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[ "Recall that the sum of the first $100$ positive integers is $\\sum^{100}_{k=1} k = \\frac{101\\cdot100}{2}=5050.$ It follows that \\begin{align*} \\sum^{100}_{i=1} \\sum^{100}_{j=1} (i+j) &= \\sum^{100}_{i=1} \\sum^{100}_{j=1}i + \\sum^{100}_{i=1} \\sum^{100}_{j=1}j \\\\ &= \\sum^{100}_{i=1} (100i) + 100 \\sum^{100}_{j=1}j \\\\ &= 100 \\sum^{100}_{i=1}i + 100 \\sum^{100}_{j=1}j \\\\ &= 100\\cdot5050 + 100\\cdot5050 \\\\ &= \\boxed{1010000} ~MRENTHUSIASM", "Recall that the sum of the first $100$ positive integers is $\\sum^{100}_{k=1} k = \\frac{101\\cdot100}{2}=5050.$ It follows that \\begin{align*} \\sum^{100}_{i=1} \\sum^{100}_{j=1} (i+j) &= \\sum^{100}_{i=1} \\left(\\sum^{100}_{j=1} i + \\sum^{100}_{j=1} j\\right) \\\\ &= \\sum^{100}_{i=1} (100i+5050) \\\\ &= 100\\sum^{100}_{i=1} i + \\sum^{100}_{i=1} 5050 \\\\ &= 100\\cdot5050+5050\\cdot100 \\\\ &= \\boxed{1010000} ~Vfire ~MRENTHUSIASM", "Recall that the sum of the first $100$ positive integers is $\\sum^{100}_{k=1} k = \\frac{101\\cdot100}{2}=5050.$ Since the nested summation is symmetric with respect to $i$ and $j,$ it follows that \\begin{align*} \\sum^{100}_{i=1} \\sum^{100}_{j=1} (i+j) &= \\sum^{100}_{i=1} \\sum^{100}_{i=1} (2i) \\\\ &= 2\\sum^{100}_{i=1} \\sum^{100}_{i=1} i \\\\ &= 2\\sum^{100}_{i=1} 5050 \\\\ &= 2\\cdot(5050\\cdot100) \\\\ &= \\boxed{1010000} ~Vfire ~MRENTHUSIASM", "The sum contains $100\\cdot100=10000$ terms, and the average value of both $i$ and $j$ is $\\frac{101}{2}.$ Therefore, the sum becomes \\[10000\\cdot\\left(\\frac{101}{2}+\\frac{101}{2}\\right)=\\boxed{1010000}.\\] ~Rejas ~MRENTHUSIASM", "We start by writing out the first few terms: \\[\\begin{array}{ccccccccc} (1+1) &+ &(1+2) &+ &(1+3) &+ &\\cdots &+ &(1+100) \\\\ (2+1) &+ &(2+2) &+ &(2+3) &+ &\\cdots &+ &(2+100) \\\\ (3+1) &+ &(3+2) &+ &(3+3) &+ &\\cdots &+ &(3+100) \\\\ [-1ex] &&&&\\vdots&&&& \\\\ (100+1) &+ &(100+2) &+ &(100+3) &+ &\\cdots &+ &(100+100). \\end{array}\\] From the first terms in the parentheses, the sum $1+2+3+\\cdots+100$ occurs $100$ times vertically.\nFrom the second terms in the parentheses, the sum $1+2+3+\\cdots+100$ occurs $100$ times horizontally.\nRecall that the sum of the first $100$ positive integers is $1+2+3+\\cdots+100=\\frac{101\\cdot100}{2}=5050.$ Therefore, the answer is \\[2\\cdot\\left(5050\\cdot100\\right)=\\boxed{1010000}.\\] ~RandomPieKevin ‎~MRENTHUSIASM", "When we expand the nested summation as shown in Solution 5, note that:\nTogether, the nested summation becomes \\begin{align*} \\sum^{100}_{k=1}\\left[(k+1)k\\right] + \\sum^{99}_{k=1}\\left[(k+101)(100-k)\\right] &= \\sum^{100}_{k=1}\\left[k^2+k\\right] + \\sum^{99}_{k=1}\\left[-k^2-k+10100\\right] \\\\ &= \\sum^{100}_{k=1}k^2 + \\sum^{100}_{k=1}k - \\sum^{99}_{k=1}k^2 - \\sum^{99}_{k=1}k + \\sum^{99}_{k=1}10100 \\\\ &= \\left(\\sum^{100}_{k=1}k^2 - \\sum^{99}_{k=1}k^2\\right) + \\left(\\sum^{100}_{k=1}k - \\sum^{99}_{k=1}k\\right) + \\sum^{99}_{k=1}10100 \\\\ &= 100^2+100+10100\\cdot99 \\\\ &= \\boxed{1010000} ~MRENTHUSIASM" ]

[ "To get from the $1^\\text{st}$ term of an arithmetic sequence to the $100^\\text{th}$ term, we must add the common difference $99$ times. The first term is $1$ and the common difference is $5-1=9-5=13-9=\\cdots = 4$ , so the $100^\\text{th}$ term is \\[1+4(99)=397 \\rightarrow \\boxed{397}\\]", "Alternatively you could create an equation for the arithmetic sequence: $a_{n}=a_{1}+4(n-1)=1+4n-4=4n-3$\n$a_{100}=4(100)-3=397$ , or $\\boxed{397}$" ]

[ "$\\frac{4}{37}=\\frac{12}{111}=\\frac{108}{999}=0.108108108...$\nSince this repeats every three digits, digit number x = digit number (x mod 3), and the 100th digit = (100 mod 3)th digit = 1st digit = $\\boxed{1}$" ]

[ "If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if $|a|=b$ , then $a$ is either $b$ or $-b$ ):\n\\begin{align*} 3x+4y=12 \\\\ -3x+4y=12 \\\\ 3x-4y=12 \\\\ -3x-4y=12 \\end{align*}\nWe can then put these equations in slope-intercept form in order to graph them.\n\\begin{align*} 3x+4y=12 \\,\\implies\\, y=-\\dfrac{3}{4}x+3\\\\ -3x+4y=12\\,\\implies\\, y=\\dfrac{3}{4}x+3\\\\ 3x-4y=12\\,\\implies\\, y=\\dfrac{3}{4}x-3\\\\ -3x-4y=12\\,\\implies\\, y=-\\dfrac{3}{4}x-3\\end{align*}\nNow you can graph the lines to find the shape of the graph:\n\nWe can easily see that it is a rhombus with diagonals of $6$ and $8$ . The area is $\\dfrac{1}{2}\\times 6\\times8$ , or $\\boxed{24}$", "The graph is symmetric with respect to both coordinate axes, and in the first quadrant it coincides with the graph of the line $3x + 4y = 12.$ Therefore the region is a rhombus, and the area is\n\\[\\text{Area} = 4\\left(\\frac{1}{2}(4\\cdot 3)\\right) = 24 \\rightarrow \\boxed{24}\\]" ]

[ "Consider the case when $x \\geq 0$ $y \\geq 0$ \\[x^2+y^2=x+y\\] \\[(x - \\frac{1}{2})^2+(y - \\frac{1}{2})^2=\\frac{1}{2}\\] Notice the circle intersect the axes at points $(0, 1)$ and $(1, 0)$ . Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of $\\frac{\\sqrt{2}}{2}$ and a triangle: \\[A = \\frac{\\pi}{4} +\\frac{1}{2}\\] Because of symmetry, the area is the same in all four quadrants.\nThe answer is $\\boxed{2}$" ]

[ "Another way to solve this problem is using cases. \nThough this may seem tedious, we only have to do one case since the area enclosed is symmetrical.\nThe equation for this figure is $x^2+y^2=|x|+|y|$ To make this as easy as possible,\nwe can make both $x$ and $y$ positive. Simplifying the equation for $x$ and $y$ being positive,\nwe get the equation $x^{2} +y^{2} -x-y = 0.$\nUsing the complete the square method, we get $\\left(x-\\frac{1}{2}\\right)^{2} + \\left(y-\\frac{1}{2}\\right)^{2}=\\frac{1}{2}$\nTherefore, the origin of this section of the shape is at $\\left(\\frac{1}{2}, \\frac{1}{2}\\right).$\nUsing the equation we can also see that the radius has a length of $\\frac{\\sqrt{2}}{2}$\nWith this shape we see that this shape can be cut into a right triangle and a semicircle.\nThe length of the hypotenuse of the triangle is $\\sqrt{2}$ so using special right triangles, we see that\nthe area of the triangle is $\\frac{1}{2}$ . The semicircle has the area of $\\frac{1}{4}\\pi$\nBut this is only $1$ case. There are $4$ cases in total so we have to multiply $\\frac{1}{2}+\\frac{1}{4}\\pi$ by $4$\nAfter multiplying, our answer is: \\[\\boxed{2}.\\]", "We solve with cases. The cases are:\nCase 1: $x\\geq0, y\\geq0.$ Case 2: $x\\geq0, y<0.$ Case 3: $x<0, y\\geq0.$ Case 4: $x<0, y<0.$\nWe can quickly realize that the whole figure is symmetrical; so when you figure out the first case, you get the first part is $\\left(x-\\dfrac12\\right)^2+\\left(x-\\dfrac12\\right)^2$ you can figure out the whole figure: (scaled 8x). The way we figure out the area is by splitting it the following way: \nWe know each of the red lines is a diameter of the circle which is $\\sqrt2$ . So the area of the red is 2. We know that the area of each semicircle is $\\dfrac14 \\pi$ so the area of the semicircles combines is $\\pi$ . Thus we get $\\boxed{2}.$" ]

[ "First consider, $|x-1|+|y-1| \\le 1.$ We can see that it is a square with a radius of $1$ (diagonal $\\sqrt{2}$ ). The area of the square is $\\sqrt{2}^2 = 2.$\nNext, we insert an absolute value sign into the equation and get $|x-1|+||y|-1| \\le 1.$ This will double the square reflecting over x-axis.\nSo now we have $2$ squares.\nFinally, we add one more absolute value and obtain $||x|-1|+||y|-1| \\le 1.$ This will double the squares as we reflect the $2$ squares we already have over the y-axis.\nConcluding, we have $4$ congruent squares. The total area is $4\\cdot2 =$ $\\boxed{8}$", "We first consider the lattice points that satisfy $||x|-1| = 0$ and $||y|-1| = 1$ . The lattice points satisfying these equations\nare $(1,0), (1,2), (1,-2), (-1,0), (-1,2),$ and $(-1,-2).$ By symmetry, we also have points $(0,1), (2,1), (-2,1), (0,-1), (2,-1),$ and $(-2,-1)$ when $||x|-1| = 1$ and $||y|-1| = 0$ . Graphing and connecting these points, we form 5 squares. However,\nwe can see that any point within the square in the middle does not satisfy the given inequality (take $(0,0)$ , for instance). As\nnoted in the above solution, each square has a diagonal $2$ for an area of $\\frac{2^2}{2} = 2$ , so the total area is $4\\cdot2 =$ $\\boxed{8}.$", "The value of $|x|$ and $|y|$ can be a maximum of 1 when the other is 0. Therefore the value of $x$ and $y$ range from -2 to 2. This forms a diamond shape which has area $4 \\times \\frac{2^2}{2}$ which is $\\boxed{8}.$", "We start by considering the graph of $|x|+|y|\\leq 1$ . To get from this graph to $||x|-1|+||y|-1| \\leq 1$ we have to translate it by $\\pm 1$ on the $x$ axis and $\\pm 1$ on the $y$ axis.\nGraphing $|x|+|y|\\leq 1$ we get a square with side length of $\\sqrt{2}$ , so the area of one of these squares is just $2$\nWe have to multiply by $4$ since there are $4$ combinations of shifting the $x$ and $y$ axis.\nSo we have $2\\times 4$ which is $\\boxed{8}$" ]

[ "First consider, $|x-1|+|y-1| \\le 1.$ We can see that it is a square with a radius of $1$ (diagonal $\\sqrt{2}$ ). The area of the square is $\\sqrt{2}^2 = 2.$\nNext, we insert an absolute value sign into the equation and get $|x-1|+||y|-1| \\le 1.$ This will double the square reflecting over x-axis.\nSo now we have $2$ squares.\nFinally, we add one more absolute value and obtain $||x|-1|+||y|-1| \\le 1.$ This will double the squares as we reflect the $2$ squares we already have over the y-axis.\nConcluding, we have $4$ congruent squares. The total area is $4\\cdot2 =$ $\\boxed{8}$", "We first consider the lattice points that satisfy $||x|-1| = 0$ and $||y|-1| = 1$ . The lattice points satisfying these equations\nare $(1,0), (1,2), (1,-2), (-1,0), (-1,2),$ and $(-1,-2).$ By symmetry, we also have points $(0,1), (2,1), (-2,1), (0,-1), (2,-1),$ and $(-2,-1)$ when $||x|-1| = 1$ and $||y|-1| = 0$ . Graphing and connecting these points, we form 5 squares. However,\nwe can see that any point within the square in the middle does not satisfy the given inequality (take $(0,0)$ , for instance). As\nnoted in the above solution, each square has a diagonal $2$ for an area of $\\frac{2^2}{2} = 2$ , so the total area is $4\\cdot2 =$ $\\boxed{8}.$", "The value of $|x|$ and $|y|$ can be a maximum of 1 when the other is 0. Therefore the value of $x$ and $y$ range from -2 to 2. This forms a diamond shape which has area $4 \\times \\frac{2^2}{2}$ which is $\\boxed{8}.$", "We start by considering the graph of $|x|+|y|\\leq 1$ . To get from this graph to $||x|-1|+||y|-1| \\leq 1$ we have to translate it by $\\pm 1$ on the $x$ axis and $\\pm 1$ on the $y$ axis.\nGraphing $|x|+|y|\\leq 1$ we get a square with side length of $\\sqrt{2}$ , so the area of one of these squares is just $2$\nWe have to multiply by $4$ since there are $4$ combinations of shifting the $x$ and $y$ axis.\nSo we have $2\\times 4$ which is $\\boxed{8}$" ]

[ "The line of symmetry divides the shaded figure into two congruent triangles, each with base $3$ and height $2.$\nTherefore, the area of the shaded figure is \\[2\\cdot\\left(\\frac12\\cdot3\\cdot2\\right)=2\\cdot3=\\boxed{6}.\\] ~MRENTHUSIASM ~Wilhelm Z", "To find the area of the shaded figure, we subtract the area of the smaller triangle (base $4$ and height $2$ ) from the area of the larger triangle (base $4$ and height $5$ ): \\[\\frac12\\cdot4\\cdot5-\\frac12\\cdot4\\cdot2=10-4=\\boxed{6}.\\] ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)", "The consecutive vertices of the shaded figure are $(1,0),(3,2),(5,0),$ and $(3,5).$ By the Shoelace Theorem , the area is \\[\\frac12\\cdot|(1\\cdot2+3\\cdot0+5\\cdot5+3\\cdot0)-(0\\cdot3+2\\cdot5+0\\cdot3+5\\cdot1)|=\\frac12\\cdot12=\\boxed{6}.\\] ~Taco12 ~I-AM-DA-KING", "We have $4$ lattice points in the interior and $6$ lattice points on the boundary. By Pick's Theorem , the area of the shaded figure is \\[4+\\frac{6}{2}-1 = 4+3-1 = \\boxed{6}.\\] ~danprathab" ]

[ "The line of symmetry divides the shaded figure into two congruent triangles, each with base $3$ and height $2.$\nTherefore, the area of the shaded figure is \\[2\\cdot\\left(\\frac12\\cdot3\\cdot2\\right)=2\\cdot3=\\boxed{6}.\\] ~MRENTHUSIASM ~Wilhelm Z", "To find the area of the shaded figure, we subtract the area of the smaller triangle (base $4$ and height $2$ ) from the area of the larger triangle (base $4$ and height $5$ ): \\[\\frac12\\cdot4\\cdot5-\\frac12\\cdot4\\cdot2=10-4=\\boxed{6}.\\] ~MRENTHUSIASM ~Steven Chen (www.professorchenedu.com)", "The consecutive vertices of the shaded figure are $(1,0),(3,2),(5,0),$ and $(3,5).$ By the Shoelace Theorem , the area is \\[\\frac12\\cdot|(1\\cdot2+3\\cdot0+5\\cdot5+3\\cdot0)-(0\\cdot3+2\\cdot5+0\\cdot3+5\\cdot1)|=\\frac12\\cdot12=\\boxed{6}.\\] ~Taco12 ~I-AM-DA-KING", "We have $4$ lattice points in the interior and $6$ lattice points on the boundary. By Pick's Theorem , the area of the shaded figure is \\[4+\\frac{6}{2}-1 = 4+3-1 = \\boxed{6}.\\] ~danprathab" ]

[ "The area of the square around the pinwheel is 25. The area of the pinwheel is equal to $\\text{the square } - \\text{ the white space.}$ Each of the four triangles have a base of 3 units and a height of 2.5 units, and so their combined area is 15 units squared. Then the unshaded space consists of the four triangles with total area of 15, and there are four white corner squares. Therefore the area of the pinwheel is $25-(15+4)$ which is $\\boxed{6}$", "We'd like to use Pick's Theorem on one of the kites, except it doesn't immediately apply since there is a single vertex (in the middle of the diagram) of each kite that does not lie on a lattice point.\nWe can remedy this be pretending the figure is twice as big:\n\nNow we can safely use Pick's Theorem:\n\\[A=\\frac{b}{2}+i-1=\\frac{6}{2}+4-1=6\\]\nHowever since we scaled the figure's dimensions by $2$ , we scaled its area by $4$ (since the area of similar shapes scales quadratically with the scaling factor). Therefore the area of each kite is $\\frac{6}{4}$ and the area of all four kites combined is $\\boxed{6}$", "The area of any kite (concave OR convex) with diagonals $p$ $q$ is $\\frac{1}{2}pq$ . Let $p$ be the smaller diagonal and $q$ be the longer diagonal. Then by Pythagorean Theorem $p=\\sqrt{2}$ . Similarly, $q$ is $\\sqrt{2}$ less than half of the diagonal of the $5 \\times 5$ grid, or $q=\\frac{5\\sqrt{2}}{2}-\\sqrt{2}=\\frac{3\\sqrt{2}}{2}$ . Therefore the area of the four kites is just:\n\\[A=4\\cdot\\frac{1}{2}pq=4\\cdot\\frac{1}{2}\\cdot\\sqrt{2}\\cdot\\frac{3\\sqrt{2}}{2}=\\boxed{6}\\]" ]

[ "First, we need to find the coordinates where the graphs intersect.\nWe want the points x and y to be the same. Thus, we set $5=x+1,$ and get $x=4.$ Plugging this into the equation, $y=1-x,$ $y=5$ , and $y=1+x$ intersect at $(4,5)$ , we call this line x.\nDoing the same thing, we get $x=-4.$ Thus, $y=5$ . Also, $y=5$ and $y=1-x$ intersect at $(-4,5)$ , and we call this line y.\nIt's apparent the only solution to $1-x=1+x$ is $0.$ Thus, $y=1.$ $y=1-x$ and $y=1+x$ intersect at $(0,1)$ , we call this line z.\nUsing the Shoelace Theorem we get: \\[\\left(\\frac{(20-4)-(-20+4)}{2}\\right)=\\frac{32}{2}\\] $=$ So, our answer is $\\boxed{16.}$", "Graphing the lines, using the intersection points we found in Solution 1, we can see that the height of the triangle is 4, and the base is 8. Using the formula for the area of a triangle, we get $\\frac{4\\cdot8}{2}$ which is equal to $\\boxed{16}$", "$y = x + 1$ and $y = -x + 1$ have $y$ -intercepts at $(0, 1)$ and slopes of $1$ and $-1$ , respectively. Since the product of these slopes is $-1$ , the two lines are perpendicular. From $y = 5$ , we see that $(-4, 5)$ and $(4, 5)$ are the other two intersection points, and they are $8$ units apart. By symmetry, this triangle is a $45-45-90$ triangle, so the legs are $4\\sqrt{2}$ each and the area is $\\frac{(4\\sqrt{2})^2}{2} = \\boxed{16}$" ]

[ "We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are $4! = 24$ ways to arrange the other numbers, so each number appears in each spot $24$ times. Therefore, the sum of all such numbers is $24 \\times (1+3+5+7+8) \\times (11111) = 24 \\times 24 \\times 11111 = 6399936.$ Since there are $5! = 120$ such numbers, we divide $6399936 \\div 120$ to get $\\boxed{53332.8}$", "We can first solve for the mean for the digits $1, 3, 5, 7,$ and $9$ since each is $2$ away from each other. The mean of the numbers than can be solved using these digits is $55555$ . The total amount of numbers that can be formed using these digits is $5! =120$ . The sum of these numbers is $55555(120) = 6666600$ . Now we can find out the total value that was gained by replacing the $8$ with a $9$ . We can start how be calculating the gain when the $8$ was in the ones digit. Since there are $4! = 24$ numbers with the $8$ in the ones digit and $1$ was gain from each of them, $24$ is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of $24+240+2400+24000+240000=266664$ as the total amount that was gained. Subtract this amount from the sum of the digits using the $9$ instead of the $8$ to get $6666600-266664=6399936$ . Finally, we divide this by $120$ to get the average. $\\frac{6399936}{120}= \\boxed{53332.8}$", "The average value of the digits is $(1 + 3 + 5 + 7 + 8)/5 = 4.8$ .\nValues occur in every place so $4.8 * 11111 = \\boxed{53332.8}$" ]

[ "Remind that $\\text{slope}=\\dfrac{\\Delta y}{\\Delta x}=\\tan \\theta$ where $\\theta$ is the angle between the slope and $x$ -axis. $k_1=2=\\tan \\alpha$ $k_2=\\dfrac{1}{3}=\\tan \\beta$ . The angle formed by the two lines is $\\alpha-\\beta$ $\\tan(\\alpha-\\beta)=\\dfrac{\\tan\\alpha-\\tan\\beta}{1+\\tan\\alpha\\tan\\beta}=\\dfrac{2-1/3}{1+2\\cdot 1/3}=1$ . Therefore, $\\alpha-\\beta=\\boxed{45}$", "Follow Solution 2 up until the lattice points section. Let's use $(0,0)$ $(2,4)$ , and $(9,3)$ . The distance between the origin and $(2,4)$ is $\\sqrt{20}$ . The distance between the origin and $(9,3)$ is $\\sqrt{90}$ . The distance between $(2,4)$ and $(9,3)$ is $\\sqrt{50}$ . Using the Law of Cosines, we see the $50 = 90 + 20 - 2\\times\\sqrt{20}$ $\\times\\sqrt{90}$ $\\times\\cos(\\theta)$ , where $\\theta$ is the angle we are looking for.\nSimplifying, we get $-60 = -2\\times(\\sqrt{20}) \\times(\\sqrt{90}) \\times\\cos(\\theta)$\n$30 = \\sqrt{1800} \\times\\cos(\\theta)$\n$30 = 30\\sqrt{2} \\times\\cos(\\theta)$\n$\\frac{1}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2}= \\cos(\\theta)$\nThus, $\\theta = \\boxed{45}$", "We can set up vectors $\\vec{a} = <1,2>$ and $\\vec{b} = <3,1>$ to represent the two lines. We know that $\\frac{\\vec{a} \\cdot \\vec{b}}{|\\vec{a}| |\\vec{b}|} = \\cos \\theta$ . Plugging the vectors in gives us $\\cos \\theta = \\frac{5}{5\\sqrt{2}} = \\frac{1}{\\sqrt{2}}$ . From this we get that $\\theta = \\boxed{45}$", "Let $Z_1 = 3 + i$ and $Z_2 = 1 + 2i$ \\begin{align*} Z_2 &= Z_1 \\cdot re^{i\\theta} \\\\ 1+2i&=(3+i) \\cdot re^{i\\theta} \\\\ 1+2i&=(3 + i) \\cdot r(\\cos\\theta + i\\sin\\theta) \\\\ 1+2i&=3r\\cos\\theta - r\\sin\\theta + 3ri\\sin\\theta + ri\\cos\\theta \\\\ \\end{align*}\nFrom this we have: \\begin{align} 1 &= 3r\\cos\\theta - r\\sin\\theta \\\\ 2 &= r\\cos\\theta + 3r\\sin\\theta \\end{align}\nTo solve this we must compute $r$ \\begin{align*} r &= \\frac{|Z_2|}{|Z_1|} \\\\ &= \\frac{\\sqrt{5}}{\\sqrt{10}} \\\\ &= \\frac{\\sqrt{2}}{2} \\end{align*}\nUsing elimination we have: $3\\cdot(2) - (1)$ \\begin{align*} 5 &= 10r\\sin\\theta \\\\ \\frac{1}{2r} &= \\sin\\theta \\\\ \\frac{1}{2\\frac{\\sqrt{2}}{2}} &= \\sin\\theta \\\\ \\frac{\\sqrt{2}}{2} &= \\sin\\theta \\\\ \\theta &= \\boxed{45}", "The lines $y = 2x, y = \\frac {1}{3}x$ , and $x = 3$ form a large right triangle and a small right triangle. Call the angle that is formed by the x-axis and the line $y = 2x$ $\\alpha$ , and call the angle that is formed by the x-axis and the line $y = \\frac {1}{3}x$ $\\beta$ . We try to find $\\sin (\\alpha - \\beta)$ first, and then try to see if any of the answer choices match up.\n$\\sin (\\alpha - \\beta)$ $\\sin \\alpha$ $\\cos \\beta$ $\\sin \\beta$ $\\cos \\alpha$\nUsing soh-cah-toa, we find that $\\sin \\alpha = \\frac {2}{\\sqrt 5}, \\sin \\beta = \\frac {1}{\\sqrt 10}, \\cos \\alpha = \\frac {1}{\\sqrt 5},$ and $\\cos \\beta = \\frac {3}{\\sqrt 10}$\nPlugging it all in, we find that $\\sin (\\alpha - \\beta) = \\frac {5}{\\sqrt {50}}$ , which is equivalent to $\\frac {\\sqrt 2}{2}$ . Since $\\sin (\\alpha - \\beta) = \\frac {\\sqrt 2}{2}$ , we get that $\\alpha - \\beta = 45^{\\circ}$ . Therefore, the answer is $\\boxed{45}$" ]

[ "We can take any two lines of this form, since the angle between them will always be the same. Let's take $y=2x$ for the line with slope of 2 and $y=\\frac{1}{3}x$ for the line with slope of 1/3. Let's take 3 lattice points and create a triangle. Let's use $(0,0)$ $(1,2)$ , and $(3,1)$ . The distance between the origin and $(1,2)$ is $\\sqrt{5}$ . The distance between the origin and $(3,1)$ is $\\sqrt{10}$ . The distance between $(1,2)$ and $(3,1)$ is $\\sqrt{5}$ . We notice that we have a triangle with 3 side lengths: $\\sqrt{5}$ $\\sqrt{5}$ , and $\\sqrt{10}$ . This forms a 45-45-90 triangle, meaning that the angle is $\\boxed{45}$" ]

[ "At $10:00$ , the hour hand will be on the $10$ while the minute hand on the $12$\nThis makes them $\\frac{1}{6}$ th of a circle apart, and $\\frac{1}{6}\\cdot360^{\\circ}=\\boxed{60}$", "We know that the full clock is a circle, and therefore has 360 degrees. Considering that there are $12$ numbers, the distance between one number will be $360\\div 12=30$ .\nIf the time is $10:00$ , then the hour hand will be on $10$ , and the minute hand will be on, $12$ , making them $2$ numbers apart, so they will be $60$ degrees apart, or $\\boxed{60}$" ]

[ "The first $2003$ even counting numbers are $2,4,6,...,4006$\nThe first $2003$ odd counting numbers are $1,3,5,...,4005$\nThus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$\n$(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$\n$= 1+1+1+...+1 = \\boxed{2003}$", "Using the sum of an arithmetic progression formula, we can write this as $\\frac{2003}{2}(2 + 4006) - \\frac{2003}{2}(1 + 4005) = \\frac{2003}{2} \\cdot 2 = \\boxed{2003}$", "The formula for the sum of the first $n$ even numbers, is $S_E=n^{2}+n$ , (E standing for even).\nSum of first $n$ odd numbers, is $S_O=n^{2}$ , (O standing for odd).\nKnowing this, plug $2003$ for $n$\n$S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \\Rightarrow$ $\\boxed{2003}$", "In the case that we don't know if $0$ is considered an even number, we note that it doesn't matter! The sum of odd numbers is $O=1+3+5+...+4005$ . And the sum of even numbers is either $E_1=0+2+4...+4004$ or $E_2=2+4+6+...+4006$ . When compared to the sum of odd numbers, we see that each of the $n$ th term in the series of even numbers differ by $1$ . For example, take series $O$ and $E_1$ . The first terms are $1$ and $0$ . Their difference is $|1-0|=1$ . Similarly, take take series $O$ and $E_2$ . The first terms are $1$ and $2$ . Their difference is $|1-2|=1$ . Since there are $2003$ terms in each set, the answer $\\boxed{2003}$", "We can pair each term of the sums - the first even number with the first odd number, the second with the second, and so forth. Then, there are 2003 pairs with a difference of 1 in each pair - 2-1 is 1, 4-3 is 1, 6-5 is 1, and so on. Then, the solution is $1 \\cdot 2003$ , and the answer is $\\boxed{2003}$" ]

[ "The first $2003$ even counting numbers are $2,4,6,...,4006$\nThe first $2003$ odd counting numbers are $1,3,5,...,4005$\nThus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$\n$(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$\n$= 1+1+1+...+1 = \\boxed{2003}$", "Using the sum of an arithmetic progression formula, we can write this as $\\frac{2003}{2}(2 + 4006) - \\frac{2003}{2}(1 + 4005) = \\frac{2003}{2} \\cdot 2 = \\boxed{2003}$", "The formula for the sum of the first $n$ even numbers, is $S_E=n^{2}+n$ , (E standing for even).\nSum of first $n$ odd numbers, is $S_O=n^{2}$ , (O standing for odd).\nKnowing this, plug $2003$ for $n$\n$S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \\Rightarrow$ $\\boxed{2003}$", "In the case that we don't know if $0$ is considered an even number, we note that it doesn't matter! The sum of odd numbers is $O=1+3+5+...+4005$ . And the sum of even numbers is either $E_1=0+2+4...+4004$ or $E_2=2+4+6+...+4006$ . When compared to the sum of odd numbers, we see that each of the $n$ th term in the series of even numbers differ by $1$ . For example, take series $O$ and $E_1$ . The first terms are $1$ and $0$ . Their difference is $|1-0|=1$ . Similarly, take take series $O$ and $E_2$ . The first terms are $1$ and $2$ . Their difference is $|1-2|=1$ . Since there are $2003$ terms in each set, the answer $\\boxed{2003}$", "We can pair each term of the sums - the first even number with the first odd number, the second with the second, and so forth. Then, there are 2003 pairs with a difference of 1 in each pair - 2-1 is 1, 4-3 is 1, 6-5 is 1, and so on. Then, the solution is $1 \\cdot 2003$ , and the answer is $\\boxed{2003}$" ]

[ "We write \\[\\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\\frac{3^{96}}{3^{96}+2^{96}}\\cdot\\frac{3^{100}}{3^{96}}+\\frac{2^{96}}{3^{96}+2^{96}}\\cdot\\frac{2^{100}}{2^{96}}=\\frac{3^{96}}{3^{96}+2^{96}}\\cdot 81+\\frac{2^{96}}{3^{96}+2^{96}}\\cdot 16.\\] Hence we see that our number is a weighted average of 81 and 16, extremely heavily weighted toward 81. Hence the number is ever so slightly less than 81, so the answer is $\\boxed{80}$", "\\[\\frac{3^{100}+2^{100}}{3^{96}+2^{96}}=\\frac{2^{96}\\left(\\frac{3^{100}}{2^{96}}\\right)+2^{96}\\left(2^{4}\\right)}{2^{96}\\left(\\frac{3}{2}\\right)^{96}+2^{96}(1)}=\\frac{\\frac{3^{100}}{2^{96}}+2^{4}}{\\left(\\frac{3}{2}\\right)^{96}+1}=\\frac{\\frac{3^{100}}{2^{100}}\\cdot2^{4}+2^{4}}{\\left(\\frac{3}{2}\\right)^{96}+1}=\\frac{2^{4}\\left(\\frac{3^{100}}{2^{100}}+1\\right)}{\\left(\\frac{3}{2}\\right)^{96}+1}.\\]\nWe can ignore the 1's on the end because they won't really affect the fraction. So, the answer is very very very close but less than the new fraction.\n\\[\\frac{2^{4}\\left(\\frac{3^{100}}{2^{100}}+1\\right)}{\\left(\\frac{3}{2}\\right)^{96}+1}<\\frac{2^{4}\\left(\\frac{3^{100}}{2^{100}}\\right)}{\\left(\\frac{3}{2}\\right)^{96}},\\]\n\\[\\frac{2^{4}\\left(\\frac{3^{100}}{2^{100}}\\right)}{\\left(\\frac{3}{2}\\right)^{96}}=\\frac{3^{4}}{2^{4}}*2^{4}=3^{4}=81.\\]\nSo, our final answer is very close but not quite 81, and therefore the greatest integer less than the number is $\\boxed{80}$", "Let $x=3^{96}$ and $y=2^{96}$ . Then our fraction can be written as $\\frac{81x+16y}{x+y}=\\frac{16x+16y}{x+y}+\\frac{65x}{x+y}=16+\\frac{65x}{x+y}$ .\nNotice that $\\frac{65x}{x+y}<\\frac{65x}{x}=65$ .\nSo , $16+\\frac{65x}{x+y}<16+65=81$ .\nAnd our only answer choice less than 81 is $\\boxed{80}$ (RegularHexagon)", "Dividing by $2^{96}$ in both numerator and denominator, this fraction can be rewritten as \\[\\frac{81 \\times (1.5)^{96} + 16}{(1.5)^{96} + 1}.\\] Notice that the $+1$ and the $+16$ will be so insignificant compared to a number such as $(1.5)^{96},$ and that thereby the fraction will be ever so slightly less than $81$ . Thereby, we see that the answer is $\\boxed{80}.$", "If you multiply $(3^{96} + 2^{96})$ by $(3^{4} + 2^{4})$ (to get the exponent up to 100), you'll get $(3^{100} + 2^{100}) + 3^{96} \\cdot 2^{4} + 2^{96} \\cdot 3^{4}$ . Thus, in the numerator, if you add and subtract by $3^{96} \\cdot 2^{4}$ and $2^{96} \\cdot 3^{4}$ , you'll get $\\frac{(3^{4} + 2^{4})(3^{100} + 2^{100}) - 3^{96} \\cdot 2^{4} - 2^{96} \\cdot 3^{4}}{3^{96}+2^{96}}$ . You can then take out out the first number to get $3^{4} + 2^{4} - \\frac{3^{96} \\cdot 2^{4} + 2^{96} \\cdot 3^{4}}{3^{96}+2^{96}}$ . This can then be written as $87 - \\frac{16 \\cdot 3^{96} + 16 \\cdot 2^{96} + 75 \\cdot 2^{96}}{3^{96}+2^{96}}$ , factoring out the 16 and splitting the fraction will give you $87 - 16 - \\frac{65 \\cdot 2^{96}}{3^{96}+2^{96}}$ , giving you $81 - \\frac{65 \\cdot 2^{96}}{3^{96}+2^{96}}$ . While you can roughly say that $\\frac{65 \\cdot 2^{96}}{3^{96}+2^{96}} < 1$ you can also notice that the only answer choice less than 81 is 80, thus the answer is $\\boxed{80}.$", "If you factor out $3^{100}$ from the numerator and $3^{96}$ from the denominator, you will get $\\frac{3^{100}\\left(1+(\\frac{2}{3}\\right)^{100})}{3^{96}\\left(1+(\\frac{2}{3}\\right)^{96})}$ . Divide the numerator and denominator by $3^{96}$ to get $\\frac{81\\left(1+(\\frac{2}{3}\\right)^{100})}{\\left(1+(\\frac{2}{3}\\right)^{96})}$ . We see that every time we multiply $\\frac{2}{3}$ by itself, it slightly decreases, so $1+(\\frac{2}{3})^{100}$ will be ever so slightly smaller than $1+(\\frac{2}{3})^{96}$ . Thus, the decimal representation of $\\frac{\\left(1+(\\frac{2}{3}\\right)^{100})}{\\left(1+(\\frac{2}{3}\\right)^{96})}$ will be extremely close to $1$ , so our solution will be the largest integer that is less than $81$ . Thus, the answer is $\\boxed{80}.$" ]

[ "We might at first think that the answer would be $9$ , because $1+2+3 \\dots +n = 45$ when $n = 9$ . But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence $-44, -43, \\cdots, 44, 45$ cancels out except $45$ . Thus, the answer is, intuitively, $\\boxed{90}$ integers.", "To maximize the number of integers, we need to make the average of them as low as possible while still being positive. The average can be $\\frac12$ if the middle two numbers are $0$ and $1$ , so the answer is $\\frac{45}{\\frac12}=\\boxed{90}$" ]

[ "We might at first think that the answer would be $9$ , because $1+2+3 \\dots +n = 45$ when $n = 9$ . But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence $-44, -43, \\cdots, 44, 45$ cancels out except $45$ . Thus, the answer is, intuitively, $\\boxed{90}$ integers.", "To maximize the number of integers, we need to make the average of them as low as possible while still being positive. The average can be $\\frac12$ if the middle two numbers are $0$ and $1$ , so the answer is $\\frac{45}{\\frac12}=\\boxed{90}$" ]

[ "Observe that $2019_{10} = 5613_7$ . To maximize the sum of the digits, we want as many $6$ s as possible (since $6$ is the highest value in base $7$ ), and this will occur with either of the numbers $4666_7$ or $5566_7$ . Thus, the answer is $4+6+6+6 = 5+5+6+6 = \\boxed{22}$", "Note that all base $7$ numbers with $5$ or more digits are in fact greater than $2019$ . Since the first answer that is possible using a $4$ digit number is $23$ , we start with the smallest base $7$ number that whose digits sum to $23$ , namely $5666_7$ . But this is greater than $2019_{10}$ , so we continue by trying $4666_7$ , which is less than 2019. So the answer is $\\boxed{22}$", "Again note that you want to maximize the number of $6$ s to get the maximum sum. Note that $666_7=342_{10}$ , so you have room to add a thousands digit base $7$ . Fix the $666$ in place and try different thousands digits, to get $4666_7$ as the number with the maximum sum of digits. The answer is $\\boxed{22}$" ]

[ "The sum of the first $n$ positive integers is $\\frac{(n)(n+1)}{2}$ , and we want this not to be a divisor of $n!$ (the product of the first $n$ positive integers). Notice that if and only if $n+1$ were composite, all of its factors would be less than or equal to $n$ , which means they would be able to cancel with the factors in $n!$ . Thus, the sum of $n$ positive integers would be a divisor of $n!$ when $n+1$ is composite. (Note: This is true for all positive integers except for 1 because 2 is not a divisor/factor of 1.) Hence in this case, $n+1$ must instead be prime. The greatest three-digit integer that is prime is $997$ , so we subtract $1$ to get $n=\\boxed{996}$", "As in Solution 1, we deduce that $n+1$ must be prime. If we can't immediately recall what the greatest three-digit prime is, we can instead use this result to eliminate answer choices as possible values of $n$ . Choices $A$ $C$ , and $E$ don't work because $n+1$ is even, and all even numbers are divisible by two, which makes choices $A$ $C$ , and $E$ composite and not prime. Choice $D$ also does not work since $999$ is divisible by $9$ , which means it's a composite number and not prime. Thus, the correct answer must be $\\boxed{996}$", "The sum of the first $n$ positive integers is $\\frac{(n)(n+1)}{2}$ and the product of the positive integers upto $n$ is $n!$ . The quotient of the two is -\n$\\frac{(2)(n!)}{(n)(n+1)}$\nwhich simplifies to $\\frac{(2)((n-1)!)}{n+1}$ . Thus, $n+1$ must be odd for the remainder to not be 0 (as $2$ will multiply with some number in $n!$ , cancelling out $n+1$ if it is even, which leaves us with the answer choices $n = 996$ and $n = 998$ . Notice that $n + 1$ must also be prime as otherwise there will be a factor of $n + 1$ in $2$ $n!$ somewhere. So either $997$ or $999$ must be prime - $999$ is obviously not prime as it is divisible by 9, so our answer should be $n$ where $n + 1 = 997$ , and so our answer is $n = 996$ or $\\boxed{996}$" ]

[ "Since $2011 \\equiv 11 \\pmod{1000},$ we know that $2011^{2011} \\equiv 11^{2011} \\pmod{1000}.$\nTo compute this, we use a clever application of the binomial theorem\n$\\begin{aligned} 11^{2011} &= (1+10)^{2011} \\\\ &= 1 + \\dbinom{2011}{1} \\cdot 10 + \\dbinom{2011}{2} \\cdot 10^2 + \\cdots \\end{aligned}$\nIn all of the other terms, the power of $10$ is greater than $2$ and so is equivalent to $0$ modulo $1000,$ which means we can ignore it. We have:\n$\\begin{aligned}11^{2011} &\\equiv 1 + 2011\\cdot 10 + \\dfrac{2011 \\cdot 2010}{2} \\cdot 100 \\\\ &\\equiv 1+20110 + \\dfrac{11\\cdot 10}{2} \\cdot 100\\\\ &= 1 + 20110 + 5500\\\\ &\\equiv 1 + 110 + 500\\\\&=611 \\pmod{1000} \\end{aligned}$\nTherefore, the hundreds digit is $\\boxed{6}.$", "We need to compute $2011^{2011} \\pmod{1000}.$ By the Chinese Remainder Theorem , it suffices to compute $2011^{2011} \\pmod{8}$ and $2011^{2011} \\pmod{125}.$\nIn modulo $8,$ we have $2011^4 \\equiv 1 \\pmod{8}$ by Euler's Theorem, and also $2011 \\equiv 3 \\pmod{8},$ so we have \\[2011^{2011} = (2011^4)^{502} \\cdot 2011^3 \\equiv 1^{502} \\cdot 3^3 \\equiv 3 \\pmod{8}.\\]\nIn modulo $125,$ we have $2011^{100} \\equiv 1 \\pmod{125}$ by Euler's Theorem, and also $2011 \\equiv 11 \\pmod{125}.$ Therefore, we have $\\begin{aligned} 2011^{2011} &= (2011^{100})^{20} \\cdot 2011^{11} \\\\ &\\equiv 1^{20} \\cdot 11^{11} \\\\ &= 121^5 \\cdot 11 \\\\ &= (-4)^5 \\cdot 11 = -1024 \\cdot 11 \\\\ &\\equiv -24 \\cdot 11 = -264 \\\\ &\\equiv 111 \\pmod{125}. \\end{aligned}$\nAfter finding the solution $2011^{2011} \\equiv 611 \\pmod{1000},$ we conclude it is the only one by the Chinese Remainder Theorem. Thus, the hundreds digit is $\\boxed{6}.$", "Notice that the hundreds digit of $2011^{2011}$ won't be affected by $2000$ . Essentially we could solve the problem by finding the hundreds digit of $11^{2011}$ . Powers of $11$ are special because they can be represented by the Pascal's Triangle. Drawing the triangle, there is a theorem that states the powers of $11$ can be found by reading rows of the triangle and adding extra numbers up. [add source] For example, the sixth row of the triangle is $1, 5, 10, 10, 5,$ and $1$ . Adding all numbers from right to left, we get $161051$ , which is also $11^5$ . In other words, each number is $10^n$ steps from the right side of the row. The hundreds digit is $0$ . We can do the same for $11^{2011}$ , but we only need to find the $3$ digits from the right. Observing, every $3$ number from the right is $1 + 2 + 3... + n$ . So to find the third number from the right on the row of $11^{2011}$ \\[f(11^n) = 1 + 2 + 3... + (n-1),\\] or $\\frac{(2010 \\cdot 2011)}{2}$ , or $2021055$ . The last digit is five, but we must remember to add the number on the right of it, which, by observing other rows is obviously $2011$ . We must carry the $1$ in $2011$ 's tens digit to the $5$ in $2021055$ 's unit digit to get $\\boxed{6}$ . The one at the very end of the row doesn't affect anything, so we can leave it alone.", "Since we are only looking at the last 3 digits and $2011^2$ has the same last 3 digits as $11^2$ , we can find $11^{2011}$ instead.\nAfter this, we can repeatedly multiply the last 3 digits by 11 and take the last 3 digits of that product. We discover that $11^{51}$ 's last 2 digits are -11, the same as $11^1$\nFrom this information, we can figure out $11^{11}$ and $11^{61}$ end in 611. Adding various multiples of 50 to the exponent gives us the fact that $11^{2011}$ 's last digits are 611. We get $\\boxed{6}$ .\n-ThisUsernameIsTaken", "We know that the hundreds digit of $2011^{2011}$ is just the hundreds digit of $11^{2011} \\equiv 011 \\pmod{1000}$ . If we actually take a look at the powers of $11 \\pmod{1000}$ , we notice a pattern:\n\\[11^{1} \\equiv 011 \\pmod{1000}\\] \\[11^{2} \\equiv 121 \\pmod{1000}\\] \\[11^{3} \\equiv 331 \\pmod{1000}\\] \\[11^{4} \\equiv 641 \\pmod{1000}\\] \\[11^{5} \\equiv 051 \\pmod{1000}\\] \\[11^{6} \\equiv 561 \\pmod{1000}\\] Notice how the units digit is always one, the tens digit is always the previous tens digit plus the ones digit (or one) and the hundreds digit is always the previous hundreds digit plus the previous tens digit. Knowing this, we confidently repeat this pattern without actually multiply the previous term by $11$ out (if you generally multiply a few numbers by $11$ , you can see why this pattern holds).\n\\[11^{7} \\equiv 171 \\pmod{1000}\\] \\[11^{8} \\equiv 881 \\pmod{1000}\\] \\[11^{9} \\equiv 691 \\pmod{1000}\\] \\[11^{10} \\equiv 601 \\pmod{1000}\\] \\[11^{11} \\equiv 611 \\pmod{1000}\\] Since $11^{11}$ ends in _ $11$ , the pattern \"cycles\" every $10$ times. $\\frac{2011}{10}=201$ remainder $1$ , meaning that $2011^{2011} \\equiv 11^{2011} \\equiv 611 \\pmod{1000}$ , giving us a hundreds digit of $\\boxed{6}$", "We know that the hundreds digit of $2011^{11}$ is just the hundreds digit of $11^{11}$ (mod 1000). If we actually take a look at the powers of 11, we notice a pattern:\n\\[11^{1} = 11\\]\n\\[11^{2} = 121\\]\n\\[11^{3} = 1331\\]\n\\[11^{4} = 14641\\]\nWe notice that the hundreds digit just increases by 1 starting from 1(the hundreds digit of $11^{1}$ ). All we have to do now is add all the numbers from 1 to 11 (since the hundreds digit increases by 1 with every power of 11 that comes after) to get the hundreds digit of $11^{11}$\n$1 + 2 + 3 + 4.... + 11 = 66$ . Thus the answer is $\\boxed{6}$" ]

[ "Expanding the binomial coefficient , we get ${200 \\choose 100}=\\frac{200!}{100!100!}$ . Let the required prime be $p$ ; then $10 \\le p < 100$ . If $p > 50$ , then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$ . The largest such prime is $\\boxed{061}$ , which is our answer.", "We know that \\[{200\\choose100}=\\frac{200!}{100!100!}\\] Since $p<100$ , there is at least $1$ factor of $p$ in each of the $100!$ in the denominator. Thus there must be at least $3$ factors of $p$ in the numerator $200!$ for $p$ to be a factor of $n=\\frac{200!}{100!100!}$ . (Note that here we assume the minimum because as $p$ goes larger in value, the number of factors of $p$ in a number decreases,)\nSo basically, $p$ is the largest prime number such that \\[\\left \\lfloor\\frac{200}{p}\\right \\rfloor>3\\] Since $p<\\frac{200}{3}=66.66...$ , the largest prime value for $p$ is $p=\\boxed{61}$" ]

[ "To maximize anything of the form $a-b$ , we maximize $a$ and minimize $b$ . The maximal element of the set is $12$ and the minimal element is $-16$ , so the maximal difference is \\[12-(-16)=28\\rightarrow \\boxed{28}.\\]" ]

[ "Take an even positive integer $x$ $x$ is either $0 \\bmod{6}$ $2 \\bmod{6}$ , or $4 \\bmod{6}$ . Notice that the numbers $9$ $15$ $21$ , ... , and in general $9 + 6n$ for nonnegative $n$ are odd composites. We now have 3 cases:\nIf $x \\ge 18$ and is $0 \\bmod{6}$ $x$ can be expressed as $9 + (9+6n)$ for some nonnegative $n$ . Note that $9$ and $9+6n$ are both odd composites.\nIf $x\\ge 44$ and is $2 \\bmod{6}$ $x$ can be expressed as $35 + (9+6n)$ for some nonnegative $n$ . Note that $35$ and $9+6n$ are both odd composites.\nIf $x\\ge 34$ and is $4 \\bmod{6}$ $x$ can be expressed as $25 + (9+6n)$ for some nonnegative $n$ . Note that $25$ and $9+6n$ are both odd composites.\nClearly, if $x \\ge 44$ , it can be expressed as a sum of 2 odd composites. However, if $x = 42$ , it can also be expressed using case 1, and if $x = 40$ , using case 3. $38$ is the largest even integer that our cases do not cover. If we examine the possible ways of splitting $38$ into two addends, we see that no pair of odd composites add to $38$ . Therefore, $\\boxed{038}$ is the largest possible number that is not expressible as the sum of two odd composite numbers.", "Let $n$ be an integer that cannot be written as the sum of two odd composite numbers. If $n>33$ , then $n-9,n-15,n-21,n-25,n-27,$ and $n-33$ must all be prime (or $n-33=1$ , which yields $n=34=9+25$ which does not work). Thus $n-9,n-15,n-21,n-27,$ and $n-33$ form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is $5,11,17,23,$ and $29$ , yielding a maximal answer of 38. Since $38-25=13$ , which is prime, the answer is $\\boxed{038}$", "The easiest method is to notice that any odd number that ends in a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35...\nFor example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9(ex. 34, 44 . . .). Hence the largest number that ends with a 4 that satisfies the conditions is 14.\nIf you list out all the numbers(15, 27, 9, 21, 33), you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and any number greater that ends with a 3 is bad(ex. 58, 68. . .), so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be $\\boxed{38}$", "Claim: The answer is $\\boxed{038}$", "All numbers that could possibly work must be $2 \\cdot p$ where $p$ is prime. As previous solutions stated, the maximum number that could possibly work by Chicken McNugget is $9 \\cdot 25 - 9 - 25 = 225-34 = 191$ . We then bash from top to bottom:\n1. $178 = 89 \\cdot 2 => 87 + 91$ - refuted\n2. $166 = 83 \\cdot 2 => 81 + 85$ - refuted\n3. $158 = 79 \\cdot 2 => 77 + 81$ - refuted\n4. $146 = 73 \\cdot 2 => 69 + 77$ - refuted\n5. $142 = 71 \\cdot 2 => 65 + 77$ - refuted\n6. $134 = 67 \\cdot 2 => 65 + 69$ - refuted\n7. $122 = 61 \\cdot 2 => 57 + 65$ - refuted\n8. $118 = 59 \\cdot 2 => 55 + 63$ - refuted\n9. $106 = 53 \\cdot 2 => 51 + 55$ - refuted\n10. $94 = 47 \\cdot 2 => 45 + 49$ - refuted\n11. $86 = 43 \\cdot 2 => 35 + 51$ - refuted\n12. $82 = 41 \\cdot 2 => 33 + 49$ - refuted\n13. $74 = 37 \\cdot 2 => 35 + 39$ - refuted\n14. $62 = 31 \\cdot 2 => 27 + 35$ - refuted\n15. $58 = 29 \\cdot 2 => 25 + 33$ - refuted\n16. $46 = 23 \\cdot 2 => 21 + 25$ - refuted\n17. $38 = 19 \\cdot 2 => = 19 + 19$ - it works!\nBecause we did a very systematic bash as shown, we are confident the answer is $\\boxed{038}$", "As stated above, all numbers that could possibly work must be $2 \\cdot p$ where $p$ is prime. If $p$ > 30, we consider $p$ by modulo 30. $p$ could be 1,7,11,13,17,19,23,29 modulo 30. $2 \\cdot p$ can be expressed as ( $p$ $q$ )+( $p$ $q$ ) for some positive, even $q$ less then $p$\nIf $p$ $1 \\bmod{30}$ , p±4 would both be composite\nIf $p$ $7 \\bmod{30}$ , p±2 would both be composite\nIf $p$ $11 \\bmod{30}$ , p±14 would both be composite\nIf $p$ $13 \\bmod{30}$ , p±8 would both be composite\nIf $p$ $17 \\bmod{30}$ , p±8 would both be composite\nIf $p$ $19 \\bmod{30}$ , p±14 would both be composite\nIf $p$ $23 \\bmod{30}$ , p±2 would both be composite\nIf $p$ $29 \\bmod{30}$ , p±4 would both be composite\nSo $p$ < 30\nFrom here, just try all possible p and find the answer is $\\boxed{038}$" ]

[ "For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is $3 \\cdot 5 = 15$ , so ${\\boxed{15}$ is the correct answer.", "We'll just test all the answer choices.\nNote that for any 3 consecutive odd integers, there will be exactly one multiple of $3.$\nLet's list all possibilities of 3 consecutive odd integers. (multiple of 3, not multiple of 3, not multiple of 3), (not multiple of 3, multiple of 3, not multiple of 3) and (not multiple of 3, not multiple of 3, multiple of 3)\nTo support this further, list the first few consecutive lists of 3 consecutive odd integers.\nWe have $(1, 3, 5), (3, 5, 7), (5, 7, 9), (7, 9, 11), (11, 13, 15), \\ldots$\nSo if the first two consecutive odd numbers aren't multiples of three, the last one must be a multiple of three.\nTherefore for five consecutive odd integers, there must be at least one multiple of three.\nIn the same fashion as we did above, note that for any 5 consecutive integers, there will also be exactly one multiple of $5.$\nTherefore, for any 5 consecutive odd integers, there must be exactly one multiple of five.\nWe can skip 7 since none of the answer choices are a multiple of 7.\nNow we try $11.$ 11 doesn't work since as we see the first set of 5 consecutive odd integers doesn't fit, namely $(1, 3, 5, 7, 9).$\nSince any 5 consecutive integers is divisible both by $3$ and $5$ , it also must be divisible by ${\\boxed{15}$ and no higher since we saw that $11$ does not work and that there is no answer choice that is multiple of $7$" ]

[ "For all consecutive odd integers, one of every five is a multiple of 5 and one of every three is a multiple of 3. The answer is $3 \\cdot 5 = 15$ , so ${\\boxed{15}$ is the correct answer.", "We'll just test all the answer choices.\nNote that for any 3 consecutive odd integers, there will be exactly one multiple of $3.$\nLet's list all possibilities of 3 consecutive odd integers. (multiple of 3, not multiple of 3, not multiple of 3), (not multiple of 3, multiple of 3, not multiple of 3) and (not multiple of 3, not multiple of 3, multiple of 3)\nTo support this further, list the first few consecutive lists of 3 consecutive odd integers.\nWe have $(1, 3, 5), (3, 5, 7), (5, 7, 9), (7, 9, 11), (11, 13, 15), \\ldots$\nSo if the first two consecutive odd numbers aren't multiples of three, the last one must be a multiple of three.\nTherefore for five consecutive odd integers, there must be at least one multiple of three.\nIn the same fashion as we did above, note that for any 5 consecutive integers, there will also be exactly one multiple of $5.$\nTherefore, for any 5 consecutive odd integers, there must be exactly one multiple of five.\nWe can skip 7 since none of the answer choices are a multiple of 7.\nNow we try $11.$ 11 doesn't work since as we see the first set of 5 consecutive odd integers doesn't fit, namely $(1, 3, 5, 7, 9).$\nSince any 5 consecutive integers is divisible both by $3$ and $5$ , it also must be divisible by ${\\boxed{15}$ and no higher since we saw that $11$ does not work and that there is no answer choice that is multiple of $7$" ]

[ "The sum of the interior angles of a hexagon is $720$ degrees. In a convex polygon, each angle must be strictly less than $180$ degrees.\nSix acute angles can only sum to less than $90\\cdot 6 = 540$ degrees, so six acute angles could not form a hexagon.\nFive acute angles and one obtuse angle can only sum to less than $90\\cdot 5 + 180 = 630$ degrees, so these angles could not form a hexagon.\nFour acute angles and two obtuse angles can only sum to less than $90\\cdot 4 + 180\\cdot 2 = 720$ degrees. This is a strict inequality, so these angles could not form a hexagon. (The limiting figure would be four right angles and two straight angles, which would really be a square with two \"extra\" points on two sides to form the straight angles.)\nThree acute angles and three obtuse angles work. For example, if you pick three acute angles of $80$ degrees, the three obtuse angles would be $160$ degrees and give a sum of $80\\cdot 3 + 160\\cdot 3 = 720$ degrees, which is a genuine hexagon. Thus, the answer is $\\boxed{3}$" ]

[ "We find that the volume of the larger block is $18$ , and the volume of the smaller block is $4$ . Dividing the two, we see that only a maximum of four $2$ by $2$ by $1$ blocks can fit inside the $3$ by $3$ by $2$ block. Drawing it out, we see that such a configuration is indeed possible. Therefore, the answer is $\\boxed{4}$" ]

[ "Flip the fractions and subtract one from all sides to yield \\[\\frac{7}{8}>\\frac{k}{n}>\\frac{6}{7}.\\] Multiply both sides by $56n$ to get \\[49n>56k>48n.\\] This is equivalent to find the largest value of $n$ such that there is only one multiple of 56 within the open interval between $48n$ and $49n$ . If $n=112,$ then $98>k>96$ and $k=97$ is the unique value. For $n\\geq 113,$ there is at least $(49\\cdot 113-48\\cdot 113)-1=112$ possible numbers for $k$ and there is one $k$ every 56 numbers. Hence, there must be at least two values of $k$ that work. So, the largest value of $n$ is $\\boxed{112}$", "Notice that in order for $k$ to be unique, $\\frac{n}{n + k+ 1} \\le \\frac{8}{15}$ and $\\frac{n}{n+ k-1} \\ge \\frac{7}{13}$ must be true. Solving these inequalities for $k$ yields $\\frac{7}{6}(k-1) \\le n \\le \\frac{8}{7}(k+1)$\nThus, we want to find $k$ such that $\\frac{7}{6}(k-1)\\le \\frac{8}{7}(k+1)$ . Solving this inequality yields $k \\le 97$ , and plugging this into $\\frac{n}{n+k} < \\frac{7}{13}$ in the original equation yields $n \\le 112$ so the answer is $\\boxed{112}$" ]

[ "Let our answer be $n$ . Write $n = 42a + b$ , where $a, b$ are positive integers and $0 \\leq b < 42$ . Then note that $b, b + 42, ... , b + 42(a-1)$ are all primes.\nIf $b$ is $0\\mod{5}$ , then $b = 5$ because $5$ is the only prime divisible by $5$ . We get $n = 215$ as our largest possibility in this case.\nIf $b$ is $1\\mod{5}$ , then $b + 2 \\times 42$ is divisible by $5$ and thus $a \\leq 2$ . Thus, $n \\leq 3 \\times 42 = 126 < 215$\nIf $b$ is $2\\mod{5}$ , then $b + 4 \\times 42$ is divisible by $5$ and thus $a \\leq 4$ . Thus, $n \\leq 5 \\times 42 = 210 < 215$\nIf $b$ is $3\\mod{5}$ , then $b + 1 \\times 42$ is divisible by $5$ and thus $a = 1$ . Thus, $n \\leq 2 \\times 42 = 84 < 215$\nIf $b$ is $4\\mod{5}$ , then $b + 3 \\times 42$ is divisible by $5$ and thus $a \\leq 3$ . Thus, $n \\leq 4 \\times 42 = 168 < 215$\nOur answer is $\\boxed{215}$" ]

[ "The distance between the two centers of the spheres can be determined via the distance formula in three dimensions: $\\sqrt{(12 - (-2))^2 + (8 - (-10))^2 + (-16 - 5)^2} = \\sqrt{14^2 + 18^2 + 21^2} = 31$ . The largest possible distance would be the sum of the two radii and the distance between the two centers, making it $19 + 87 + 31 = \\boxed{137}$" ]

[ "First, we use difference of squares on $13^4 - 11^4 = (13^2)^2 - (11^2)^2$ to get $13^4 - 11^4 = (13^2 + 11^2)(13^2 - 11^2)$ . Using difference of squares again and simplifying, we get $(169 + 121)(13+11)(13-11) = 290 \\cdot 24 \\cdot 2 = (2\\cdot 8 \\cdot 2) \\cdot (3 \\cdot 145)$ . Realizing that we don't need the right-hand side because it doesn't contain any factor of 2, we see that the greatest power of $2$ that is a divisor $13^4 - 11^4$ is $\\boxed{32}$", "Just like in the above solution, we use the difference-of-squares factorization, but only once to get $13^4-11^4=(13^2-11^2)(13^2+11^2).$ We can then compute that this is equal to $48\\cdot290.$ Note that $290=2\\cdot145$ (we don't need to factorize any further as $145$ is already odd) thus the largest power of $2$ that divides $290$ is only $2^1=2,$ while $48=2^4\\cdot3,$ so the largest power of $2$ that divides $48$ is $2^4=16.$ Hence, the largest power of $2$ that is a divisor of $13^4-11^4$ is $2\\cdot16=\\boxed{32}.$", "Let $n=13^4-11^4.$ We wish to find the largest power of $2$ that divides $n$\nDenote $v_p(k)$ as the largest exponent of $p$ in the prime factorization of $n$ . In this problem, we have $p=2$\nBy the Lifting the Exponent Lemma on $n$\n\\[v_2(13^4-11^4)=v_2(13-11)+v_2(4)+v_2(13+11)-1\\] \\[=v_2(2)+v_2(4)+v_2(24)-1\\] \\[=1+2+3-1=5.\\]\nTherefore, exponent of the largest power of $2$ that divids $13^4-11^4$ is $5,$ so the largest power of $2$ that divides this number is $2^5=\\boxed{32}$", "We can simply take 13 to the 4th power, which is 28561. We subtract that by 11 to the 4th power, which is 14641 (You can use Pascal's Triangle to find this). Finally, subtract the numbers to get 13920.\nTo test the options, since we need the largest one, we can go from top down. Testing, we see that both D and E are decimals,\nand 32 works. So, our answer is $\\boxed{32}.$" ]

[ "Let the two chosen numbers be $a$ and $b$ . To maximize the quotient, we first have either $a,b>0$ or $a,b<0$ , and from there we maximize $|a|$ and minimize $|b|$\nFor the case $a,b<0$ , we have $a=-24$ and $b=-2$ , which gives us $(-24)/(-2)=12$ . For the case $a,b>0$ , we have $a=8$ and $b=1$ , which gives us $8/1=8$\nSince $12>8$ , our answer is $\\boxed{12}$" ]

[ "Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$ . By the Trivial Inequality (all squares are nonnegative) the minimum value for this is $\\boxed{1}$ , which can be achieved at $x=y=0$", "We expand the original expression, then factor the result by grouping: \\begin{align*} (xy-1)^2+(x+y)^2&=\\left(x^2y^2-2xy+1\\right)+\\left(x^2+2xy+y^2\\right) \\\\ &=x^2y^2+x^2+y^2+1 \\\\ &=x^2\\left(y^2+1\\right)+\\left(y^2+1\\right) \\\\ &=\\left(x^2+1\\right)\\left(y^2+1\\right). \\end{align*} Clearly, both factors are positive. By the Trivial Inequality, we have \\[\\left(x^2+1\\right)\\left(y^2+1\\right)\\geq\\left(0+1\\right)\\left(0+1\\right)=\\boxed{1}.\\] Note that the least possible value of $(xy-1)^2+(x+y)^2$ occurs at $x=y=0.$", "Like solution 1, expand and simplify the original equation to $x^2+y^2+x^2y^2+1$ and let $f(x, y) = x^2+y^2+x^2y^2+1$ . To find local extrema, find where $\\nabla f(x, y) = \\boldsymbol{0}$ . First, find the first partial derivative with respect to x and y and find where they are $0$ \\[\\frac{\\partial f}{\\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \\implies x = 0\\] \\[\\frac{\\partial f}{\\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \\implies y = 0\\] Thus, there is a local extremum at $(0, 0)$ . Because this is the only extremum, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning $f(0, 0)$ is the minimum of $f(x, y)$ . Plugging $(0, 0)$ into $f(x, y)$ , we find 1 $\\implies \\boxed{1}$" ]

[ "Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$ . By the Trivial Inequality (all squares are nonnegative) the minimum value for this is $\\boxed{1}$ , which can be achieved at $x=y=0$", "We expand the original expression, then factor the result by grouping: \\begin{align*} (xy-1)^2+(x+y)^2&=\\left(x^2y^2-2xy+1\\right)+\\left(x^2+2xy+y^2\\right) \\\\ &=x^2y^2+x^2+y^2+1 \\\\ &=x^2\\left(y^2+1\\right)+\\left(y^2+1\\right) \\\\ &=\\left(x^2+1\\right)\\left(y^2+1\\right). \\end{align*} Clearly, both factors are positive. By the Trivial Inequality, we have \\[\\left(x^2+1\\right)\\left(y^2+1\\right)\\geq\\left(0+1\\right)\\left(0+1\\right)=\\boxed{1}.\\] Note that the least possible value of $(xy-1)^2+(x+y)^2$ occurs at $x=y=0.$", "Like solution 1, expand and simplify the original equation to $x^2+y^2+x^2y^2+1$ and let $f(x, y) = x^2+y^2+x^2y^2+1$ . To find local extrema, find where $\\nabla f(x, y) = \\boldsymbol{0}$ . First, find the first partial derivative with respect to x and y and find where they are $0$ \\[\\frac{\\partial f}{\\partial x} = 2x + 2xy^{2} = 2x(1 + y^{2}) = 0 \\implies x = 0\\] \\[\\frac{\\partial f}{\\partial y} = 2y + 2yx^{2} = 2y(1 + x^{2}) = 0 \\implies y = 0\\] Thus, there is a local extremum at $(0, 0)$ . Because this is the only extremum, we can assume that this is a minimum because the problem asks for the minimum (though this can also be proven using the partial second derivative test) and the global minimum since it's the only minimum, meaning $f(0, 0)$ is the minimum of $f(x, y)$ . Plugging $(0, 0)$ into $f(x, y)$ , we find 1 $\\implies \\boxed{1}$" ]

[ "Grouping the first and last terms and two middle terms gives $(x^2+5x+4)(x^2+5x+6)+2019$ , which can be simplified to $(x^2+5x+5)^2-1+2019$ . Noting that squares are nonnegative, and verifying that $x^2+5x+5=0$ for some real $x$ , the answer is $\\boxed{2018}$", "Let $a=x+\\tfrac{5}{2}$ . Then the expression $(x+1)(x+2)(x+3)(x+4)$ becomes $\\left(a-\\tfrac{3}{2}\\right)\\left(a-\\tfrac{1}{2}\\right)\\left(a+\\tfrac{1}{2}\\right)\\left(a+\\tfrac{3}{2}\\right)$\nWe can now use the difference of two squares to get $\\left(a^2-\\tfrac{9}{4}\\right)\\left(a^2-\\tfrac{1}{4}\\right)$ , and expand this to get $a^4-\\tfrac{5}{2}a^2+\\tfrac{9}{16}$\nRefactor this by completing the square to get $\\left(a^2-\\tfrac{5}{4}\\right)^2-1$ , which has a minimum value of $-1$ . The answer is thus $2019-1=\\boxed{2018}$", "Similar to Solution 1, grouping the first and last terms and the middle terms, we get $(x^2+5x+4)(x^2+5x+6)+2019$\nLetting $y=x^2+5x$ , we get the expression $(y+4)(y+6)+2019$ . Now, we can find the critical points of $(y+4)(y+6)$ to minimize the function:\n$\\frac{d}{dx}(y^2+10y+24)=0$\n$2y+10=0$\n$2y=-5$\n$y=-5,0$\nTo minimize the result, we use $y=-5$ . Hence, the minimum is $(-5+4)(-5+6)=-1$ , so $-1+2019 = \\boxed{2018}$", "The expression is negative when an odd number of the factors are negative. This happens when $-2 < x < -1$ or $-4 < x < -3$ . Plugging in $x = -\\frac32$ or $x = -\\frac72$ yields $-\\frac{15}{16}$ , which is very close to $-1$ . Thus the answer is $-1 + 2019 = \\boxed{2018}$", "Answer choices $C$ $D$ , and $E$ are impossible, since $(x+1)(x+2)(x+3)(x+4)$ can be negative (as seen when e.g. $x = -\\frac{3}{2}$ ). Plug in $x = -\\frac{3}{2}$ to see that it becomes $2019 - \\frac{15}{16}$ , so round this to $\\boxed{2018}$", "We can ignore the $2019$ and consider it later, as it is a constant. By difference of squares, we can group this into $\\left((x+2.5)^2-0.5^2\\right)\\left((x+2.5)^2-1.5^2\\right)$ . We pull a factor of $4$ into each term to avoid dealing with decimals:\n\\[\\dfrac{\\left((2x+5)^2-1\\right)\\left((2x+5)^2-9\\right)}{16}.\\]\nNow, we let $a=2x+5$ . Our expression becomes:\n\\[\\dfrac{(a-1)(a-9)}{16}=\\dfrac{a^2-10a+9}{16}.\\]\nTaking the derivative, we get $\\dfrac{2a-10}{16}=\\dfrac{a-5}8.$ This is equal to $0$ when $a=5$ , and plugging in $a=5$ , we get the expression is equal to $-1$ and therefore our answer is $2019-1=\\boxed{2018}.$" ]

[ "The volume of the cube is $V_{\\text{cube}}=6^3=216,$ and the volume of a clay ball is $V_{\\text{ball}}=\\frac43\\cdot\\pi\\cdot2^3=\\frac{32}{3}\\pi.$\nSince the balls can be reshaped but not compressed, the maximum number of balls that can completely fit inside a cube is \\[\\left\\lfloor\\frac{V_{\\text{cube}}}{V_{\\text{ball}}}\\right\\rfloor=\\left\\lfloor\\frac{81}{4\\pi}\\right\\rfloor.\\] Approximating with $\\pi\\approx3.14,$ we have $12<4\\pi<13,$ or $\\left\\lfloor\\frac{81}{13}\\right\\rfloor \\leq \\left\\lfloor\\frac{81}{4\\pi}\\right\\rfloor \\leq \\left\\lfloor\\frac{81}{12}\\right\\rfloor.$ We simplify to get \\[6 \\leq \\left\\lfloor\\frac{81}{4\\pi}\\right\\rfloor \\leq 6,\\] from which $\\left\\lfloor\\frac{81}{4\\pi}\\right\\rfloor=\\boxed{6}.$", "As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is $\\left\\lfloor\\frac{81}{4\\pi}\\right\\rfloor.$\nBy an underestimation $\\pi\\approx3,$ we have $4\\pi>12,$ or $\\frac{81}{4\\pi}<6\\frac34.$\nBy an overestimation $\\pi\\approx\\frac{22}{7},$ we have $4\\pi<\\frac{88}{7},$ or $\\frac{81}{4\\pi}>6\\frac{39}{88}.$\nTogether, we get \\[6 < 6\\frac{39}{88} < \\frac{81}{4\\pi} < 6\\frac34 < 7,\\] from which $\\left\\lfloor\\frac{81}{4\\pi}\\right\\rfloor=\\boxed{6}.$", "As shown in Solution 1, we conclude that the maximum number of balls that can completely fit inside a cube is $\\left\\lfloor\\frac{81}{4\\pi}\\right\\rfloor.$\nApproximating with $\\pi\\approx3,$ we have $\\frac{81}{4\\pi}\\approx6\\frac34.$ Since $\\pi$ is about $5\\%$ greater than $3,$ it is safe to claim that $\\left\\lfloor\\frac{81}{4\\pi}\\right\\rfloor=\\boxed{6}.$" ]

[ "The intersections of the two polynomials, $p(x)$ and $q(x)$ , are precisely the roots of the equation $p(x)=q(x) \\rightarrow p(x) - q(x) = 0$ . Since the leading coefficients of both polynomials are $1$ , the degree of $p(x) - q(x) = 0$ is at most three, and the maximum point of intersection is three, because a third degree polynomial can have at most three roots. Thus, the answer is $\\boxed{3}$" ]

[ "Circle-triangle problem.PNG\nWe can draw a circle and a triangle, such that each side is tangent to the circle. This means that each side would intersect the circle at one point.\nYou would then have $3$ points, but what if the circle was bigger? Then, each side would intersect the circle at 2 points.\nTherefore, $2 \\times 3 = \\boxed{6}$", "We know that the maximum amount of points that a circle and a line segment can intersect is $2$ . Therefore, because there are $3$ line segments in a triangle, the maximum amount of points of intersection is $2 \\times 3 = \\boxed{6}$" ]

[ "Note that $k^2_1 + k^2_2 + ... + k^2_n = 2002 \\leq \\frac{n(n+1)(2n+1)}{6}$\nWhen $n = 17$ $\\frac{n(n+1)(2n+1)}{6} = \\frac{(17)(18)(35)}{6} = 1785 < 2002$\nWhen $n = 18$ $\\frac{n(n+1)(2n+1)}{6} = 1785 + 18^2 = 2109 > 2002$\nTherefore, we know $n \\leq 17$\nNow we must show that $n = 17$ works. We replace one of $1, 2, ... 17$ with an integer $a > 17$ to account for the amount under $2002$ , which is $2002-1785 = 217$\nEssentially, this boils down to writing $217$ as a difference of squares. We know $217 = (7)(31)$ , so we assume there exist positive integers $a$ and $b$ where $a > 17$ and $b \\leq 17$ such that $a^2 - b^2 = 217$\nWe can rewrite this as $(a+b)(a-b) = 217$ , so either $a+b = 217$ and $a-b = 1$ or $a+b = 31$ and $a-b = 7$ . We analyze each case separately.\nCase 1: $a+b = 217$ and $a-b = 1$\nSolving this system of equations gives $a = 109$ and $b = 108$ . However, $108 > 17$ , so this case does not yield a solution.\nCase 2: $a+b = 31$ and $a-b = 7$\nSolving this system of equations gives $a = 19$ and $b = 12$ . This satisfies all the requirements of the problem.\nThe list $1, 2 ... 11, 13, 14 ... 17, 19$ has $17$ terms whose sum of squares equals $2002$ . Therefore, the answer is $\\boxed{17}$" ]

[ "Imagine the clock as a circle. The minute hand will be at the 4 at 20 minutes past the hour. The central angle formed between $4$ and $5$ is $30$ degrees (since it is 1/12 of a full circle, 360). By $4:20$ , the hour hand would have moved $\\frac{1}{3}$ way from 4 to 5 since $\\frac{20}{60}$ is reducible to $\\frac{1}{3}$ . One third of the way from 4 to 5 is one third of 30 degrees, which is 10 degrees past the 4. Recall that the minute hand is at the 4, so the angle between them is $\\boxed{10}$" ]

[ "We can see that $44^2=1936$ which is less than 2020. Therefore, there are $2020-44=1976$ of the $4040$ numbers greater than $2020$ . Also, there are $2020+44=2064$ numbers that are less than or equal to $2020$\nSince there are $44$ duplicates/extras, it will shift up our median's placement down $44$ . Had the list of numbers been $1,2,3, \\dots, 4040$ , the median of the whole set would be $\\dfrac{1+4040}{2}=2020.5$\nThus, our answer is $2020.5-44=\\boxed{1976.5}$", "As we are trying to find the median of a $4040$ -term set, we must find the average of the $2020$ th and $2021$ st terms.\nSince $45^2 = 2025$ is slightly greater than $2020$ , we know that the $44$ perfect squares $1^2$ through $44^2$ are less than $2020$ , and the rest are greater. Thus, from the number $1$ to the number $2020$ , there are $2020 + 44 = 2064$ terms. Since $44^2$ is $44 + 45 = 89$ less than $45^2 = 2025$ and $84$ less than $2020$ , we will only need to consider the perfect square terms going down from the $2064$ th term, $2020$ , after going down $84$ terms. Since the $2020$ th and $2021$ st terms are only $44$ and $43$ terms away from the $2064$ th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$ . Averaging the two, we get $\\boxed{1976.5}.$", "We want to know the $2020$ th term and the $2021$ st term to get the median.\nWe know that $44^2=1936$ . So, numbers $1^2, 2^2, \\ldots,44^2$ are in between $1$ and $1936$\nSo, the sum of $44$ and $1936$ will result in $1980$ , which means that $1936$ is the $1980$ th number.\nAlso, notice that $45^2=2025$ , which is larger than $2021$\nThen the $2020$ th term will be $1936+40 = 1976$ , and similarly the $2021$ th term will be $1977$\nSolving for the median of the two numbers, we get $\\boxed{1976.5}$", "We note that $44^2 = 1936$ , which is the first square less than $2020$ , which means that there are $44$ additional terms before $2020$ . This makes $2020$ the $2064$ th term. To find the median, we need the $2020$ th and $2021$ st term. We note that every term before $2020$ is one less than the previous term (That is, we subtract $1$ to get the previous term.). If $2020$ is the $2064$ th term, than $2020 - 44$ is the $(2064 - 44)$ th term. So, the $2020$ th term is $1976$ , and the $2021$ st term is $1977$ , and the average of these two terms is the median, or $\\boxed{1976.5}$", "To find the median, we sort the $4040$ numbers in decreasing order, then average the $2020$ th and the $2021$ st numbers of the sorted list.\nSince $45^2=2025$ and $44^2=1936,$ the first $2021$ numbers of the sorted list are \\[\\underbrace{2020^2,2019^2,2018^2,\\ldots,46^2,45^2}_{1976\\mathrm{ \\ numbers}}\\phantom{ },\\phantom{ }\\underbrace{2020,2019,2018,\\ldots,1977,1976}_{45\\mathrm{ \\ numbers}}\\phantom{ },\\] from which the answer is $\\frac{1977+1976}{2}=\\boxed{1976.5}.$" ]

[ "We can see that $44^2=1936$ which is less than 2020. Therefore, there are $2020-44=1976$ of the $4040$ numbers greater than $2020$ . Also, there are $2020+44=2064$ numbers that are less than or equal to $2020$\nSince there are $44$ duplicates/extras, it will shift up our median's placement down $44$ . Had the list of numbers been $1,2,3, \\dots, 4040$ , the median of the whole set would be $\\dfrac{1+4040}{2}=2020.5$\nThus, our answer is $2020.5-44=\\boxed{1976.5}$", "As we are trying to find the median of a $4040$ -term set, we must find the average of the $2020$ th and $2021$ st terms.\nSince $45^2 = 2025$ is slightly greater than $2020$ , we know that the $44$ perfect squares $1^2$ through $44^2$ are less than $2020$ , and the rest are greater. Thus, from the number $1$ to the number $2020$ , there are $2020 + 44 = 2064$ terms. Since $44^2$ is $44 + 45 = 89$ less than $45^2 = 2025$ and $84$ less than $2020$ , we will only need to consider the perfect square terms going down from the $2064$ th term, $2020$ , after going down $84$ terms. Since the $2020$ th and $2021$ st terms are only $44$ and $43$ terms away from the $2064$ th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$ . Averaging the two, we get $\\boxed{1976.5}.$", "We want to know the $2020$ th term and the $2021$ st term to get the median.\nWe know that $44^2=1936$ . So, numbers $1^2, 2^2, \\ldots,44^2$ are in between $1$ and $1936$\nSo, the sum of $44$ and $1936$ will result in $1980$ , which means that $1936$ is the $1980$ th number.\nAlso, notice that $45^2=2025$ , which is larger than $2021$\nThen the $2020$ th term will be $1936+40 = 1976$ , and similarly the $2021$ th term will be $1977$\nSolving for the median of the two numbers, we get $\\boxed{1976.5}$", "We note that $44^2 = 1936$ , which is the first square less than $2020$ , which means that there are $44$ additional terms before $2020$ . This makes $2020$ the $2064$ th term. To find the median, we need the $2020$ th and $2021$ st term. We note that every term before $2020$ is one less than the previous term (That is, we subtract $1$ to get the previous term.). If $2020$ is the $2064$ th term, than $2020 - 44$ is the $(2064 - 44)$ th term. So, the $2020$ th term is $1976$ , and the $2021$ st term is $1977$ , and the average of these two terms is the median, or $\\boxed{1976.5}$", "To find the median, we sort the $4040$ numbers in decreasing order, then average the $2020$ th and the $2021$ st numbers of the sorted list.\nSince $45^2=2025$ and $44^2=1936,$ the first $2021$ numbers of the sorted list are \\[\\underbrace{2020^2,2019^2,2018^2,\\ldots,46^2,45^2}_{1976\\mathrm{ \\ numbers}}\\phantom{ },\\phantom{ }\\underbrace{2020,2019,2018,\\ldots,1977,1976}_{45\\mathrm{ \\ numbers}}\\phantom{ },\\] from which the answer is $\\frac{1977+1976}{2}=\\boxed{1976.5}.$" ]

[ "We can rewrite the fraction as $\\frac{123456789}{2^{22} \\cdot 10^4} = \\frac{12345.6789}{2^{22}}$ . Since the last digit of the numerator is odd, a $5$ is added to the right if the numerator is divided by $2$ , and this will continuously happen because $5$ , itself, is odd. Indeed, this happens twenty-two times since we divide by $2$ twenty-two times, so we will need $22$ more digits. Hence, the answer is $4 + 22 = \\boxed{26}$", "Multiply the numerator and denominator of the fraction by $5^{22}$ (which is the same as multiplying by 1) to give $\\frac{5^{22} \\cdot 123456789}{10^{26}}$ . Now, instead of thinking about this as a fraction, think of it as the division calculation $(5^{22} \\cdot 123456789) \\div 10^{26}$ . The dividend is a huge number, but we know it doesn't have any digits to the right of the decimal point. Also, the dividend is not a multiple of 10 (it's not a multiple of 2), so these 26 divisions by 10 will each shift the entire dividend one digit to the right of the decimal point. Thus, $\\boxed{26}$ is the minimum number of digits to the right of the decimal point needed.", "The denominator is $10^4 \\cdot 2^{22}$ . Each $10$ adds one digit to the right of the decimal, and each additional $2$ adds another digit. The answer is $4 + 22 = \\boxed{26}$" ]

[ "Rotating square $ABCD$ counterclockwise $45^\\circ$ so that the line of symmetry $BD$ is a vertical line makes it easier to see that $\\boxed{4}$ squares need to be colored to match its corresponding square." ]

[ "The only way to get a negative product using three numbers is to multiply one negative number and two positives or three negatives. Only two reasonable choices\nexist: $(-8)\\times(-6)\\times(-4) = (-8)\\times(24) = -192$ and $(-8)\\times5\\times7 = (-8)\\times35 = -280$ .\nThe latter is smaller, so $\\boxed{280}$" ]

[ "If we graph each term separately, we will notice that all of the zeros occur at $\\frac{1}{m}$ , where $m$ is any integer from $1$ to $119$ , inclusive: $|mx-1|=0\\implies mx=1\\implies x=\\frac{1}{m}$\nThe minimum value of $f(x)$ occurs where the absolute value of the sum of the slopes is at a minimum $\\ge 0$ , since it is easy to see that the value will be increasing on either side. That means the minimum must happen at some $\\frac{1}{m}$\nThe sum of the slopes at $x = \\frac{1}{m}$ is\n\\begin{align*}&\\sum_{i=m+1}^{119}i - \\sum_{i=1}^{m}i\\\\ &=\\sum_{i=1}^{119}i - 2\\sum_{i=1}^{m}i\\\\ &=-m^2-m+7140\\end{align*}\nNow we want to minimize $-m^2-m+7140$ . The zeros occur at $-85$ and $84$ , which means the slope is $0$ where $m = 84, 85$\nWe can now verify that both $x=\\frac{1}{84}$ and $x=\\frac{1}{85}$ yield $\\boxed{49}$" ]

[ "By the triangle inequality, $|x-1|+|2x-1|+|3x-1|+\\cdots + |119x-1| \\geq |(x-1)+(2x-1)+\\cdots+(119x)-1|.$ However, we may change signs of some of these terms to cancel out the $x$ 's.\nSince the minimum exists, we want all the $x$ s to cancel out. Thus, we want to find some $n$ such that \\[1+2+3+...+n=(n+1)+(n+2)+(n+3)+...+119\\] \\[\\frac{n(n+1)}{2}=\\frac{119\\cdot120}{2}-\\frac{n(n+1)}{2}\\] \\[n^2+n-7140=0\\] \\[n=84\\]\nThen, $x=\\frac{1}{n}= \\frac{1}{84}$ . The answer(expression's value) is then $84*1+(119-85+1)*(-1)$ , which becomes $84-35=\\boxed{49}$" ]

[ "The smaller angle makes up $5/12$ of the circle which is the clock. A circle is $360^\\circ$ , so the measure of the smaller angle is \\[\\frac{5}{12}\\cdot 360^\\circ = 150^\\circ \\rightarrow \\boxed{150}\\]" ]

[ "By the Binomial Theorem, each term in the expansion is of the form \\[\\binom{1000}{k}\\left(x\\sqrt[3]{2}\\right)^k\\left(y\\sqrt{3}\\right)^{1000-k}=\\binom{1000}{k}2^{\\frac k3}3^{\\frac{1000-k}{2}}x^k y^{1000-k},\\] where $k\\in\\{0,1,2,\\ldots,1000\\}.$\nThis problem is equivalent to counting the values of $k$ such that both $\\frac k3$ and $\\frac{1000-k}{2}$ are integers. Note that $k$ must be a multiple of $3$ and a multiple of $2,$ so $k$ must be a multiple of $6.$ There are $\\boxed{167}$ such values of $k:$ \\[6(0), 6(1), 6(2), \\ldots, 6(166).\\]" ]

[ "We can rewrite the expression as \\[222,222-(22,222+2,222+222+22+2).\\]\nWe note that the units digit of the addition is $0$ because all the units digits of the five numbers are $2$ and $5*2=10$ , which has a units digit of $0$\nNow, we have something with a units digit of $0$ subtracted from $222,222$ . The units digit of this expression is obviously $2$ , and we get $\\boxed{2}$ as our answer.", "222,222-22,222 = 200,000\n200,000 - 2,222 = 197778\n197778 - 222 = 197556\n197556 - 22 = 197534\n197534 - 2 = 1957532\nSo our answer is $\\boxed{2}$", "We only care about the unit's digits.\nThus, $2-2$ ends in $0$ $0-2$ ends in $8$ $8-2$ ends in $6$ $6-2$ ends in $4$ , and $4-2$ ends in $\\boxed{2}$", "We just take the units digit of each and subtract, or you can do it this way by adding an extra ten to the first number (so we don't get a negative number): \\[12-2-(2+2+2+2)=10-8=2\\] Thus, we get the answer $\\boxed{2}$" ]

[ " Notice that we are given a parametric form of the region, and $w$ is used in both $x$ and $y$ . We first fix $u$ and $v$ to $0$ , and graph $(-3w,4w)$ from $0\\le w\\le1$ . When $w$ is $0$ , we have the point $(0,0)$ , and when $w$ is $1$ , we have the point $(-3,4)$ . We see that since this is a directly proportional function, we can just connect the dots like this:\n\nNow, when we vary $u$ from $0$ to $2$ , this line is translated to the right $2$ units:\n\nWe know that any points in the region between the line (or rather segment) and its translation satisfy $w$ and $u$ , so we shade in the region:\n\nWe can also shift this quadrilateral one unit up, because of $v$ . Thus, this is our figure:\n\n\nThe length of the boundary is simply $1+2+5+1+2+5$ $5$ can be obtained by Pythagorean theorem since we have side lengths $3$ and $4$ .). This equals $\\boxed{16.}$", "We can find the \"boundary points\" and work with our intuition to solve the problem. We set each of $u, v, w$ equal to $0, 1$ for a total of $8$ combinations in $u, v, w$ . We now test each one.\nCase 1: $u = 0, v = 0, w = 0 \\implies (0, 0)$\nCase 2: $u = 0, v = 0, w = 1 \\implies (-3, 4)$\nCase 3: $u = 0, v = 1, w = 0 \\implies (0, 1)$\nCase 4: $u = 0, v = 1, w = 1 \\implies (-3, 5)$\nCase 5: $u = 1, v = 0, w = 0 \\implies (2, 0)$\nCase 6: $u = 1, v = 0, w = 1 \\implies (-1, 4)$\nCase 7: $u = 1, v = 1, w = 0 \\implies (2, 1)$\nCase 8: $u = 1, v = 1, w = 1 \\implies (-1, 5)$\nWhen graphed on a coordinate plane, the points appear as follows.\n\nNotice how there are two distinct rectangles visible in the figure. This leads us to believe that the region tracks the motion of this region as it travels in space. To understand why this is true, we can imagine a fixed $w$ (as it is present in both the $x$ and $y$ coordinates). Then if we hold one of $u$ or $v$ fixed and let the other vary, we get a straight line parallel to the $x$ or $y$ axis respectively. If we let the other vary, we get the other type of straight line. Together, they form a rectangular region. In addition, $w$ serves as a diagonal translation, so if we now let $w$ vary, it traces out the motion of the rectangle. Keeping this in mind, we connect the dots.\n\nEach of the diagonal sides have length $5$ by the distance formula on $(0,0)$ and $(-3,4)$ (the other diagonal side is congruent), so our total area is $2 + 1 + 5 + 2 + 1 + 5 = \\boxed{16}$" ]

[ "Draw altitudes from $B$ and $C$ to base $AD$ to create a rectangle and two right triangles. The side opposite $BC$ is equal to $50$ . The bases of the right triangles can be found using Pythagorean or special triangles to be $18$ and $7$ . Add it together to get $AD=18+50+7=75$ . The perimeter is $75+30+50+25=\\boxed{180}$" ]

[ "The numerator in each fraction cancels out with the denominator of the next fraction. There are only two numbers that didn't cancel: $\\frac{2006}{2}=\\boxed{1003}$" ]

[ "First, notice that there must be two such numbers: one greater than $\\log_69$ and one less than it. Furthermore, they both have to be the same distance away, namely $2(\\log_610 - 1)$ . Let these two numbers be $\\log_6a$ and $\\log_6b$ . Because they are equidistant from $\\log_69$ , we have $\\frac{\\log_6a + \\log_6b}{2} = \\log_69$ . Using log properties, this simplifies to $\\log_6{\\sqrt{ab}} = \\log_69$ . We then have $\\sqrt{ab} = 9$ , so $ab = \\boxed{81}$", "\\[\\log_6 9 - \\log_6 x_2 = -2d\\] Subbing in for $-2d$ and using log rules, \\[\\log_6 \\frac{9}{x_2} = \\log_6 \\frac{9}{25}\\] From this we conclude that \\[\\frac{9}{x_2} = \\frac{9}{25} \\implies x_2 = 25\\]\nFinding the product of the distinct values, $x_1x_2 = \\boxed{81}$" ]

[ "Let $a = 2 \\cdot |\\log_6 10 - 1| = |\\log_6 9 - \\log_6 x| = |\\log_6 \\frac{9}{x}|$\n$\\pm a = \\log_6 \\frac{9}{x} \\implies 6^{\\pm a} = b^{\\pm 1} = \\frac{9}{x} \\implies x = 9 \\cdot b^{\\pm 1}$\n$9b^1 \\cdot 9b^{-1} = \\boxed{81}$" ]

[ "For $\\log_{7x}2023\\cdot \\log_{289x}2023=\\log_{2023x}2023$ , transform it into $\\dfrac{\\ln 289+\\ln 7}{\\ln 7 + \\ln x}\\cdot \\dfrac{\\ln 289+\\ln 7}{\\ln 289 + \\ln x}=\\dfrac{\\ln 289+\\ln 7}{\\ln 289+\\ln 7+\\ln x}$ . Replace $\\ln x$ with $y$ . Because we want to find the product of all solutions of $x$ , it is equivalent to finding the exponential of the sum of all solutions of $y$ . Change the equation to standard quadratic equation form, the term with 1 power of $y$ is canceled. By using Vieta, we see that since there does not exist a $by$ term, $\\sum y=0$ and $\\prod x=e^0=\\boxed{1}$", "\\[\\log_{7x}2023\\cdot \\log_{289x}2023=\\log_{2023x}2023\\]\nRearranging it give us:\n\\[\\log_{2023}7x\\cdot \\log_{2023}289x=\\log_{2023}2023x\\]\n\\[(\\log_{2023}7+\\log_{2023}x)(\\log_{2023}289+\\log_{2023}x)=(\\log_{2023}2023+\\log_{2023}x)\\]\nlet $\\log_{2023}x$ be $a$ , we get\n\\[(\\log_{2023}7+a)(\\log_{2023}289+a)=1+a\\]\n\\[a^2+(\\log_{2023}7+\\log_{2023}289)a+\\log_{2023}7 \\cdot \\log_{2023}289=1+a\\]\n\\[a^2+\\log_{2023}7 \\cdot \\log_{2023}289-1=0\\]\nby Vieta's Formulas,\n\\[a_1+a_2=0\\]\n\\[\\log_{2023}{x_1}+\\log_{2023}{x_2}=0\\]\n\\[\\log_{2023}{x_1x_2}=0\\]\n\\[x_1x_2=\\boxed{1}\\]", "Similar to solution 1, change the bases first \\[\\frac{\\ln 289+\\ln 7}{\\ln7 + \\ln{x}} \\cdot \\frac{\\ln 289+\\ln 7}{2\\ln17 + \\ln{x}} = \\frac{\\ln 289+\\ln 7}{\\ln7 + 2\\ln17 + \\ln{x}}\\] Cancel and cross multiply to get \\[(\\ln7 + 2\\ln17)(\\ln7 + 2\\ln17 + \\ln{x}) = (\\ln7 + \\ln{x})(2\\ln17 + \\ln{x})\\] Simplify to get \\[(\\ln{x})^2 = 4(\\ln17)^2 + 2\\ln17\\ln7 + (\\ln7)^2\\] \\[\\ln{x} = \\pm \\sqrt{4(\\ln17)^2 + 2\\ln17\\ln7 + (\\ln7)^2}\\] The sum of all possible $\\ln{x}$ is 0, thus the product of all solutions of $x$ is $\\boxed{1}$", "We take the reciprocal of both sides: \\[\\frac{1}{\\log_{7x}2023}\\cdot \\frac{1}{\\log_{289x}2023}=\\frac{1}{\\log_{2023x}2023}.\\] Using logarithm properties, we have \\[\\log_{2023}7x\\cdot \\log_{2023}289x=\\log_{2023}2023x.\\] Simplify to obtain \\[2023x^2=2023x,\\] from which we have $x=\\boxed{1}$" ]

[ "First, square both sides, and isolate the absolute value. \\begin{align*} 5|x|+8&=x^2-16\\\\ 5|x|&=x^2-24\\\\ |x|&=\\frac{x^2-24}{5}. \\\\ \\end{align*} Solve for the absolute value and factor.\nCase 1: $x=\\frac{x^2-24}{5}$\nMultiplying both sides by $5$ gives us \\[5x=x^2-24.\\] Rearranging and factoring, we have \\begin{align*} x^2-5x-24 &=0, \\\\ (x-8)(x+3) &= 0.\\\\ \\end{align*}\nCase 2: $x=\\frac{-x^2+24}{5}$\nAs above, we multiply both sides by $5$ to find \\[5x=-x^2+24.\\] Rearranging and factoring gives us \\begin{align*} x^2+5x-24 &=0, \\\\ (x+8)(x-3) &= 0. \\\\ \\end{align*}\nCombining these cases, we have $x= -8, -3, 3, 8$ . Because our first step of squaring is not reversible, however, we need to check for extraneous solutions. Plug each solution for $x$ back into the original equation to ensure it works. Whether the number is positive or negative does not matter since the absolute value or square will cancel it out anyways.\nTrying $|x|=|3|$ , we have \\begin{align*} \\sqrt{5|3|+8}&=\\sqrt{3^2-16}, \\\\ \\sqrt{15+8}&=\\sqrt{9-16}, \\\\ \\sqrt{23} &\\not= \\sqrt{-7}.\\\\ \\end{align*} Therefore, $x = 3$ and $x= -3$ are extraneous.\nChecking $|x|=|8|$ , we have \\begin{align*} \\sqrt{5|8|+8}&=\\sqrt{8^2-16}, \\\\ \\sqrt{40+8}&=\\sqrt{64-16}, \\\\ \\sqrt{48}&=\\sqrt{48}.\\\\ \\end{align*}\nThe roots of our original equation are $-8$ and $8$ and product is $-8 \\times 8 = \\boxed{64}$", "Square both sides, to get $5|x| + 8 = x^2-16$ . Rearrange to get $x^2 - 5|x| - 24 = 0$ . Seeing that $x^2 = |x|^2$ , substitute to get $|x|^2 - 5|x| - 24 = 0$ . We see that this is a quadratic in $|x|$ . Factoring, we get $(|x|-8)(|x|+3) = 0$ , so $|x| = \\{8,-3\\}$ . Since the radicand of the equation can't be negative, the sole solution is $|x| = 8$ . Therefore, $x$ can be $8$ or $-8$ . The product is then $-8 \\times 8 = \\boxed{64}$", "First we note that $x \\in (-\\infty,-4] \\cup [4,\\infty]$ . This will help us later with finding extraneous solutions. \nNext, we have two cases: \\[\\text{IF } x\\leq 4:\\] \\[\\text{ } \\sqrt{-5x+8} = \\sqrt{x^2-16} \\implies x^2+5x-24=0 \\rightarrow x = -8,3\\] .\nWe note that $3$ is not in the range of possible $x$ 's and thus is not a solution.\n\\[\\text{IF } x\\geq 4:\\] \\[\\text{ } \\sqrt{5x+8} = \\sqrt{x^2-16} \\implies x^2-5x-24=0 \\rightarrow x = -3,8\\] .\nWe again not that $-3$ is an extraneous solution.\nThus, we have the two solutions $-8$ and $8$ . Therefore product is $-8 \\cdot 8 = \\boxed{64}$" ]

[ "If we were to expand by squaring, we would get a quartic polynomial , which isn't always the easiest thing to deal with.\nInstead, we substitute $y$ for $x^2+18x+30$ , so that the equation becomes $y=2\\sqrt{y+15}$\nNow we can square; solving for $y$ , we get $y=10$ or $y=-6$ . The second root is extraneous since $2\\sqrt{y+15}$ is always non-negative (and moreover, plugging in $y=-6$ , we get $-6=6$ , which is obviously false). Hence we have $y=10$ as the only solution for $y$ . Substituting $x^2+18x+30$ back in for $y$\nBoth of the roots of this equation are real, since its discriminant is $18^2 - 4 \\cdot 1 \\cdot 20 = 244$ , which is positive. Thus by Vieta's formulas , the product of the real roots is simply $\\boxed{020}$", "We begin by noticing that the polynomial on the left is $15$ less than the polynomial under the radical sign. Thus: \\[(x^2+ 18x + 45) - 2\\sqrt{x^2+18x+45} - 15 = 0.\\] Letting $n = \\sqrt{x^2+18x+45}$ , we have $n^2-2n-15 = 0 \\Longrightarrow (n-5)(n+3) = 0$ . Because the square root of a real number can't be negative, the only possible $n$ is $5$\nSubstituting that in, we have \\[\\sqrt{x^2+18x+45} = 5 \\Longrightarrow x^2 + 18x + 45 = 25 \\Longrightarrow x^2+18x+20=0.\\]\nReasoning as in Solution 1, the product of the roots is $\\boxed{020}$", "Begin by completing the square on both sides of the equation, which gives \\[(x+9)^2-51=2\\sqrt{(x+3)(x+15)}\\] Now by substituting $y=x+9$ , we get $y^2-51=2\\sqrt{(y-6)(y+6)}$ , or \\[y^4-106y^2+2745=0\\] The solutions in $y$ are then \\[y=x+9=\\pm3\\sqrt{5},\\pm\\sqrt{61}\\] Turns out, $\\pm3\\sqrt{5}$ are a pair of extraneous solutions. Thus, our answer is then \\[\\left(\\sqrt{61}-9\\right)\\left(-\\sqrt{61}-9\\right)=81-61=\\boxed{020}\\] By difference of squares.", "We are given the equation \\[x^2+18x+30=2\\sqrt{x^2+18x+45}\\] Squaring both sides yields \\[(x^2+18x+30)^2=4(x^2+18x+45)\\] \\[(x^2+18x+30)^2=4(x^2+18x+30+15)\\] \\[(x^2+18x+30)^2=4(x^2+18x+30)+60\\] \\[(x^2+18x+30)^2-4(x^2+18x+30)-60=0\\] Substituting $y=x^2+18x+30$ yields \\[y^2-4y-60=0\\] \\[(y+6)(y-10)=0\\] Thus $y=x^2+18x+30=-6,10$ . However if $y=-6$ , the left side of the equation \\[x^2+18x+30=2\\sqrt{x^2+18x+45}\\] would be negative while the right side is negative. Thus $y=10$ is the only possible value and we have \\[x^2+18x+30=10\\] \\[x^2+18x+20=0\\] Since the discriminant $\\sqrt{18^2-4\\cdot20}$ is real, both the roots are real. Thus by Vieta's Formulas, the product of the roots is the constant, $\\boxed{20}$" ]

[ "To find either the LCM or the GCF of two numbers, always prime factorize first.\nThe prime factorization of $180 = 3^2 \\times 5 \\times 2^2$\nThe prime factorization of $594 = 3^3 \\times 11 \\times 2$\nThen, to find the LCM, we have to find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is $3^3, 5, 11, 2^2$ ). Multiply all of these to get 5940.\nFor the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. $3^2 \\times 2$ = 18.\nThus the answer = $\\frac{5940}{18}$ $\\boxed{330}$", "We start off with a similar approach as the original solution. From the prime factorizations, the GCF is $18$\nIt is a well known fact that $\\gcd(m,n)\\times \\operatorname{lcm}(m,n)=|mn|$ . So we have, $18\\times \\operatorname{lcm} (180,594)=594\\times 180$\nDividing by $18$ yields $\\operatorname{lcm} (180,594)=594\\times 10=5940$\nTherefore, $\\frac{\\operatorname{lcm} (180,594)}{\\operatorname{gcf}(180,594)}=\\frac{5940}{18}=\\boxed{330}$", "From Solution 1, \nthe prime factorization of $180 = 2^2 \\cdot 3^2 \\cdot 5$ .\nThe prime factorization of $594 = 2 \\cdot 3^3 \\cdot 11$ .\nHence, $\\operatorname{lcm} (180,594) = 2^2 \\cdot 3^3 \\cdot 5 \\cdot 11$ , and $\\operatorname{gcf} (180,594) = 2 \\cdot 3^2$ .\nTherefore, $\\frac{\\operatorname{lcm} (180,594)}{\\operatorname{gcf} (180,594)} = \\frac{2^2 \\cdot 3^3 \\cdot 5 \\cdot 11}{2 \\cdot 3^2} = 2 \\cdot 3 \\cdot 5 \\cdot 11 = 330 \\Longrightarrow \\boxed{330}$" ]

[ "Completing the square, then difference of squares:\n\\begin{align*} 2^{202} + 202 &= (2^{101})^2 + 2\\cdot 2^{101} + 1 - 2\\cdot 2^{101} + 201\\\\ &= (2^{101} + 1)^2 - 2^{102} + 201\\\\ &= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201. \\end{align*}\nThus, we see that the remainder is surely $\\boxed{201}$", "Let $x=2^{50}$ . We are now looking for the remainder of $\\frac{4x^4+202}{2x^2+2x+1}$\nWe could proceed with polynomial division, but the numerator looks awfully similar to the Sophie Germain Identity , which states that \\[a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)\\]\nLet's use the identity, with $a=1$ and $b=x$ , so we have\n\\[1+4x^4=(1+2x^2+2x)(1+2x^2-2x)\\]\nRearranging, we can see that this is exactly what we need:\n\\[\\frac{4x^4+1}{2x^2+2x+1}=2x^2-2x+1\\]\nSo \\[\\frac{4x^4+202}{2x^2+2x+1} = \\frac{4x^4+1}{2x^2+2x+1} +\\frac{201}{2x^2+2x+1}\\]\nSince the first half divides cleanly as shown earlier, the remainder must be $\\boxed{201}$", "We let \\[x = 2^{50}\\] and \\[2^{202} + 202 = 4x^{4} + 202\\] .\nNext we write \\[2^{101} + 2^{51} + 1 = 2x^{2} + 2x + 1\\] .\nWe know that \\[4x^{4} + 1 = (2x^{2} + 2x + 1)(2x^{2} - 2x + 1)\\] by the Sophie Germain identity so to find \\[4x^{4} + 202,\\] we find that \\[4x^{4} + 202 = 4x^{4} + 201 + 1\\] which shows that the remainder is $\\boxed{201}$", "We let $x=2^{50.5}$ . That means $2^{202}+202=x^{4}+202$ and $2^{101}+2^{51}+1=x^{2}+x\\sqrt{2}+1$ . Then, we simply do polynomial division, and find that the remainder is $\\boxed{201}$", "Let $n=2^{101}+2^{51}+1$ . Then, mod $n$\n$2^{202}+202 \\equiv (-2^{51}-1)^2 + 202$\n$\\equiv 2^{102}+2^{52}+203$\n$= 2(n-1)+203 \\equiv 201 \\pmod{n}$\nThus, the remainder is $\\boxed{201}$", "We can repeatedly manipulate the numerator to make parts of it divisible by the denominator:\n\\[\\frac{2^{202}+202}{2^{101}+2^{51}+1}\\] \\[= \\frac{2^{202} + 2^{152} + 2^{101}}{2^{101}+2^{51}+1} - \\frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}\\] \\[= 2^{101} - \\frac{2^{152} + 2^{101} - 202}{2^{101}+2^{51}+1}\\] \\[=2^{101} - \\frac{2^{152}+2^{101}+2^{101}+2^{51} - 2^{101} - 2^{51} - 202}{2^{101}+2^{51}+1}\\] \\[=2^{101} - 2^{51} + \\frac{2^{101}+2^{51}+202}{2^{101}+2^{51}+1}\\] \\[= 2^{101} - 2^{51} + \\frac{2^{101}+2^{51}+1+201}{2^{101}+2^{51}+1}\\] \\[= 2^{101} - 2^{51} + 1 + \\frac{201}{2^{101} + 2^{51} + 1}.\\]\nClearly, $201 < 2^{201} + 2^{51} + 1$ , hence, we can not manipulate the numerator further to make the denominator divide into one of its parts. This concludes, that the remainder is $\\boxed{201}$" ]

[ "The sum of any four consecutive powers of 3 is divisible by $3^0 + 3^1 + 3^2 +3^3 = 40$ and hence is divisible by 8. Therefore\nis divisible by 8. So the required remainder is $3^0 + 3^1 = \\boxed{4}$" ]

[ "We have $3^2 = 9 \\equiv 1 \\pmod 8$ . Hence for any $k$ we have $3^{2k}\\equiv 1^k = 1 \\pmod 8$ , and then $3^{2k+1} = 3\\cdot 3^{2k} \\equiv 3\\cdot 1 = 3 \\pmod 8$\nTherefore our sum gives the same remainder modulo $8$ as $1 + 3 + 1 + 3 + 1 + \\cdots + 1 + 3$ . There are $2010$ terms in the sum, hence there are $2010/2 = 1005$ pairs of $1+3$ , and thus the sum is $1005 \\cdot 4 = 4020 \\equiv 20 \\equiv \\boxed{4} \\pmod 8$" ]

[ "From the Remainder Theorem , the remainder when $x^{51}+51$ is divided by $x+1$ is $(-1)^{51}+51=-1+51=50, \\boxed{50}$" ]

[ "Let's first obtain an algebraic expression for the root mean square of the first $n$ integers, which we denote $I_n$ . By repeatedly using the identity $(x+1)^3 = x^3 + 3x^2 + 3x + 1$ , we can write \\[1^3 + 3\\cdot 1^2 + 3 \\cdot 1 + 1 = 2^3,\\] \\[1^3 + 3 \\cdot(1^2 + 2^2) + 3 \\cdot (1 + 2) + 1 + 1 = 3^3,\\] and \\[1^3 + 3\\cdot(1^2 + 2^2 + 3^2) + 3 \\cdot (1 + 2 + 3) + 1 + 1 + 1 = 4^3.\\] We can continue this pattern indefinitely, and thus for any\npositive integer $n$ \\[1 + 3\\sum_{j=1}^n j^2 + 3 \\sum_{j=1}^n j^1 + \\sum_{j=1}^n j^0 = (n+1)^3.\\] Since $\\sum_{j=1}^n j = n(n+1)/2$ , we obtain \\[\\sum_{j=1}^n j^2 = \\frac{2n^3 + 3n^2 + n}{6}.\\] Therefore, \\[I_n = \\left(\\frac{1}{n} \\sum_{j=1}^n j^2\\right)^{1/2} = \\left(\\frac{2n^2 + 3n + 1}{6}\\right)^{1/2}.\\] Requiring that $I_n$ be an integer, we find that \\[(2n+1 ) (n+1) = 6k^2,\\] where $k$ is an integer. Using the Euclidean algorithm, we see that $\\gcd(2n+1, n+1) = \\gcd(n+1,n) = 1$ , and so $2n+1$ and $n+1$ share no\nfactors greater than 1. The equation above thus implies that $2n+1$ and $n+1$ is each proportional to a perfect square. Since $2n+1$ is\nodd, there are only two possible cases:\nCase 1: $2n+1 = 3 a^2$ and $n+1 = 2b^2$ , where $a$ and $b$ are integers.\nCase 2: $2n+1 = a^2$ and $n+1 = 6b^2$\nIn Case 1, $2n+1 = 4b^2 -1 = 3a^2$ . This means that $(4b^2 -1)/3 = a^2$ for some integers $a$ and $b$ . We proceed by checking whether $(4b^2-1)/3$ is a perfect square for $b=2, 3, 4, \\dots$ . (The solution $b=1$ leads to $n=1$ , and we are asked to find a value of $n$ greater than 1.) The smallest positive integer $b$ greater than 1 for\nwhich $(4b^2-1)/3$ is a perfect square is $b=13$ , which results in $n=337$\nIn Case 2, $2n+1 = 12b^2 - 1 = a^2$ . Note that $a^2 = 2n+1$ is an odd square, and hence is congruent to $1 \\pmod 4$ . But $12b^2 -1 \\equiv 3 \\pmod 4$ for any $b$ , so Case 2 has no solutions.\nAlternatively, one can proceed by checking whether $12b^2 -1$ is a perfect square for $b=1, 2 ,3 ,\\dots$ . We find that $12b^2 -1$ is not a perfect square for $b = 1,2, 3, ..., 7, 8$ , and $n= 383$ when $b=8$ . Thus the smallest positive integers $a$ and $b$ for which $12b^2- 1 = a^2$ result in a value of $n$ exceeding the value found in Case 1, which was 337.\nIn summary, the smallest value of $n$ greater than 1 for which $I_n$ is an integer is $\\boxed{337}$" ]

[ "$(\\sqrt{3}+\\sqrt{2})^6=(5+2\\sqrt{6})^3=(5+2\\sqrt{6})(49+20\\sqrt{6})=(485+198\\sqrt{6})$ Then, find that $\\sqrt{6}$ is about $2.449$ . Finally, multiply and add to find that the smallest integer higher is $\\boxed{970}$", "Let's evaluate $(\\sqrt{3}+\\sqrt{2})^6 + (\\sqrt{3}-\\sqrt{2})^6$ . We see that all the irrational terms cancel. Then, using binomial theorem, we evaluate all the rational terms in the first expression to get 485. Then, the sum of the rational parts of the 2nd term will be 485 as well. Then, we get a total of 970 and since $(\\sqrt{3}-\\sqrt{2})^6) < 1$ , the greatest integer greater than our original expression is $\\boxed{970}$" ]

[ "Expanding, we have $2kx^2-8x-x^2+6=0$ , or $(2k-1)x^2-8x+6=0$ . For this quadratic not to have real roots, it must have a negative discriminant. Therefore, $(-8)^2-4(2k-1)(6)<0\\implies 64-48k+24<0\\implies k>\\frac{11}{6}$ . From here, we can easily see that the smallest integral value of $k$ is $2, \\boxed{2}$" ]

[ "Adding $\\sqrt{n - 1}$ to both sides, we get \\[\\sqrt{n} < \\sqrt{n - 1} + 0.01.\\] Squaring both sides, we get \\[n < n - 1 + 0.02 \\sqrt{n - 1} + 0.0001,\\] which simplifies to \\[0.9999 < 0.02 \\sqrt{n - 1},\\] or \\[\\sqrt{n - 1} > 49.995.\\] Squaring both sides again, we get \\[n - 1 > 2499.500025,\\] so $n > 2500.500025$ . The smallest positive integer $n$ that satisfies this inequality is $\\boxed{2501}$" ]

[ "Let the desired integer be $n$ . From the information given, it can be determined that, for positive integers $a, \\ b, \\ c$\n$n = 9a + 36 = 10b + 45 = 11c + 55$\nThis can be rewritten as the following congruences:\n$n \\equiv 0 \\pmod{9}$\n$n \\equiv 5 \\pmod{10}$\n$n \\equiv 0 \\pmod{11}$\nSince 9 and 11 are relatively prime, n is a multiple of 99. It can then easily be determined that the smallest multiple of 99 with a units digit 5 (this can be interpreted from the 2nd congruence) is $\\boxed{495}$", "Let $n$ be the desired integer. From the given information, we have \\begin{align*}9x &= a \\\\ 11y &= a \\\\ 10z + 5 &= a, \\end{align*} here, $x,$ and $y$ are the middle terms of the sequence of 9 and 11 numbers, respectively. Similarly, we have $z$ as the 4th term of the sequence. Since, $a$ is a multiple of $9$ and $11,$ it is also a multiple of $\\text{lcm}[9,11]=99.$ Hence, $a=99m,$ for some $m.$ So, we have $10z + 5 = 99m.$ It follows that $99(5) = \\boxed{495}$ is the smallest integer that can be represented in such a way.", "By the method in Solution 1, we find that the number $n$ can be written as $9a+36=10b+45=11c+55$ for some integers $a,b,c$ . From this, we can see that $n$ must be divisible by 9, 5, and 11. This means $n$ must be divisible by 495. The only multiples of 495 that are small enough to be AIME answers are 495 and 990. From the second of the three expressions above, we can see that $n$ cannot be divisible by 10, so $n$ must equal $\\boxed{495}$ . Solution by Zeroman.", "First note that the integer clearly must be divisible by $9$ and $11$ since we can use the \"let the middle number be x\" trick. Let the number be $99k$ for some integer $k.$ Now let the $10$ numbers be $x,x+1, \\cdots x+9.$ We have $10x+45 = 99k.$ Taking mod $5$ yields $k \\equiv 0 \\pmod{5}.$ Since $k$ is positive, we take $k=5$ thus obtaining $99 \\cdot 5 = \\boxed{495}$ as our answer." ]

[ "The problem states that the answer cannot be a perfect square or have prime factors less than $50$ . Therefore, the answer will be the product of at least two different primes greater than $50$ . The two smallest primes greater than $50$ are $53$ and $59$ . Multiplying these two primes, we obtain the number $3127$ , which is also the smallest number on the list of answer choices.\nSo we are done, and the answer is $\\boxed{3127}$" ]

[ "We use the fact that the number of divisors of a number $n = p_1^{e_1}p_2^{e_2} \\cdots p_k^{e_k}$ is $(e_1 + 1)(e_2 + 1) \\cdots (e_k + 1)$ . If a number has $18 = 2 \\cdot 3 \\cdot 3$ factors, then it can have at most $3$ distinct primes in its factorization.\nDividing the greatest power of $2$ from $n$ , we have an odd integer with six positive divisors, which indicates that it either is ( $6 = 2 \\cdot 3$ ) a prime raised to the $5$ th power, or two primes, one of which is squared. The smallest example of the former is $3^5 = 243$ , while the smallest example of the latter is $3^2 \\cdot 5 = 45$\nSuppose we now divide all of the odd factors from $n$ ; then we require a power of $2$ with $\\frac{18}{6} = 3$ factors, namely $2^{3-1} = 4$ . Thus, our answer is $2^2 \\cdot 3^2 \\cdot 5 = \\boxed{180}$", "Somewhat similar to the first solution, we see that the number $n$ has two even factors for every odd factor. Thus, if $x$ is an odd factor of $n$ , then $2x$ and $4x$ must be the two corresponding even factors. So, the prime factorization of $n$ is $2^2 3^a 5^b 7^c...$ for some set of integers $a, b, c, ...$\nSince there are $18$ factors of $n$ , we can write:\n$(2+1)(a+1)(b+1)(c+1)... = 18$\n$(a+1)(b+1)(c+1)... = 6$\nSince $6$ only has factors from the set $1, 2, 3, 6$ , either $a=5$ and all other variables are $0$ , or $a=3$ and $b=2$ , with again all other variables equal to $0$ . This gives the two numbers $2^2 \\cdot 3^5$ and $2^2 \\cdot 3^2 \\cdot 5$ . The latter number is smaller, and is equal to $\\boxed{180}$" ]

[ "Combine the terms in the product to get $2^{\\frac{1+3+5+ \\dots +(2n-1)+(2n+1)}{7}}$\nThe exponent can be simplified to \\[\\frac{1+3+5+ \\dots +(2n-1)+(2n+1)}{7} \\Rightarrow \\frac{\\frac{n(1+(2n+1))}{2}}{7} \\Rightarrow \\frac{n^2}{7}.\\]\nWe want this inequality to be true with the smallest positive odd integer value of $n$ \\[2^{\\frac{n^2}{7}} > 1000.\\]\nNow, let's test the answer choices. For $n=7$ , we have $2^{49/7}=2^{7}<1000$ . For $n=9$ , we have $2^{81/7}>2^{10}>1000$\nSo our answer is $\\boxed{9}$ . ~ jiang147369" ]

[ "In order to get the smallest possible average, we want the 4 even numbers to be as small as possible. The first 4 positive even numbers are 2, 4, 6, and 8. Their average is $\\frac{2+4+6+8}{4}=\\boxed{5}$" ]

[ "Since we want the smallest possible result, and we are only adding and multiplying positive numbers over $1$ , we can \"prune\" the set to the three smallest numbers $\\{3,5,7\\}$ . Using bigger numbers will create bigger sums and bigger products.\nFrom there, compute the $3$ ways you can do the two operations:\n$(3+5)7 = 8\\cdot 7 = 56$\n$(3 + 7)5 = 10\\cdot 5 = 50$\n$(7+5)3 = 12\\cdot 3 = 36$\nThe smallest number is 36, giving an answer of $\\boxed{36}$" ]

[ "Let the two three-digit numbers be $\\overline{abc}$ and $\\overline{def}$ . Their sum is equal to $100(a+d)+10(b+e)+(c+f)$\nTo minimize this, we need to minimize the contribution of the $100$ factor, so we let $a=4$ and $d=5$ . Similarly, we let $b=6$ $e=7$ , and then $c=8$ and $f=9$ . The sum is \\[100(9)+10(13)+(17)=1047 \\rightarrow \\boxed{1047}\\]" ]

[ "We know from the Triangle Inequality that the last side, $s$ , fulfills $s<5+19=24$ . Adding $5+19$ to both sides of the inequality, we get $s+5+19<48$ , and because $s+5+19$ is the perimeter of our triangle, $\\boxed{48}$ is our answer." ]

[ "$(x-2)^2$ = perfect square.\n1 < perfect square < 25\nPerfect square can equal: 4, 9, or 16\nSolve for $x$\n$(x-2)^2=4$\n$x=4,0$\nand\n$(x-2)^2=9$\n$x=5,-1$\nand\n$(x-2)^2=16$\n$x=6,-2$\nThe sum of all integer solutions is\n$4+5+6+0+(-1)+(-2)=\\boxed{12}$" ]

[ "We factor $x^2-3x+2$ into $(x-1)(x-2)$ . Thus, either $1$ or $2$ is a root of $x^2-5x+k$ . If $1$ is a root, then $1^2-5\\cdot1+k=0$ , so $k=4$ . If $2$ is a root, then $2^2-5\\cdot2+k=0$ , so $k=6$ . The sum of all possible values of $k$ is $\\boxed{10}$" ]

[ "Let $A = (\\cos 40^{\\circ}, \\sin 40^{\\circ}), B = (\\cos 60^{\\circ}, \\sin 60^{\\circ}),$ and $C = (\\cos t^{\\circ}, \\sin t^{\\circ}).$ We apply casework to the legs of isosceles $\\triangle ABC:$\nTogether, the sum of all such possible values of $t$ is $20+80+50+230=\\boxed{380}.$" ]

[ "Split the equation into two cases, where the value inside the absolute value is positive and nonpositive.\nCase 1:\nThe equation yields $x^2-12x+34=2$ , which is equal to $(x-4)(x-8)=0$ . Therefore, the two values for the positive case is $4$ and $8$\nCase 2:\nSimilarly, taking the nonpositive case for the value inside the absolute value notation yields $x^2-12x+34=-2$ . Factoring and simplifying gives $(x-6)^2=0$ , so the only value for this case is $6$\nSumming all the values results in $4+8+6=\\boxed{18}$", "We have the equations $x^2-12x+32=0$ and $x^2-12x+36=0$\nNotice that the second is a perfect square with a double root at $x=6$ , and the first has two distinct real roots. By Vieta's, the sum of the roots of the first equation is $-(-12)$ or $12$ $12+6=\\boxed{18}$", "Completing the square gives \\begin{align*} \\left|(x-6)^2-2\\right|&=2 \\\\ (x-6)^2-2&=\\pm2. \\hspace{15mm}(\\bigstar) \\end{align*} Note that the graph of $y=(x-6)^2-2$ is an upward parabola with the vertex $(6,-2)$ and the axis of symmetry $x=6;$ the graphs of $y=\\pm2$ are horizontal lines.\nWe apply casework to $(\\bigstar):$\nFinally, the sum of all solutions is $12+6=\\boxed{18}.$" ]

[ "The mean is $\\frac{4+6+8+17+x}{5}=\\frac{35+x}{5}$\nThere are three possibilities for the median: it is either $6$ $8$ , or $x$\nLet's start with $6$\n$\\frac{35+x}{5}=6$ has solution $x=-5$ , and the sequence is $-5, 4, 6, 8, 17$ , which does have median $6$ , so this is a valid solution.\nNow let the median be $8$\n$\\frac{35+x}{5}=8$ gives $x=5$ , so the sequence is $4, 5, 6, 8, 17$ , which has median $6$ , so this is not valid.\nFinally we let the median be $x$\n$\\frac{35+x}{5}=x \\implies 35+x=5x \\implies x=\\frac{35}{4}=8.75$ , and the sequence is $4, 6, 8, 8.75, 17$ , which has median $8$ . This case is therefore again not valid.\nHence the only possible value of $x$ is $\\boxed{5}.$" ]

[ "The mean is $\\frac{4+6+8+17+x}{5}=\\frac{35+x}{5}$\nThere are three possibilities for the median: it is either $6$ $8$ , or $x$\nLet's start with $6$\n$\\frac{35+x}{5}=6$ has solution $x=-5$ , and the sequence is $-5, 4, 6, 8, 17$ , which does have median $6$ , so this is a valid solution.\nNow let the median be $8$\n$\\frac{35+x}{5}=8$ gives $x=5$ , so the sequence is $4, 5, 6, 8, 17$ , which has median $6$ , so this is not valid.\nFinally we let the median be $x$\n$\\frac{35+x}{5}=x \\implies 35+x=5x \\implies x=\\frac{35}{4}=8.75$ , and the sequence is $4, 6, 8, 8.75, 17$ , which has median $8$ . This case is therefore again not valid.\nHence the only possible value of $x$ is $\\boxed{5}.$" ]

[ "We evaluate this in cases:\nCase 1 $x<30$\nWhen $x<30$ we are going to have $60-2x>0$ . When $x>0$ we are going to have $|x|>0\\implies x>0$ and when $-x>0$ we are going to have $|x|>0\\implies -x>0$ . Therefore we have $x=|2x-(60-2x)|$ $x=|2x-60+2x|\\implies x=|4x-60|$\nSubcase 1 $30>x>15$\nWhen $30>x>15$ we are going to have $4x-60>0$ . When this happens, we can express $|4x-60|$ as $4x-60$ .\nTherefore we get $x=4x-60\\implies -3x=-60\\implies x=20$ . We check if $x=20$ is in the domain of the numbers that we put into this subcase, and it is, since $30>20>15$ . Therefore $20$ is one possible solution.\nSubcase 2 $x<15$\nWhen $x<15$ we are going to have $4x-60<0$ , therefore $|4x-60|$ can be expressed in the form $60-4x$ .\nWe have the equation $x=60-4x\\implies 5x=60\\implies x=12$ . Since $12$ is less than $15$ $12$ is another possible solution. $x=|2x-|60-2x||$\nCase 2 $x>30$\nWhen $x>30$ $60-2x<0$ . When $x<0$ we can express this in the form $-x$ . Therefore we have $-(60-2x)=2x-60$ . This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we have $x=|2x-(2x-60)|$\n$x=|2x-2x+60|$\n$x=|60|$\n$x=60$\nWe have now evaluated all the cases, and found the solution to be $\\{60,12,20\\}$ which have a sum of $\\boxed{92}$", "From the equation $x = \\left|2x-|60-2x|\\right|$ , we have $x = 2x-|60-2x|$ , or $-x = 2x-|60-2x|$ . Therefore, $x=|60-2x|$ , or $3x=|60-2x|$ . From here we have four possible cases:\n1. $x=60-2x$ ; this simplifies to $3x=60$ , so $x=20$\n2. $-x=60-2x$ ; this simplifies to $x=60$\n3. $3x=60-2x$ ; this simplifies to $5x=60$ , so $x=12$\n4. $-3x=60-2x$ ; this simplifies to $-x=60$ , so $x=-60$ . However, this solution is extraneous because the absolute value of $2x-|60-2x|$ cannot be negative.\nThe sum of all of the solutions of $x$ is $20+60+12=\\boxed{92}$" ]

[ "$2^{1999}\\cdot5^{2001}=2^{1999}\\cdot5^{1999}\\cdot5^{2}=25\\cdot10^{1999}$ , a number with the digits \"25\" followed by 1999 zeros. The sum of the digits in the decimal form would be $2+5=7$ , thus making the answer $\\boxed{7}$" ]

[ "Using the standard multiplication algorithm, $111,111,111^2=12,345,678,987,654,321,$ whose digit sum is $\\boxed{81.}$", "We note that\n$1^2 = 1$\n$11^2 = 121$\n$111^2 = 12321$\nand $1,111^2 = 1234321$\nWe can clearly see the pattern: If $X$ is $111\\cdots111$ , with $n$ ones (and for the sake of simplicity, assume that $n<10$ ), then the sum of the digits of $X^2$ is\n$1+2+3+4+5\\cdots n+(n-1)+(n-2)\\cdots+1$\n$=(1+2+3\\cdots n)+(1+2+3+\\cdots n-1)$\n$=\\dfrac{n(n+1)}{2}+\\dfrac{(n-1)n}{2}$\n$=\\dfrac{n(n+1+n-1)}{2}=\\dfrac{2n^2}{2}=n^2.$\nAha! We know that $111,111,111$ has $9$ digits, so its digit sum is $9^2=\\boxed{81}$", "We see that $111^2$ can be written as $111(100+10+1)=11100+1110+111=12321$\nWe can apply this strategy to find $111,111,111^2$ , as seen below.\n$111111111^2=111111111(100000000+10000000\\cdots+10+1)$\n$=11111111100000000+1111111110000000+\\cdots+111111111$\n$=12,345,678,987,654,321$\nThe digit sum is thus $1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \\boxed{81}$" ]

[ "The prime factorization is $2016=2^5\\times3^2\\times7$ . Since the problem is only asking us for the distinct prime factors, we have $2,3,7$ . Their desired sum is then $\\boxed{12}$", "We notice that $9 \\mid 2016$ , since $2+0+1+6 = 9$ , and $9 \\mid 9$ . We can divide $2016$ by $9$ to get $224$ . This is divisible by $4$ , as $4 \\mid 24$ . Dividing $224$ by $4$ , we have $56$ . This is clearly divisible by $7$ , leaving $8$ . We have $2016 = 9\\cdot 4\\cdot 7\\cdot 8$ . We know that $4$ and $8$ are both multiples of $2$ $9$ is $3^2$ , and $7$ is prime. This means that the distinct prime factors are $2,3,$ and $7$ . Their sum is $\\boxed{12}$" ]

[ "Looking at the prime numbers under $12$ , we see that there are $\\left\\lfloor\\frac{12}{2}\\right\\rfloor+\\left\\lfloor\\frac{12}{2^2}\\right\\rfloor+\\left\\lfloor\\frac{12}{2^3}\\right\\rfloor=6+3+1=10$ factors of $2$ $\\left\\lfloor\\frac{12}{3}\\right\\rfloor+\\left\\lfloor\\frac{12}{3^2}\\right\\rfloor=4+1=5$ factors of $3$ , and $\\left\\lfloor\\frac{12}{5}\\right\\rfloor=2$ factors of $5$ . All greater primes are represented once or none in $12!$ , so they cannot be part of the square. Since we are looking for a perfect square, the exponents on its prime factors must be even, so we can only use $4$ of the $5$ factors of $3$ . Therefore, the prime factorization of the square is $2^{10}\\cdot3^4\\cdot5^2$ . To find the square root of this, we halve the exponents, leaving $2^5\\cdot3^2\\cdot5$ . The sum of the exponents is $\\boxed{8}$" ]

[ "Putting the numbers in numerical order we get the list $0,0,1,2,3,3,3,4.$ The mode is $3.$ The median is $\\frac{2+3}{2}=2.5.$ The average is $\\frac{0+0+1+2+3+3+3+4}{8}=\\frac{16}{8}=2.$ The sum of all three is $3+2.5+2=\\boxed{7.5}$" ]

[ "First, we must find the prime factorization of $2010$ $2010=2\\cdot 3 \\cdot 5 \\cdot 67$ . We add the factors up to get $\\boxed{77}$" ]

[ "Multiplying both sides by $x$\n$\\frac{2003}{2004}x^{2}+1x+1=0$\nLet the roots be $a$ and $b$\nThe problem is asking for $\\frac{1}{a}+\\frac{1}{b}= \\frac{a+b}{ab}$\nBy Vieta's formulas\n$a+b=(-1)^{1}\\frac{1}{\\frac{2003}{2004}}=-\\frac{2004}{2003}$\n$ab=(-1)^{2}\\frac{1}{\\frac{2003}{2004}}=\\frac{2004}{2003}$\nSo the answer is $\\frac{a+b}{ab}=\\frac{-\\frac{2004}{2003}}{\\frac{2004}{2003}}=-1 \\Rightarrow\\boxed{1}$", "Dividing both sides by $x$\n$\\frac{2003}{2004}+\\frac 1x+\\frac{1}{x^2}=0$\nwe see by Vieta's formulas that the sum of the roots is $-1 \\Rightarrow\\boxed{1}$", "Re-stating and multiplying $\\frac{1}{x}$ by $\\frac{2004}{2004}$ , we have $\\frac{2003x}{2004} + \\frac{2004}{2004x} + 1 = 0$\nSimplifying, we have $\\frac{2003}{2004}+\\frac{\\frac{2004}{x}}{2004}=-1.$\nPutting this together, we have $\\frac{2003+\\frac{2004}{x}}{2004}=-1.$\nSimplifying, we have $2004x + \\frac{2004}{x} = -2004.$\nMultiplying both sides by $x,$ we have $2003x^2+2004=-2004x.$\nMoving $-2004x$ to the left side, we have $2003x^2 + 2004x + 2004 = 0.$\nLet's go back and see what we want.\nWe want the sum of the reciprocals of the roots, in which if the roots were to be called $r_1$ and $r_2,$ we would want $\\frac{1}{r_1}+\\frac{1}{r_2}=\\frac{r_1 + r_2}{r_1 \\cdot r_2}$\nWe can use Vieta's formulas to solve the sum of the roots and the product of the roots.\nWe have that the sum of the two roots is $\\frac{-b}{a}$ where the quadratic is $ax^2+bx+c=0.$\nIn this case, $a=2003$ and $b=2004$\nTherefore, $r_1+r_2=-\\frac{2004}{2003}.$\nWe have that the product of the two roots is $\\frac{c}{a}$ where the quadratic is $ax^2+bx+c=0.$\nIn this case, $c=2004$ and $a=2003.$\nTherefore, $r_1 \\cdot r_2 = \\frac{2004}{2003}.$\nNow that we have the sum and product of the roots, we can substitute back into the expression of what we want.\nWe have $\\frac{-\\frac{2004}{2003}}{\\frac{2004}{2003}}=-\\frac{2004}{2003} \\cdot \\frac{2003}{2004} = \\boxed{1}.$" ]

[ "Let $y = \\sqrt[4]{x}$ . Then we have $y(7 - y) = 12$ , or, by simplifying, \\[y^2 - 7y + 12 = (y - 3)(y - 4) = 0.\\]\nThis means that $\\sqrt[4]{x} = y = 3$ or $4$\nThus the sum of the possible solutions for $x$ is $4^4 + 3^4 = \\boxed{337}$" ]

[ "Since $10!$ is divisible by $100$ , any factorial greater than $10!$ is also divisible by $100$ . The last two digits of all factorials greater than $10!$ are $00$ , so the last two digits of $10!+11!+...+2006!$ are $00$ . \n(*)\nSo all that is needed is the tens digit of the sum $7!+8!+9!$\n$7!+8!+9!=5040+40320+362880=408240$\nSo the tens digit is $\\boxed{4}$" ]

[ "Since we only need the tens digits, we only need to care about the multiplication of tens and ones. (If you want to use mathematical terms then we only need to look at the exponents in $mod 100$ .) We will use the \" $\\equiv$ \" sign to denote congruence in modulus, basically taking the last two digits and ignoring everything else.\n\\[\\begin{split} 15^{1}\\equiv15 \\\\ 15^{2}\\equiv25 \\end{split}\\] From here we only need to multiply $15\\cdot25$ and we can ignore the hundreds digits. \\[\\begin{split} 15^{3}\\equiv75 \\\\ 15^{4}\\equiv25 \\\\ 15^{5}\\equiv75 \\end{split}\\] Notice that for every $x\\neq1$ $15^{x}\\equiv25$ if $x$ is even, and $15^{x}\\equiv75$ if $x$ is odd. Since $2015^{2016}$ has an even exponent, we conclude that the last two digits will be $25$ , and subtracting $25-17=0 \\Longrightarrow \\boxed{0}$ .\n~JH. L" ]

[ "Notice that, for $n\\ge 2$ $2015^n\\equiv 15^n$ is congruent to $25\\pmod{100}$ when $n$ is even and $75\\pmod{100}$ when $n$ is odd. (Check for yourself). Since $2016$ is even, $2015^{2016} \\equiv 25\\pmod{100}$ and $2015^{2016}-2017 \\equiv 25 - 17 \\equiv \\underline{0}8\\pmod{100}$\nSo the answer is $\\textbf{(A)}\\ \\boxed{0}$" ]

[ "Since we want the tens digit, we can find the last two digits of $7^{2011}$ . We can do this by using modular arithmetic. \\[7^1\\equiv 07 \\pmod{100}.\\] \\[7^2\\equiv 49 \\pmod{100}.\\] \\[7^3\\equiv 43 \\pmod{100}.\\] \\[7^4\\equiv 01 \\pmod{100}.\\] We can write $7^{2011}$ as $(7^4)^{502}\\times 7^3$ . Using this, we can say: \\[7^{2011}\\equiv (7^4)^{502}\\times 7^3\\equiv 7^3\\equiv 343\\equiv 43\\pmod{100}.\\] From the above, we can conclude that the last two digits of $7^{2011}$ are 43. Since they have asked us to find the tens digit, our answer is $\\boxed{4}$", "We can use patterns to figure out the answer.\n7 to the power of 2 is 49. So, the tens digit is 4.\n7 to the power of 3 is 343. So, the tens digit is 4. \n7 to the power of 4 is 2401.So, the tens digit is 0.\n7 to the power of 5 is 16807. So, the tens digit is 0.\nBy now, we can notice the pattern. The tens digit for 7 to the power of 2 is 4, then 4, then 0, then 0. It keeps on going, 2 fours, and then 2 zeros in 4 numbers. If we round up for 2011/4, we get 503. 503 * 4 is 2012. So 7 to the 2012 power has a tens digit of 0, since 2012 is a mutiple of 4, and 7 to the power of 4 has a tens digit of 0. We have to subtract a power from 7 to the 2012 power, so the tens digit goes back from 0 to 4 because if we subtract a power from 7 to the power of 4, we have 7 to the power of 3, which has a tens digit of 4. Hence the answer is $\\boxed{4}$" ]

[ "$13^{2003}\\equiv 3^{2003}\\pmod{10}$\nSince $3^4=81\\equiv1\\pmod{10}$\n$3^{2003}=(3^{4})^{500}\\cdot3^{3}\\equiv1^{500}\\cdot27\\equiv7\\pmod{10}$\nTherefore, the units digit is $7 \\Rightarrow\\boxed{7}$", "Since we are looking for the units digit of $13^{2003}$ , we only have to focus on the units digit of the base (13) as none of the other digits of the base affect the units digit of the resulting value.\nBy testing the first few values or through previous knowledge, you might see that the units digit of exponents with base 3 follow this pattern: \\[3^1=3\\] \\[3^2=9\\] \\[3^3=27\\] \\[3^4=81,\\] giving us the rotation $3-9-7-1.$\nAs this cycle resets every time the index increases by 4, we know that this cycle ends on 2000, and starts once again on 2001. As our expression is raised to the power of 2003, we know that the units digit of our expression must end with the third term of our pattern: $7$\nTherefore, the units digit of our expression is $7 \\Rightarrow\\boxed{7}$" ]

[ "$2022^{2023} + 2023^{2022} \\equiv 2^3 + 3^2 \\equiv 17 \\equiv 7$ (mod 10). $\\boxed{7}$ ~andliu766", "When looking at the units digit patterns of the powers of $2$ , we see that\n$2^1=$ , units digit $2$\n$2^2=$ , units digit $4$\n$2^3=$ , units digit $8$\n$2^4=$ , units digit $6$\n$2^5=$ , units digit $2$\nAnd the pattern repeats. This pattern will apply for the powers of $2022$ as well, since the units digit of $2022$ is $2$ . We can find the pattern for the powers of $3$ too. The pattern follows with units digits, $3$ $9$ $7$ $1$ $3$ $9$ , ...\nSimilarly, the units digit of $2023$ will follow the same pattern as the powers of $3$\nBoth of these powers cycle in groups of $4$ . When diving $2023$ by $4$ , we get $505$ remainder $3$ , meaning $505$ complete cycles; or the power being a multiple of $4$ $505$ times, and $3$ extra. So the units digit of $2022^{2023}$ is $8$ $2022$ divided by $4$ is $505$ reminder $2$ , which means $505$ complete cycles, or the power being a multiple of $4$ $505$ times, and $2$ extra. So the units digit of $2023^{2022}$ is $9$\nWe only need to find the units digit in the end, so we just add those $2$ already found units digits, to get a new units digit of $7$ . Therefore the answer is $\\boxed{7}$", "Note that the units digit will be the same regardless of the tens, hundreds, and thousands digits, so we can simplify this problem to finding the last digit of $2^{2023} + 3^{2022}$ . We can find the units digit of $2^{2023}$ , by listing the units digits of the first few powers of two, and trying to find a pattern.\n$2^1=2$\n$2^2=4$\n$2^3=8$\n$2^4=6+10$\n$2^5=2+30$\n$2^6=4+60$\nAs we can see the units digits of powers of two repeat after every four iterations. Now we know the units digit of $2^{2020}$ is $6$ and the units digit of $2^{2023} \\Rightarrow 2^3\\cdot 2^{2020} \\Rightarrow 6\\cdot 8 \\Rightarrow 8$ . Similarly we can find the last digits of powers of three repeat after every four, so the units digit of $3^{2022}$ is $1\\cdot 3^2 = 9$ . Adding these together, the ones digit is the same as the ones digit of $9+8$ which is $7$ $\\boxed{7}$" ]