The smallest whole number that satisfies this is $\\boxed{4}$" ]

[ "Since $6$ and $4$ are fixed sides, the smallest possible side has to be larger than $6-4=2$ and the largest possible side has to be smaller than $6+4=10$ . This gives us the triangle inequality $2<x<10$ and $2<y<10$ $7$ can be attained by letting $x=9.1$ and $y=2.1$ . However, $8=10-2$ cannot be attained. Thus, the answer is $\\boxed{8}$" ]

[ "Consider points on the complex plane $A (0,0),\\ B (11,0),\\ C (11,10),\\ D (0,10)$ . Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one vertex of the triangle at $A$ , and the other two points $E$ and $F$ on $BC$ and $CD$ , respectively. Let $E (11,a)$ and $F (b, 10)$ . Since it's equilateral, then $E\\cdot\\text{cis}60^{\\circ} = F$ , so $(11 + ai)\\left(\\frac {1}{2} + \\frac {\\sqrt {3}}{2}i\\right) = b + 10i$ , and expanding we get $\\left(\\frac {11}{2} - \\frac {a\\sqrt {3}}{2}\\right) + \\left(\\frac {11\\sqrt {3}}{2} + \\frac {a}{2}\\right)i = b + 10i$\nWe can then set the real and imaginary parts equal, and solve for $(a,b) = (20 - 11\\sqrt {3},22 - 10\\sqrt {3})$ . Hence a side $s$ of the equilateral triangle can be found by $s^2 = AE^2 = a^2 + AB^2 = 884 - 440\\sqrt{3}$ . Using the area formula $\\frac{s^2\\sqrt{3}}{4}$ , the area of the equilateral triangle is $\\frac{(884-440\\sqrt{3})\\sqrt{3}}{4} = 221\\sqrt{3} - 330$ . Thus $p + q + r = 221 + 3 + 330 = \\boxed{554}$", "Since $\\angle{BAD}=90$ and $\\angle{EAF}=60$ , it follows that $\\angle{DAF}+\\angle{BAE}=90-60=30$ . Rotate triangle $ADF$ $60$ degrees clockwise. Note that the image of $AF$ is $AE$ . Let the image of $D$ be $D'$ . Since angles are preserved under rotation, $\\angle{DAF}=\\angle{D'AE}$ . It follows that $\\angle{D'AE}+\\angle{BAE}=\\angle{D'AB}=30$ . Since $\\angle{ADF}=\\angle{ABE}=90$ , it follows that quadrilateral $ABED'$ is cyclic with circumdiameter $AE=s$ and thus circumradius $\\frac{s}{2}$ . Let $O$ be its circumcenter. By Inscribed Angles, $\\angle{BOD'}=2\\angle{BAD'}=60$ . By the definition of circle, $OB=OD'$ . It follows that triangle $OBD'$ is equilateral. Therefore, $BD'=r=\\frac{s}{2}$ . Applying the Law of Cosines to triangle $ABD'$ $\\frac{s}{2}=\\sqrt{10^2+11^2-(2)(10)(11)(\\cos{30})}$ . Squaring and multiplying by $\\sqrt{3}$ yields $\\frac{s^2\\sqrt{3}}{4}=221\\sqrt{3}-330\\implies{p+q+r=221+3+330=\\boxed{554}$" ]

[ "By the triangle inequality in $\\triangle ABC$ , we find that $BC$ and $CA$ must sum to greater than $41$ , so they must be (in some order) $7$ and $36$ $13$ and $36$ $18$ and $27$ $18$ and $36$ , or $27$ and $36$ . We try $7$ and $36$ , and now by the triangle inequality in $\\triangle ABD$ , we must use the remaining numbers $13$ $18$ , and $27$ to get a sum greater than $41$ , so the only possibility is $18$ and $27$ . This works as we can put $BC = 36$ $AC = 7$ $AD = 18$ $BD = 27$ $CD = 13$ , so that $\\triangle ADC$ and $\\triangle BDC$ also satisfy the triangle inequality. Hence we have found a solution that works, and it can be verified that the other possibilities don't work, though as this is a multiple-choice competition, you probably wouldn't do that in order to save time. In any case, the answer is $CD = 13$ , which is $\\boxed{13}$" ]

[ "First, modulo $2$ or $5$ $\\underline{20210A} \\equiv A$ .\nHence, $A \\neq 0, 2, 4, 5, 6, 8$\nSecond modulo $3$ $\\underline{20210A} \\equiv 2 + 0 + 2 + 1 + 0 + A \\equiv 5 + A$ .\nHence, $A \\neq 1, 4, 7$\nThird, modulo $11$ $\\underline{20210A} \\equiv A + 1 + 0 - 0 - 2 - 2 \\equiv A - 3$ .\nHence, $A \\neq 3$\nTherefore, the answer is $\\boxed{9}$", "Any number ending in $5$ is divisible by $5$ . So we can eliminate option $\\textbf{(C)}$\nIf the sum of the digits of a number is divisible by $3$ , the number is divisible by $3$ . The sum of the digits of this number is $2 + 0 + 2 + 1 + 0 + A = 5 + A$ . If $5 + A$ is divisible by $3$ , the number is divisible by $3$ . Thus we can eliminate options $\\textbf{(A)}$ and $\\textbf{(D)}$\nSo the correct option is either $\\textbf{(B)}$ or $\\textbf{(E)}$ . Let's try dividing the number with some integers.\n$20210A/7 = 2887x$ , where $x$ is $1A/7$ . Since $13$ and $19$ are both indivisible by $7$ , this does not help us narrow the choices down.\n$20210A/11 = 1837x$ , where $x$ is $3A/11$ . Since $33/11 = 3$ , option $\\textbf{(B)}$ would make $20210A$ divisible by $11$ . Thus, by elimination, the correct choice must be option $\\boxed{9}$", "$202100 \\implies$ divisible by $2$\n$202101 \\implies$ divisible by $3$\n$202102 \\implies$ divisible by $2$\n$202103 \\implies$ divisible by $11$\n$202104 \\implies$ divisible by $2$\n$202105 \\implies$ divisible by $5$\n$202106 \\implies$ divisible by $2$\n$202107 \\implies$ divisible by $3$\n$202108 \\implies$ divisible by $2$\nThis leaves only $A=\\boxed{9}$" ]

[ "First, modulo $2$ or $5$ $\\underline{20210A} \\equiv A$ .\nHence, $A \\neq 0, 2, 4, 5, 6, 8$\nSecond modulo $3$ $\\underline{20210A} \\equiv 2 + 0 + 2 + 1 + 0 + A \\equiv 5 + A$ .\nHence, $A \\neq 1, 4, 7$\nThird, modulo $11$ $\\underline{20210A} \\equiv A + 1 + 0 - 0 - 2 - 2 \\equiv A - 3$ .\nHence, $A \\neq 3$\nTherefore, the answer is $\\boxed{9}$", "Any number ending in $5$ is divisible by $5$ . So we can eliminate option $\\textbf{(C)}$\nIf the sum of the digits of a number is divisible by $3$ , the number is divisible by $3$ . The sum of the digits of this number is $2 + 0 + 2 + 1 + 0 + A = 5 + A$ . If $5 + A$ is divisible by $3$ , the number is divisible by $3$ . Thus we can eliminate options $\\textbf{(A)}$ and $\\textbf{(D)}$\nSo the correct option is either $\\textbf{(B)}$ or $\\textbf{(E)}$ . Let's try dividing the number with some integers.\n$20210A/7 = 2887x$ , where $x$ is $1A/7$ . Since $13$ and $19$ are both indivisible by $7$ , this does not help us narrow the choices down.\n$20210A/11 = 1837x$ , where $x$ is $3A/11$ . Since $33/11 = 3$ , option $\\textbf{(B)}$ would make $20210A$ divisible by $11$ . Thus, by elimination, the correct choice must be option $\\boxed{9}$", "$202100 \\implies$ divisible by $2$\n$202101 \\implies$ divisible by $3$\n$202102 \\implies$ divisible by $2$\n$202103 \\implies$ divisible by $11$\n$202104 \\implies$ divisible by $2$\n$202105 \\implies$ divisible by $5$\n$202106 \\implies$ divisible by $2$\n$202107 \\implies$ divisible by $3$\n$202108 \\implies$ divisible by $2$\nThis leaves only $A=\\boxed{9}$" ]

[ "To get the smallest possible product, we want to multiply the smallest negative number by the largest positive number. These are $-7$ and $3$ , respectively, and their product is $-21$ , which is $\\boxed{21}$" ]

[ "To find the smallest sum, we just have to find the smallest 3 numbers and add them together.\nObviously, the numbers are $-3, -1, 7$ , and adding them gets us $3$\n$\\boxed{3}$" ]

[ "First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$ . The hypotenuse of the triangle formed is the median of equilateral triangle $ADE$ , and one of the legs is $3\\sqrt{2}$ . We apply the Pythagorean Theorem to deduce that the height is $6$\nNext, we complete t he figure into a triangular prism, and find its volume, which is $\\frac{6\\sqrt{2}\\cdot 12\\sqrt{2}\\cdot 6}{2}=432$\nNow, we subtract off the two extra pyramids that we included, whose combined volume is $2\\cdot \\left( \\frac{6\\sqrt{2}\\cdot 3\\sqrt{2} \\cdot 6}{3} \\right)=144$\nThus, our answer is $432-144=\\boxed{288}$", "Extend $EA$ and $FB$ to meet at $G$ , and $ED$ and $FC$ to meet at $H$ . Now, we have a regular tetrahedron $EFGH$ , which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s = 12\\sqrt{2}$ . Using the formula for the volume of a regular tetrahedron, which is $V = \\frac{\\sqrt{2}S^3}{12}$ , where S is the side length of the tetrahedron, the volume of our original solid is:\n$V = \\frac{1}{2} \\cdot \\frac{\\sqrt{2} \\cdot (12\\sqrt{2})^3}{12} = \\boxed{288}$", "We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$ ; thus, we will integrate with respect to height from $0$ to $6$ , noting that each cross section of height $dh$ is a rectangle. The volume is then $\\int_0^h(wl) \\ \\text{d}h$ , where $w$ is the width of the rectangle and $l$ is the length. We can express $w$ in terms of $h$ as $w=6\\sqrt{2}-\\sqrt{2}h$ since it decreases linearly with respect to $h$ , and $l=6\\sqrt{2}+\\sqrt{2}h$ since it similarly increases linearly with respect to $h$ . Now we solve: \\[\\int_0^6(6\\sqrt{2}-\\sqrt{2}h)(6\\sqrt{2}+\\sqrt{2}h)\\ \\text{d}h =\\int_0^6(72-2h^2)\\ \\text{d}h=72(6)-2\\left(\\frac{1}{3}\\right)\\left(6^3\\right)=\\boxed{288}\\]", "Draw an altitude from a vertex of the square base to the top edge. By using $30,60, 90$ triangle ratios, we obtain that the altitude has a length of $3 \\sqrt{6}$ , and that little portion that hangs out has a length of $3\\sqrt2$ . This is a triangular pyramid with a base of $3\\sqrt6, 3\\sqrt6, 3\\sqrt2$ , and a height of $3\\sqrt{2}$ . Since there are two of these, we can compute the sum of the volumes of these two to be $72$ . Now we are left with a triangular prism with a base of dimensions $3\\sqrt6, 3\\sqrt6, 3\\sqrt2$ and a height of $6\\sqrt2$ . We can compute the volume of this to be 216, and thus our answer is $\\boxed{288}$" ]

[ "First, square both sides. This gives us\n\\[\\sqrt{5x-1}^2+2\\cdot\\sqrt{5x-1}\\cdot\\sqrt{x-1}+\\sqrt{x-1}^2=4 \\Longrightarrow 5x-1+2\\cdot\\sqrt{(5x-1)\\cdot(x-1)}+x-1=4 \\Longrightarrow 2\\cdot\\sqrt{5x^2-6x+1}+6x-2=4\\] Then, adding $-6x$ to both sides gives us\n\\[2\\cdot\\sqrt{5x^2-6x+1}+6x-2-6x=4-6x \\Longrightarrow 2\\cdot\\sqrt{5x^2-6x+1}-2 =-6x+4\\] After that, adding $2$ to both sides will give us\n\\[2\\cdot\\sqrt{5x^2-6x+1}-2+2=-6x+4+2 \\Longrightarrow 2\\cdot\\sqrt{5x^2-6x+1}=-6x+6\\] Next, we divide both sides by 2 which gives us\n\\[\\frac{2\\cdot\\sqrt{5x^2-6x+1}-2+2}{2}=\\frac{-6x+4+2}{2} \\Longrightarrow \\sqrt{5x^2-6x+1}=-3x+3\\] Finally, solving the equation, we get\n\\[5x^2-6x+1=(-3x+3)^2 \\Longrightarrow 5x^2-6x+1=9x^2-18x+9\\] \\[\\Longrightarrow 5x^2-6x+1-(9x^2-18x+9)=9x^2-18x+9-(9x^2-18x+9)\\] \\[\\Longrightarrow -4x^2+12x-8=0 \\Longrightarrow -4(x-1)(x-2)=0\\] \\[\\Longrightarrow x-1=0 \\text{or}\\ x-2=0 \\Longrightarrow x=1 \\text{or}\\ x=2\\] Plugging 1 and 2 into the original equation, $\\sqrt{5x-1}+\\sqrt{x-1}=2$ , we see that when $x=1$\n\\[\\sqrt{5x-1}+\\sqrt{x-1}=2 \\Longrightarrow \\sqrt{5\\cdot1-1}+\\sqrt{1-1}=2 \\Longrightarrow \\sqrt4+\\sqrt0=2 \\Longrightarrow 2+0=2 \\Longrightarrow 2=2\\] the equation is true. On the other hand, we note that when $x=2$\n\\[\\sqrt{5x-1}+\\sqrt{x-1}=2 \\Longrightarrow \\sqrt{5\\cdot2-1}+\\sqrt{2-1}=2 \\Longrightarrow \\sqrt9+\\sqrt1=2 \\Longrightarrow 3+1=2 \\Longrightarrow 4=0\\] the equation is false. \nTherefore the answer is $\\boxed{1}$", "Let us test the answer choices, for it is in this case simpler and quicker. $x=0$ and $x=2/3$ obviously doesn't work, since square roots of negative numbers are not real. $x=2$ doesn't work either, because the first term is already bigger than 2, and the second term cannot be negative. By the process of elimination, the answer is $\\boxed{1}$" ]

[ "We apply the Change of Base Formula, then rearrange: \\begin{align*} \\frac{\\log_2{4}}{\\log_2{(3x)}}&=\\frac{\\log_2{8}}{\\log_2{(2x)}} \\\\ \\frac{2}{\\log_2{(3x)}}&=\\frac{3}{\\log_2{(2x)}} \\\\ 3\\log_2{(3x)}&=2\\log_2{(2x)}. \\\\ \\end{align*} By the logarithmic identity $n\\log_b{a}=\\log_b{\\left(a^n\\right)},$ it follows that \\begin{align*} \\log_2{\\left[(3x)^3\\right]}&=\\log_2{\\left[(2x)^2\\right]} \\\\ (3x)^3&=(2x)^2 \\\\ 27x^3&=4x^2 \\\\ x&=\\frac{4}{27}, \\end{align*} from which the answer is $4+27=\\boxed{31}.$", "We will apply the following logarithmic identity: \\[\\log_{p^n}{\\left(q^n\\right)}=\\log_{p}{q},\\] which can be proven by the Change of Base Formula: \\[\\log_{p^n}{\\left(q^n\\right)}=\\frac{\\log_{p}{\\left(q^n\\right)}}{\\log_{p}{\\left(p^n\\right)}}=\\frac{n\\log_{p}{q}}{n}=\\log_{p}{q}.\\] We rewrite the original equation as $\\log_{(3x)^3} 64 = \\log_{(2x)^2} 64,$ from which \\begin{align*} (3x)^3&=(2x)^2 \\\\ 27x^3&=4x^2 \\\\ x&=\\frac{4}{27}. \\end{align*} Therefore, the answer is $4+27=\\boxed{31}.$", "By the logarithmic identity $n\\log_b{a}=\\log_b{\\left(a^n\\right)},$ the original equation becomes \\[2\\log_{3x} 2 = 3\\log_{2x} 2.\\] By the logarithmic identity $\\log_b{a}\\cdot\\log_a{b}=1,$ we multiply both sides by $\\log_2{(2x)},$ then apply the Change of Base Formula to the left side: \\begin{align*} 2\\left[\\log_{3x}2\\right]\\left[\\log_2{(2x)}\\right] &= 3 \\\\ 2\\left[\\frac{\\log_2 2}{\\log_2{(3x)}}\\right]\\left[\\frac{\\log_2{(2x)}}{\\log_2 2}\\right] &= 3 \\\\ 2\\left[\\frac{\\log_2{(2x)}}{\\log_2{(3x)}}\\right] &=3 \\\\ 2\\left[\\log_{3x}{(2x)}\\right] &= 3 \\\\ \\log_{3x}{\\left[(2x)^2\\right]} &= 3 \\\\ (3x)^3&=(2x)^2 \\\\ 27x^3&=4x^2 \\\\ x&=\\frac{4}{27}. \\end{align*} Therefore, the answer is $4+27=\\boxed{31}.$", "We can convert both $4$ and $8$ into $2^2$ and $2^3,$ respectively: \\[2\\log_{3x} (2) = 3\\log_{2x} (2).\\] Converting the bases of the right side, we get \\begin{align*} \\log_{2x} 2 &= \\frac{\\ln 2}{\\ln (2x)} \\\\ \\frac{2}{3}\\cdot\\log_{3x} (2) &= \\frac{\\ln 2}{\\ln (2x)} \\\\ 2^\\frac{2}{3} &= (3x)^\\frac{\\ln 2}{\\ln (2x)} \\\\ \\frac{2}{3} \\cdot \\ln 2 &= \\frac{\\ln 2}{\\ln (2x)} \\cdot \\ln (3x). \\end{align*} Dividing both sides by $\\ln 2,$ we get $\\frac{2}{3} = \\frac{\\ln (3x)}{\\ln (2x)},$ from which \\[2\\ln (2x) = 3\\ln (3x).\\] Expanding this equation gives \\begin{align*} 2\\ln 2 + 2\\ln (x) &= 3\\ln 3 + 3\\ln (x) \\\\ \\ln (x) &= 2\\ln 2 - 3\\ln 3. \\end{align*} Thus, we have \\[x = e^{2\\ln 2 - 3\\ln 3} = \\frac{e^{2\\ln 2}}{e^{3\\ln 3}} = \\frac{2^2}{3^3} = \\frac{4}{27},\\] from which the answer is $4+27=\\boxed{31}.$", "Let $y=\\log_{3x} 4 = \\log_{2x} 8.$ We convert the equations with $y$ to the exponential form: \\begin{align*} (3x)^y&=4, \\\\ (2x)^y&=8. \\end{align*} Cubing the first equation and squaring the second equation, we have \\begin{align*} (3x)^{3y}&=64, \\\\ (2x)^{2y}&=64. \\end{align*} Applying the Transitive Property, we get \\begin{align*} (3x)^{3y}&=(2x)^{2y} \\\\ (3x)^3&=(2x)^2 \\\\ 27x^3&=4x^2 \\\\ x&=\\frac{4}{27}, \\end{align*} from which the answer is $4+27=\\boxed{31}.$" ]

[ "We solve each equation separately:\nNote that the problem is equivalent to finding the area of a parallelogram with consecutive vertices $(x_1,y_1)=\\left(\\sqrt{10}, \\sqrt{6}\\right),(x_2,y_2)=\\left(\\sqrt{3},1\\right),(x_3,y_3)=\\left(-\\sqrt{10},-\\sqrt{6}\\right),$ and $(x_4,y_4)=\\left(-\\sqrt{3}, -1\\right)$ in the coordinate plane. By the Shoelace Theorem, the area we seek is \\[\\frac{1}{2} \\left|(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1)\\right| = 6\\sqrt2-2\\sqrt{10},\\] so the answer is $6+2+2+10=\\boxed{20}.$", "We solve each equation separately:\nWe continue with the last paragraph of Solution 1 to get the answer $\\boxed{20}.$", "Let $z_1$ and $z_2$ be the solutions to the equation $z^2=4+4\\sqrt{15}i,$ and $z_3$ and $z_4$ be the solutions to the equation $z^2=2+2\\sqrt 3i.$ Clearly, $z_1$ and $z_2$ are opposite complex numbers, so are $z_3$ and $z_4.$ This solution refers to the results of De Moivre's Theorem in Solution 2.\nFrom Solution 2, let $z_1=4\\operatorname{cis}\\phi$ for some $0<\\phi<\\frac{\\pi}{4}.$ It follows that $z_2=4\\operatorname{cis}(\\phi+\\pi).$ On the other hand, we have $z_3=2\\operatorname{cis}\\frac{\\pi}{6}$ and $z_4=2\\operatorname{cis}\\frac{7\\pi}{6}$ without the loss of generality. Since $\\tan(2\\phi)>\\tan\\frac{\\pi}{3},$ we deduce that $2\\phi>\\frac{\\pi}{3},$ from which $\\phi>\\frac{\\pi}{6}.$\nIn the complex plane, the positions of $z_1,z_2,z_3,$ and $z_4$ are shown below: Note that the diagonals of every parallelogram partition the shape into four triangles with equal areas. Therefore, to find the area of the parallelogram with vertices $z_1,z_2,z_3,$ and $z_4,$ we find the area of the triangle with vertices $0,z_1,$ and $z_3,$ then multiply by $4.$\nRecall that $|z_1|=4, |z_2|=2, \\sin\\phi=\\frac{\\sqrt6}{4},$ and $\\cos\\phi=\\frac{\\sqrt{10}}{4}$ from Solution 2. The area of the parallelogram is \\begin{align*} 4\\cdot\\left[\\frac12\\cdot|z_1|\\cdot|z_3|\\cdot\\sin\\left(\\phi-\\frac{\\pi}{6}\\right)\\right] &= 16\\sin\\left(\\phi-\\frac{\\pi}{6}\\right) \\\\ &= 16\\left[\\sin\\phi\\cos\\frac{\\pi}{6}-\\cos\\phi\\sin\\frac{\\pi}{6}\\right] \\\\ &= 16\\left[\\frac{\\sqrt3}{2}\\sin\\phi-\\frac12\\cos\\phi\\right] \\\\ &= 6\\sqrt2-2\\sqrt{10}, \\end{align*} so the answer is $6+2+2+10=\\boxed{20}.$", "Rather than thinking about this with complex numbers, notice that if we take two solutions and think of them as vectors, the area of the parallelogram they form is half the desired area. Also, notice that the area of a parallelogram is $ab\\sin \\theta$ where $a$ and $b$ are the side lengths.\nThe side lengths are easily found since we are given the squares of $z$ . Thus, the magnitude of $z$ in the first equation is just $\\sqrt{16} = 4$ and in the second equation is just $\\sqrt{4} = 2$ . Now, we need $\\sin \\theta$\nTo find $\\theta$ , think about what squaring is in complex numbers. The angle between the squares of the two solutions is twice the angle between the two solutions themselves. In addition, we can find $\\cos$ of this angle by taking the dot product of those two complex numbers and dividing by their magnitudes. The vectors are $\\Bigl\\langle 4, 4\\sqrt{15}\\Bigr\\rangle$ and $\\Bigl\\langle 2, 2\\sqrt{3}\\Bigr\\rangle$ , so their dot product is $8 + 24\\sqrt{5}$ . Dividing by the magnitudes yields: $\\dfrac{8+24\\sqrt{5}}{4 \\cdot 16} = \\dfrac{1 + 3\\sqrt{5}}{8}$ . This is $\\cos 2\\theta$ , and recall the identity $\\cos 2\\theta = 1 - 2\\sin^2 \\theta$ . This means that $\\sin^2 \\theta = \\dfrac{7 - 3\\sqrt{5}}{16}$ , so $\\sin \\theta = \\dfrac{\\sqrt{7- 3\\sqrt{5}}}{4}$ . Now, notice that $\\sqrt{7- 3\\sqrt{5}} = \\dfrac{3\\sqrt{2}-\\sqrt{10}}{2}$ (which is not too hard to discover) so $\\sin \\theta = \\dfrac{3\\sqrt{2}-\\sqrt{10}}{8}$ . Finally, putting everything together yields: $2\\cdot 4 \\cdot \\dfrac{3\\sqrt{2}-\\sqrt{10}}{8} = 3\\sqrt{2} - \\sqrt{10}$ as the area of the parallelogram found by treating two of the solutions as vectors. However, drawing a picture out shows that we actually want twice this (each fourth of the parallelogram from the problem is one half of the parallelogram whose area was found above) so the desired area is actually $6\\sqrt{2} - 2\\sqrt{10}$ . Then, the answer is $\\boxed{20}$" ]

[ "Let $A=\\log_{225}x$ and let $B=\\log_{64}y$\nFrom the first equation: $A+B=4 \\Rightarrow B = 4-A$\nPlugging this into the second equation yields $\\frac{1}{A}-\\frac{1}{B}=\\frac{1}{A}-\\frac{1}{4-A}=1 \\Rightarrow A = 3\\pm\\sqrt{5}$ and thus, $B=1\\pm\\sqrt{5}$\nSo, $\\log_{225}(x_1x_2)=\\log_{225}(x_1)+\\log_{225}(x_2)=(3+\\sqrt{5})+(3-\\sqrt{5})=6$ $\\Rightarrow x_1x_2=225^6=15^{12}$\nAnd $\\log_{64}(y_1y_2)=\\log_{64}(y_1)+\\log_{64}(y_2)=(1+\\sqrt{5})+(1-\\sqrt{5})=2$ $\\Rightarrow y_1y_2=64^2=2^{12}$\nThus, $\\log_{30}\\left(x_1y_1x_2y_2\\right) = \\log_{30}\\left(15^{12}\\cdot2^{12} \\right) = \\log_{30}\\left(30^{12} \\right) = \\boxed{012}$" ]

[ "All the unknown entries can be expressed in terms of $b$ .\nSince $100e = beh = ceg = def$ , it follows that $h = \\frac{100}{b}, g = \\frac{100}{c}$ ,\nand $f = \\frac{100}{d}$ . Comparing rows $1$ and $3$ then gives $50bc = 2 \\cdot \\frac{100}{b} \\cdot \\frac{100}{c}$ ,\nfrom which $c = \\frac{20}{b}$ .\nComparing columns $1$ and $3$ gives $50d \\cdot \\frac{100}{c}= 2c \\cdot \\frac{100}{d}$ ,\nfrom which $d = \\frac{c}{5} = \\frac{4}{b}$ .\nFinally, $f = 25b, g = 5b$ , and $e = 10$ . All the entries are positive integers\nif and only if $b = 1, 2,$ or $4$ . The corresponding values for $g$ are $5, 10,$ and $20$ , and their sum is $\\boxed{35}$", "We know because this is a multiplicative magic square that each of the following are equal to each other: $100e=ceg=50dg=beh=2cf=50bc=def=2gh$\nFrom this we know that $50dg=2hg$ , thus $h=25d$ .\nThus $beh=be(25d)$ and $be(25d)=100e$ . Thus $b=\\frac{4}{d}$ From this we know that $50bc=(50)(\\frac{4}{d})(c)=50dg$ . Thus $c=\\frac{d^2g}{4}$ .\nNow we know from the very beginning that $100e=ceg$ or $100=cg$ or $100=\\frac{d^2g}{4}(g)$ or $\\frac{d^2g^2}{4}$ . Rearranging the equation $100=\\frac{d^2g^2}{4}$ we have $(d^2)(g^2)=400$ or $dg=20$ due to $d$ and $g$ both being positive. Now that $dg=20$ we find all pairs of positive integers that multiply to $20$ . There is $(d,g)= (20,1);(10,2);(5,4);(4,5);(2,10);(1,20)$ . Now we know that $b=\\frac{4}{d}$ and b has to be a positive integer. Thus $d$ can only be $1$ $2$ , or $4$ . Thus $g$ can only be $20$ $10$ ,or $5$ . Thus sum of $20+10+5$ $35$ . The answer is $\\boxed{35}$" ]

[ "We can guess and check small primes, subtract it from $126$ , and see if the result is a prime because the further away the two numbers are, the greater the difference will be. Since $126 = 2 \\cdot 3^2 \\cdot 7$ , we can eliminate $2$ $3$ , and $7$ as an option because subtracting these would result in a composite number.\nIf we subtract $5$ , then the resulting number is $121$ , which is not prime. If we subtract $11$ , then the resulting number is $115$ , which is also not prime. But when we subtract $13$ , the resulting number is $113$ , a prime number. The largest possible difference is $113-13=\\boxed{100}$" ]

[ "Since all answers are over $2000$ , work backwards and find the cost of the first $1999$ lockers. The first $9$ lockers cost $0.18$ dollars, while the next $90$ lockers cost $0.04\\cdot 90 = 3.60$ . Lockers $100$ through $999$ cost $0.06\\cdot 900 = 54.00$ , and lockers $1000$ through $1999$ inclusive cost $0.08\\cdot 1000 = 80.00$\nThis gives a total cost of $0.18 + 3.60 + 54.00 + 80.00 = 137.78$ . There are $137.94 - 137.78 = 0.16$ dollars left over, which is enough for $8$ digits, or $2$ more four digit lockers. These lockers are $2000$ and $2001$ , leading to answer $\\boxed{2001}$" ]

[ "Let $b$ be the number of boys and $g$ be the number of girls.\n\\[\\frac23 b = \\frac34 g \\Rightarrow b = \\frac98 g\\]\nFor $g$ and $b$ to be integers, $g$ must cancel out with the denominator, and the smallest possible value is $8$ . This yields $9$ boys. The minimum number of students is $8+9=\\boxed{17}$", "We know that $\\frac23 B = \\frac34 G$ or $\\frac69 B = \\frac68 G$ . So, the ratio of the number of boys to girls is 9:8. The smallest total number of students is $9 + 8 = \\boxed{17}$ . ~DY" ]

[ "From the bar graph, we can see that $5$ students chose candy E. There are $6+8+4+2+5=25$ total students in Mrs. Sawyers class. The percent that chose E is $\\frac{5}{25} \\cdot 100 = \\boxed{20}$" ]

[ "Note that $\\frac{n}{(n+1)!} = \\frac{1}{n!} - \\frac{1}{(n+1)!}$ , and therefore this sum is a telescoping sum, which is equivalent to $1 - \\frac{1}{2022!}$ . Our answer is $1 + 2022 = \\boxed{2023}$", "We add $\\frac{1}{2022!}$ to the original expression \\[\\left(\\frac{1}{2!}+\\frac{2}{3!}+\\frac{3}{4!}+\\cdots+\\frac{2021}{2022!}\\right)+\\frac{1}{2022!}=\\left(\\frac{1}{2!}+\\frac{2}{3!}+\\frac{3}{4!}+\\dots+\\frac{2020}{2021!}\\right)+\\frac{1}{2021!}.\\] This sum clearly telescopes, thus we end up with $\\left(\\frac{1}{2!}+\\frac{2}{3!}\\right)+\\frac{1}{3!}=\\frac{2}{2!}=1$ . Thus the original expression is equal to $1-\\frac{1}{2022!}$ , and $1+2022=\\boxed{2023}$", "By looking for a pattern, we see that $\\tfrac{1}{2!} = 1 - \\tfrac{1}{2!}$ and $\\tfrac{1}{2!} + \\tfrac{2}{3!} = \\tfrac{5}{6} = 1 - \\tfrac{1}{3!}$ , so we can conclude by engineer's induction that the sum in the problem is equal to $1 - \\tfrac{1}{2022!}$ , for an answer of $\\boxed{2023}$ , completing the proof.\n~eibc", "Let $x=\\frac{1}{1!}+\\frac{1}{2!}+\\frac{1}{3!}+\\dots+\\frac{1}{2022!}.$\nNote that \\begin{align*} \\left(\\frac{1}{2!}+\\frac{2}{3!}+\\frac{3}{4!}+\\dots+\\frac{2021}{2022!}\\right)+\\left(\\frac{1}{1!}+\\frac{1}{2!}+\\frac{1}{3!}+\\dots+\\frac{1}{2022!}\\right)&=\\frac{1}{1!}+\\frac{2}{2!}+\\frac{3}{3!}+\\dots+\\frac{2022}{2022!}\\\\ \\left(\\frac{1}{2!}+\\frac{2}{3!}+\\frac{3}{4!}+\\dots+\\frac{2021}{2022!}\\right)+x&=\\frac{1}{0!}+\\frac{1}{1!}+\\frac{1}{2!}+\\dots+\\frac{1}{2021!}\\\\ \\left(\\frac{1}{2!}+\\frac{2}{3!}+\\frac{3}{4!}+\\dots+\\frac{2021}{2022!}\\right)+x&=x+1-\\frac{1}{2022!}\\\\ \\left(\\frac{1}{2!}+\\frac{2}{3!}+\\frac{3}{4!}+\\dots+\\frac{2021}{2022!}\\right)&=1-\\frac{1}{2022!}. \\end{align*} Therefore, the answer is $1+2022=\\boxed{2023}.$", "Because the fractions get smaller, it is obvious that the answer is less than $1$ , so we can safely assume that $a=1$ (this can also be guessed by intuition using similar math problems). Looking at the answer choices, $2018<b<2024$ . Because the last term consists of $2022!$ (and the year is $2022$ ) we can guess that $b=2022$ . Adding them yields $1+2022=\\boxed{2023}$", "Knowing that the answer will be in the form $a-\\frac{1}{b!}$ , we can guess that the sum telescopes. Using partial fractions, we can hope to rewrite $\\frac{n-1}{n!}$ as $\\frac{A}{(n-1)!}-\\frac{B}{n}$ . Setting these equal and multiplying by $n!$ , we get $n-1=An-B(n-1)!$ . Since $An$ is the only term with $n$ with degree $1$ , we can conclude that $A=1$ . This means that $B=\\frac{1}{(n-1)!}$ . Substituting, we find that $\\frac{n-1}{n!}=\\frac{1}{(n-1)!}-\\frac{1}{n!}$ . This sum clearly telescopes and we obtain $1-\\frac{1}{2022!}$ . This means that our desired answer is $1+2022=\\boxed{2023}.$" ]

[ "Let $n$ be the 13th consecutive even integer that's being added up. By doing this, we can see that the sum of all 25 even numbers will simplify to $25n$ since $(n-2k)+\\dots+(n-4)+(n-2)+(n)+(n+2)+(n+4)+ \\dots +(n+2k)=25n$ . Now, $25n=10000 \\rightarrow n=400$ . Remembering that this is the 13th integer, we wish to find the 25th, which is $400+2(25-13)=\\boxed{424}$", "Let $x$ be the largest number. Then, $x+(x-2)+(x-4)+\\cdots +(x-48)=10000$ . Factoring this gives $2\\left(\\frac{x}{2} + \\left(\\frac{x}{2} - 1\\right) + \\left(\\frac{x}{2} - 2\\right) +\\cdots + \\left(\\frac{x}{2} - 24\\right)\\right)=10,000$ . Grouping like terms gives $25\\left(\\frac{x}{2}\\right) - 300=5000$ , and continuing down the line, we find $x=\\boxed{424}$", "Let $x$ be the smallest number. The equation will become, $x+(x+2)+(x+4)+\\cdots +(x+48)=10,000$ . After you combine like terms, you get $25x+(50*12)=10,000$ which turns into $10,000-600=25x$ $25x=9400$ , so $x=376$ . Then, you add $376+48 = \\boxed{424}$", "Dividing the series by $2$ , we get that the sum of $25$ consecutive integers is $5000$ . Let the middle number be $k$ we know that the sum is $25k$ , so $25k=5000$ . Solving, $k=200$ $2k=400$ is the middle term of the original sequence, so the original last term is $400+\\frac{25-1}{2}\\cdot 2=424$ . So the answer is $\\boxed{424}$" ]

[ "It is a well-known fact that the sum of the first $n$ odd numbers is $n^2$ . This makes the sum of the first $8$ odd numbers equal to $64$\nLet $n$ be equal to the smallest of the $5$ even integers. Then $(n+2)$ is the next highest, $(n+4)$ even higher, and so on.\nThis sets up the equation $n+(n+2)+(n+4)+(n+6)+(n+8)+4=64$\nNow we solve:\n\\[5n+24=64\\] \\[5n=40\\] \\[n=8\\]\nThus, the smallest integer is $\\boxed{8}$" ]

[ "Let the progression start at $a$ , have common difference $2$ , and end at $a + 2(n-1)$\nThe average term is $\\frac{a + (a + 2(n-1))}{2}$ , or $a + n - 1$ . Since the number of terms is $n$ , and the sum of the terms is $153$ , we have:\n$n(a+n-1) = 153$\nSince $n$ is a positive integer, it must be a factor of $153$ . This means $n = 1, 3, 9, 17, 51, 153$ are the only possibilities. We are given $n>1$ , leaving the other five factors.\nWe now must check if $a$ is an integer. We have $a = \\frac{153}{n} + 1 - n$ . If $n$ is a factor of $153$ , then $\\frac{153}{n}$ will be an integer. Adding $1-n$ wil keep it an integer.\nThus, there are $5$ possible values for $n$ , which is answer $\\boxed{5}$" ]

[ "Note that the sum of the interior angles of a convex polygon of $n$ sides is $180(n-2)^\\circ$ , and each interior angle belongs to $[0, 180^\\circ)$ . Therefore, we must have $n - 2 = \\lfloor \\frac{2570^\\circ}{180^\\circ} \\rfloor = 15$ . Then the missing angle must be $180*15^\\circ - 2570^\\circ = 130^\\circ$ , so our answer is $\\boxed{130}$ and we are done." ]

[ "The numbers that we are adding are $51,61,71 \\cdots 341$ . The numbers are part of an arithmetic series with first term $51$ , last term $341$ , common difference $10$ , and $30$ terms. Using the arithmetic series formula, the sum of the terms is $\\tfrac{30 \\cdot 392}{2} = \\boxed{5880}$" ]

[ "The sum of the odd integers $2k-1$ from $1$ to $n$ is $n^2$ . However, in this problem, the sum is instead $2k+1$ , starting at $3$ rather than $1$ . To rewrite this, we note that $2k-1$ is $2$ less than $2k+1$ for every $k$ we add, so for $n$ $k$ 's, we subtract $2n$ , giving us $n^2+2n$ ,which factors as $n(n+2) \\implies \\boxed{2}$" ]

[ "Note that $(2^x-4)+(4^x-2)=4^x+2^x-6,$ so we let $a=2^x-4$ and $b=4^x-2.$ The original equation becomes \\[a^3+b^3=(a+b)^3.\\] We expand the right side, then rearrange: \\begin{align*} a^3+b^3 &= a^3+3a^2b+3ab^2+b^3 \\\\ 0 &= 3a^2b+3ab^2 \\\\ 0 &= 3ab(a+b). \\end{align*}\nTogether, the answer is $2+\\frac12+1=\\boxed{72}.$" ]

[ "The sum of an infinite geometric series is of the form: \\[\\begin{split} S & = \\frac{a_1}{1-r} \\end{split}\\] where $a_1$ is the first term and $r$ is the ratio whose absolute value is less than 1.\nWe know that the second term is the first term multiplied by the ratio. \nIn other words: \\[\\begin{split} a_1 \\cdot r & = 1 \\\\ a_1 & = \\frac{1}{r} \\end{split}\\]\nThus, the sum is the following: \\[\\begin{split} S & = \\frac{\\frac{1}{r}}{1-r} \\\\\\\\ S & =\\frac{1}{r-r^2} \\end{split}\\]\nSince we want the minimum value of this expression, we want the maximum value for the denominator, $-r^2+r$ .\nThe maximum x-value of a quadratic with leading coefficient $-a$ is $\\frac{-b}{2a}$ \\[\\begin{split} r & = \\frac{-(1)}{2(-1)} \\\\\\\\ r & = \\frac{1}{2} \\end{split}\\]\nPlugging $r$ $=$ $\\frac{1}{2}$ into the quadratic yields: \\[\\begin{split} S & = \\frac{1}{\\frac{1}{2} -\\left(\\frac{1}{2}\\right)^2} \\\\\\\\ S & = \\frac{1}{\\frac{1}{4}} \\end{split}\\]\nTherefore, the minimum sum of our infinite geometric sequence is $\\boxed{4}$ .\n(Solution by akaashp11)", "After observation we realize that in order to minimize our sum $\\frac{a}{1-r}$ with $a$ being the reciprocal of r. The common ratio $r$ has to be in the form of $\\frac{1}{x}$ with $x$ being an integer as anything more than $1$ divided by $x$ would give a larger sum than a ratio in the form of $\\frac{1}{x}$\nThe first term has to be $x$ , so then in order to minimize the sum, we have minimize $x$\nThe smallest possible value for $x$ such that it is an integer that's greater than $1$ is $2$ . So our first term is $2$ and our common ratio is $\\frac{1}{2}$ . Thus the sum is $\\frac{2}{\\frac {1}{2}}$ or $\\boxed{4}$ .\nSolution 2 by No_One", "We can see that if $a$ is the first term, and $r$ is the common ratio between each of the terms, then we can get \\[S=\\frac{a}{1-r} \\implies S-Sr=a\\] Also, we know that the second term can be expressed as $a\\cdot r$ notice if we multiply $S-Sr=a$ by $r$ , we would get \\[r(S-Sr)=ar \\implies Sr-Sr^2=1 \\implies Sr^2-Sr+1=0\\] This quadratic has real solutions if the discriminant is greater than or equal to zero, or \\[S^2-4\\cdot S \\cdot 1 \\ge 0\\] This yields that $S\\le 0$ or $S\\ge 4$ . \nHowever, since we know that $S$ has to be positive, we can safely conclude that the minimum possible value of $S$ is $\\boxed{4}$", "Let the first term of the geometric series $x$ . Since it must be decreasing, we have $x>1$ and the third term is $\\frac{1}{x}$ . Realize that by AM-GM inequality $x+\\frac{1}{x} \\ge 2$ with equality if $x = 1$ . However, we established that $x>1$ so that means $x+\\frac{1}{x} > 2$ . So the sum of the first three terms of the sequence $x + \\frac{1}{x} + 1$ is greater than $3$ , and the geometric series keeps continuing infinitely. This means the sum continues increasing. The only answer choice greater than $3$ is $\\boxed{4}$ . ~skyscraper", "Let the first term be $k.$ The sum of the series is $\\frac{k}{1- \\frac{1}{k}} =\\frac{k^2}{k-1}.$ Rewrite this as $\\frac{k^2-2k+1}{k-1} +\\frac{2k-1}{k-1} = k-1+\\frac{2k-2}{k-1} +\\frac{1}{k-1} = (k-1) + \\left(\\frac{1}{k-1}\\right) + 2.$ By AM-GM we know that $(k-1) + \\left(\\frac{1}{k-1}\\right) \\ge 2$ so the minimum is $2+2 = \\boxed{4}.$", "Set the first term is $a.$ , the common ratio should be $\\frac{1}{a}.$\nThe sum to infinity of the series is $S=\\frac{a}{1-\\frac{1}{a}}=\\frac{a^2}{a-1}.$\nSince $S$ is positive, we have $a>1.$ Define the function $f(a)=\\frac{a^2}{a-1}$ , the domain of this function is $a>1.$\nLet $f^{'}(a)=\\frac{2a^2-2a-a^2}{(a-1)^2}=\\frac{a(a-2)}{(a-1)^2}=0.$ We solve that $a=2.$\nIt's easy to find that when $1<a<2, f^{'}(a)<0,$ when $a>2, f^{'}(a)>0.$ Thus $f(a)$ has a minimum value when $a=2.$ , which is $f(2)=4.$ Choose $\\boxed{4}.$" ]

[ "The sum of the geometric sequence is $\\frac{a}{1 - r}$ where $a$ is the first term and $r$ is the common ratio. We know the second term, $ar,$ is equal to $1.$ Thus $ar = 1 \\Rightarrow a = \\frac{1}{r}.$ This means, \\[S = \\frac{a}{1 - r} = \\frac{1/r}{1 - r} = \\frac{1}{r(1 - r)}.\\] In order to minimize $S,$ we maximize the denominator. By AM-GM, \\[\\frac{(r) + (1 - r)}{2} \\ge \\sqrt{r(1-r)} \\Rightarrow \\frac{1}{4} \\ge r(1-r).\\] Equality occurs at $r = 1-r \\Rightarrow r = \\frac{1}{2}.$ This gives the minimum value of $S$ as $\\frac{\\frac{1}{1/2}}{1 - \\frac{1}{2}} = \\boxed{4}.$", "A geometric sequence always looks like\n\\[a,ar,ar^2,ar^3,\\dots\\]\nand they say that the second term $ar=1$ . You should know that the sum of an infinite geometric series (denoted by $S$ here) is $\\frac{a}{1-r}$ . We now have a system of equations which allows us to find $S$ in one variable.\n\\begin{align*} ar&=1 \\\\ S&=\\frac{a}{1-r} \\end{align*}\n$\\textbf{Solving in terms of \\textit{a} then graphing}$\n\\[S=\\frac{a^2}{a-1}\\]\nWe seek the smallest positive value of $S$ . We proceed by graphing in the $aS$ plane (if you want to be rigorous, only stop graphing when you know all the rest you didn't graph is just the approaching of asymptotes and infinities) and find the answer is $\\boxed{4}.$", "\\[\\textbf{Completing the Square and Quadratics}\\] Let $r$ be the common ratio. If the second term is $1$ , the first must be $\\frac{1}{r}$ . By the infinite geometric series formula, the sum must be \\[S=\\frac{\\frac{1}{r}}{1-r}\\] This equals $\\frac{1}{r(1-r)}$ . To find the minimum value of S, we must find the maximum value of the denominator, $r(1-r)$ , which is $\\frac{1}{4}$ , completing the square. Thus, the minimum value of $S$ is $\\boxed{4}$", "Our sequence is $a_1,1,...$ . Since we know this is a converging series, our ratio is in $(0,1)$ . Because the 2nd term in the sequence is a 1, the ratio must be $\\frac{1}{a_1}$ , so we can write $S$ as $\\frac{a_1}{1-\\frac{1}{a_1}}$ . With some manipulation we get $S=\\frac{1}{a_1-1}$ . Since S has to be a \"positive number,\" we come to think $a_1$ is $2$ (makes S positive & we know a sequence/series of a ratio $1/2$ is definitely convergent). So our sequence is $2,1,1/2,1,4,...$ $2+1+1=\\boxed{4}$ .\n-thedodecagon" ]

[ "We know that the formula for the sum of an infinite geometric series is $S = \\frac{a}{1-r}$\nSo we can apply this to the conditions given by the problem.\nWe have two equations:\n\\begin{align*} 15 &= \\frac{a}{1-r} \\\\ 45 &= \\frac{a^{2}}{1-r^{2}} \\end{align*}\nWe get\n\\begin{align*} a &= 15 - 15r \\\\ a^{2} &= 45 - 45r^{2} \\\\ \\\\ (15 - 15r)^{2} &= 45 - 45r^{2} \\\\ 270r^{2} - 450r + 180 &= 0 \\\\ 3r^{2} - 5r + 2 &= 0 \\\\ (3r - 2)(r - 1) &= 0 \\end{align*}\nSince $|r| < 1$ $r = \\frac{2}{3}$ , so plug this into the equation above and we get $a = 15 - 15r = 15 - 10 = \\boxed{5}$" ]

[ "$221$ can be written as the sum of four two-digit numbers, let's say $\\overline{ae}$ $\\overline{bf}$ $\\overline{cg}$ , and $\\overline{dh}$ . Then $221= 10(a+b+c+d)+(e+f+g+h)$ . The last digit of $221$ is $1$ , and $10(a+b+c+d)$ won't affect the units digits, so $(e+f+g+h)$ must end with $1$ . The smallest value $(e+f+g+h)$ can have is $(1+2+3+4)=10$ , and the greatest value is $(6+7+8+9)=30$ . Therefore, $(e+f+g+h)$ must equal $11$ or $21$\nCase 1: $(e+f+g+h)=11$\nThe only distinct positive integers that can add up to $11$ is $(1+2+3+5)$ . So, $a$ $b$ $c$ , and $d$ must include four of the five numbers $(4,6,7,8,9)$ . We have $10(a+b+c+d)=221-11=210$ , or $a+b+c+d=21$ . We can add all of $4+6+7+8+9=34$ , and try subtracting one number to get to $21$ , but none of them work. Therefore, $(e+f+g+h)$ cannot add up to $11$\nCase 2: $(e+f+g+h)=21$\nChecking all the values for $e$ $f$ $g$ ,and $h$ each individually may be time-consuming, instead of only having $1$ solution like Case 1. We can try a different approach by looking at $(a+b+c+d)$ first. If $(e+f+g+h)=21$ $10(a+b+c+d)=221-21=200$ , or $(a+b+c+d)=20$ . That means $(a+b+c+d)+(e+f+g+h)=21+20=41$ . We know $(1+2+3+4+5+6+7+8+9)=45$ , so the missing digit is $45-41=\\boxed{4}$" ]

[ "Alternatively, we know that a number is congruent to the sum of its digits mod 9, so $221 \\equiv 5 \\equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \\equiv -d$ , where $d$ is some digit. Clearly, $\\boxed{4}$" ]

[ "If the first six integers are $14$ , the last number can be $(-14\\cdot 6) - 1 = -85$ . The sum of all seven integers will be $-1$\nHowever, if all seven integers are over $13$ , the smallest possible sum is $14\\cdot 7 = 98$\nThus, the answer is $6$ , which is option $\\boxed{6}$" ]

[ "The arithmetic mean of these numbers is $\\frac{\\frac{2013}{3}}{2}=\\frac{671}{2}=335.5$ . Therefore the numbers are $333$ $334$ $335$ $336$ $337$ $338$ , so the answer is $\\boxed{338}$", "Let the $4^{\\text{th}}$ number be $x$ . Then our desired number is $x+2$\nOur integers are $x-3,x-2,x-1,x,x+1,x+2$ , so we have that $6x-3 = 2013 \\implies x = \\frac{2016}{6} = 336 \\implies x+2 = \\boxed{338}$", "Let the first term be $x$ . Our integers are $x,x+1,x+2,x+3,x+4,x+5$ . We have, $6x+15=2013\\implies x=333\\implies x+5=\\boxed{338}$", "Let the $6th$ number be $x$ . Then our list is: $x-6+x-5+x-4+x-3+x-x-1=2013$ . Simplifying this gets you $6x-21=2013\\implies 6x=2034$ , which means that $x = \\boxed{338}$" ]

[ "Since there are $6$ numbers, we divide $2013$ by $6$ to find the mean of the numbers. $\\frac{2013}{6} = 335 \\frac{1}{2}$ .\nThen, $335 \\frac{1}{2} + \\frac{1}{2} = 336$ (the fourth number). Fifth: $337$ ; Sixth: $\\boxed{338}$" ]

[ "Since there are $8$ vertices of a cube , there are ${8 \\choose 3} = 56$ total triangles to consider. They fall into three categories: there are those which are entirely contained within a single face of the cube (whose sides are two edges and one face diagonal ), those which lie in a plane perpendicular to one face of the cube (whose sides are one edge, one face diagonal and one space diagonal of the cube) and those which lie in a plane oblique to the edges of the cube, whose sides are three face diagonals of the cube.\nEach face of the cube contains ${4\\choose 3} = 4$ triangles of the first type, and there are $6$ faces, so there are $24$ triangles of the first type. Each of these is a right triangle with legs of length $1$ , so each triangle of the first type has area $\\frac 12$\nEach edge of the cube is a side of exactly $2$ of the triangles of the second type, and there are $12$ edges, so there are $24$ triangles of the second type. Each of these is a right triangle with legs of length $1$ and $\\sqrt 2$ , so each triangle of the second type has area $\\frac{\\sqrt{2}}{2}$\nEach vertex of the cube is associated with exactly one triangle of the third type (whose vertices are its three neighbors), and there are $8$ vertices of the cube, so there are $8$ triangles of the third type. Each of the these is an equilateral triangle with sides of length $\\sqrt 2$ , so each triangle of the third type has area $\\frac{\\sqrt 3}2$\nThus the total area of all these triangles is $24 \\cdot \\frac12 + 24\\cdot\\frac{\\sqrt2}2 + 8\\cdot\\frac{\\sqrt3}2 = 12 + 12\\sqrt2 + 4\\sqrt3 = 12 + \\sqrt{288} + \\sqrt{48}$ and the answer is $12 + 288 + 48 = \\boxed{348}$" ]

[ "Every factor of $10^n$ will be of the form $2^a \\times 5^b , a\\leq n , b\\leq n$ . Not all of these base ten logarithms will be rational, but we can add them together in a certain way to make it rational. Recall the logarithmic property $\\log(a \\times b) = \\log(a)+\\log(b)$ . For any factor $2^a \\times 5^b$ , there will be another factor $2^{n-a} \\times 5^{n-b}$ . Note this is not true if $10^n$ is a perfect square. When these are multiplied, they equal $2^{a+n-a} \\times 5^{b+n-b} = 10^n$ $\\log 10^n=n$ so the number of factors divided by 2 times n equals the sum of all the factors, 792.\nThere are $n+1$ choices for the exponent of 5 in each factor, and for each of those choices, there are $n+1$ factors (each corresponding to a different exponent of 2), yielding $(n+1)^2$ total factors. $\\frac{(n+1)^2}{2}*n = 792$ . We then plug in answer choices and arrive at the answer $\\boxed{11}$" ]

[ "We see that the result of this expression will always be in the form $(100\\text{ some number of zeros }001)^2.$ Multiplying these together yields: \\[110\\text{ some number of zeros }011.\\] This works because of the way they are multiplied. Therefore, the answer is $\\boxed{4}$" ]

[ "Let the number be $10a+b$ where $a$ and $b$ are the tens and units digits of the number.\nSo $(10a+b)-(a+b)=9a$ must have a units digit of $6$\nThis is only possible if $9a=36$ , so $a=4$ is the only way this can be true.\nSo the numbers that have this property are $40, 41, 42, 43, 44, 45, 46, 47, 48, 49$\nTherefore the answer is $\\boxed{10}$", "Let a two-digit number equal $10a+b$ , where $a$ and $b$ are the tens and units digits of the number.\nFrom the problem, we have $10a+b-(a+b)=9a$\nNow let $9a=10x+y$ , where $x$ and $y$ are the tens and units digits of the number. Then it must be that $y=6$ as stated in the problem.\nNote that $10a$ ends in $0$ , but $9a$ ends in $6$ , so $a=4$ . We need not to care about $b$ , since it cancels out in the calculation.\nSo the answer is $\\boxed{10}$ , since there are $10$ numbers that have $a=4$" ]

[ "Since the sum of the first $2011$ terms is $200$ , and the sum of the first $4022$ terms is $380$ , the sum of the second $2011$ terms is $180$ .\nThis is decreasing from the first 2011, so the common ratio is less than one.\nBecause it is a geometric sequence and the sum of the first 2011 terms is $200$ , second $2011$ is $180$ , the ratio of the second $2011$ terms to the first $2011$ terms is $\\frac{9}{10}$ . Following the same pattern, the sum of the third $2011$ terms is $\\frac{9}{10}*180 = 162$\nThus, $200+180+162=542$ , so the sum of the first $6033$ terms is $\\boxed{542}$", "Solution by e_power_pi_times_i\nThe sum of the first $2011$ terms can be written as $\\dfrac{a_1(1-k^{2011})}{1-k}$ , and the first $4022$ terms can be written as $\\dfrac{a_1(1-k^{4022})}{1-k}$ . Dividing these equations, we get $\\dfrac{1-k^{2011}}{1-k^{4022}} = \\dfrac{10}{19}$ . Noticing that $k^{4022}$ is just the square of $k^{2011}$ , we substitute $x = k^{2011}$ , so $\\dfrac{1}{x+1} = \\dfrac{10}{19}$ . That means that $k^{2011} = \\dfrac{9}{10}$ . Since the sum of the first $6033$ terms can be written as $\\dfrac{a_1(1-k^{6033})}{1-k}$ , dividing gives $\\dfrac{1-k^{2011}}{1-k^{6033}}$ . Since $k^{6033} = \\dfrac{729}{1000}$ , plugging all the values in gives $\\boxed{542}$", "The sum of the first 2011 terms of the sequence is expressible as $a_1 + a_1r + a_1r^2 + a_1r^3$ .... until $a_1r^{2010}$ . The sum of the 2011 terms following the first 2011 is expressible as $a_1r^{2011} + a_1r^{2012} + a_1r^{2013}$ .... until $a_1r^{4021}$ . Notice that the latter sum of terms can be expressed as $(r^{2011})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})$ . We also know that the latter sum of terms can be obtained by subtracting 200 from 380, which then means that $r^{2011} = 9/10$ . The terms from 4023 to 6033 can be expressed as $(r^{4022})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})$ , which is equivalent to $((9/10)^2)(200) = 162$ . Adding 380 and 162 gives the answer of $\\boxed{542}$" ]

[ "The sum of the first $m$ odd integers is given by $m^2$ . The sum of the first $n$ even integers is given by $n(n+1)$\nThus, $m^2 = n^2 + n + 212$ . Since we want to solve for n, rearrange as a quadratic equation: $n^2 + n + (212 - m^2) = 0$\nUse the quadratic formula: $n = \\frac{-1 + \\sqrt{1 - 4(212 - m^2)}}{2}$ . Since $n$ is clearly an integer, $1 - 4(212 - m^2) = 4m^2 - 847$ must be not only a perfect square, but also an odd perfect square for $n$ to be an integer.\nLet $x = \\sqrt{4m^2 - 847}$ ; note that this means $n = \\frac{-1 + x}{2}$ . It can be rewritten as $x^2 = 4m^2 - 847$ , so $4m^2 - x^2 = 847$ . Factoring the left side by using the difference of squares, we get $(2m + x)(2m - x) = 847 = 7\\cdot11^2$\nOur goal is to find possible values for $x$ , then use the equation above to find $n$ . The difference between the factors is $(2m + x) - (2m - x) = 2m + x - 2m + x = 2x.$ We have three pairs of factors, $847\\cdot1, 121\\cdot 7,$ and $77\\cdot 11$ . The differences between these factors are $846$ $114$ , and $66$ - those are all possible values for $2x$ . Thus the possibilities for $x$ are $423$ $57$ , and $33$\nNow plug in these values into the equation $n = \\frac{-1 + x}{2}$ , so $n$ can equal $211$ $28$ , or $16$ , hence the answer is $\\boxed{255}$", "As above, start off by noting that the sum of the first $m$ odd integers $= m^2$ and the sum of the first $n$ even integers $= n(n+1)$ . Clearly $m > n$ , so let $m = n + a$ , where $a$ is some positive integer. We have:\n$(n+a)^2 = n(n+1) + 212$ .\nExpanding, grouping like terms and factoring, we get: $n = \\frac{(212 - a^2)}{(2a - 1)}$\nWe know that $n$ and $a$ are both positive integers, so we need only check values of $a$ from $1$ to $14$ $14^2 = 196 < 212 < 15^2 = 225$ ). Plugging in, the only values of $a$ that give integral solutions are $1, 4,$ and $6$ . These gives $n$ values of $211, 28,$ and $16$ , respectively. $211 + 28 + 16 = 255$ . Hence, the answer is $\\boxed{255}$", "Using the closed forms for the sums, we get $m^2=n(n+1)+212$ , or $m^2=n^2+n+212$ . We would like to factor this equation, but the current expressions don't allow for this. So we multiply both sides by 4 to let us complete the square. Our equation is now $4m^2=4n^2+4n+848$ . Complete the square on the right hand side: $4m^2=(4n^2+4n+1)+848-1=(2n+1)^2+847$ . Move over the $(2n+1)^2$ and factor to get $(2m-2n-1)(2m+2n+1)=847=7\\cdot11\\cdot11$ . The second factor is clearly greater than the first, and the only possible factor pairs are $1$ and $847$ $7$ and $121$ $11$ and $77$ . In each of these cases, solve for $m$ and $n$ and we find the solutions $(m,n)=(212,211), (32,28), (22,16)$ . The sum of all possible values of $n$ is $211+28+16=\\boxed{255}$" ]

[ "Solution by e_power_pi_times_i\nWhen the $n$ th odd positive integer is subtracted from the $n$ th even positive integer, the result is $1$ . Therefore the sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is $80\\cdot1 = \\boxed{80}$" ]

[ "If any of the approximations $A_i$ is less than 2 or more than 3, the error associated with that term will be larger than 1, so the largest error will be larger than 1. However, if all of the $A_i$ are 2 or 3, the largest error will be less than 1. So in the best case, we write 19 as a sum of 7 numbers, each of which is 2 or 3. Then there must be five 3s and two 2s. It is clear that in the best appoximation, the two 2s will be used to approximate the two smallest of the $a_i$ , so our approximations are $A_1 = A_2 = 2$ and $A_3 = A_4 = A_5 = A_6 = A_7 = 3$ and the largest error is $|A_2 - a_2| = 0.61$ , so the answer is $\\boxed{061}$" ]

[ "Let $x, y$ , and $z$ be the unique lengths of the edges of the box. Each box has $4$ edges of each length, so: \\[4x + 4y + 4z = 140 \\ \\Longrightarrow \\ x + y + z = 35.\\] The spacial diagonal (longest distance) is given by $\\sqrt{x^2 + y^2 + z^2}$ . Thus, we have $\\sqrt{x^2 + y^2 + z^2} = 21$ , so $x^2 + y^2 + z^2 = 21^2$\nOur target expression is the surface area of the box:\n\\[S = 2xy + 2xz + 2yz.\\]\nSince $S$ is a symmetric polynomial of degree $2$ , we try squaring the first equation to get:\n\\[35^2 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy +2yz + 2xz = 35^2.\\]\nSubstituting in our long diagonal and surface area expressions, we get: $21^2 + S = 35^2$ , so $S = (35 + 21)(35 - 21) = 56\\cdot 14 = 784$ , which is option $\\boxed{784}$" ]

[ "For any polynomial, even a polynomial with more than one variable, the sum of all the coefficients (including the constant, which is the coefficient of $x^0y^0$ ) is found by setting all variables equal to $1$ . Note that we don't have to worry about whether a constant is a coefficient of an \"invisible\" $x^0y^0$ term, because there is no such term here.\nSetting $x=y=1$ gives $(-1)^{18}$ , which is equal to $1$ , which is answer $\\boxed{1}$" ]

[ "$AMC10+AMC12=123422$\n$AMC00+AMC00=123400$\n$AMC+AMC=1234$\n$2\\cdot AMC=1234$\n$AMC=\\frac{1234}{2}=617$\nSince $A$ $M$ , and $C$ are digits, $A=6$ $M=1$ $C=7$\nTherefore, $A+M+C = 6+1+7 = \\boxed{14}$", "We know that $AMC12$ is $2$ more than $AMC10$ . We set up $AMC10=x$ and $AMC12=x+2$ . We have $x+x+2=123422$ . Solving for $x$ , we get $x=61710$ . Therefore, the sum $A+M+C= \\boxed{14}$", "Consider the place values of the digits of $AMC10$ and $AMC12$\nWhen we add $AMC10$ and $AMC12$ $C + C$ must result in a units digit of $4$ , meaning $C$ is either $2$ or $7$ . Since $M$ is odd, this means a ten was carried over to the next place value from $C + C$ , and thus $C = 7$ (as $7 + 7 = 14$ and the ten is carried over). Now, we know 3 is the units digit of $M + M + 1$ , so $M$ is either $1$ or $6$ . Again, we must look at the digit before $M$ , or $A$ $A$ is even, so $M$ must be less than $5$ , or else the ten would be carried over. Ergo, $M$ is $1$ . Nothing is carried over, so we have $A + A = 12$ , and $A = 6$ . Therefore, the sum of $A$ $M$ , and $C$ is $6 + 1 + 7 = \\boxed{14}$" ]

[ "$AMC10+AMC12=123422$\n$AMC00+AMC00=123400$\n$AMC+AMC=1234$\n$2\\cdot AMC=1234$\n$AMC=\\frac{1234}{2}=617$\nSince $A$ $M$ , and $C$ are digits, $A=6$ $M=1$ $C=7$\nTherefore, $A+M+C = 6+1+7 = \\boxed{14}$", "We know that $AMC12$ is $2$ more than $AMC10$ . We set up $AMC10=x$ and $AMC12=x+2$ . We have $x+x+2=123422$ . Solving for $x$ , we get $x=61710$ . Therefore, the sum $A+M+C= \\boxed{14}$", "Consider the place values of the digits of $AMC10$ and $AMC12$\nWhen we add $AMC10$ and $AMC12$ $C + C$ must result in a units digit of $4$ , meaning $C$ is either $2$ or $7$ . Since $M$ is odd, this means a ten was carried over to the next place value from $C + C$ , and thus $C = 7$ (as $7 + 7 = 14$ and the ten is carried over). Now, we know 3 is the units digit of $M + M + 1$ , so $M$ is either $1$ or $6$ . Again, we must look at the digit before $M$ , or $A$ $A$ is even, so $M$ must be less than $5$ , or else the ten would be carried over. Ergo, $M$ is $1$ . Nothing is carried over, so we have $A + A = 12$ , and $A = 6$ . Therefore, the sum of $A$ $M$ , and $C$ is $6 + 1 + 7 = \\boxed{14}$" ]

[ "Let the numbers be $x$ $y$ , and $z$ in that order. The given tells us that\n\\begin{eqnarray*}y&=&7z\\\\ x&=&4(y+z)=4(7z+z)=4(8z)=32z\\\\ x+y+z&=&32z+7z+z=40z=20\\\\ z&=&\\frac{20}{40}=\\frac{1}{2}\\\\ y&=&7z=7\\cdot\\frac{1}{2}=\\frac{7}{2}\\\\ x&=&32z=32\\cdot\\frac{1}{2}=16 \\end{eqnarray*}\nTherefore, the product of all three numbers is $xyz=16\\cdot\\frac{7}{2}\\cdot\\frac{1}{2}=28 \\Rightarrow \\boxed{28}$", "Alternatively, we can set up the system in equation form:\n\\begin{eqnarray*}1x+1y+1z&=&20\\\\ 1x-4y-4z&=&0\\\\ 0x+1y-7z&=&0\\\\ \\end{eqnarray*}\nOr, in matrix form $\\begin{bmatrix} 1 & 1 & 1 \\\\ 1 & -4 & -4 \\\\ 0 & 1 & -7 \\end{bmatrix} \\begin{bmatrix} x \\\\ y \\\\ z \\\\ \\end{bmatrix} =\\begin{bmatrix} 20 \\\\ 0 \\\\ 0 \\\\ \\end{bmatrix}$\nTo solve this matrix equation, we can rearrange it thus:\n$\\begin{bmatrix} x \\\\ y \\\\ z \\\\ \\end{bmatrix} = \\begin{bmatrix} 1 & 1 & 1 \\\\ 1 & -4 & -4 \\\\ 0 & 1 & -7 \\end{bmatrix} ^{-1} \\begin{bmatrix} 20 \\\\ 0 \\\\ 0 \\\\ \\end{bmatrix}$\nSolving this matrix equation by using inverse matrices and matrix multiplication yields\n$\\begin{bmatrix} x \\\\ y \\\\ z \\\\ \\end{bmatrix} = \\begin{bmatrix} \\frac{1}{2} \\\\ \\frac{7}{2} \\\\ 16 \\\\ \\end{bmatrix}$\nWhich means that $x = \\frac{1}{2}$ $y = \\frac{7}{2}$ , and $z = 16$ . Therefore, $xyz = \\frac{1}{2}\\cdot\\frac{7}{2}\\cdot16 = 28 \\Rightarrow \\boxed{28}$" ]

[ "Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$\nWe have \\[6x+(x+40)+x=8x+40=96,\\] from which $x=7.$\nTherefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\\boxed{5}.$", "Solve this using a system of equations. Let $x,y,$ and $z$ be the three numbers, respectively. We get three equations: \\begin{align*} x+y+z&=96, \\\\ x&=6z, \\\\ z&=y-40. \\end{align*} Rewriting the third equation gives us $y=z+40,$ so we can substitute $x$ as $6z$ and $y$ as $z+40.$\nTherefore, we get \\begin{align*} 6z+(z+40)+z&=96 \\\\ 8z+40&=96 \\\\ 8z&=56 \\\\ z&=7. \\end{align*} Substituting 7 in for $z$ gives us $x=6z=6(7)=42$ and $y=z+40=7+40=47.$\nSo, the answer is $|x-y|=|42-47|=\\boxed{5}.$", "In accordance with Solution 2, \\[y = z+40, x = 6z \\implies |x-y| = |6z - z - 40| = 5|z - 8| \\implies \\boxed{5}.\\] [email protected], vvsss" ]

[ "Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$\nWe have \\[6x+(x+40)+x=8x+40=96,\\] from which $x=7.$\nTherefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\\boxed{5}.$", "Solve this using a system of equations. Let $x,y,$ and $z$ be the three numbers, respectively. We get three equations: \\begin{align*} x+y+z&=96, \\\\ x&=6z, \\\\ z&=y-40. \\end{align*} Rewriting the third equation gives us $y=z+40,$ so we can substitute $x$ as $6z$ and $y$ as $z+40.$\nTherefore, we get \\begin{align*} 6z+(z+40)+z&=96 \\\\ 8z+40&=96 \\\\ 8z&=56 \\\\ z&=7. \\end{align*} Substituting 7 in for $z$ gives us $x=6z=6(7)=42$ and $y=z+40=7+40=47.$\nSo, the answer is $|x-y|=|42-47|=\\boxed{5}.$", "In accordance with Solution 2, \\[y = z+40, x = 6z \\implies |x-y| = |6z - z - 40| = 5|z - 8| \\implies \\boxed{5}.\\] [email protected], vvsss" ]

[ "Let the $3$ numbers be $a,$ $b,$ and $c$ . We see that \\[a+b+c = 98\\] and \\[\\frac{a}{b} = \\frac{2}{3} \\Rrightarrow 3a = 2b\\] \\[\\frac{b}{c} = \\frac{5}{8} \\Rrightarrow 8b = 5c\\] Writing $a$ and $c$ in terms of $b$ we have $a = \\frac{2}{3} b$ and $c = \\frac{8}{5} b$ . \nSubstituting in the sum, we have \\[\\frac{2}{3} b + b + \\frac{8}{5} b = 98\\] \\[\\frac{49}{15} b = 98\\] \\[b = 98 \\cdot \\frac{15}{49} \\Rrightarrow b = 30\\] $\\boxed{30}$" ]

[ "The sum of two angles in a triangle is $\\frac{6}{5}$ of a right angle $\\longrightarrow \\frac{6}{5} \\times 90 = 108$\nIf $x$ is the measure of the first angle, then the measure of the second angle is $x+30$ \\[x + x + 30 = 108 \\longrightarrow 2x = 78 \\longrightarrow x = 39\\]\nNow we know the measure of two angles are $39^{\\circ}$ and $69^{\\circ}$ . By the Triangle Sum Theorem, the sum of all angles in a triangle is $180^{\\circ},$ so the final angle is $72^{\\circ}$ . Therefore, the largest angle in the triangle is $\\boxed{72}$" ]

[ "The units digit of a multiple of $10$ will always be $0$ . We add a $0$ whenever we multiply by $10$ . So, removing the units digit is equal to dividing by $10$\nLet the smaller number (the one we get after removing the units digit) be $a$ . This means the bigger number would be $10a$\nWe know the sum is $10a+a = 11a$ so $11a=17402$ . So $a=1582$ . The difference is $10a-a = 9a$ . So, the answer is $9(1582) = \\boxed{14238}$", "Since the unit's place of a multiple of $10$ is $0$ , the other integer must end with a $2$ , for both integers sum up to a number ending in a $2$ . Thus, the unit's place of the difference must be $10-2=8$ , and the only answer choice that ends with an $8$ is $\\boxed{14238}$", "Let the larger number be $\\underline{ABCD0}.$ It follows that the smaller number is $\\underline{ABCD}.$ Adding vertically, we have \\[\\begin{array}{cccccc} & A & B & C & D & 0 \\\\ +\\quad & & A & B & C & D \\\\ \\hline & & & & & \\\\ [-2.5ex] & 1 & 7 & 4 & 0 & 2 \\\\ \\end{array}\\] Working from right to left, we get \\[D=2\\implies C=8 \\implies B=5 \\implies A=1.\\] The larger number is $15820$ and the smaller number is $1582.$ Their difference is $15820-1582=\\boxed{14238}.$", "We know that the larger number has a units digit of $0$ since it is divisible by 10. If $D$ is the ten's digit of the larger number, then $D$ is the units digit of the smaller number. Since the sum of the natural numbers has a unit's digit of $2$ $D=2$\nThe units digit of the larger number is $0$ and the units digit of the smaller number is $2$ , so the positive difference between the numbers is 8. There is only one answer choice that has this units digit, and that is $\\boxed{14238}.$" ]

[ "The units digit of a multiple of $10$ will always be $0$ . We add a $0$ whenever we multiply by $10$ . So, removing the units digit is equal to dividing by $10$\nLet the smaller number (the one we get after removing the units digit) be $a$ . This means the bigger number would be $10a$\nWe know the sum is $10a+a = 11a$ so $11a=17402$ . So $a=1582$ . The difference is $10a-a = 9a$ . So, the answer is $9(1582) = \\boxed{14238}$", "Since the unit's place of a multiple of $10$ is $0$ , the other integer must end with a $2$ , for both integers sum up to a number ending in a $2$ . Thus, the unit's place of the difference must be $10-2=8$ , and the only answer choice that ends with an $8$ is $\\boxed{14238}$", "Let the larger number be $\\underline{ABCD0}.$ It follows that the smaller number is $\\underline{ABCD}.$ Adding vertically, we have \\[\\begin{array}{cccccc} & A & B & C & D & 0 \\\\ +\\quad & & A & B & C & D \\\\ \\hline & & & & & \\\\ [-2.5ex] & 1 & 7 & 4 & 0 & 2 \\\\ \\end{array}\\] Working from right to left, we get \\[D=2\\implies C=8 \\implies B=5 \\implies A=1.\\] The larger number is $15820$ and the smaller number is $1582.$ Their difference is $15820-1582=\\boxed{14238}.$", "We know that the larger number has a units digit of $0$ since it is divisible by 10. If $D$ is the ten's digit of the larger number, then $D$ is the units digit of the smaller number. Since the sum of the natural numbers has a unit's digit of $2$ $D=2$\nThe units digit of the larger number is $0$ and the units digit of the smaller number is $2$ , so the positive difference between the numbers is 8. There is only one answer choice that has this units digit, and that is $\\boxed{14238}.$" ]

[ "Let the two real numbers be $x,y$ . We are given that $x+y=4xy,$ and dividing both sides by $xy$ $\\frac{x}{xy}+\\frac{y}{xy}=4.$\n\\[\\frac{1}{y}+\\frac{1}{x}=\\boxed{4}.\\]", "Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction.\nSee for yourself. And by looking into fractions, we immediately see that $\\frac{1}{3}$ and $1$ would fit the rule. $\\boxed{4}.$", "Notice that from the information given above, $x+y=4xy$\nBecause the sum of the reciprocals of two numbers is just the sum of the two numbers over the product of the two numbers or $\\frac{x+y}{xy}$\nWe can solve this by substituting $x+y\\implies 4xy$\nOur answer is simply $\\frac{4xy}{xy}\\implies4$\nTherefore, the answer is $\\boxed{4}$" ]

[ "Let $x, y$ be our two numbers. Then $x+y = 4xy$ . Thus,\n$\\frac{1}{x} + \\frac{1}{y} = \\frac{x+y}{xy} = 4$\n$\\boxed{4}$", "We can let $x=y.$ Then, we have that $2x=4x^2$ making $x=\\tfrac{1}{2}.$ The answer is $\\dfrac{2}{x}=4=\\boxed{4}.$ Solasky talk" ]

[ "$x+y=10$\n$xy=20$\n$\\frac1x+\\frac1y=\\frac{y}{xy}+\\frac{x}{xy}=\\frac{x+y}{xy}=\\frac{10}{20}=\\boxed{12}$" ]

[ "Let $a$ be the bigger number and $b$ be the smaller.\n$a + b = 5(a - b)$\nMultiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives\n$\\frac{a}{b} = \\frac32$ , so the answer is $\\boxed{32}$", "Without loss of generality, let the two numbers be $3$ and $2$ , as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\\boxed{32}$" ]