https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Relay400&feedformat=atom AoPS Wiki - User contributions [en] 2021-09-20T21:32:26Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_18&diff=122378 2004 AMC 12A Problems/Problem 18 2020-05-14T01:31:48Z <p>Relay400: This is not a solution.</p> <hr /> <div>{{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #18]] and [[2004 AMC 10A Problems/Problem 22|2004 AMC 10A #22]]}}<br /> <br /> ==Problem==<br /> [[Square]] &lt;math&gt;ABCD&lt;/math&gt; has side length &lt;math&gt;2&lt;/math&gt;. A [[semicircle]] with [[diameter]] &lt;math&gt;\overline{AB}&lt;/math&gt; is constructed inside the square, and the [[tangent (geometry)|tangent]] to the semicircle from &lt;math&gt;C&lt;/math&gt; intersects side &lt;math&gt;\overline{AD}&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt;. What is the length of &lt;math&gt;\overline{CE}&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(100);<br /> defaultpen(fontsize(10));<br /> pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2);<br /> draw(A--B--C--D--cycle);draw(C--E);<br /> draw(Arc((1,0),1,0,180));<br /> label(&quot;$A$&quot;,A,(-1,-1));<br /> label(&quot;$B$&quot;,B,( 1,-1));<br /> label(&quot;$C$&quot;,C,( 1, 1));<br /> label(&quot;$D$&quot;,D,(-1, 1));<br /> label(&quot;$E$&quot;,E,(-1, 0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{2+\sqrt{5}}{2} \qquad \mathrm{(B) \ } \sqrt{5} \qquad \mathrm{(C) \ } \sqrt{6} \qquad \mathrm{(D) \ } \frac{5}{2} \qquad \mathrm{(E) \ } 5-\sqrt{5} &lt;/math&gt;<br /> <br /> __TOC__<br /> == Solution 1 ==<br /> <br /> &lt;asy&gt;<br /> size(150);<br /> defaultpen(fontsize(10));<br /> pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2;<br /> draw(A--B--C--D--cycle);draw(C--E);<br /> draw(Arc((1,0),1,0,180));draw((A+B)/2--F);<br /> label(&quot;$A$&quot;,A,(-1,-1));<br /> label(&quot;$B$&quot;,B,( 1,-1));<br /> label(&quot;$C$&quot;,C,( 1, 1));<br /> label(&quot;$D$&quot;,D,(-1, 1));<br /> label(&quot;$E$&quot;,E,(-1, 0));<br /> label(&quot;$F$&quot;,F,( 0, 1));<br /> label(&quot;$x$&quot;,(A+E)/2,(-1, 0));<br /> label(&quot;$x$&quot;,(E+F)/2,( 0, 1));<br /> label(&quot;$2$&quot;,(F+C)/2,( 0, 1));<br /> label(&quot;$2$&quot;,(D+C)/2,( 0, 1));<br /> label(&quot;$2$&quot;,(B+C)/2,( 1, 0));<br /> label(&quot;$2-x$&quot;,(D+E)/2,(-1, 0));<br /> &lt;/asy&gt;<br /> Let the point of tangency be &lt;math&gt;F&lt;/math&gt;. By the [[Two Tangent Theorem]] &lt;math&gt;BC = FC = 2&lt;/math&gt; and &lt;math&gt;AE = EF = x&lt;/math&gt;. Thus &lt;math&gt;DE = 2-x&lt;/math&gt;. The [[Pythagorean Theorem]] on &lt;math&gt;\triangle CDE&lt;/math&gt; yields<br /> <br /> &lt;cmath&gt;\begin{align*}<br /> DE^2 + CD^2 &amp;= CE^2\\<br /> (2-x)^2 + 2^2 &amp;= (2+x)^2\\<br /> x^2 - 4x + 8 &amp;= x^2 + 4x + 4\\<br /> x &amp;= \frac{1}{2}\end{align*}&lt;/cmath&gt;<br /> <br /> Hence &lt;math&gt;CE = FC + x = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}&lt;/math&gt;.<br /> <br /> <br /> == Solution 2 ==<br /> Call the point of tangency point &lt;math&gt;F&lt;/math&gt; and the midpoint of &lt;math&gt;AB&lt;/math&gt; as &lt;math&gt;G&lt;/math&gt;. &lt;math&gt;CF=2&lt;/math&gt; by Tangent Theorem. Notice that &lt;math&gt;\angle EGF=\frac{180-2\cdot\angle CGF}{2}=90-\angle CGF&lt;/math&gt;. Thus, &lt;math&gt;\angle EGF=\angle FCG&lt;/math&gt; and &lt;math&gt;tanEGF=tanFCG=\frac{1}{2}&lt;/math&gt;. Solving &lt;math&gt;EF=\frac{1}{2}&lt;/math&gt;. Adding, the answer is &lt;math&gt;\frac{5}{2}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> [[Image:2004_AMC12A-18.png]]<br /> <br /> Clearly, &lt;math&gt;EA = EF = BG&lt;/math&gt;. Thus, the sides of [[right triangle]] &lt;math&gt;CDE&lt;/math&gt; are in arithmetic progression. Thus it is [[similar triangles|similar]] to the triangle &lt;math&gt;3 - 4 - 5&lt;/math&gt; and since &lt;math&gt;DC = 2&lt;/math&gt;, &lt;math&gt;CE = 5/2&lt;/math&gt;.<br /> <br /> == Solution 4 ==<br /> &lt;asy&gt;<br /> size(150);<br /> defaultpen(fontsize(10));<br /> pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0);<br /> draw(A--B--C--D--cycle);draw(C--E);<br /> draw(Arc((1,0),1,0,180));draw((A+B)/2--F);<br /> label(&quot;$A$&quot;,A,(-1,-1));<br /> label(&quot;$B$&quot;,B,( 1,-1));<br /> label(&quot;$C$&quot;,C,( 1, 1));<br /> label(&quot;$D$&quot;,D,(-1, 1));<br /> label(&quot;$E$&quot;,E,(-1, 0));<br /> label(&quot;$F$&quot;,F,( 0, 1));<br /> label(&quot;$x$&quot;,(A+E)/2,(-1, 0));<br /> label(&quot;$x$&quot;,(E+F)/2,( 0, 1));<br /> label(&quot;$2$&quot;,(F+C)/2,( 0, 1));<br /> label(&quot;$2$&quot;,(D+C)/2,( 0, 1));<br /> label(&quot;$2$&quot;,(B+C)/2,( 1, 0));<br /> label(&quot;$2-x$&quot;,(D+E)/2,(-1, 0));<br /> label(&quot;$G$&quot;,G,(0,-1));<br /> dot(G);<br /> draw(G--C);<br /> label(&quot;$\sqrt{5}$&quot;,(G+C)/2,(-1,0));<br /> &lt;/asy&gt;<br /> <br /> Let us call the midpoint of side &lt;math&gt;AB&lt;/math&gt;, point &lt;math&gt;G&lt;/math&gt;. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides &lt;math&gt;GB, BC, GC&lt;/math&gt;. We get &lt;math&gt;GC=\sqrt{5}&lt;/math&gt;. We then know that &lt;math&gt;CF=2&lt;/math&gt; by Pythagorean theorem. Then by connecting &lt;math&gt;EG&lt;/math&gt;, we get similar triangles &lt;math&gt;EFG&lt;/math&gt; and &lt;math&gt;GFC&lt;/math&gt;. Solving the ratios, we get &lt;math&gt;x=\frac{1}{2}&lt;/math&gt;, so the answer is &lt;math&gt; \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}&lt;/math&gt;.<br /> <br /> == Solution 5 ==<br /> Using the diagram as drawn in Solution 5, let the total area of square &lt;math&gt;ABCD&lt;/math&gt; be divided into the triangles &lt;math&gt;DCE&lt;/math&gt;, &lt;math&gt;EAG&lt;/math&gt;, &lt;math&gt;CGB&lt;/math&gt;, and &lt;math&gt;EGC&lt;/math&gt;. Let x be the length of AE. Thus, the area of each triangle can be determined as follows: <br /> <br /> &lt;cmath&gt;DCE = \frac{DC\cdot{DE}}{2} = \frac{2\cdot(2-x)}{2} = 1-x&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;EAG= \frac{AE\cdot{AG}}{2} = \frac{1\cdot{x}}{2} = \frac{x}{2}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;CGB = \frac{GB\cdot{CB}}{2} = \frac{1\cdot(2)}{2} = 1&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;EGC= \frac{EG\cdot{GC}}{2} = \frac{\sqrt{4+(2-x)^2}}{2}&lt;/cmath&gt; (the length of CE is calculated with the [[Pythagorean Theorem]], lines GE and <br /> CE are perpendicular by definition of tangent)<br /> <br /> Adding up the areas and equating to the area of the total square (2*2=4), we get<br /> <br /> &lt;cmath&gt;1-x+\frac{x}{2}+1+ \frac{\sqrt{4+(2-x)^2}}{2} = 4 &lt;/cmath&gt;<br /> <br /> Solving for x:<br /> <br /> &lt;cmath&gt; 2-2x+x+1+\sqrt{x^2-4x+8}=8 &lt;/cmath&gt;<br /> &lt;cmath&gt; \sqrt{x^2-4x+8}=x+2 &lt;/cmath&gt;<br /> &lt;cmath&gt; x^2-4x+8=x^2+4x+4 &lt;/cmath&gt;<br /> &lt;cmath&gt; 8-4x=4x+4 \rightarrow x=\frac{1}{2} &lt;/cmath&gt;<br /> <br /> Solving for length of CE with the value we have for x:<br /> &lt;cmath&gt; \sqrt{4+(2-x)^2} = \sqrt{4+(3/2)^2} = \sqrt{25/4} = \boxed{\frac{5}{2}} &lt;/cmath&gt;</div> Relay400 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_25&diff=115136 2012 AMC 10A Problems/Problem 25 2020-01-22T05:29:53Z <p>Relay400: Be proper.</p> <hr /> <div>== Problem ==<br /> <br /> Real numbers &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are chosen independently and at random from the interval &lt;math&gt;[0,n]&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;. The probability that no two of &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are within 1 unit of each other is greater than &lt;math&gt;\frac {1}{2}&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since &lt;math&gt;x,y,z&lt;/math&gt; are all reals located in &lt;math&gt;[0, n]&lt;/math&gt;, the number of choices for each one is continuous.<br /> <br /> WLOG([[Without loss of generality]]), assume that &lt;math&gt;n\geq x \geq y \geq z \geq 0&lt;/math&gt;. Then the set of points &lt;math&gt;(x,y,z)&lt;/math&gt; is a tetrahedron, or a triangular pyramid. The point &lt;math&gt;(x,y,z)&lt;/math&gt; distributes uniformly in this region. If this is not easy to understand, read Solution II.<br /> <br /> The altitude of the tetrahedron is &lt;math&gt;n&lt;/math&gt; and the base is an isosceles right triangle with a leg length &lt;math&gt;n&lt;/math&gt;. The volume is &lt;math&gt;V_1=\dfrac{n^3}{6}&lt;/math&gt;, as shown in the first figure in red.<br /> <br /> &lt;asy&gt;<br /> import three;<br /> unitsize(10cm);<br /> size(150);<br /> currentprojection=orthographic(1/2,-1,2/3); <br /> // three - currentprojection, orthographic<br /> draw((1,1,0)--(0,1,0)--(0,0,0),dashed+green);<br /> draw((0,0,0)--(0,0,1),green);<br /> draw((0,1,0)--(0,1,1),dashed+green);<br /> draw((1,1,0)--(1,1,1),green);<br /> draw((1,0,0)--(1,0,1),green);<br /> draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle,green);<br /> <br /> draw((0,0,0)--(1,0,0)--(1,1,0)--(1,1,1), red);<br /> draw((1,1,0)--(0,0,0)--(1,1,1), dashed+red);<br /> draw((1,1,1)--(1,0,0), red);<br /> &lt;/asy&gt;<br /> <br /> <br /> Now we will find the region with points satisfying &lt;math&gt;|x-y|\geq1&lt;/math&gt;, &lt;math&gt;|y-z|\geq1&lt;/math&gt;, &lt;math&gt;|z-x|\geq1&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;n\geq x \geq y \geq z \geq 0&lt;/math&gt;, we have &lt;math&gt;x-y\geq1&lt;/math&gt;, &lt;math&gt;y-z\geq1&lt;/math&gt;.<br /> <br /> The region of points &lt;math&gt;(x,y,z)&lt;/math&gt; satisfying the condition is shown in the second figure in black. It is a tetrahedron, too.<br /> <br /> &lt;asy&gt;<br /> import three;<br /> unitsize(10cm);<br /> size(150);<br /> currentprojection=orthographic(1/2, -1, 2/3); <br /> // three - currentprojection, orthographic<br /> draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green);<br /> draw((0, 0, 0)--(0, 0, 1), green);<br /> draw((0, 1, 0)--(0, 1, 1), dashed+green);<br /> <br /> draw((1, 0, 0)--(1, 0, 1), green);<br /> draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green);<br /> <br /> <br /> <br /> draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red);<br /> draw((0,0,0)--(1,0,0)--(1,1,1), red);<br /> draw((1,1,1)--(1,1,0)--(1,0.9,0), red);<br /> <br /> draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle);<br /> draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed);<br /> draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed);<br /> <br /> &lt;/asy&gt;<br /> <br /> The volume of this region is &lt;math&gt;V_2=\dfrac{(n-2)^3}{6}&lt;/math&gt;.<br /> <br /> So the probability is &lt;math&gt;p=\dfrac{V_2}{V_1}=\dfrac{(n-2)^3}{n^3}&lt;/math&gt;.<br /> <br /> Substituting &lt;math&gt;n&lt;/math&gt; with the values in the choices, we find that when &lt;math&gt;n=10&lt;/math&gt;, &lt;math&gt;p=\frac{512}{1000}&gt;\frac{1}{2}&lt;/math&gt;, when &lt;math&gt;n=9&lt;/math&gt;, &lt;math&gt;p=\frac{343}{729}&lt;\frac{1}{2}&lt;/math&gt;. So &lt;math&gt;n\geq 10&lt;/math&gt;.<br /> <br /> So the answer is &lt;math&gt; \boxed{\textbf{(D)}\ 10} &lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Because &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are chosen independently and at random from the interval &lt;math&gt;[0,n]&lt;/math&gt;, which means that &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; distributes uniformly and independently in the interval &lt;math&gt;[0,n]&lt;/math&gt;. So the point &lt;math&gt;(x, y, z)&lt;/math&gt; distributes uniformly in the cubic &lt;math&gt;0\leqslant x, y, z \leqslant n&lt;/math&gt;, as shown in the figure below. The volume of this cubic is &lt;math&gt;V_0=n^3&lt;/math&gt;.<br /> <br /> [[File:Cubic.png]]<br /> <br /> As we want to find the probablity of the incident <br /> &lt;math&gt;A=\big\{ |x-y| \geq 1, |y-z| \geq1, |z-x| \geq 1 \big\}&lt;/math&gt;, <br /> we should find the volume of the region of points such that &lt;math&gt;|x-y|\geq 1&lt;/math&gt;, &lt;math&gt;|y-z|\geq 1&lt;/math&gt;, &lt;math&gt;|z-x|\geq 1&lt;/math&gt; and &lt;math&gt;0\leq x, y, z \leq n&lt;/math&gt;.<br /> <br /> Now we will find the region &lt;math&gt;\big\{ (x,y,z)\ | \ 0\leq x, y, z \leq n, |x-y|\geq 1, |y-z|\geq 1, |z-x|\geq 1 \big\} &lt;/math&gt;.<br /> <br /> The region can be generated by cutting off 3 slices corresponding to &lt;math&gt;|x-y|&lt; 1&lt;/math&gt;, &lt;math&gt;|y-z|&lt; 1&lt;/math&gt;, and &lt;math&gt;|z-x|&lt; 1&lt;/math&gt;, respectively, from the cubic.<br /> <br /> After cutting off a slice corresponding to &lt;math&gt;|x-y|&lt; 1&lt;/math&gt;, we get two triangular prisms, as shown in the figure.<br /> <br /> [[File:2.png]]<br /> <br /> In order to observe the object clearly, we rotate the object by the &lt;math&gt;z&lt;/math&gt; axis, as shown.<br /> <br /> [[File:3.png]]<br /> <br /> We can draw the slice corresponding to &lt;math&gt;|y-z|&lt; 1&lt;/math&gt; on the object.<br /> <br /> [[File:4B.png]]<br /> <br /> After cutting off the slice corresponding to &lt;math&gt;|y-z|&lt; 1&lt;/math&gt;, we have 4 pieces left.<br /> <br /> [[File:5.png]]<br /> <br /> After cutting off the slice corresponding to &lt;math&gt;|z-x|&lt; 1&lt;/math&gt;, we have 6 congruent triangular prisms. <br /> <br /> [[File:6B.png]]<br /> <br /> Here we draw all the pictures in colors in order to explain the solution clearly. That does not mean that the students should do it in the examination. They can draw a figure with lines only, as shown below.<br /> <br /> [[File:7.png]]<br /> <br /> Every triangular pyramid has an altitude &lt;math&gt;n-2&lt;/math&gt; and a base of isoceless right triangle with leg length &lt;math&gt;n-2&lt;/math&gt;, so the volume is &lt;math&gt;(n-2)^3/6&lt;/math&gt;.<br /> Then the volume of the region &lt;math&gt;\big\{ (x,y,z)\ | \ 0\leqslant x, y, z \leqslant n, |x-y|\geqslant 1, |y-z|\geqslant 1, |z-x|\geqslant 1 \big\}&lt;/math&gt; is &lt;math&gt;V_A=6\times(n-2)^3/6&lt;/math&gt;=&lt;math&gt;(n-2)^3&lt;/math&gt;.<br /> <br /> So the probability of the incident &lt;math&gt;A&lt;/math&gt; is &lt;math&gt;P(A)=\dfrac{V_A}{V_0}&lt;/math&gt;=&lt;math&gt;\dfrac{(n-2)^3}{n^3}&lt;/math&gt;.<br /> <br /> Then we can get the answer the same way as Solution I.<br /> <br /> The answer is &lt;math&gt; \boxed{\textbf{(D)}\ 10} &lt;/math&gt;.<br /> <br /> <br /> ----<br /> <br /> <br /> If there is no choice for selection, we can also find the minimum value of the integer &lt;math&gt;n&lt;/math&gt; if we do not substitute &lt;math&gt;n&lt;/math&gt; by the possible values one by one.<br /> <br /> Let &lt;math&gt;P(A)&gt;1/2&lt;/math&gt;, i.e., &lt;math&gt;\dfrac{(n-2)^3}{n^3}&gt;\dfrac{1}{2}&lt;/math&gt;, so &lt;math&gt;\dfrac{n-2}{n}&gt;\dfrac{1}{\sqrt[^3\!]{2}}&lt;/math&gt;, or &lt;math&gt;1-\dfrac{2}{n}&gt;\dfrac{1}{\sqrt[^3\!]{2}}&lt;/math&gt;, hence &lt;math&gt;n&gt;\dfrac{2\sqrt[^3\!]{2}}{\sqrt[^3\!]{2}-1}&lt;/math&gt;.<br /> <br /> Now we will estimate the value of &lt;math&gt;\dfrac{2\sqrt[^3\!]{2}}{\sqrt[^3\!]{2}-1}&lt;/math&gt; without a calculator.<br /> <br /> Since &lt;math&gt;a^3-1&lt;/math&gt;=&lt;math&gt;(a-1)(a^2+a+1)&lt;/math&gt;, so<br /> &lt;math&gt;\dfrac{2\sqrt[^3\!]{2}}{\sqrt[^3\!]{2}-1}&lt;/math&gt;<br /> =&lt;math&gt;\dfrac{2\sqrt[^3\!]{2}\times\left( \sqrt[^3\!]{2}^2+\sqrt[^3\!]{2}+1\right)}{\left( \sqrt[^3\!]{2}-1\right)\left( \sqrt[^3\!]{2}^2+\sqrt[^3\!]{2}+1\right)}&lt;/math&gt;<br /> =&lt;math&gt;\dfrac{2\times\left( 2+\sqrt[^3\!]{2}^2+\sqrt[^3\!]{2}\right)}{ \sqrt[^3\!]{2}^3-1}&lt;/math&gt;<br /> =&lt;math&gt;2\times\left( 2+\sqrt[^3\!]{4}+\sqrt[^3\!]{2}\right)&lt;/math&gt;.<br /> <br /> Now we would get the approximation of &lt;math&gt;\sqrt[^3\!]{4}&lt;/math&gt; and &lt;math&gt;\sqrt[^3\!]{2}&lt;/math&gt;.<br /> <br /> In order to avoid compicated computation, we get the approximation with one decimal digit only.<br /> <br /> Estimation of &lt;math&gt;\sqrt[^3\!]{2}&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;1.5^3=2.25\times1.5&gt;2&lt;/math&gt;, so &lt;math&gt;1&lt;\sqrt[^3\!]{2}&lt;1.5&lt;/math&gt;.<br /> <br /> The mean of 1 and 1.5 with one decimal digit is about 1.3 .<br /> <br /> As &lt;math&gt;1.3^3=1.69\times 1.3=2.197&gt;2&lt;/math&gt;, so &lt;math&gt;1&lt;\sqrt[^3\!]{2}&lt;1.3&lt;/math&gt;.<br /> <br /> The mean of 1 and 1.3 with one decimal digit is about 1.2.<br /> <br /> As &lt;math&gt;1.2^3=1.44\times 1.2=1.728&lt;2&lt;/math&gt;, so &lt;math&gt;1.2&lt;\sqrt[^3\!]{2}&lt;1.3&lt;/math&gt;.<br /> <br /> Estimation of &lt;math&gt;\sqrt[^3\!]{4}&lt;/math&gt;.<br /> <br /> As &lt;math&gt;\sqrt[^3\!]{4}=\sqrt[^3\!]{2}^2&lt;/math&gt;, so &lt;math&gt;1.2^2&lt;\sqrt[^3\!]{4}&lt;1.3^2&lt;/math&gt;,<br /> then &lt;math&gt;1.24&lt;\sqrt[^3\!]{4}&lt;1.69&lt;/math&gt;.<br /> <br /> As &lt;math&gt;1.5^3=2.25\times 1.5=3.375&lt;4&lt;/math&gt;, so &lt;math&gt;1.5&lt;\sqrt[^3\!]{4}&lt;1.69&lt;/math&gt;.<br /> <br /> The mean of 1.5 and 1.69 with one decimal digit is about 1.6.<br /> <br /> As &lt;math&gt;1.6^3=(16/10)^3=(2^4/10)^3=2^{12}/10^3=4\times 2^10/10^3=4\times 1.024&gt;4&lt;/math&gt;, so &lt;math&gt;1.5&lt;\sqrt[^3\!]{4}&lt;1.6&lt;/math&gt;.<br /> <br /> <br /> Then &lt;math&gt;2\times(2+1.5+1.2)&lt;2\times\left(2+\sqrt[^3\!]{4}+\sqrt[^3\!]{2}\right)&lt;2\times(2+1.6+1.3)&lt;/math&gt;, i.e., <br /> &lt;math&gt;9.4&lt;2\times\left(2+\sqrt[^3\!]{4}+\sqrt[^3\!]{2}\right)&lt;9.8&lt;/math&gt;,<br /> <br /> As &lt;math&gt;n&gt;2\times\left(2+\sqrt[^3\!]{4}+\sqrt[^3\!]{2}\right)&lt;/math&gt;, So the minimal value of integer &lt;math&gt;n&lt;/math&gt; is 10.<br /> <br /> ===Appendix===<br /> This solution is motivated by the suggestive formula &lt;math&gt;\frac{(n-2)^{3}}{n^{3}}&lt;/math&gt;.<br /> <br /> The problem generalizes easily to &lt;math&gt;k&lt;/math&gt;-dimensional real space &lt;math&gt;\mathbb{R}^{k}&lt;/math&gt;. In the general &lt;math&gt;k&lt;/math&gt;-dimensional case, we are asked to find the probability that a randomly chosen &lt;math&gt;k&lt;/math&gt;-tuple &lt;math&gt;(x_{1},\dotsc,x_{k}) \in [0,n]^{k}&lt;/math&gt; satisfies &lt;math&gt;|x_{i} - x_{j}| &gt; 1&lt;/math&gt; for all &lt;math&gt;i \ne j&lt;/math&gt;. To avoid repetition, let us say that &lt;math&gt;(x_{1},\dotsc,x_{k})&lt;/math&gt; is &lt;i&gt;spaced-out&lt;/i&gt; if &lt;math&gt;|x_{i} - x_{j}| &gt; 1&lt;/math&gt; for all &lt;math&gt;i \ne j&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C&lt;/math&gt; be the &lt;math&gt;k&lt;/math&gt;-dimensional hyper-cube of side length &lt;math&gt;n&lt;/math&gt;: <br /> &lt;cmath&gt; C = [0,n]^{k} = \big\{(x_{1},\dotsc,x_{k}) \in \mathbb{R}^{k} \;:\; 0 \leqslant x_{i} \leqslant n \text{ for all }i \big\} \;.&lt;/cmath&gt; <br /> Then &lt;math&gt;C&lt;/math&gt; has volume &lt;math&gt;n^{k}&lt;/math&gt;. Let &lt;math&gt;S&lt;/math&gt; be the set of spaced-out &lt;math&gt;k&lt;/math&gt;-tuples &lt;math&gt;(x_{1},\dotsc,x_{k})&lt;/math&gt;. The desired probability is Vol&lt;math&gt;(S)/n^{k}&lt;/math&gt;. <br /> <br /> The set of &lt;math&gt;k&lt;/math&gt;-tuples &lt;math&gt;(x_{1},\dotsc,x_{k})&lt;/math&gt; such that there exist distinct indices &lt;math&gt;i, j&lt;/math&gt; such that &lt;math&gt;x_{i} = x_{j}&lt;/math&gt; has volume &lt;math&gt;0&lt;/math&gt;, so we may restrict our attention to &lt;math&gt;k&lt;/math&gt;-tuples such that &lt;math&gt;x_{i} \ne x_{j}&lt;/math&gt; for all &lt;math&gt;i \ne j&lt;/math&gt;. <br /> <br /> Further, the condition that &lt;math&gt;(x_{1},\dotsc,x_{k})&lt;/math&gt; is spaced-out is &quot;invariant upon permuting the indices&quot;; in other words, if &lt;math&gt;\sigma&lt;/math&gt; is a permutation of the set of indices &lt;math&gt;\{1,\dotsc,k\}&lt;/math&gt;, then &lt;math&gt;(x_{1},\dotsc,x_{k})&lt;/math&gt; is spaced-out if and only if &lt;math&gt;(x_{\sigma(1)},\dotsc,x_{\sigma(k)})&lt;/math&gt; is spaced-out. Therefore, we may consider the set of spaced-out &lt;math&gt;k&lt;/math&gt;-tuples &lt;math&gt;(x_{1},\dotsc,x_{k})&lt;/math&gt; which additionally satisfy &lt;math&gt;x_{1} &lt; \dotsb &lt; x_{k}&lt;/math&gt;. Let us denote this set by &lt;math&gt;T&lt;/math&gt;. This condition is equivalent to &lt;cmath&gt;0 \leqslant x_{1} &lt; x_{2} - 1 &lt; \dotsb &lt; x_{i}-(i-1) &lt; \dotsb &lt; x_{k}-(k-1) \leqslant n-(k-1) \;.&lt;/cmath&gt;<br /> Let us choose new variables &lt;math&gt;y_{i} = x_{i} - (i-1)&lt;/math&gt; for &lt;math&gt;i = 1,\dotsc,k&lt;/math&gt;. This change of variables is just a translation of each &lt;math&gt;(x_{1},\dotsc,x_{k})&lt;/math&gt; by the vector &lt;math&gt;(0,1,\dots,k-1)&lt;/math&gt;; in the above solutions, it corresponds to taking the 6 tetrahedrons and gluing them together to form a cube. <br /> <br /> We now compute the volume of the set of &lt;math&gt;(y_{1},\dotsc,y_{k}) \in [0,n-(k-1)]^{k}&lt;/math&gt; which satisfy &lt;math&gt;y_{1} &lt; \dotsb &lt; y_{k}&lt;/math&gt;. As above, we can disregard any &lt;math&gt;(y_{1},\dotsc,y_{k})&lt;/math&gt; such that &lt;math&gt;y_{i} = y_{j}&lt;/math&gt; for some &lt;math&gt;i \ne j&lt;/math&gt;. Given any &lt;math&gt;(y_{1},\dotsc,y_{k})&lt;/math&gt; such that &lt;math&gt;y_{i} \ne y_{j}&lt;/math&gt; for all &lt;math&gt;i \ne j&lt;/math&gt;, there exists exactly one permutation &lt;math&gt;\sigma&lt;/math&gt; of the indices &lt;math&gt;\{1,\dotsc,k\}&lt;/math&gt; such that &lt;math&gt;y_{\sigma(1)} &lt; \dotsb &lt; y_{\sigma(k)}&lt;/math&gt;. Since there are &lt;math&gt;k!&lt;/math&gt; permutations of &lt;math&gt;\{1,\dotsc,k\}&lt;/math&gt;, the desired volume is equal to &lt;math&gt;\frac{1}{k!}&lt;/math&gt; times the volume of the &lt;math&gt;k&lt;/math&gt;-dimensional hyper-cube of side length &lt;math&gt;n-(k-1)&lt;/math&gt;, which is &lt;math&gt;\frac{1}{k!}(n-(k-1))^{k}&lt;/math&gt;. Hence &lt;math&gt;T&lt;/math&gt; has volume &lt;math&gt;\frac{1}{k!}(n-(k-1))^{k}&lt;/math&gt; as well and &lt;math&gt;S&lt;/math&gt; has volume &lt;math&gt;(n-(k-1))^{k}&lt;/math&gt;. Hence the desired probability is &lt;math&gt;\frac{(n-(k-1))^{k}}{n^{k}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> If &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are separated by at least one, then by subtracting the minimum space between the three variables (which is &lt;math&gt;2&lt;/math&gt;), &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; can be chosen randomly in the interval &lt;math&gt;[0,n-2]&lt;/math&gt;. The probability is hence &lt;math&gt;\dfrac{(n-2)^3}{n^3} &gt; \dfrac{1}{2}&lt;/math&gt;, which simplifies to &lt;math&gt;n^3-12n^2+24n-16&gt;0&lt;/math&gt;. &lt;math&gt;\boxed{\textbf{(D)} 10}&lt;/math&gt; is the minimum which satisfies this inequality.<br /> <br /> ==Solution 4==<br /> Imagine Points &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, &lt;math&gt;z&lt;/math&gt; as the &quot;starting points&quot; of three &quot;blocks&quot; of real numbers that have length &lt;math&gt;1&lt;/math&gt;. We are just trying to find the probability that those three &quot;blocks&quot; do not overlap. To do this we can set each unit of &lt;math&gt;1&lt;/math&gt; into &lt;math&gt;\mu&lt;/math&gt; equal little increments, and take the limit of the probability as &lt;math&gt;\mu&lt;/math&gt; approaches &lt;math&gt;\infty&lt;/math&gt;. (This is because there are indefinitely many real numbers in a given interval.) We see that the total number of arrangements for the three blocks of &lt;math&gt;1&lt;/math&gt;, without considering the rule that no two blocks shall overlap is &lt;math&gt;(n\mu + 1)^3&lt;/math&gt;. We see that the number of ways to arrange the three blocks such that no three of them are overlapping is simply &lt;math&gt;\dbinom{(n-2)\mu + 3}{3}&lt;/math&gt;. Taking the limit as &lt;math&gt;\mu&lt;/math&gt; approaches infinity, we obtain our closed form to be &lt;math&gt;\lim_{\mu \to \infty} \frac{\dbinom{(n-2)\mu + 3}{3}}{ (n\mu + 1)^3 }. &lt;/math&gt; Dividing leading coefficients, we obtain &lt;math&gt;\lim_{\mu \to \infty} \frac{\dbinom{(n-2)\mu + 3}{3}}{ (n\mu + 1)^3 } = <br /> \frac{\mu(n-2)^3}{\mu n^3} = \frac{(n-2)^3}{n^3}&lt;/math&gt;. Solving the inequality &lt;math&gt;\frac{(n-2)^3}{n^3} &gt; \frac{1}{2}&lt;/math&gt;, we get the least value of &lt;math&gt;n&lt;/math&gt; as &lt;math&gt;n=\boxed{10}&lt;/math&gt;.<br /> <br /> -fidgetboss_4000<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=24|after=Last Problem.}}<br /> {{MAA Notice}}</div> Relay400 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_25&diff=115135 2013 AMC 10B Problems/Problem 25 2020-01-22T05:15:22Z <p>Relay400: small tweaks to make solution more cohesive &amp; understandable</p> <hr /> <div>{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #23]] and [[2013 AMC 10B Problems|2013 AMC 10B #25]]}}<br /> <br /> ==Problem==<br /> <br /> Bernardo chooses a three-digit positive integer &lt;math&gt;N&lt;/math&gt; and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer &lt;math&gt;S&lt;/math&gt;. For example, if &lt;math&gt;N = 749&lt;/math&gt;, Bernardo writes the numbers &lt;math&gt;10,\!444&lt;/math&gt; and &lt;math&gt;3,\!245&lt;/math&gt;, and LeRoy obtains the sum &lt;math&gt;S = 13,\!689&lt;/math&gt;. For how many choices of &lt;math&gt;N&lt;/math&gt; are the two rightmost digits of &lt;math&gt;S&lt;/math&gt;, in order, the same as those of &lt;math&gt;2N&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.<br /> <br /> Say that &lt;math&gt;N \equiv a \pmod{6}&lt;/math&gt;<br /> <br /> also that &lt;math&gt;N \equiv b \pmod{5}&lt;/math&gt;<br /> <br /> Substituting these equations into the question and setting the units digits of &lt;math&gt;2N&lt;/math&gt; and &lt;math&gt;S&lt;/math&gt; equal to each other, it can be seen that &lt;math&gt;b &lt; 5&lt;/math&gt; (because otherwise &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; will have different parities), and thus &lt;math&gt;a=b&lt;/math&gt;.<br /> &lt;math&gt;N \equiv a \pmod{6}&lt;/math&gt;,<br /> &lt;math&gt;N \equiv a \pmod{5}&lt;/math&gt;,<br /> &lt;math&gt;\implies N=a \pmod{30}&lt;/math&gt;,<br /> &lt;math&gt;0 \le a \le 4 &lt;/math&gt;<br /> <br /> Therefore, &lt;math&gt;N&lt;/math&gt; can be written as &lt;math&gt;30x+y&lt;/math&gt;<br /> and &lt;math&gt;2N&lt;/math&gt; can be written as &lt;math&gt;60x+2y&lt;/math&gt; <br /> <br /> Just keep in mind that &lt;math&gt;y&lt;/math&gt; can be one of five choices: &lt;math&gt;0, 1, 2, 3,&lt;/math&gt; or &lt;math&gt;4&lt;/math&gt;, ;<br /> Also, we have already found which digits of &lt;math&gt;y&lt;/math&gt; will add up into the units digits of &lt;math&gt;2N&lt;/math&gt;.<br /> <br /> Now, examine the tens digit, &lt;math&gt;x&lt;/math&gt; by using &lt;math&gt;\mod{25}&lt;/math&gt; and &lt;math&gt;\mod{36}&lt;/math&gt; to find the tens digit (units digits can be disregarded because &lt;math&gt;y=0,1,2,3,4&lt;/math&gt; will always work)<br /> Then we take &lt;math&gt;N=30x+y&lt;/math&gt; &lt;math&gt;\mod{25}&lt;/math&gt; and &lt;math&gt;\mod{36}&lt;/math&gt; to find the last two digits in the base &lt;math&gt;5&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt; representation.<br /> &lt;cmath&gt;N \equiv 30x \pmod{36}&lt;/cmath&gt;<br /> &lt;cmath&gt;N \equiv 30x \equiv 5x \pmod{25}&lt;/cmath&gt; <br /> Both of those must add up to <br /> &lt;cmath&gt;2N\equiv60x \pmod{100}&lt;/cmath&gt;<br /> <br /> (&lt;math&gt;33 \ge x \ge 4&lt;/math&gt;)<br /> <br /> Now, since &lt;math&gt;y=0,1,2,3,4&lt;/math&gt; will always work if &lt;math&gt;x&lt;/math&gt; works, then we can treat &lt;math&gt;x&lt;/math&gt; as a units digit instead of a tens digit in the respective bases and decrease the mods so that &lt;math&gt;x&lt;/math&gt; is now the units digit.<br /> &lt;cmath&gt;N \equiv 6x \equiv x \pmod{5}&lt;/cmath&gt; <br /> &lt;cmath&gt;N \equiv 5x \pmod{6}&lt;/cmath&gt; <br /> &lt;cmath&gt;2N\equiv 6x \pmod{10}&lt;/cmath&gt;<br /> <br /> Say that &lt;math&gt;x=5m+n&lt;/math&gt; (m is between 0-6, n is 0-4 because of constraints on x)<br /> Then <br /> <br /> &lt;cmath&gt;N \equiv 5m+n \pmod{5}&lt;/cmath&gt; <br /> &lt;cmath&gt;N \equiv 25m+5n \pmod{6}&lt;/cmath&gt; <br /> &lt;cmath&gt;2N\equiv30m + 6n \pmod{10}&lt;/cmath&gt;<br /> <br /> and this simplifies to <br /> <br /> &lt;cmath&gt;N \equiv n \pmod{5}&lt;/cmath&gt; <br /> &lt;cmath&gt;N \equiv m+5n \pmod{6}&lt;/cmath&gt;<br /> &lt;cmath&gt;2N\equiv 6n \pmod{10}&lt;/cmath&gt;<br /> <br /> From careful inspection, this is true when<br /> <br /> &lt;math&gt;n=0, m=6&lt;/math&gt;<br /> <br /> &lt;math&gt;n=1, m=6&lt;/math&gt;<br /> <br /> &lt;math&gt;n=2, m=2&lt;/math&gt;<br /> <br /> &lt;math&gt;n=3, m=2&lt;/math&gt;<br /> <br /> &lt;math&gt;n=4, m=4&lt;/math&gt;<br /> <br /> This gives you &lt;math&gt;5&lt;/math&gt; choices for &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; choices for &lt;math&gt;y&lt;/math&gt;, so the answer is <br /> &lt;math&gt;5* 5 = \boxed{\textbf{(E) }25}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Notice that there are exactly &lt;math&gt;1000-100=900=5^2\cdot 6^2&lt;/math&gt; possible values of &lt;math&gt;N&lt;/math&gt;. This means, in &lt;math&gt;100\le N\le 999&lt;/math&gt;, every possible combination of &lt;math&gt;2&lt;/math&gt; digits will happen exactly once. We know that &lt;math&gt;N=900,901,902,903,904&lt;/math&gt; works because &lt;math&gt;900\equiv\dots00_5\equiv\dots00_6&lt;/math&gt;.<br /> <br /> We know for sure that the units digit will add perfectly every &lt;math&gt;30&lt;/math&gt; added or subtracted, because &lt;math&gt;\text{lcm }5,6=30&lt;/math&gt;. So we only have to care about cases of &lt;math&gt;N&lt;/math&gt; every &lt;math&gt;30&lt;/math&gt; subtracted. In each case, &lt;math&gt;2N&lt;/math&gt; subtracts &lt;math&gt;6&lt;/math&gt;/adds &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;N_5&lt;/math&gt; subtracts &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;N_6&lt;/math&gt; adds &lt;math&gt;1&lt;/math&gt; for the &lt;math&gt;10&lt;/math&gt;'s digit.<br /> <br /> &lt;cmath&gt;\textbf{5 }\textcolor{red}{\text{ 0}}\text{ 4 3 2 1 0 }\textcolor{red}{\text{4}}\text{ 3 2 1 0 4 3 2 1 0 4 }\textcolor{red}{\text{3 2}}\text{ 1 0 4 3 2 1 0 4 3 2 }\textcolor{red}{\text{1}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\textbf{6 }\textcolor{red}{\text{ 0}}\text{ 1 2 3 4 5 }\textcolor{red}{\text{0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5 0}}\text{ 1 2 3 4 5 0 1 2 3 4 }\textcolor{red}{\text{5}}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\textbf{10}\textcolor{red}{\text{ 0}}\text{ 4 8 2 6 0 }\textcolor{red}{\text{4}}\text{ 8 2 6 0 4 8 2 6 0 4 }\textcolor{red}{\text{8 2}}\text{ 6 0 4 8 2 6 0 4 8 2 }\textcolor{red}{\text{6}}&lt;/cmath&gt;<br /> <br /> As we can see, there are &lt;math&gt;5&lt;/math&gt; cases, including the original, that work. These are highlighted in &lt;math&gt;\textcolor{red}{\text{red}}&lt;/math&gt;. So, thus, there are &lt;math&gt;5&lt;/math&gt; possibilities for each case, and &lt;math&gt;5\cdot 5=\boxed{\textbf{(E) }25}&lt;/math&gt;.<br /> <br /> == Solution 3 ==<br /> <br /> Notice that &lt;math&gt;N_5&lt;/math&gt; ranges from &lt;math&gt;3&lt;/math&gt; to &lt;math&gt;5&lt;/math&gt; digits and &lt;math&gt;N_6&lt;/math&gt; ranges from &lt;math&gt;3&lt;/math&gt; to &lt;math&gt;4&lt;/math&gt; digits.<br /> <br /> Then let &lt;math&gt;a_i&lt;/math&gt;, &lt;math&gt;b_i&lt;/math&gt; denotes the digits of &lt;math&gt;N_5&lt;/math&gt;, &lt;math&gt;N_6&lt;/math&gt;, respectively such that &lt;cmath&gt;0\le a_i&lt;5,0\le b_i&lt;6&lt;/cmath&gt; Thus we have &lt;cmath&gt;N=5^4a_1+5^3a_2+5^2a_3+5a_4+a_5=6^3b_1+6^2b_2+6b_3+b_4&lt;/cmath&gt;<br /> &lt;cmath&gt;625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4&lt;/cmath&gt; <br /> Now we are given &lt;cmath&gt;2N \equiv S \equiv N_5+N_6\pmod{100}&lt;/cmath&gt;<br /> &lt;cmath&gt;2(625a_1+125a_2+25a_3+5a_4+a_5) \equiv (10000a_1+1000a_2+100a_3+10a_4+a_5)+(1000b_1+100b_2+10b_3+b_4)\pmod{100}&lt;/cmath&gt;<br /> &lt;cmath&gt;1250a_1+250a_2+50a_3+10a_4+2a_5 \equiv 10000a_1+1000a_2+1000b_1+100a_3+100b_2+10a_4+10b_3+a_5+b_4\pmod{100}&lt;/cmath&gt;<br /> &lt;cmath&gt;50a_1+50a_2+50a_3+10a_4+2a_5 \equiv 10a_4+10b_3+a_5+b_4\pmod{100}&lt;/cmath&gt;<br /> Canceling out &lt;math&gt;a_5&lt;/math&gt; left with<br /> &lt;cmath&gt;50a_1+50a_2+50a_3+10a_4+a_5 \equiv 10a_4+10b_3+b_4\pmod{100}&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;a_5&lt;/math&gt;, &lt;math&gt;b_4&lt;/math&gt; determine the unit digits of the two sides of the congruence equation, we have &lt;math&gt;a_5=b_4=0,1,2,3,4&lt;/math&gt;. Thus,<br /> <br /> &lt;cmath&gt;50a_1+50a_2+50a_3+10a_4 \equiv 10a_4+10b_3\pmod{100}&lt;/cmath&gt; canceling out &lt;math&gt;10a_4&lt;/math&gt;, we have <br /> &lt;cmath&gt;50a_1+50a_2+50a_3 \equiv 10b_3\pmod{100}&lt;/cmath&gt;<br /> &lt;cmath&gt;5a_1+5a_2+5a_3 \equiv b_3\pmod{10}&lt;/cmath&gt;<br /> &lt;cmath&gt;5(a_1+a_2+a_3) \equiv b_3\pmod{10}&lt;/cmath&gt;<br /> <br /> Thus &lt;math&gt;b_3&lt;/math&gt; is a multiple of &lt;math&gt;5&lt;/math&gt;.<br /> <br /> Now going back to our original equation <br /> &lt;cmath&gt;625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4&lt;/cmath&gt; <br /> Since &lt;math&gt;a_5=b_4&lt;/math&gt;,<br /> &lt;cmath&gt;625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3&lt;/cmath&gt; <br /> &lt;cmath&gt;5(125a_1+25a_2+5a_3+a_4)=6(36b_1+6b_2+b_3)&lt;/cmath&gt; <br /> &lt;cmath&gt;5(125a_1+25a_2+5a_3+a_4)=6[6(6b_1+b_2)+b_3]&lt;/cmath&gt; <br /> <br /> Since the left side is a multiple of &lt;math&gt;5&lt;/math&gt;, then so does the right side. Thus &lt;math&gt;5\mid6(6b_1+b_2)+b_3&lt;/math&gt;.<br /> <br /> Since we already know that &lt;math&gt;5\mid b_3&lt;/math&gt;, then &lt;math&gt;5\mid6(6b_1+b_2)&lt;/math&gt;, from where we also know that &lt;math&gt;5\mid6b_1+b_2&lt;/math&gt;.<br /> <br /> For &lt;math&gt;b_1,b_2&lt;6&lt;/math&gt;, there is a total of 7 ordered pairs that satisfy the condition. Namely,<br /> <br /> &lt;cmath&gt;(b_1,b_2)=(0,0),(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;N_6&lt;/math&gt; has at least &lt;math&gt;3&lt;/math&gt; digits, &lt;math&gt;(b_1,b_2)=(0,0)&lt;/math&gt; doesn't work. Furthermore, when &lt;math&gt;b_1=5&lt;/math&gt;, &lt;math&gt;216b_1&lt;/math&gt; exceeds &lt;math&gt;1000&lt;/math&gt; which is not possible as &lt;math&gt;N&lt;/math&gt; is a three digit number, thus &lt;math&gt;(b_1,b_2)=(5,0)&lt;/math&gt; won't work as well. <br /> <br /> Since we know that &lt;math&gt;a_i&lt;5&lt;/math&gt;, for each of the ordered pairs &lt;math&gt;(b_1,b_2)&lt;/math&gt;, there is respectively one and only one solution &lt;math&gt;(a_1,a_2,a_3,a_5)&lt;/math&gt; that satisfies the equation <br /> <br /> &lt;cmath&gt;625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3&lt;/cmath&gt;<br /> <br /> Thus there are five solutions to the equation. Also since we have 5 possibilities for &lt;math&gt;a_5=b_4&lt;/math&gt;, we have a total of &lt;math&gt;5\cdot5=25&lt;/math&gt; values for &lt;math&gt;N&lt;/math&gt;. &lt;math&gt;\boxed{\textbf{(E) }25}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> ==Solution 4==<br /> <br /> Observe that the maximum possible value of the sum of the last two digits of the base &lt;math&gt;5&lt;/math&gt; number and the base &lt;math&gt;6&lt;/math&gt; number is &lt;math&gt;44+55=99&lt;/math&gt;. <br /> Let &lt;math&gt;N \equiv a \mod 25&lt;/math&gt; and &lt;math&gt;N \equiv b \mod 36&lt;/math&gt;. <br /> <br /> If &lt;math&gt;a &lt; \frac{25}{2}&lt;/math&gt;, &lt;math&gt;2N \equiv 2a \mod 25&lt;/math&gt; and if &lt;math&gt;a &gt; \frac{25}{2}&lt;/math&gt;, &lt;math&gt;2N \equiv 2a - 25 \mod 25&lt;/math&gt;.<br /> <br /> Using the same logic for &lt;math&gt;b&lt;/math&gt;, if &lt;math&gt;b &lt; 18&lt;/math&gt;, &lt;math&gt;2N \equiv 2b \mod 36&lt;/math&gt;, and in the other case &lt;math&gt;2N \equiv 2b - 36 \mod 36&lt;/math&gt;.<br /> <br /> We can do four cases:<br /> <br /> Case 1: &lt;math&gt;a + b = 2a - 25 + 2b - 36 \implies a + b = 61&lt;/math&gt;.<br /> <br /> For this case, there is trivially only one possible solution, &lt;math&gt;(a, b) = (25, 36)&lt;/math&gt;, which is equivalent to &lt;math&gt;(a, b) = (0, 0)&lt;/math&gt;. <br /> <br /> Case 2: &lt;math&gt;a + b = 2a - 25 + 2b \implies a + b = 25&lt;/math&gt;.<br /> <br /> Note that in this case, &lt;math&gt;a \geq 13&lt;/math&gt; must hold, and &lt;math&gt;b &lt; 18&lt;/math&gt; must hold. <br /> We find the possible ordered pairs to be: &lt;math&gt;(13, 12), (14, 11), (15, 10), ..., (24, 1)&lt;/math&gt; for a total of &lt;math&gt;12&lt;/math&gt; ordered pairs.<br /> <br /> Case 3: &lt;math&gt;a + b = 2a + 2b - 36 \implies a + b = 36&lt;/math&gt;.<br /> <br /> Note that in this case, &lt;math&gt;b \geq 18&lt;/math&gt; must hold, and &lt;math&gt;a &lt; 13&lt;/math&gt; must hold.<br /> We find the possible ordered pairs to be: &lt;math&gt;(24, 12), (25, 11), (26, 10), ..., (35, 1)&lt;/math&gt; for a total of &lt;math&gt;12&lt;/math&gt; ordered pairs.<br /> <br /> Case 4: &lt;math&gt;a + b = 2a + 2b&lt;/math&gt;.<br /> <br /> Trivially no solutions except &lt;math&gt;(a, b) = (0, 0)&lt;/math&gt;, which matches the solution in Case 1, which makes this an overcount.<br /> <br /> By CRT, each solution &lt;math&gt;(a, b)&lt;/math&gt; corresponds exactly one positive integer in a set of exactly &lt;math&gt;\text{lcm} (25, 36) = 900&lt;/math&gt; consecutive positive integers, and since there are &lt;math&gt;900&lt;/math&gt; positive integers between &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;999&lt;/math&gt;, our induction is complete, and our answer is &lt;math&gt;1 + 12 + 12 = \boxed{\textbf{(E) }25}&lt;/math&gt;.<br /> <br /> -fidgetboss_4000<br /> <br /> {{AMC12 box|year=2013|ab=B|num-b=22|num-a=24}}<br /> <br /> {{AMC10 box|year=2013|ab=B|num-b=24|after=Last Question}}<br /> <br /> {{MAA Notice}}</div> Relay400 https://artofproblemsolving.com/wiki/index.php?title=Kevin_Zhao&diff=106098 Kevin Zhao 2019-06-03T21:55:44Z <p>Relay400: ... Definitely not appropriate.</p> <hr /> <div></div> Relay400 https://artofproblemsolving.com/wiki/index.php?title=2019_USAJMO_Problems/Problem_4&diff=105623 2019 USAJMO Problems/Problem 4 2019-05-04T18:46:07Z <p>Relay400: /* See also */</p> <hr /> <div>==Problem==<br /> &lt;math&gt;(*)&lt;/math&gt; Let &lt;math&gt;ABC&lt;/math&gt; be a triangle with &lt;math&gt;\angle ABC&lt;/math&gt; obtuse. The &lt;math&gt;A&lt;/math&gt;''-excircle'' is a circle in the exterior of &lt;math&gt;\triangle ABC&lt;/math&gt; that is tangent to side &lt;math&gt;\overline{BC}&lt;/math&gt; of the triangle and tangent to the extensions of the other two sides. Let &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; be the feet of the altitudes from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; to lines &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;, respectively. Can line &lt;math&gt;EF&lt;/math&gt; be tangent to the &lt;math&gt;A&lt;/math&gt;-excircle?<br /> <br /> ==Solution==<br /> <br /> Instead of trying to find a synthetic way to describe &lt;math&gt;EF&lt;/math&gt; being tangent to the &lt;math&gt;A&lt;/math&gt;-excircle (very hard), we instead consider the foot of the perpendicular from the &lt;math&gt;A&lt;/math&gt;-excircle to &lt;math&gt;EF&lt;/math&gt;, hoping to force something via the length of the perpendicular. It would be nice if there were an easier way to describe &lt;math&gt;EF&lt;/math&gt;, something more closely related to the &lt;math&gt;A&lt;/math&gt;-excircle; as we are considering perpendicularity, if we could generate a line parallel to &lt;math&gt;EF&lt;/math&gt;, that would be good. <br /> <br /> So we recall that it is well known that triangle &lt;math&gt;AEF&lt;/math&gt; is similar to &lt;math&gt;ABC&lt;/math&gt;. This motivates reflecting &lt;math&gt;BC&lt;/math&gt; over the angle bisector at &lt;math&gt;A&lt;/math&gt; to obtain &lt;math&gt;B'C'&lt;/math&gt;, which is parallel to &lt;math&gt;EF&lt;/math&gt; for obvious reasons. <br /> <br /> Furthermore, as reflection preserves intersection, &lt;math&gt;B'C'&lt;/math&gt; is tangent to the reflection of the &lt;math&gt;A&lt;/math&gt;-excircle over the &lt;math&gt;A&lt;/math&gt;-angle bisector. But it is well-known that the &lt;math&gt;A&lt;/math&gt;-excenter lies on the &lt;math&gt;A&lt;/math&gt;-angle bisector, so the &lt;math&gt;A&lt;/math&gt;-excircle must be preserved under reflection over the &lt;math&gt;A&lt;/math&gt;-excircle. Thus &lt;math&gt;B'C'&lt;/math&gt; is tangent to the &lt;math&gt;A&lt;/math&gt;-excircle.Yet for all lines parallel to &lt;math&gt;EF&lt;/math&gt;, there are only two lines tangent to the &lt;math&gt;A&lt;/math&gt;-excircle, and only one possibility for &lt;math&gt;EF&lt;/math&gt;, so &lt;math&gt;EF = B'C'&lt;/math&gt;. <br /> <br /> Thus as &lt;math&gt;ABB'&lt;/math&gt; is isoceles, &lt;cmath&gt;[ABC] = \frac{1}{2} \cdot AC \cdot BE = \frac{AC}{2} \cdot \sqrt{AB^2 - AE^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB'^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB^2} = 0,&lt;/cmath&gt; contradiction. -alifenix-<br /> <br /> ==Solution 2 ==<br /> <br /> The answer is no.<br /> <br /> Suppose otherwise. Consider the reflection over the bisector of &lt;math&gt;\angle BAC&lt;/math&gt;. This swaps rays &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;; suppose &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are sent to &lt;math&gt;E'&lt;/math&gt; and &lt;math&gt;F'&lt;/math&gt;. Note that the &lt;math&gt;A&lt;/math&gt;-excircle is fixed, so line &lt;math&gt;E'F'&lt;/math&gt; must also be tangent to the &lt;math&gt;A&lt;/math&gt;-excircle.<br /> <br /> Since &lt;math&gt;BEFC&lt;/math&gt; is cyclic, we obtain &lt;math&gt;\measuredangle ECB = \measuredangle EFB = \measuredangle EF'E'&lt;/math&gt;, so &lt;math&gt;\overline{E'F'} \parallel \overline{BC}&lt;/math&gt;. However, as &lt;math&gt;\overline{EF}&lt;/math&gt; is a chord in the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;EF \le BC&lt;/math&gt;.<br /> <br /> If &lt;math&gt;EF &lt; BC&lt;/math&gt; then &lt;math&gt;E'F' &lt; BC&lt;/math&gt; too, so then &lt;math&gt;\overline{E'F'}&lt;/math&gt; lies inside &lt;math&gt;\triangle ABC&lt;/math&gt; and cannot be tangent to the excircle.<br /> <br /> The remaining case is when &lt;math&gt;EF = BC&lt;/math&gt;. In this case, &lt;math&gt;\overline{EF}&lt;/math&gt; is also a diameter, so &lt;math&gt;BECF&lt;/math&gt; is a rectangle. In particular &lt;math&gt;\overline{BE} \parallel \overline{CF}&lt;/math&gt;. However, by the existence of the orthocenter, the lines &lt;math&gt;BE&lt;/math&gt; and &lt;math&gt;CF&lt;/math&gt; must intersect, contradiction.<br /> <br /> ==See also==<br /> {{MAA Notice}}<br /> {{USAJMO newbox|year=2019|num-b=3|num-a=5}}</div> Relay400 https://artofproblemsolving.com/wiki/index.php?title=2019_USAJMO_Problems/Problem_4&diff=105622 2019 USAJMO Problems/Problem 4 2019-05-04T18:45:51Z <p>Relay400: /* Solution */</p> <hr /> <div>==Problem==<br /> &lt;math&gt;(*)&lt;/math&gt; Let &lt;math&gt;ABC&lt;/math&gt; be a triangle with &lt;math&gt;\angle ABC&lt;/math&gt; obtuse. The &lt;math&gt;A&lt;/math&gt;''-excircle'' is a circle in the exterior of &lt;math&gt;\triangle ABC&lt;/math&gt; that is tangent to side &lt;math&gt;\overline{BC}&lt;/math&gt; of the triangle and tangent to the extensions of the other two sides. Let &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt; be the feet of the altitudes from &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; to lines &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt;, respectively. Can line &lt;math&gt;EF&lt;/math&gt; be tangent to the &lt;math&gt;A&lt;/math&gt;-excircle?<br /> <br /> ==Solution==<br /> <br /> Instead of trying to find a synthetic way to describe &lt;math&gt;EF&lt;/math&gt; being tangent to the &lt;math&gt;A&lt;/math&gt;-excircle (very hard), we instead consider the foot of the perpendicular from the &lt;math&gt;A&lt;/math&gt;-excircle to &lt;math&gt;EF&lt;/math&gt;, hoping to force something via the length of the perpendicular. It would be nice if there were an easier way to describe &lt;math&gt;EF&lt;/math&gt;, something more closely related to the &lt;math&gt;A&lt;/math&gt;-excircle; as we are considering perpendicularity, if we could generate a line parallel to &lt;math&gt;EF&lt;/math&gt;, that would be good. <br /> <br /> So we recall that it is well known that triangle &lt;math&gt;AEF&lt;/math&gt; is similar to &lt;math&gt;ABC&lt;/math&gt;. This motivates reflecting &lt;math&gt;BC&lt;/math&gt; over the angle bisector at &lt;math&gt;A&lt;/math&gt; to obtain &lt;math&gt;B'C'&lt;/math&gt;, which is parallel to &lt;math&gt;EF&lt;/math&gt; for obvious reasons. <br /> <br /> Furthermore, as reflection preserves intersection, &lt;math&gt;B'C'&lt;/math&gt; is tangent to the reflection of the &lt;math&gt;A&lt;/math&gt;-excircle over the &lt;math&gt;A&lt;/math&gt;-angle bisector. But it is well-known that the &lt;math&gt;A&lt;/math&gt;-excenter lies on the &lt;math&gt;A&lt;/math&gt;-angle bisector, so the &lt;math&gt;A&lt;/math&gt;-excircle must be preserved under reflection over the &lt;math&gt;A&lt;/math&gt;-excircle. Thus &lt;math&gt;B'C'&lt;/math&gt; is tangent to the &lt;math&gt;A&lt;/math&gt;-excircle.Yet for all lines parallel to &lt;math&gt;EF&lt;/math&gt;, there are only two lines tangent to the &lt;math&gt;A&lt;/math&gt;-excircle, and only one possibility for &lt;math&gt;EF&lt;/math&gt;, so &lt;math&gt;EF = B'C'&lt;/math&gt;. <br /> <br /> Thus as &lt;math&gt;ABB'&lt;/math&gt; is isoceles, &lt;cmath&gt;[ABC] = \frac{1}{2} \cdot AC \cdot BE = \frac{AC}{2} \cdot \sqrt{AB^2 - AE^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB'^2} = \frac{AC}{2} \cdot \sqrt{AB^2 - AB^2} = 0,&lt;/cmath&gt; contradiction. -alifenix-<br /> <br /> ==Solution 2 ==<br /> <br /> The answer is no.<br /> <br /> Suppose otherwise. Consider the reflection over the bisector of &lt;math&gt;\angle BAC&lt;/math&gt;. This swaps rays &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;AC&lt;/math&gt;; suppose &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are sent to &lt;math&gt;E'&lt;/math&gt; and &lt;math&gt;F'&lt;/math&gt;. Note that the &lt;math&gt;A&lt;/math&gt;-excircle is fixed, so line &lt;math&gt;E'F'&lt;/math&gt; must also be tangent to the &lt;math&gt;A&lt;/math&gt;-excircle.<br /> <br /> Since &lt;math&gt;BEFC&lt;/math&gt; is cyclic, we obtain &lt;math&gt;\measuredangle ECB = \measuredangle EFB = \measuredangle EF'E'&lt;/math&gt;, so &lt;math&gt;\overline{E'F'} \parallel \overline{BC}&lt;/math&gt;. However, as &lt;math&gt;\overline{EF}&lt;/math&gt; is a chord in the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;EF \le BC&lt;/math&gt;.<br /> <br /> If &lt;math&gt;EF &lt; BC&lt;/math&gt; then &lt;math&gt;E'F' &lt; BC&lt;/math&gt; too, so then &lt;math&gt;\overline{E'F'}&lt;/math&gt; lies inside &lt;math&gt;\triangle ABC&lt;/math&gt; and cannot be tangent to the excircle.<br /> <br /> The remaining case is when &lt;math&gt;EF = BC&lt;/math&gt;. In this case, &lt;math&gt;\overline{EF}&lt;/math&gt; is also a diameter, so &lt;math&gt;BECF&lt;/math&gt; is a rectangle. In particular &lt;math&gt;\overline{BE} \parallel \overline{CF}&lt;/math&gt;. However, by the existence of the orthocenter, the lines &lt;math&gt;BE&lt;/math&gt; and &lt;math&gt;CF&lt;/math&gt; must intersect, contradiction.<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2019|num-b=3|num-a=5}}</div> Relay400 https://artofproblemsolving.com/wiki/index.php?title=AMC_historical_results&diff=104073 AMC historical results 2019-03-06T04:10:56Z <p>Relay400: /* AMC 12A */</p> <hr /> <div>&lt;!-- Post AMC statistics and lists of high scorers here so that the AMC page doesn't get cluttered. --&gt;<br /> This is the '''AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]].<br /> ==2019==<br /> ===AMC 10A===<br /> *Average score: TBD<br /> *AIME floor: TBD<br /> *DHR: TBD<br /> <br /> ===AMC 10B===<br /> *Average score: TBD<br /> *AIME floor: TBD <br /> *DHR: TBD<br /> <br /> ===AMC 12A===<br /> *Average score: TBD <br /> *AIME floor: TBD<br /> *DHR: TBD<br /> <br /> ===AMC 12B===<br /> *Average score: TBD<br /> *AIME floor: TBD <br /> *DHR: TBD<br /> <br /> ==2018==<br /> ===AMC 10A===<br /> *Average score:53.84<br /> *AIME floor: 111<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 57.81<br /> *AIME floor: 108<br /> *DHR: 123<br /> <br /> ===AMC 12A===<br /> *Average score: 56.36<br /> *AIME floor: 93<br /> *DHR: 120<br /> <br /> ===AMC 12B===<br /> *Average score: 57.85<br /> *AIME floor: 99<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.09<br /> *Median score: 5<br /> *USAMO cutoff: 215 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.48<br /> *Median score: 5<br /> *USAMO cutoff: 216 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 222 (AMC 10A), 212 (AMC 10B)<br /> <br /> ===AMC 8===<br /> *Honor roll: 15<br /> *DHR: 19<br /> <br /> ==2017==<br /> ===AMC 10A===<br /> *Average score: 59.33<br /> *AIME floor: 112.5<br /> *DHR: 127.5<br /> <br /> ===AMC 10B===<br /> *Average score: 66.56<br /> *AIME floor: 120<br /> *DHR: 136.5<br /> <br /> ===AMC 12A===<br /> *Average score: 60.32<br /> *AIME floor: 96<br /> *DHR: 115.5<br /> <br /> ===AMC 12B===<br /> *Average score: 58.35<br /> *AIME floor: 100.5<br /> *DHR: 129<br /> <br /> ===AIME I===<br /> *Average score:5.70<br /> *Median score: 6<br /> *USAMO cutoff: 225 (AMC 12A), 235 (AMC 12B)<br /> *USAJMO cutoff: 224.5 (AMC 10A), 233 (AMC 10B)<br /> <br /> ===AIME II===<br /> *Average score: 5.64<br /> *Median score: 6<br /> *USAMO cutoff: 221 (AMC 12A), 230.5 (AMC 12B)<br /> *USAJMO cutoff: 219 (AMC 10A), 225 (AMC 10B)<br /> ===AMC 8===<br /> *Honor roll: 17<br /> *DHR: 20<br /> <br /> ==2016==<br /> ===AMC 10A===<br /> *Average score: 65.3<br /> *AIME floor: 110<br /> *DHR: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 65.4<br /> *AIME floor: 111<br /> *DHR: 124.5<br /> <br /> ===AMC 12A===<br /> *Average score: 59.06<br /> *AIME floor: 93<br /> *DHR: 111<br /> <br /> ===AMC 12B===<br /> *Average score: 67.96<br /> *AIME floor: 100.5<br /> *DHR: 127.5<br /> <br /> ===AIME I===<br /> *Average score: 5.83<br /> *Median score: 6<br /> *USAMO cutoff: 220<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 4.33<br /> *Median score: 4<br /> *USAMO cutoff: 205<br /> *USAJMO cutoff: 200<br /> <br /> ===AMC 8===<br /> *Honor roll: 18<br /> *DHR: 22<br /> <br /> ==2015==<br /> ===AMC 10A===<br /> *Average score: 73.39<br /> *AIME floor: 106.5<br /> *DHR: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.10<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 69.90<br /> *AIME floor: 99<br /> *DHR: 117<br /> <br /> ===AMC 12B===<br /> *Average score: 66.92<br /> *AIME floor: 100.5<br /> *DHR: 126<br /> <br /> ===AIME I===<br /> *Average score: 5.29<br /> *Median score: 5<br /> *USAMO cutoff: 219.0<br /> *USAJMO cutoff: 213.0<br /> <br /> ===AIME II===<br /> *Average score: 6.63<br /> *Median score: 6<br /> *USAMO cutoff: 229.0<br /> *USAJMO cutoff: 223.5<br /> <br /> ==2014==<br /> ===AMC 10A===<br /> *Average score: 63.83<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 10B===<br /> *Average score: 71.44<br /> *AIME floor: 120<br /> *DHR: 132<br /> <br /> ===AMC 12A===<br /> *Average score: 64.01<br /> *AIME floor: 93<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.11<br /> *AIME floor: 100.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 4.88<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ===AIME II===<br /> *Average score: 5.49<br /> *Median score: 5<br /> *USAMO cutoff: 211.5<br /> *USAJMO cutoff: 211<br /> <br /> ==2013==<br /> ===AMC 10A===<br /> *Average score: 72.50<br /> *AIME floor: 108<br /> *DHR: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 72.62<br /> *AIME floor: 120<br /> *DHR: 129<br /> <br /> ===AMC 12A===<br /> *Average score: 65.06<br /> *AIME floor: 88.5<br /> *DHR: 106.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.21<br /> *AIME floor: 93<br /> *DHR: 108<br /> <br /> ===AIME I===<br /> *Average score: 4.69<br /> *Median score: 4<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ===AIME II===<br /> *Average score: 6.56<br /> *Median score: 6<br /> *USAMO cutoff: 209<br /> *USAJMO cutoff: 210.5<br /> <br /> ==2012==<br /> ===AMC 10A===<br /> *Average score: 72.51<br /> *AIME floor: 115.5<br /> *DHR: 121.5<br /> <br /> ===AMC 10B===<br /> *Average score: 76.59<br /> *AIME floor: 120<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 64.62<br /> *AIME floor: 94.5<br /> *DHR: 109.5<br /> <br /> ===AMC 12B===<br /> *Average score: 70.08<br /> *AIME floor: 99<br /> *DHR: 114<br /> <br /> ===AIME I===<br /> *Average score: 5.13<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ===AIME II===<br /> *Average score: 4.94<br /> *Median score: <br /> *USAMO cutoff: 204.5<br /> *USAJMO cutoff: 204<br /> <br /> ==2011==<br /> ===AMC 10A===<br /> *Average score: 64.24<br /> *AIME floor: 117<br /> *DHR: 129<br /> <br /> ===AMC 10B===<br /> *Average score: 71.78<br /> *AIME floor: 117<br /> *DHR: 133.5<br /> <br /> ===AMC 12A===<br /> *Average score: 65.38<br /> *AIME floor: 93<br /> *DHR: 112.5<br /> <br /> ===AMC 12B===<br /> *Average score: 64.71<br /> *AIME floor: 97.5<br /> *DHR: 121.5<br /> <br /> ===AIME I===<br /> *Average score: 2.23<br /> *Median score: <br /> *USAMO cutoff: 188<br /> *USAJMO cutoff: 179<br /> <br /> ===AIME II===<br /> *Average score: 5.47<br /> *Median score: <br /> *USAMO cutoff: 215.5<br /> *USAJMO cutoff: 196.5<br /> <br /> ==2010==<br /> ===AMC 10A===<br /> *Average score: 68.11<br /> *AIME floor: 115.5<br /> <br /> ===AMC 10B===<br /> *Average score: 68.57<br /> *AIME floor: 118.5<br /> <br /> ===AMC 12A===<br /> *Average score: 61.02<br /> *AIME floor: 88.5<br /> <br /> ===AMC 12B===<br /> *Average score: 59.58<br /> *AIME floor: 88.5<br /> <br /> ===AIME I===<br /> *Average score: 5.90<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ===AIME II===<br /> *Average score: 3.39<br /> *Median score: <br /> *USAMO cutoff: 208.5 (204.5 for non juniors and seniors)<br /> *USAJMO cutoff: 188.5<br /> <br /> ==2009==<br /> ===AMC 10A===<br /> *Average score: 67.41<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 74.73<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 66.37<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 71.88<br /> *AIME floor: 100 (Top 5% (1.00))<br /> <br /> ===AIME I===<br /> *Average score: 4.17<br /> *Median score: 4<br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score: 3.27<br /> *Median score: 3<br /> *USAMO floor:<br /> <br /> ==2008==<br /> ===AMC 10A===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 10B===<br /> *Average score: <br /> *AIME floor: <br /> <br /> ===AMC 12A===<br /> *Average score: 65.6<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 68.9<br /> *AIME floor: 97.5<br /> <br /> ===AIME I===<br /> *Average score: <br /> *Median score: <br /> *USAMO floor: <br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2007==<br /> <br /> ===AMC 10A===<br /> *Average score: 67.9<br /> *AIME floor: 117<br /> <br /> ===AMC 10B===<br /> *Average score: 61.5<br /> *AIME floor: 115.5<br /> <br /> ===AMC 12A===<br /> *Average score: 66.8<br /> *AIME floor: 97.5<br /> <br /> ===AMC 12B===<br /> *Average score: 73.1<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score: 5<br /> *Median score: 3<br /> *USAMO floor: 6<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2006==<br /> ===AMC 10A===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 68.5<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 85.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 85.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score: <br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2005==<br /> ===AMC 10A===<br /> *Average score: 74.0<br /> *AIME floor: 120<br /> <br /> ===AMC 10B===<br /> *Average score: 79.0<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 78.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 83.4<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2004==<br /> ===AMC 10A===<br /> *Average score: 69.1<br /> *AIME floor: 110<br /> <br /> ===AMC 10B===<br /> *Average score: 80.4<br /> *AIME floor: 120<br /> <br /> ===AMC 12A===<br /> *Average score: 73.9<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 84.5<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2003==<br /> ===AMC 10A===<br /> *Average score: 74.4<br /> *AIME floor: 119<br /> <br /> ===AMC 10B===<br /> *Average score: 79.6<br /> *AIME floor: 121<br /> <br /> ===AMC 12A===<br /> *Average score: 77.8<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 76.6<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2002==<br /> ===AMC 10A===<br /> *Average score: 68.5<br /> *AIME floor: 115<br /> <br /> ===AMC 10B===<br /> *Average score: 74.9<br /> *AIME floor: 118<br /> <br /> ===AMC 12A===<br /> *Average score: 72.7<br /> *AIME floor: 100<br /> <br /> ===AMC 12B===<br /> *Average score: 80.8<br /> *AIME floor: 100<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2001==<br /> ===AMC 10===<br /> *Average score: 67.8<br /> *AIME floor: 116<br /> <br /> ===AMC 12===<br /> *Average score: 56.6<br /> *AIME floor: 84<br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==2000==<br /> ===AMC 10===<br /> *Average score: 64.2<br /> *AIME floor: 120<br /> <br /> ===AMC 12===<br /> *Average score: 64.9<br /> *AIME floor: <br /> <br /> ===AIME I===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ===AIME II===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:<br /> <br /> ==1999==<br /> ===AHSME===<br /> *Average score: 68.8<br /> *AIME floor:<br /> <br /> ===AIME===<br /> *Average score:<br /> *Median score:<br /> *USAMO floor:</div> Relay400 https://artofproblemsolving.com/wiki/index.php?title=1966_AHSME_Answer_Key&diff=101978 1966 AHSME Answer Key 2019-02-13T03:29:16Z <p>Relay400: #1 is incorrect, I have verified this and answer page agrees</p> <hr /> <div>#B<br /> #E<br /> #D<br /> #B<br /> #A<br /> #C<br /> #A<br /> #B<br /> #A<br /> #E&lt;!-- # 10 --&gt;<br /> #C<br /> #E<br /> #E<br /> #C<br /> #D<br /> #B<br /> #C<br /> #A<br /> #B<br /> #C&lt;!-- # 20 --&gt;<br /> #E<br /> #A<br /> #A<br /> #B<br /> #D<br /> #C<br /> #A<br /> #B<br /> #B<br /> #D&lt;!-- # 30 --&gt;<br /> #D<br /> #B<br /> #D<br /> #B<br /> #C<br /> #E<br /> #C<br /> #D<br /> #E<br /> #A&lt;!-- # 40 --&gt;</div> Relay400 https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=95668 Gmaas 2018-06-29T01:15:46Z <p>Relay400: /* gmaas in Popular Culture */</p> <hr /> <div> :) :O :P {{{=== Gmaas Facts ===}}} :) :O :P<br /> <br /> -Gmaas founded Target, but then Gmaas sued them for making the mascot look like a dog when it was supposed to look like Gmaas. <br /> <br /> -Everyone has a bit of Gmaas inside them. EDIT: Except you. You are pure Gmaas. <br /> <br /> -Gmaas likes to eat popsicles. Especially the fudge ones that get him hyper. When he is hyper he runs across Washington D.C. grabbing unsuspecting pedestrians and steals their phone, hacks into them and downloads Fortnite onto their poor phone.<br /> <br /> -Gmaas' favorite cereal is fruit loops. Gmaas thinks it tastes like unicorns jumping on rainbows<br /> <br /> -Gmaas is a champion pillow-fighter.<br /> <br /> -Gmaas colonized Mars.<br /> <br /> -He also colonized Jupiter, Pluto, and several other galaxies. He cloned some little Gmaas robots (with his amazingly robotic skill of coding) and put them all over a galaxy called Gmaasalaxy. <br /> <br /> -Gmaas has the ability to make every device play &quot;The Duck Song&quot; at will.<br /> <br /> -Gmaas once caught the red dot and ate it.<br /> <br /> -Gmaas' favorite color is neon klfhsadkhfd.<br /> <br /> -Gmaas is a champion pvp Minecraft player<br /> <br /> -Gmaas founded Costco.<br /> <br /> -Gmaas can solve any puzzle instantly.<br /> <br /> -When Gmaas flips coins, they always land tails.<br /> <br /> -On Gmaas's math tests, he always gets 101.777777777777777777777777777777777777777777777777777701%<br /> <br /> -Gmaas' favorite number is pi. It's also one of his favorite foods. <br /> <br /> -Gmaas' burps created all gaseous planets.<br /> <br /> -Gmass beat Luke Robatille in an epic showdown of catnip consumption.<br /> <br /> - Gmaas' wealth is unknown, but it is estimated to be way more than Scrooge's.<br /> <br /> - Gmaas has a summer house on Mars. <br /> <br /> -Gmaas attended Harvard, Yale, Stanford, MIT, UC Berkeley, Princeton, Columbia, and Caltech at the same time, using a time-turner.<br /> <br /> -Gmaas also attended Hogwarts and was a prefect. <br /> <br /> -Mrs.Norris is Gmaas's archenemy.<br /> <br /> -Gmaas is a demigod and attends Camp Half-Blood over summer. He is the counselor for the Apollo cabin. Because cats are demigod counselors too.<br /> <br /> -Gmaas has completed over 2,000 quests, and is very popular throughout Camp Half-Blood. He has also been to Camp Jupiter.<br /> <br /> -Gmaas painted the Mona Lisa, The Last Supper, and A Starry Night.<br /> <br /> -Gmass knows that the real names are Gmassa Lisa, The Last Domestic Meal, and Far-away Light.<br /> <br /> -Gmaas actually attended all the Ivy Leagues.<br /> <br /> - I am Gmaas<br /> <br /> - I too am Gmaas<br /> <br /> - But it is I who is Gmaas<br /> <br /> - Gmaas is us all<br /> <br /> - Gmaas is all of us yet none of us<br /> <br /> - Gmaas was captured by the infamous j3370 in 2017 but was released due to sympathy. EDIT: j3370 only captured his concrete form, his abstract form cannot be processed by a feeble human brain<br /> <br /> - Gmaas's fur is purple and yellow and red and green and orange and blue and brown and pink all at the same time. <br /> <br /> - Gmaas crossed the event horizon of a black hole and ended up in the AoPS universe.<br /> <br /> - Gmaas crossed the Delaware River with Washington.<br /> <br /> - Gmaas also crossed the Atlantic with the pilgrims.<br /> <br /> - if you are able to capture a Gmaas hair, he will give you some of his gmaas power.<br /> <br /> - Chuck norris makes Gmaas jokes.<br /> <br /> - Gmaas is also the ruler of Oceania, Eastasia, and Eurasia (1984 reference)<br /> <br /> - Gmaas killed Big Brother by farting on him. Though he was caught by the Ministry of Love, he escaped easily.<br /> <br /> - Everyone thinks that Gmaas is a god<br /> <br /> - Gmaas also owns Animal Farm. Napoleon was his servant.<br /> <br /> - Gmaas is the only one who knows where Amelia Earhart is.<br /> <br /> - Gmaas is the only cat that has been proven transcendental.<br /> <br /> - Gmaas happened to notice http://artofproblemsolving.com/community/c402403h1598015p9983782 and is not very happy about it.<br /> <br /> - Grumpy cat reads Gmaas memes. EDIT: Grumpy cat then steals them and claims they're his. Gmass in't very happy about that, either.<br /> <br /> - The real reason why AIME cutoffs aren't out yet is because Gmaas refused to grade them due to too much problem misplacement.<br /> <br /> - Gmaas dueled Grumpy Cat and lost. He wasn't trying.<br /> <br /> - Gmaas sits on the statue of pallas and says forevermore (the Raven refrence )<br /> <br /> - Gmaas is a big fan of Edgar Allan Poe, because he is actually Poe. <br /> <br /> - Gmaas does merely not use USD. He owns it.<br /> <br /> - Gmaas really knows that Roblox is awful and does not play it seriously, thank god our lord is sane<br /> <br /> - In 2003, Gmaas used elliptical curves to force his reign over AoPS.<br /> <br /> - &quot;Actually, my name is spelled GMAAS&quot;<br /> <br /> - Gmaas is the smartest living being in the universe.<br /> <br /> - It was Gmaas who helped Sun Wukong on the Journey to the West.<br /> <br /> - Gmaas is the real creator of Wikipedia.<br /> <br /> -It is said Gmaas could hack any website he desires.<br /> <br /> -Gmaas is the basis of Greek mythology.<br /> <br /> -Gmaas once sold Google to a man for around &lt;math&gt;12&lt;/math&gt; dollars!<br /> <br /> -Gmaas uses a HP printer.<br /> <br /> -Gmaas owns all AoPS staff including Richard Rusczyk.<br /> <br /> -Gmaas was there when Yoda was born.<br /> <br /> - Gmaas's true number of lives left is unknown; however, Gmaas recently confirmed that he had at least one left. Why doesn't Gmaas have so many more lives than other cats? The power of Gmaas.<br /> <br /> - sseraj once spelled Gmaas as gmASS on accident in Introduction to Geometry (1532).<br /> <br /> -Gmaas actively plays Roblox, and is a globally ranked professional gamer: https://www.roblox.com/users/29708533/profile...but he hates Roblox.<br /> <br /> -Gmaas has beaten Chuck Norris and The Rock and John Cena all together in a fight.<br /> <br /> -Gmaas is a South Korean, North Korean, Palestinian, Israeli, U.S., and Soviet citizen at the same time. EDIT: Gmaas has now been found out to be a citizen of every country in the world. Gmaas seems to enjoy the country of AOPS best, however.<br /> <br /> -&quot;i am sand&quot; destroyed Gmaas in FTW<br /> <br /> - sseraj posted a picture of gmaas with a game controller in Introduction to Geometry (1532).<br /> <br /> -Gmaas plays roblox mobile edition and likes Minecraft, Candy Crush, and Club Penguin Rewritten. He also &lt;math&gt;\boxed{\text{loves}}&lt;/math&gt; &lt;b&gt;Cat&lt;/b&gt;ch that fish.<br /> <br /> -Gmaas is Roy Moore's horse in the shape of a cat<br /> <br /> -Gmaas is a known roblox/club penguin rewritten player and is a legend at it. He has over &lt;math&gt;289547987693&lt;/math&gt; robux and &lt;math&gt;190348&lt;/math&gt; in CPR.<br /> <br /> -This is all hypothetical.<br /> <br /> - EDIT: This is all factual <br /> <br /> -Gmaas's real name is Princess. He has a sibling named Rusty/Fireheart/Firestar<br /> (Warrior cats reference)<br /> <br /> - He is capable of salmon powers, according to PunSpark (ask him)<br /> <br /> -The Gmaas told Richard Rusczyk to make AoPS<br /> <br /> -The Gmaas is everything. Yes, you are part of the Gmaas-Dw789<br /> <br /> -The Gmaas knows every dimension up to 9999999999999999999999999999999999999999999999999999999999999999999999999999999999th dimension.<br /> <br /> -Certain theories provide evidence that he IS darth plagueis the wise.<br /> <br /> -Gmaas is &quot;TIRED OF PEOPLE ADDING TO HIS PAGE!!&quot; (Maas 45).<br /> <br /> -Gmaas has multiple accounts; some of them are pifinity, cyumi, srej, squareman, Electro3.0, sturdywill, g1zq, alleycat, Equinox8, anc3, batta, beastgert, and lakecomo224. His best account, though, is wlm7, later renamed to &quot;gmail.com&quot;.<br /> <br /> -Gmaas has a penguin servant named sm24136 who runs GMAASINC. The penguin may or may not be dead. <br /> <br /> -Gmaas owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br /> <br /> -Gmaas knows how to hack into top secret aops community pages.<br /> <br /> -Gmaas was a river clant cat who crossed the event horizon of a black hole and came out the other end!<br /> <br /> -Gmaas is king of the first men, the anduls.<br /> <br /> -Gmaas is a well known professor at MEOWston Academy.<br /> <br /> -Gmaas is a Tuna addict, along with other, more potent fish such as Salmon and Trout.<br /> <br /> -Gmaas won the reward of being cutest and fattest cat ever--he surpassed grumpy cat. (He also out-grumped grumpy cat!!!)<br /> <br /> -Last sighting 1665 Algebra-A 3/9/18 at 9:08 PM.<br /> <br /> - owner of sseraj, not pet.<br /> <br /> - embodiment of life and universe and beyond .<br /> <br /> - Watches memes.<br /> <br /> -After Death became the GOD OF HYPERDEATH and obtained over 9000 souls.<br /> <br /> -Gmaas's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)]<br /> <br /> -Gmaas is a certified Slytherin.<br /> <br /> -Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom <br /> <br /> -Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom)<br /> <br /> - Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br /> <br /> -Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br /> <br /> -Gmaas is a supreme overlord who must be given &lt;math&gt;10^{1000000000000000000000^{1000000000000000000000}}&lt;/math&gt; minecraft DIAMONDS<br /> <br /> - gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br /> <br /> - Gmaas is everyone's favorite animal. <br /> <br /> - He lives with sseraj. <br /> <br /> -Gmaas is my favorite pokemon<br /> <br /> -Gmaas dislikes number theory but enjoys geometry.<br /> <br /> - Gmaas is cool<br /> <br /> - He is often overfed (with probability &lt;math&gt;\frac{3972}{7891}&lt;/math&gt;), or malnourished (with probability &lt;math&gt;\frac{3919}{7891}&lt;/math&gt;) by sseraj.<br /> <br /> - He has &lt;cmath&gt;\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAS&lt;/cmath&gt; supercars, excluding the Purrari and the 138838383 Teslas. <br /> <br /> - He is an employee of AoPS.<br /> <br /> - He is a gmaas with yellow fur and white hypnotizing eyes.<br /> <br /> - He has the ability to divide by zero.<br /> <br /> - He was born with a tail that is a completely different color from the rest of his fur.<br /> <br /> - His stare is very hypnotizing and effective at getting table scraps.<br /> <br /> - He sometimes appears several minutes before certain classes start as an admin. <br /> <br /> - He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br /> <br /> - It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br /> <br /> - Actually, he is a cat. He said so. And science also says so.<br /> <br /> - He is distant relative of Mathcat1234.<br /> <br /> - He is very famous now, and mods always talk about him before class starts.<br /> <br /> - His favorite food is AoPS textbooks because they help him digest problems.<br /> <br /> - Gmaas tends to reside in sseraj's fridge.<br /> <br /> - Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br /> <br /> - The fur of Gmaas can protect him from the harsh conditions of a freezer.<br /> <br /> - Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br /> <br /> - Gmaas is a sage omniscient cat.<br /> <br /> - He is looking for suitable places other than sseraj's fridge to live in.<br /> <br /> - Places where gmaas sightings have happened: <br /> ~The Royal Scoop ice cream store in Bonita Beach Florida<br /> <br /> ~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br /> <br /> -Prealgebra 2 (1440)<br /> <br /> ~Alligator Swamp A 1072 <br /> <br /> ~Alligator Swamp B 1073<br /> <br /> ~Prealgebra A (1488)<br /> <br /> ~Introduction to Algebra A (1170)<br /> <br /> ~Introduction to Algebra B (1529)<br /> <br /> ~Welcome to Panda Town Gate 1076<br /> <br /> ~Welcome to Gmaas Town Gate 1221<br /> <br /> ~Welcome to Gmaas Town Gate 1125<br /> <br /> ~33°01'17.4&quot;N 117°05'40.1&quot;W (Rancho Bernardo Road, San Diego, CA)<br /> <br /> ~The other side of the ice in Antarctica<br /> <br /> ~Feisty Alligator Swamp 1115<br /> <br /> ~Introduction to Geometry 1221 (Taught by sseraj)<br /> <br /> ~Introduction to Counting and Probability 1142 <br /> <br /> ~Feisty-ish Alligator Swamp 1115 (AGAIN)<br /> <br /> ~Intermediate Counting and Probability 1137<br /> <br /> ~Intermediate Counting and Probability 1207<br /> <br /> ~Posting student surveys<br /> <br /> ~USF Castle Walls - Elven Tribe 1203<br /> <br /> ~Dark Lord's Hut 1210<br /> <br /> ~AMC 10 Problem Series 1200<br /> <br /> ~Intermediate Number Theory 1138<br /> <br /> ~Intermediate Number Theory 1476<br /> <br /> ~Introduction To Number Theory 1204. Date:7/27/16.<br /> <br /> ~Algebra B 1112<br /> <br /> ~Intermediate Algebra 1561 7:17 PM 12/11/16<br /> <br /> ~Nowhere Else, Tasmania<br /> <br /> ~Earth Dimension C-137<br /> ~Geometry 1694 at 1616 PST military time. There was a boy riding him, and he seemed extremely miffed.<br /> <br /> <br /> <br /> <br /> - These have all been designated as the most glorious sections of Aopsland now (especially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&amp;B).<br /> <br /> - Gmaas has also been sighted in Olympiad Geometry 1148.<br /> <br /> - Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br /> <br /> - Gmaas once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<br /> <br /> - Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br /> <br /> EDIT: Nobody has yet seen him atop a tribal base yet.<br /> <br /> - Gmaas are often under the disguise of a penguin or cat. Look out for them.<br /> <br /> EDIT: Gmaas rarely disguises himself as a penguin.<br /> <br /> - Many know that leafy stole dream island. In truth, After leafy stole it, Gmaas stole it himself. (BFDI Reference)<br /> <br /> - He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br /> <br /> EDIT: He IS an AoPS site admin.<br /> <br /> - If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br /> <br /> - Is this the real life? Is this just fantasy? No. This is Gmaas, the legend.<br /> <br /> -Aha!! An impostor!! <br /> http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br /> (look at the acronym).<br /> <br /> -EDIT. The above fact is slightly irrelevant.<br /> <br /> - Gmaas might have been viewing (with a &lt;math&gt;\frac{99999.\overline{9}}{100000}&lt;/math&gt; chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br /> <br /> - EDIT: Gmaas is a he.<br /> <br /> -Gmaas is love, Gmaas is life<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mew.<br /> <br /> - Gmaas is on the list of &quot;Elusive Creatures.&quot; If you have questions or want the full list, contact moab33.<br /> <br /> - Gmaas can be summoned using the &lt;math&gt;\tan(90)&lt;/math&gt; ritual. Draw a pentagram and write the numerical value of &lt;math&gt;\tan(90)&lt;/math&gt; in the middle, and he will be summoned.<br /> <br /> - EDIT: The above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br /> https://artofproblemsolving.com/community/c287916h1291232<br /> <br /> - EDIT EDIT: The above 'proof' is non-conclusive. math101010 had only put an approximation.<br /> <br /> - Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br /> <br /> - EDIT: That has never happened and thus it does not contain the singularity of a black hole.<br /> <br /> - Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br /> <br /> -Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br /> <br /> - The original owner of Gmaas is Gmaas.<br /> <br /> - Gmaas was not the fourth Peverell brother, but he ascended into a higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.) EDIT: he wasn't the fourth Peverell brother, but he was a cousin of theirs, and he was the one who advised Ignotus to give up his cloak.<br /> <br /> - It is suspected that Gmaas may be ordering his cyber hairballs to take the forums, along with microbots.<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mu.<br /> <br /> - Gmaas rarely frequents the headquarters of the Illuminati. He was their symbol for one yoctosecond, but soon decided that the job was too low for his power to be wasted on.<br /> <br /> - It has been wondered if Gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br /> <br /> - Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br /> <br /> - It has been confirmed that gmaas uses gmewal as his email service<br /> <br /> - Gmaas enjoys wearing gmean shorts<br /> <br /> - Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br /> <br /> - Gmaas is well known behind his stage name, Michael Stevens (also known as Vsauce XD), or his page name, Purrshanks. EDIT: Crookshanks was his brother.<br /> <br /> - Gmaas rekt sseraj at 12:54 June 4, 2016 UTC time zone. And then the Doctor chased him.<br /> <br /> - Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br /> <br /> - Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br /> <br /> - In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seen the Dark Lord's hut knows that both Gmaas and the DL (USF code name of the Dark Lord) love BBC. How Gmaas gave him a TV may be lost to history. And it has been lost.<br /> <br /> - The TV has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br /> <br /> -The reason why Dacammel left the USF is that gmaas entrusted his TV to him, and not wanting to be discovered by LF, Cobra, or Z9, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br /> <br /> - EDIT: The above fact is somewhat irrelevant.<br /> <br /> -EDIT EDIT. Dacammel gave the TV back to gmaas, and he left the dark side and their cookies alone. <br /> <br /> - Gmaas is a Super Duper Uper Cat Time Lord. He has &lt;math&gt;57843504&lt;/math&gt; regenerations and has used &lt;math&gt;3&lt;/math&gt;. &lt;cmath&gt;9\cdot12\cdot2\cdot267794=57843504&lt;/cmath&gt;. <br /> <br /> -Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part.<br /> <br /> - Gmaas loves to eat turnips. At &lt;math&gt;\frac{13}{32}&lt;/math&gt; of the sites he was spotted at, he was seen with a turnip.<br /> <br /> -Gmaas has a secret hidden garden full of turnips under sseraj's house.<br /> <br /> - Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br /> <br /> -Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br /> <br /> -Gmaas is in alliance with the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything in sight? Nobody knows (except him), not even the leader of the Cult of Skaro.<br /> <br /> -Gmaas lives in Gallifrey and in Gotham City (he has sleepovers with Batman).<br /> <br /> -Gmaas is an excellent driver. EDIT: he was to one who designed the driver's license test, although he didn't bother with the permit test.<br /> <br /> -The native location of Gmaas is the twilight zone.<br /> <br /> -Donald Trump once sang &quot;All Hail the Chief&quot; to Gmaas, 3 days after being sworn in as US President.<br /> <br /> - Gmaas likes to talk with rrusczyk from time to time.<br /> <br /> - Gmaas can shoot fire from his smelly butt.<br /> <br /> - Gmaas is the reason why the USF has the longest thread on AoPS.<br /> <br /> - Gmass is an avid watcher of the popular T.V. show &quot;Bernie Sanders and the Gauntlet of DOOM&quot;<br /> <br /> - sseraj, in 1521 Introduction to Number Theory, posted an image of Gmaas after saying &quot;Who wants to see 5space?&quot; at around 5:16 PM Mountain Time, noting Gmaas was &quot;also 5space&quot;<br /> <br /> -EDIT: he also did it in Introduction to Algebra A once.<br /> <br /> - Gmaas is now my HD background on my Mac.<br /> <br /> - In 1521 Into to Number Theory, sseraj posted an image of a 5space Gmaas fusion. (First sighting) <br /> <br /> - Also confirmed that Gmaas doesn't like ketchup because it was the only food left the photo.<br /> <br /> - In 1447 Intro to Geometry, sseraj posted a picture of Gmaas with a rubik's cube suggesting that Gmaas's has an average solve time of &lt;math&gt;-GMAAS&lt;/math&gt; seconds.<br /> <br /> -Gmaas beat Superman in a fight with ease<br /> <br /> -Gmaas was an admin of Roblox<br /> <br /> -Gmaas traveled around the world, paying so much &lt;math&gt;MONEY&lt;/math&gt; just to eat :D<br /> <br /> -Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br /> Summary:<br /> <br /> -When Gmaas subtracts &lt;math&gt;0.\overline{99}&lt;/math&gt; from &lt;math&gt;1&lt;/math&gt;, the difference is greater than &lt;math&gt;0&lt;/math&gt;.<br /> <br /> -Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br /> <br /> -Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br /> <br /> -The results of the revival are top secret, and nobody knows what happened.<br /> <br /> -sseraj, in 1496 Prealgebra 2, said that Gmaas is Santacat.<br /> <br /> -sseraj likes to post a picture of gmaas in every class he passes by.<br /> <br /> -sseraj posted a picture of gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Endor is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought gmaas is now wandering space in search for a home.<br /> EDIT: What evidence is there Endor was affected? Other Ewoks still live there.<br /> EDIT EDIT: also, glass doesn't care. He can live there no matter what the climate is.<br /> <br /> -Gmaas is the lord of the pokemans<br /> <br /> -Gmaas can communicate with, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will.<br /> <br /> -Picture of Gmaas http://i.imgur.com/PP9xi.png<br /> <br /> -Known by Mike Miller<br /> <br /> -Gmaas got mad at sseraj once, so he locked him in his own freezer<br /> <br /> -Then, sseraj decided to eat all of Gmaas's hidden turnips in the freezer as punishment<br /> <br /> -Gmaas is an obviously omnipotent cat.<br /> <br /> -ehawk11 met him<br /> <br /> -sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br /> <br /> -sseraj has posted pictures of gmaas in '&quot;intro to algebra&quot;, before class started, with the title, &quot;caption contest&quot;. anyone who posted a caption mysteriously vanished in the middle of the night. <br /> EDIT: This has happened many times, including in Introduction to Geometry 1533, among other active classes. The person writing this (Poeshuman) did participate, and did not disappear. (You could argue Gmaas is typing this through his/her account...)<br /> <br /> - gmaas has once slept in your bed and made it wet<br /> <br /> -It is rumored that rrusczyk is actually Gmaas in disguise<br /> <br /> -Gmaas is suspected to be a Mewtwo in disguise<br /> <br /> -Gmaas is a cat but has characteristics of every other animal on Earth.<br /> <br /> -Gmaas is the ruler of the universe and has been known to be the creator of the species &quot;Gmaasians&quot;.<br /> <br /> -There is a rumor that Gmaas is starting a poll<br /> <br /> -Gmaas is a rumored past ThunderClan cat who ran away, founded GmaasClan, then became a kittypet.<br /> <br /> -There is a rumored sport called &quot;Gmaas Hunting&quot; where people try to successfully capture gmaas in the wild with video/camera/eyes. Strangely, no one has been able to do this, and those that have have mysteriously disappeared into the night. Nobody knows why. The person who is writing this(g1zq) has tried Gmaas Hunting, but has never been successful.<br /> <br /> - Gmaas burped and caused an earthquake.<br /> <br /> - Gmaas once drank from your pretty teacup.<br /> <br /> -GMAAS IS HERE.... PURRRRRRRRRRRRRRRRRRRR<br /> <br /> -Gmass made, and currently owns the Matrix.<br /> <br /> -The above fact is true. Therefore, this is an illusion.<br /> <br /> -Gmaas is the reason Salah will become better than Ronaldo.<br /> <br /> -Who is Gmaas, really?<br /> -Gmass is a heavenly being<br /> <br /> -Illuminati was a manifestation of Gmaas, but Gmass decided illuminati was not great enough for his godly self.<br /> <br /> -jlikemath has met Gmaas and Gmaas is his best friend.<br /> <br /> -Gmaas loves Kpop<br /> <br /> -Gmaas read Twilight EDIT: ...and SURVIVED<br /> <br /> -there is a secret code when put into super smash, Gmass would be a playible character. Too bad he didn't say it.<br /> <br /> -Gmaas was a tribute to one of the Hunger Games and came out a Victor and now lives in District 4<br /> <br /> === Gmaas photos ===<br /> http://cdn.artofproblemsolving.com/images/f/f/8/ff8efef3a0d2eb51254634e54bec215b948a1bba.jpg<br /> <br /> === gmaas in Popular Culture ===<br /> <br /> - &lt;s&gt;Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, &quot;The Adventures of gmaas&quot;.&lt;/s&gt;Sorry, this was a rick roll troll.<br /> <br /> - BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,&quot;What are gwali?&quot; the customary answer &quot;This is gwali&quot; is returned. Scientist 5space is now looking into it.<br /> <br /> - Sullymath and themoocow are also writing a book about Gmaas<br /> <br /> -Sighting of Gmaas: https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project<br /> <br /> -Oryx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed<br /> <br /> - Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br /> <br /> - Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br /> <br /> - Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles<br /> <br /> - gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br /> <br /> -oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here<br /> <br /> -Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br /> <br /> -No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br /> <br /> -In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br /> https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br /> https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br /> https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br /> https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br /> <br /> <br /> - Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br /> <br /> -&lt;math&gt;Another&lt;/math&gt; sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br /> - Yet Another Gmaas sighting ? [https://supportforums.cisco.com/t5/user/viewprofilepage/user-id/45046]<br /> <br /> -Gmaas has been sighted several times on the Global Announcements forum<br /> <br /> -Gmaas uses the following transportation: &lt;img&gt; http://cdn.artofproblemsolving.com/images/3/6/8/368da4e615ea3476355ee3388b39f30a48b8dd48.jpg &lt;/img&gt;<br /> <br /> - When Gmaas was mad, he started world wars 1 &amp; 2. It is only because of Gmaas that we have not had World War 3.<br /> <br /> - Gmaas is the only cat to have been proved irrational and transcendental, though we suspect all cats fall in the first category.<br /> <br /> - Gmaas plays Geometry Dash and shares an account with Springhill, his username is D3m0nG4m1n9.<br /> <br /> -Gmaas likes to whiz on the wilzo<br /> <br /> -Gmaas has been spotted in AMC 8 Basics<br /> <br /> -Gmaas is cool<br /> <br /> -Gmaas hemoon card that does over 9000000 dmg<br /> <br /> -Gmaas is a skilled swordsman who should not to be mistaken for Puss in Boots. Some say he even trained the mysterious and valiant Meta Knight.<br /> <br /> -Kirby once swallowed Gmaas. Gmaas had to spit him out.<br /> <br /> -Gmaas was the creator of pokemon, and his pokemon card can OHKO anyone in one turn. He is invisible and he will always move first.<br /> <br /> -Gmaas beat Dongmin in The Genius Game Seasons 1, 2, 3, 4, 5, 6, and 7.</div> Relay400 https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=95667 Gmaas 2018-06-29T01:14:39Z <p>Relay400: /* gmaas in Popular Culture */</p> <hr /> <div> :) :O :P {{{=== Gmaas Facts ===}}} :) :O :P<br /> <br /> -Gmaas founded Target, but then Gmaas sued them for making the mascot look like a dog when it was supposed to look like Gmaas. <br /> <br /> -Everyone has a bit of Gmaas inside them. EDIT: Except you. You are pure Gmaas. <br /> <br /> -Gmaas likes to eat popsicles. Especially the fudge ones that get him hyper. When he is hyper he runs across Washington D.C. grabbing unsuspecting pedestrians and steals their phone, hacks into them and downloads Fortnite onto their poor phone.<br /> <br /> -Gmaas' favorite cereal is fruit loops. Gmaas thinks it tastes like unicorns jumping on rainbows<br /> <br /> -Gmaas is a champion pillow-fighter.<br /> <br /> -Gmaas colonized Mars.<br /> <br /> -He also colonized Jupiter, Pluto, and several other galaxies. He cloned some little Gmaas robots (with his amazingly robotic skill of coding) and put them all over a galaxy called Gmaasalaxy. <br /> <br /> -Gmaas has the ability to make every device play &quot;The Duck Song&quot; at will.<br /> <br /> -Gmaas once caught the red dot and ate it.<br /> <br /> -Gmaas' favorite color is neon klfhsadkhfd.<br /> <br /> -Gmaas is a champion pvp Minecraft player<br /> <br /> -Gmaas founded Costco.<br /> <br /> -Gmaas can solve any puzzle instantly.<br /> <br /> -When Gmaas flips coins, they always land tails.<br /> <br /> -On Gmaas's math tests, he always gets 101.777777777777777777777777777777777777777777777777777701%<br /> <br /> -Gmaas' favorite number is pi. It's also one of his favorite foods. <br /> <br /> -Gmaas' burps created all gaseous planets.<br /> <br /> -Gmass beat Luke Robatille in an epic showdown of catnip consumption.<br /> <br /> - Gmaas' wealth is unknown, but it is estimated to be way more than Scrooge's.<br /> <br /> - Gmaas has a summer house on Mars. <br /> <br /> -Gmaas attended Harvard, Yale, Stanford, MIT, UC Berkeley, Princeton, Columbia, and Caltech at the same time, using a time-turner.<br /> <br /> -Gmaas also attended Hogwarts and was a prefect. <br /> <br /> -Mrs.Norris is Gmaas's archenemy.<br /> <br /> -Gmaas is a demigod and attends Camp Half-Blood over summer. He is the counselor for the Apollo cabin. Because cats are demigod counselors too.<br /> <br /> -Gmaas has completed over 2,000 quests, and is very popular throughout Camp Half-Blood. He has also been to Camp Jupiter.<br /> <br /> -Gmaas painted the Mona Lisa, The Last Supper, and A Starry Night.<br /> <br /> -Gmass knows that the real names are Gmassa Lisa, The Last Domestic Meal, and Far-away Light.<br /> <br /> -Gmaas actually attended all the Ivy Leagues.<br /> <br /> - I am Gmaas<br /> <br /> - I too am Gmaas<br /> <br /> - But it is I who is Gmaas<br /> <br /> - Gmaas is us all<br /> <br /> - Gmaas is all of us yet none of us<br /> <br /> - Gmaas was captured by the infamous j3370 in 2017 but was released due to sympathy. EDIT: j3370 only captured his concrete form, his abstract form cannot be processed by a feeble human brain<br /> <br /> - Gmaas's fur is purple and yellow and red and green and orange and blue and brown and pink all at the same time. <br /> <br /> - Gmaas crossed the event horizon of a black hole and ended up in the AoPS universe.<br /> <br /> - Gmaas crossed the Delaware River with Washington.<br /> <br /> - Gmaas also crossed the Atlantic with the pilgrims.<br /> <br /> - if you are able to capture a Gmaas hair, he will give you some of his gmaas power.<br /> <br /> - Chuck norris makes Gmaas jokes.<br /> <br /> - Gmaas is also the ruler of Oceania, Eastasia, and Eurasia (1984 reference)<br /> <br /> - Gmaas killed Big Brother by farting on him. Though he was caught by the Ministry of Love, he escaped easily.<br /> <br /> - Everyone thinks that Gmaas is a god<br /> <br /> - Gmaas also owns Animal Farm. Napoleon was his servant.<br /> <br /> - Gmaas is the only one who knows where Amelia Earhart is.<br /> <br /> - Gmaas is the only cat that has been proven transcendental.<br /> <br /> - Gmaas happened to notice http://artofproblemsolving.com/community/c402403h1598015p9983782 and is not very happy about it.<br /> <br /> - Grumpy cat reads Gmaas memes. EDIT: Grumpy cat then steals them and claims they're his. Gmass in't very happy about that, either.<br /> <br /> - The real reason why AIME cutoffs aren't out yet is because Gmaas refused to grade them due to too much problem misplacement.<br /> <br /> - Gmaas dueled Grumpy Cat and lost. He wasn't trying.<br /> <br /> - Gmaas sits on the statue of pallas and says forevermore (the Raven refrence )<br /> <br /> - Gmaas is a big fan of Edgar Allan Poe, because he is actually Poe. <br /> <br /> - Gmaas does merely not use USD. He owns it.<br /> <br /> - Gmaas really knows that Roblox is awful and does not play it seriously, thank god our lord is sane<br /> <br /> - In 2003, Gmaas used elliptical curves to force his reign over AoPS.<br /> <br /> - &quot;Actually, my name is spelled GMAAS&quot;<br /> <br /> - Gmaas is the smartest living being in the universe.<br /> <br /> - It was Gmaas who helped Sun Wukong on the Journey to the West.<br /> <br /> - Gmaas is the real creator of Wikipedia.<br /> <br /> -It is said Gmaas could hack any website he desires.<br /> <br /> -Gmaas is the basis of Greek mythology.<br /> <br /> -Gmaas once sold Google to a man for around &lt;math&gt;12&lt;/math&gt; dollars!<br /> <br /> -Gmaas uses a HP printer.<br /> <br /> -Gmaas owns all AoPS staff including Richard Rusczyk.<br /> <br /> -Gmaas was there when Yoda was born.<br /> <br /> - Gmaas's true number of lives left is unknown; however, Gmaas recently confirmed that he had at least one left. Why doesn't Gmaas have so many more lives than other cats? The power of Gmaas.<br /> <br /> - sseraj once spelled Gmaas as gmASS on accident in Introduction to Geometry (1532).<br /> <br /> -Gmaas actively plays Roblox, and is a globally ranked professional gamer: https://www.roblox.com/users/29708533/profile...but he hates Roblox.<br /> <br /> -Gmaas has beaten Chuck Norris and The Rock and John Cena all together in a fight.<br /> <br /> -Gmaas is a South Korean, North Korean, Palestinian, Israeli, U.S., and Soviet citizen at the same time. EDIT: Gmaas has now been found out to be a citizen of every country in the world. Gmaas seems to enjoy the country of AOPS best, however.<br /> <br /> -&quot;i am sand&quot; destroyed Gmaas in FTW<br /> <br /> - sseraj posted a picture of gmaas with a game controller in Introduction to Geometry (1532).<br /> <br /> -Gmaas plays roblox mobile edition and likes Minecraft, Candy Crush, and Club Penguin Rewritten. He also &lt;math&gt;\boxed{\text{loves}}&lt;/math&gt; &lt;b&gt;Cat&lt;/b&gt;ch that fish.<br /> <br /> -Gmaas is Roy Moore's horse in the shape of a cat<br /> <br /> -Gmaas is a known roblox/club penguin rewritten player and is a legend at it. He has over &lt;math&gt;289547987693&lt;/math&gt; robux and &lt;math&gt;190348&lt;/math&gt; in CPR.<br /> <br /> -This is all hypothetical.<br /> <br /> - EDIT: This is all factual <br /> <br /> -Gmaas's real name is Princess. He has a sibling named Rusty/Fireheart/Firestar<br /> (Warrior cats reference)<br /> <br /> - He is capable of salmon powers, according to PunSpark (ask him)<br /> <br /> -The Gmaas told Richard Rusczyk to make AoPS<br /> <br /> -The Gmaas is everything. Yes, you are part of the Gmaas-Dw789<br /> <br /> -The Gmaas knows every dimension up to 9999999999999999999999999999999999999999999999999999999999999999999999999999999999th dimension.<br /> <br /> -Certain theories provide evidence that he IS darth plagueis the wise.<br /> <br /> -Gmaas is &quot;TIRED OF PEOPLE ADDING TO HIS PAGE!!&quot; (Maas 45).<br /> <br /> -Gmaas has multiple accounts; some of them are pifinity, cyumi, srej, squareman, Electro3.0, sturdywill, g1zq, alleycat, Equinox8, anc3, batta, beastgert, and lakecomo224. His best account, though, is wlm7, later renamed to &quot;gmail.com&quot;.<br /> <br /> -Gmaas has a penguin servant named sm24136 who runs GMAASINC. The penguin may or may not be dead. <br /> <br /> -Gmaas owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br /> <br /> -Gmaas knows how to hack into top secret aops community pages.<br /> <br /> -Gmaas was a river clant cat who crossed the event horizon of a black hole and came out the other end!<br /> <br /> -Gmaas is king of the first men, the anduls.<br /> <br /> -Gmaas is a well known professor at MEOWston Academy.<br /> <br /> -Gmaas is a Tuna addict, along with other, more potent fish such as Salmon and Trout.<br /> <br /> -Gmaas won the reward of being cutest and fattest cat ever--he surpassed grumpy cat. (He also out-grumped grumpy cat!!!)<br /> <br /> -Last sighting 1665 Algebra-A 3/9/18 at 9:08 PM.<br /> <br /> - owner of sseraj, not pet.<br /> <br /> - embodiment of life and universe and beyond .<br /> <br /> - Watches memes.<br /> <br /> -After Death became the GOD OF HYPERDEATH and obtained over 9000 souls.<br /> <br /> -Gmaas's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)]<br /> <br /> -Gmaas is a certified Slytherin.<br /> <br /> -Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom <br /> <br /> -Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom)<br /> <br /> - Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br /> <br /> -Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br /> <br /> -Gmaas is a supreme overlord who must be given &lt;math&gt;10^{1000000000000000000000^{1000000000000000000000}}&lt;/math&gt; minecraft DIAMONDS<br /> <br /> - gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br /> <br /> - Gmaas is everyone's favorite animal. <br /> <br /> - He lives with sseraj. <br /> <br /> -Gmaas is my favorite pokemon<br /> <br /> -Gmaas dislikes number theory but enjoys geometry.<br /> <br /> - Gmaas is cool<br /> <br /> - He is often overfed (with probability &lt;math&gt;\frac{3972}{7891}&lt;/math&gt;), or malnourished (with probability &lt;math&gt;\frac{3919}{7891}&lt;/math&gt;) by sseraj.<br /> <br /> - He has &lt;cmath&gt;\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAS&lt;/cmath&gt; supercars, excluding the Purrari and the 138838383 Teslas. <br /> <br /> - He is an employee of AoPS.<br /> <br /> - He is a gmaas with yellow fur and white hypnotizing eyes.<br /> <br /> - He has the ability to divide by zero.<br /> <br /> - He was born with a tail that is a completely different color from the rest of his fur.<br /> <br /> - His stare is very hypnotizing and effective at getting table scraps.<br /> <br /> - He sometimes appears several minutes before certain classes start as an admin. <br /> <br /> - He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br /> <br /> - It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br /> <br /> - Actually, he is a cat. He said so. And science also says so.<br /> <br /> - He is distant relative of Mathcat1234.<br /> <br /> - He is very famous now, and mods always talk about him before class starts.<br /> <br /> - His favorite food is AoPS textbooks because they help him digest problems.<br /> <br /> - Gmaas tends to reside in sseraj's fridge.<br /> <br /> - Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br /> <br /> - The fur of Gmaas can protect him from the harsh conditions of a freezer.<br /> <br /> - Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br /> <br /> - Gmaas is a sage omniscient cat.<br /> <br /> - He is looking for suitable places other than sseraj's fridge to live in.<br /> <br /> - Places where gmaas sightings have happened: <br /> ~The Royal Scoop ice cream store in Bonita Beach Florida<br /> <br /> ~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br /> <br /> -Prealgebra 2 (1440)<br /> <br /> ~Alligator Swamp A 1072 <br /> <br /> ~Alligator Swamp B 1073<br /> <br /> ~Prealgebra A (1488)<br /> <br /> ~Introduction to Algebra A (1170)<br /> <br /> ~Introduction to Algebra B (1529)<br /> <br /> ~Welcome to Panda Town Gate 1076<br /> <br /> ~Welcome to Gmaas Town Gate 1221<br /> <br /> ~Welcome to Gmaas Town Gate 1125<br /> <br /> ~33°01'17.4&quot;N 117°05'40.1&quot;W (Rancho Bernardo Road, San Diego, CA)<br /> <br /> ~The other side of the ice in Antarctica<br /> <br /> ~Feisty Alligator Swamp 1115<br /> <br /> ~Introduction to Geometry 1221 (Taught by sseraj)<br /> <br /> ~Introduction to Counting and Probability 1142 <br /> <br /> ~Feisty-ish Alligator Swamp 1115 (AGAIN)<br /> <br /> ~Intermediate Counting and Probability 1137<br /> <br /> ~Intermediate Counting and Probability 1207<br /> <br /> ~Posting student surveys<br /> <br /> ~USF Castle Walls - Elven Tribe 1203<br /> <br /> ~Dark Lord's Hut 1210<br /> <br /> ~AMC 10 Problem Series 1200<br /> <br /> ~Intermediate Number Theory 1138<br /> <br /> ~Intermediate Number Theory 1476<br /> <br /> ~Introduction To Number Theory 1204. Date:7/27/16.<br /> <br /> ~Algebra B 1112<br /> <br /> ~Intermediate Algebra 1561 7:17 PM 12/11/16<br /> <br /> ~Nowhere Else, Tasmania<br /> <br /> ~Earth Dimension C-137<br /> ~Geometry 1694 at 1616 PST military time. There was a boy riding him, and he seemed extremely miffed.<br /> <br /> <br /> <br /> <br /> - These have all been designated as the most glorious sections of Aopsland now (especially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&amp;B).<br /> <br /> - Gmaas has also been sighted in Olympiad Geometry 1148.<br /> <br /> - Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br /> <br /> - Gmaas once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<br /> <br /> - Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br /> <br /> EDIT: Nobody has yet seen him atop a tribal base yet.<br /> <br /> - Gmaas are often under the disguise of a penguin or cat. Look out for them.<br /> <br /> EDIT: Gmaas rarely disguises himself as a penguin.<br /> <br /> - Many know that leafy stole dream island. In truth, After leafy stole it, Gmaas stole it himself. (BFDI Reference)<br /> <br /> - He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br /> <br /> EDIT: He IS an AoPS site admin.<br /> <br /> - If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br /> <br /> - Is this the real life? Is this just fantasy? No. This is Gmaas, the legend.<br /> <br /> -Aha!! An impostor!! <br /> http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br /> (look at the acronym).<br /> <br /> -EDIT. The above fact is slightly irrelevant.<br /> <br /> - Gmaas might have been viewing (with a &lt;math&gt;\frac{99999.\overline{9}}{100000}&lt;/math&gt; chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br /> <br /> - EDIT: Gmaas is a he.<br /> <br /> -Gmaas is love, Gmaas is life<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mew.<br /> <br /> - Gmaas is on the list of &quot;Elusive Creatures.&quot; If you have questions or want the full list, contact moab33.<br /> <br /> - Gmaas can be summoned using the &lt;math&gt;\tan(90)&lt;/math&gt; ritual. Draw a pentagram and write the numerical value of &lt;math&gt;\tan(90)&lt;/math&gt; in the middle, and he will be summoned.<br /> <br /> - EDIT: The above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br /> https://artofproblemsolving.com/community/c287916h1291232<br /> <br /> - EDIT EDIT: The above 'proof' is non-conclusive. math101010 had only put an approximation.<br /> <br /> - Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br /> <br /> - EDIT: That has never happened and thus it does not contain the singularity of a black hole.<br /> <br /> - Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br /> <br /> -Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br /> <br /> - The original owner of Gmaas is Gmaas.<br /> <br /> - Gmaas was not the fourth Peverell brother, but he ascended into a higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.) EDIT: he wasn't the fourth Peverell brother, but he was a cousin of theirs, and he was the one who advised Ignotus to give up his cloak.<br /> <br /> - It is suspected that Gmaas may be ordering his cyber hairballs to take the forums, along with microbots.<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mu.<br /> <br /> - Gmaas rarely frequents the headquarters of the Illuminati. He was their symbol for one yoctosecond, but soon decided that the job was too low for his power to be wasted on.<br /> <br /> - It has been wondered if Gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br /> <br /> - Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br /> <br /> - It has been confirmed that gmaas uses gmewal as his email service<br /> <br /> - Gmaas enjoys wearing gmean shorts<br /> <br /> - Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br /> <br /> - Gmaas is well known behind his stage name, Michael Stevens (also known as Vsauce XD), or his page name, Purrshanks. EDIT: Crookshanks was his brother.<br /> <br /> - Gmaas rekt sseraj at 12:54 June 4, 2016 UTC time zone. And then the Doctor chased him.<br /> <br /> - Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br /> <br /> - Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br /> <br /> - In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seen the Dark Lord's hut knows that both Gmaas and the DL (USF code name of the Dark Lord) love BBC. How Gmaas gave him a TV may be lost to history. And it has been lost.<br /> <br /> - The TV has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br /> <br /> -The reason why Dacammel left the USF is that gmaas entrusted his TV to him, and not wanting to be discovered by LF, Cobra, or Z9, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br /> <br /> - EDIT: The above fact is somewhat irrelevant.<br /> <br /> -EDIT EDIT. Dacammel gave the TV back to gmaas, and he left the dark side and their cookies alone. <br /> <br /> - Gmaas is a Super Duper Uper Cat Time Lord. He has &lt;math&gt;57843504&lt;/math&gt; regenerations and has used &lt;math&gt;3&lt;/math&gt;. &lt;cmath&gt;9\cdot12\cdot2\cdot267794=57843504&lt;/cmath&gt;. <br /> <br /> -Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part.<br /> <br /> - Gmaas loves to eat turnips. At &lt;math&gt;\frac{13}{32}&lt;/math&gt; of the sites he was spotted at, he was seen with a turnip.<br /> <br /> -Gmaas has a secret hidden garden full of turnips under sseraj's house.<br /> <br /> - Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br /> <br /> -Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br /> <br /> -Gmaas is in alliance with the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything in sight? Nobody knows (except him), not even the leader of the Cult of Skaro.<br /> <br /> -Gmaas lives in Gallifrey and in Gotham City (he has sleepovers with Batman).<br /> <br /> -Gmaas is an excellent driver. EDIT: he was to one who designed the driver's license test, although he didn't bother with the permit test.<br /> <br /> -The native location of Gmaas is the twilight zone.<br /> <br /> -Donald Trump once sang &quot;All Hail the Chief&quot; to Gmaas, 3 days after being sworn in as US President.<br /> <br /> - Gmaas likes to talk with rrusczyk from time to time.<br /> <br /> - Gmaas can shoot fire from his smelly butt.<br /> <br /> - Gmaas is the reason why the USF has the longest thread on AoPS.<br /> <br /> - Gmass is an avid watcher of the popular T.V. show &quot;Bernie Sanders and the Gauntlet of DOOM&quot;<br /> <br /> - sseraj, in 1521 Introduction to Number Theory, posted an image of Gmaas after saying &quot;Who wants to see 5space?&quot; at around 5:16 PM Mountain Time, noting Gmaas was &quot;also 5space&quot;<br /> <br /> -EDIT: he also did it in Introduction to Algebra A once.<br /> <br /> - Gmaas is now my HD background on my Mac.<br /> <br /> - In 1521 Into to Number Theory, sseraj posted an image of a 5space Gmaas fusion. (First sighting) <br /> <br /> - Also confirmed that Gmaas doesn't like ketchup because it was the only food left the photo.<br /> <br /> - In 1447 Intro to Geometry, sseraj posted a picture of Gmaas with a rubik's cube suggesting that Gmaas's has an average solve time of &lt;math&gt;-GMAAS&lt;/math&gt; seconds.<br /> <br /> -Gmaas beat Superman in a fight with ease<br /> <br /> -Gmaas was an admin of Roblox<br /> <br /> -Gmaas traveled around the world, paying so much &lt;math&gt;MONEY&lt;/math&gt; just to eat :D<br /> <br /> -Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br /> Summary:<br /> <br /> -When Gmaas subtracts &lt;math&gt;0.\overline{99}&lt;/math&gt; from &lt;math&gt;1&lt;/math&gt;, the difference is greater than &lt;math&gt;0&lt;/math&gt;.<br /> <br /> -Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br /> <br /> -Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br /> <br /> -The results of the revival are top secret, and nobody knows what happened.<br /> <br /> -sseraj, in 1496 Prealgebra 2, said that Gmaas is Santacat.<br /> <br /> -sseraj likes to post a picture of gmaas in every class he passes by.<br /> <br /> -sseraj posted a picture of gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Endor is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought gmaas is now wandering space in search for a home.<br /> EDIT: What evidence is there Endor was affected? Other Ewoks still live there.<br /> EDIT EDIT: also, glass doesn't care. He can live there no matter what the climate is.<br /> <br /> -Gmaas is the lord of the pokemans<br /> <br /> -Gmaas can communicate with, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will.<br /> <br /> -Picture of Gmaas http://i.imgur.com/PP9xi.png<br /> <br /> -Known by Mike Miller<br /> <br /> -Gmaas got mad at sseraj once, so he locked him in his own freezer<br /> <br /> -Then, sseraj decided to eat all of Gmaas's hidden turnips in the freezer as punishment<br /> <br /> -Gmaas is an obviously omnipotent cat.<br /> <br /> -ehawk11 met him<br /> <br /> -sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br /> <br /> -sseraj has posted pictures of gmaas in '&quot;intro to algebra&quot;, before class started, with the title, &quot;caption contest&quot;. anyone who posted a caption mysteriously vanished in the middle of the night. <br /> EDIT: This has happened many times, including in Introduction to Geometry 1533, among other active classes. The person writing this (Poeshuman) did participate, and did not disappear. (You could argue Gmaas is typing this through his/her account...)<br /> <br /> - gmaas has once slept in your bed and made it wet<br /> <br /> -It is rumored that rrusczyk is actually Gmaas in disguise<br /> <br /> -Gmaas is suspected to be a Mewtwo in disguise<br /> <br /> -Gmaas is a cat but has characteristics of every other animal on Earth.<br /> <br /> -Gmaas is the ruler of the universe and has been known to be the creator of the species &quot;Gmaasians&quot;.<br /> <br /> -There is a rumor that Gmaas is starting a poll<br /> <br /> -Gmaas is a rumored past ThunderClan cat who ran away, founded GmaasClan, then became a kittypet.<br /> <br /> -There is a rumored sport called &quot;Gmaas Hunting&quot; where people try to successfully capture gmaas in the wild with video/camera/eyes. Strangely, no one has been able to do this, and those that have have mysteriously disappeared into the night. Nobody knows why. The person who is writing this(g1zq) has tried Gmaas Hunting, but has never been successful.<br /> <br /> - Gmaas burped and caused an earthquake.<br /> <br /> - Gmaas once drank from your pretty teacup.<br /> <br /> -GMAAS IS HERE.... PURRRRRRRRRRRRRRRRRRRR<br /> <br /> -Gmass made, and currently owns the Matrix.<br /> <br /> -The above fact is true. Therefore, this is an illusion.<br /> <br /> -Gmaas is the reason Salah will become better than Ronaldo.<br /> <br /> -Who is Gmaas, really?<br /> -Gmass is a heavenly being<br /> <br /> -Illuminati was a manifestation of Gmaas, but Gmass decided illuminati was not great enough for his godly self.<br /> <br /> -jlikemath has met Gmaas and Gmaas is his best friend.<br /> <br /> -Gmaas loves Kpop<br /> <br /> -Gmaas read Twilight EDIT: ...and SURVIVED<br /> <br /> -there is a secret code when put into super smash, Gmass would be a playible character. Too bad he didn't say it.<br /> <br /> -Gmaas was a tribute to one of the Hunger Games and came out a Victor and now lives in District 4<br /> <br /> === Gmaas photos ===<br /> http://cdn.artofproblemsolving.com/images/f/f/8/ff8efef3a0d2eb51254634e54bec215b948a1bba.jpg<br /> <br /> === gmaas in Popular Culture ===<br /> <br /> - &lt;s&gt;Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, &quot;The Adventures of gmaas&quot;.&lt;/s&gt;Sorry, this was a rick roll troll.<br /> <br /> - BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,&quot;What are gwali?&quot; the customary answer &quot;This is gwali&quot; is returned. Scientist 5space is now looking into it.<br /> <br /> - Sullymath and themoocow are also writing a book about Gmaas<br /> <br /> -Sighting of Gmaas: https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project<br /> <br /> -Oryx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed<br /> <br /> - Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br /> <br /> - Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br /> <br /> - Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles<br /> <br /> - gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br /> <br /> -oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here<br /> <br /> -Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br /> <br /> -No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br /> <br /> -In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br /> https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br /> https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br /> https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br /> https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br /> <br /> <br /> - Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br /> <br /> -&lt;math&gt;Another&lt;/math&gt; sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br /> - Yet Another Gmaas sighting ? [https://supportforums.cisco.com/t5/user/viewprofilepage/user-id/45046]<br /> <br /> -Gmaas has been sighted several times on the Global Announcements forum<br /> <br /> -Gmaas uses the following transportation: &lt;img&gt; http://cdn.artofproblemsolving.com/images/3/6/8/368da4e615ea3476355ee3388b39f30a48b8dd48.jpg &lt;/img&gt;<br /> <br /> - When Gmaas was mad, he started world wars 1 &amp; 2. It is only because of Gmaas that we have not had World War 3.<br /> <br /> - Gmaas is the only cat to have been proved irrational and transcendental, though we suspect all cats fall in the first category.<br /> <br /> - Gmaas plays Geometry Dash and shares an account with Springhill, his username is D3m0nG4m1n9.<br /> <br /> -Gmaas likes to whiz on the wilzo<br /> <br /> -Gmaas has been spotted in AMC 8 Basics<br /> <br /> -Gmaas is cool<br /> <br /> -Gmaas hemoon card that does over 9000000 dmg<br /> <br /> -Gmaas is a skilled swordsman who should not to be mistaken for Puss in Boots. Some say he even trained the mysterious and valiant Meta Knight.<br /> <br /> -Kirby once swallowed Gmaas. Gmaas had to spit him out.<br /> <br /> -Gmaas was the creator of pokemon, and his pokemon card can OHKO anyone in one turn. He is invisible and he will always move first.<br /> <br /> -Gmass beat Dongmin in [url=The Genius Game]https://en.wikipedia.org/wiki/The_Genius_(TV_series)[/url] Season 1, 2, 3, 4, and 5.</div> Relay400 https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=95666 Gmaas 2018-06-29T01:14:05Z <p>Relay400: /* gmaas in Popular Culture */</p> <hr /> <div> :) :O :P {{{=== Gmaas Facts ===}}} :) :O :P<br /> <br /> -Gmaas founded Target, but then Gmaas sued them for making the mascot look like a dog when it was supposed to look like Gmaas. <br /> <br /> -Everyone has a bit of Gmaas inside them. EDIT: Except you. You are pure Gmaas. <br /> <br /> -Gmaas likes to eat popsicles. Especially the fudge ones that get him hyper. When he is hyper he runs across Washington D.C. grabbing unsuspecting pedestrians and steals their phone, hacks into them and downloads Fortnite onto their poor phone.<br /> <br /> -Gmaas' favorite cereal is fruit loops. Gmaas thinks it tastes like unicorns jumping on rainbows<br /> <br /> -Gmaas is a champion pillow-fighter.<br /> <br /> -Gmaas colonized Mars.<br /> <br /> -He also colonized Jupiter, Pluto, and several other galaxies. He cloned some little Gmaas robots (with his amazingly robotic skill of coding) and put them all over a galaxy called Gmaasalaxy. <br /> <br /> -Gmaas has the ability to make every device play &quot;The Duck Song&quot; at will.<br /> <br /> -Gmaas once caught the red dot and ate it.<br /> <br /> -Gmaas' favorite color is neon klfhsadkhfd.<br /> <br /> -Gmaas is a champion pvp Minecraft player<br /> <br /> -Gmaas founded Costco.<br /> <br /> -Gmaas can solve any puzzle instantly.<br /> <br /> -When Gmaas flips coins, they always land tails.<br /> <br /> -On Gmaas's math tests, he always gets 101.777777777777777777777777777777777777777777777777777701%<br /> <br /> -Gmaas' favorite number is pi. It's also one of his favorite foods. <br /> <br /> -Gmaas' burps created all gaseous planets.<br /> <br /> -Gmass beat Luke Robatille in an epic showdown of catnip consumption.<br /> <br /> - Gmaas' wealth is unknown, but it is estimated to be way more than Scrooge's.<br /> <br /> - Gmaas has a summer house on Mars. <br /> <br /> -Gmaas attended Harvard, Yale, Stanford, MIT, UC Berkeley, Princeton, Columbia, and Caltech at the same time, using a time-turner.<br /> <br /> -Gmaas also attended Hogwarts and was a prefect. <br /> <br /> -Mrs.Norris is Gmaas's archenemy.<br /> <br /> -Gmaas is a demigod and attends Camp Half-Blood over summer. He is the counselor for the Apollo cabin. Because cats are demigod counselors too.<br /> <br /> -Gmaas has completed over 2,000 quests, and is very popular throughout Camp Half-Blood. He has also been to Camp Jupiter.<br /> <br /> -Gmaas painted the Mona Lisa, The Last Supper, and A Starry Night.<br /> <br /> -Gmass knows that the real names are Gmassa Lisa, The Last Domestic Meal, and Far-away Light.<br /> <br /> -Gmaas actually attended all the Ivy Leagues.<br /> <br /> - I am Gmaas<br /> <br /> - I too am Gmaas<br /> <br /> - But it is I who is Gmaas<br /> <br /> - Gmaas is us all<br /> <br /> - Gmaas is all of us yet none of us<br /> <br /> - Gmaas was captured by the infamous j3370 in 2017 but was released due to sympathy. EDIT: j3370 only captured his concrete form, his abstract form cannot be processed by a feeble human brain<br /> <br /> - Gmaas's fur is purple and yellow and red and green and orange and blue and brown and pink all at the same time. <br /> <br /> - Gmaas crossed the event horizon of a black hole and ended up in the AoPS universe.<br /> <br /> - Gmaas crossed the Delaware River with Washington.<br /> <br /> - Gmaas also crossed the Atlantic with the pilgrims.<br /> <br /> - if you are able to capture a Gmaas hair, he will give you some of his gmaas power.<br /> <br /> - Chuck norris makes Gmaas jokes.<br /> <br /> - Gmaas is also the ruler of Oceania, Eastasia, and Eurasia (1984 reference)<br /> <br /> - Gmaas killed Big Brother by farting on him. Though he was caught by the Ministry of Love, he escaped easily.<br /> <br /> - Everyone thinks that Gmaas is a god<br /> <br /> - Gmaas also owns Animal Farm. Napoleon was his servant.<br /> <br /> - Gmaas is the only one who knows where Amelia Earhart is.<br /> <br /> - Gmaas is the only cat that has been proven transcendental.<br /> <br /> - Gmaas happened to notice http://artofproblemsolving.com/community/c402403h1598015p9983782 and is not very happy about it.<br /> <br /> - Grumpy cat reads Gmaas memes. EDIT: Grumpy cat then steals them and claims they're his. Gmass in't very happy about that, either.<br /> <br /> - The real reason why AIME cutoffs aren't out yet is because Gmaas refused to grade them due to too much problem misplacement.<br /> <br /> - Gmaas dueled Grumpy Cat and lost. He wasn't trying.<br /> <br /> - Gmaas sits on the statue of pallas and says forevermore (the Raven refrence )<br /> <br /> - Gmaas is a big fan of Edgar Allan Poe, because he is actually Poe. <br /> <br /> - Gmaas does merely not use USD. He owns it.<br /> <br /> - Gmaas really knows that Roblox is awful and does not play it seriously, thank god our lord is sane<br /> <br /> - In 2003, Gmaas used elliptical curves to force his reign over AoPS.<br /> <br /> - &quot;Actually, my name is spelled GMAAS&quot;<br /> <br /> - Gmaas is the smartest living being in the universe.<br /> <br /> - It was Gmaas who helped Sun Wukong on the Journey to the West.<br /> <br /> - Gmaas is the real creator of Wikipedia.<br /> <br /> -It is said Gmaas could hack any website he desires.<br /> <br /> -Gmaas is the basis of Greek mythology.<br /> <br /> -Gmaas once sold Google to a man for around &lt;math&gt;12&lt;/math&gt; dollars!<br /> <br /> -Gmaas uses a HP printer.<br /> <br /> -Gmaas owns all AoPS staff including Richard Rusczyk.<br /> <br /> -Gmaas was there when Yoda was born.<br /> <br /> - Gmaas's true number of lives left is unknown; however, Gmaas recently confirmed that he had at least one left. Why doesn't Gmaas have so many more lives than other cats? The power of Gmaas.<br /> <br /> - sseraj once spelled Gmaas as gmASS on accident in Introduction to Geometry (1532).<br /> <br /> -Gmaas actively plays Roblox, and is a globally ranked professional gamer: https://www.roblox.com/users/29708533/profile...but he hates Roblox.<br /> <br /> -Gmaas has beaten Chuck Norris and The Rock and John Cena all together in a fight.<br /> <br /> -Gmaas is a South Korean, North Korean, Palestinian, Israeli, U.S., and Soviet citizen at the same time. EDIT: Gmaas has now been found out to be a citizen of every country in the world. Gmaas seems to enjoy the country of AOPS best, however.<br /> <br /> -&quot;i am sand&quot; destroyed Gmaas in FTW<br /> <br /> - sseraj posted a picture of gmaas with a game controller in Introduction to Geometry (1532).<br /> <br /> -Gmaas plays roblox mobile edition and likes Minecraft, Candy Crush, and Club Penguin Rewritten. He also &lt;math&gt;\boxed{\text{loves}}&lt;/math&gt; &lt;b&gt;Cat&lt;/b&gt;ch that fish.<br /> <br /> -Gmaas is Roy Moore's horse in the shape of a cat<br /> <br /> -Gmaas is a known roblox/club penguin rewritten player and is a legend at it. He has over &lt;math&gt;289547987693&lt;/math&gt; robux and &lt;math&gt;190348&lt;/math&gt; in CPR.<br /> <br /> -This is all hypothetical.<br /> <br /> - EDIT: This is all factual <br /> <br /> -Gmaas's real name is Princess. He has a sibling named Rusty/Fireheart/Firestar<br /> (Warrior cats reference)<br /> <br /> - He is capable of salmon powers, according to PunSpark (ask him)<br /> <br /> -The Gmaas told Richard Rusczyk to make AoPS<br /> <br /> -The Gmaas is everything. Yes, you are part of the Gmaas-Dw789<br /> <br /> -The Gmaas knows every dimension up to 9999999999999999999999999999999999999999999999999999999999999999999999999999999999th dimension.<br /> <br /> -Certain theories provide evidence that he IS darth plagueis the wise.<br /> <br /> -Gmaas is &quot;TIRED OF PEOPLE ADDING TO HIS PAGE!!&quot; (Maas 45).<br /> <br /> -Gmaas has multiple accounts; some of them are pifinity, cyumi, srej, squareman, Electro3.0, sturdywill, g1zq, alleycat, Equinox8, anc3, batta, beastgert, and lakecomo224. His best account, though, is wlm7, later renamed to &quot;gmail.com&quot;.<br /> <br /> -Gmaas has a penguin servant named sm24136 who runs GMAASINC. The penguin may or may not be dead. <br /> <br /> -Gmaas owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br /> <br /> -Gmaas knows how to hack into top secret aops community pages.<br /> <br /> -Gmaas was a river clant cat who crossed the event horizon of a black hole and came out the other end!<br /> <br /> -Gmaas is king of the first men, the anduls.<br /> <br /> -Gmaas is a well known professor at MEOWston Academy.<br /> <br /> -Gmaas is a Tuna addict, along with other, more potent fish such as Salmon and Trout.<br /> <br /> -Gmaas won the reward of being cutest and fattest cat ever--he surpassed grumpy cat. (He also out-grumped grumpy cat!!!)<br /> <br /> -Last sighting 1665 Algebra-A 3/9/18 at 9:08 PM.<br /> <br /> - owner of sseraj, not pet.<br /> <br /> - embodiment of life and universe and beyond .<br /> <br /> - Watches memes.<br /> <br /> -After Death became the GOD OF HYPERDEATH and obtained over 9000 souls.<br /> <br /> -Gmaas's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)]<br /> <br /> -Gmaas is a certified Slytherin.<br /> <br /> -Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom <br /> <br /> -Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom)<br /> <br /> - Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br /> <br /> -Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br /> <br /> -Gmaas is a supreme overlord who must be given &lt;math&gt;10^{1000000000000000000000^{1000000000000000000000}}&lt;/math&gt; minecraft DIAMONDS<br /> <br /> - gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br /> <br /> - Gmaas is everyone's favorite animal. <br /> <br /> - He lives with sseraj. <br /> <br /> -Gmaas is my favorite pokemon<br /> <br /> -Gmaas dislikes number theory but enjoys geometry.<br /> <br /> - Gmaas is cool<br /> <br /> - He is often overfed (with probability &lt;math&gt;\frac{3972}{7891}&lt;/math&gt;), or malnourished (with probability &lt;math&gt;\frac{3919}{7891}&lt;/math&gt;) by sseraj.<br /> <br /> - He has &lt;cmath&gt;\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAS&lt;/cmath&gt; supercars, excluding the Purrari and the 138838383 Teslas. <br /> <br /> - He is an employee of AoPS.<br /> <br /> - He is a gmaas with yellow fur and white hypnotizing eyes.<br /> <br /> - He has the ability to divide by zero.<br /> <br /> - He was born with a tail that is a completely different color from the rest of his fur.<br /> <br /> - His stare is very hypnotizing and effective at getting table scraps.<br /> <br /> - He sometimes appears several minutes before certain classes start as an admin. <br /> <br /> - He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br /> <br /> - It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br /> <br /> - Actually, he is a cat. He said so. And science also says so.<br /> <br /> - He is distant relative of Mathcat1234.<br /> <br /> - He is very famous now, and mods always talk about him before class starts.<br /> <br /> - His favorite food is AoPS textbooks because they help him digest problems.<br /> <br /> - Gmaas tends to reside in sseraj's fridge.<br /> <br /> - Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br /> <br /> - The fur of Gmaas can protect him from the harsh conditions of a freezer.<br /> <br /> - Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br /> <br /> - Gmaas is a sage omniscient cat.<br /> <br /> - He is looking for suitable places other than sseraj's fridge to live in.<br /> <br /> - Places where gmaas sightings have happened: <br /> ~The Royal Scoop ice cream store in Bonita Beach Florida<br /> <br /> ~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br /> <br /> -Prealgebra 2 (1440)<br /> <br /> ~Alligator Swamp A 1072 <br /> <br /> ~Alligator Swamp B 1073<br /> <br /> ~Prealgebra A (1488)<br /> <br /> ~Introduction to Algebra A (1170)<br /> <br /> ~Introduction to Algebra B (1529)<br /> <br /> ~Welcome to Panda Town Gate 1076<br /> <br /> ~Welcome to Gmaas Town Gate 1221<br /> <br /> ~Welcome to Gmaas Town Gate 1125<br /> <br /> ~33°01'17.4&quot;N 117°05'40.1&quot;W (Rancho Bernardo Road, San Diego, CA)<br /> <br /> ~The other side of the ice in Antarctica<br /> <br /> ~Feisty Alligator Swamp 1115<br /> <br /> ~Introduction to Geometry 1221 (Taught by sseraj)<br /> <br /> ~Introduction to Counting and Probability 1142 <br /> <br /> ~Feisty-ish Alligator Swamp 1115 (AGAIN)<br /> <br /> ~Intermediate Counting and Probability 1137<br /> <br /> ~Intermediate Counting and Probability 1207<br /> <br /> ~Posting student surveys<br /> <br /> ~USF Castle Walls - Elven Tribe 1203<br /> <br /> ~Dark Lord's Hut 1210<br /> <br /> ~AMC 10 Problem Series 1200<br /> <br /> ~Intermediate Number Theory 1138<br /> <br /> ~Intermediate Number Theory 1476<br /> <br /> ~Introduction To Number Theory 1204. Date:7/27/16.<br /> <br /> ~Algebra B 1112<br /> <br /> ~Intermediate Algebra 1561 7:17 PM 12/11/16<br /> <br /> ~Nowhere Else, Tasmania<br /> <br /> ~Earth Dimension C-137<br /> ~Geometry 1694 at 1616 PST military time. There was a boy riding him, and he seemed extremely miffed.<br /> <br /> <br /> <br /> <br /> - These have all been designated as the most glorious sections of Aopsland now (especially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&amp;B).<br /> <br /> - Gmaas has also been sighted in Olympiad Geometry 1148.<br /> <br /> - Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br /> <br /> - Gmaas once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<br /> <br /> - Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br /> <br /> EDIT: Nobody has yet seen him atop a tribal base yet.<br /> <br /> - Gmaas are often under the disguise of a penguin or cat. Look out for them.<br /> <br /> EDIT: Gmaas rarely disguises himself as a penguin.<br /> <br /> - Many know that leafy stole dream island. In truth, After leafy stole it, Gmaas stole it himself. (BFDI Reference)<br /> <br /> - He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br /> <br /> EDIT: He IS an AoPS site admin.<br /> <br /> - If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br /> <br /> - Is this the real life? Is this just fantasy? No. This is Gmaas, the legend.<br /> <br /> -Aha!! An impostor!! <br /> http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br /> (look at the acronym).<br /> <br /> -EDIT. The above fact is slightly irrelevant.<br /> <br /> - Gmaas might have been viewing (with a &lt;math&gt;\frac{99999.\overline{9}}{100000}&lt;/math&gt; chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br /> <br /> - EDIT: Gmaas is a he.<br /> <br /> -Gmaas is love, Gmaas is life<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mew.<br /> <br /> - Gmaas is on the list of &quot;Elusive Creatures.&quot; If you have questions or want the full list, contact moab33.<br /> <br /> - Gmaas can be summoned using the &lt;math&gt;\tan(90)&lt;/math&gt; ritual. Draw a pentagram and write the numerical value of &lt;math&gt;\tan(90)&lt;/math&gt; in the middle, and he will be summoned.<br /> <br /> - EDIT: The above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br /> https://artofproblemsolving.com/community/c287916h1291232<br /> <br /> - EDIT EDIT: The above 'proof' is non-conclusive. math101010 had only put an approximation.<br /> <br /> - Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br /> <br /> - EDIT: That has never happened and thus it does not contain the singularity of a black hole.<br /> <br /> - Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br /> <br /> -Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br /> <br /> - The original owner of Gmaas is Gmaas.<br /> <br /> - Gmaas was not the fourth Peverell brother, but he ascended into a higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.) EDIT: he wasn't the fourth Peverell brother, but he was a cousin of theirs, and he was the one who advised Ignotus to give up his cloak.<br /> <br /> - It is suspected that Gmaas may be ordering his cyber hairballs to take the forums, along with microbots.<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mu.<br /> <br /> - Gmaas rarely frequents the headquarters of the Illuminati. He was their symbol for one yoctosecond, but soon decided that the job was too low for his power to be wasted on.<br /> <br /> - It has been wondered if Gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br /> <br /> - Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br /> <br /> - It has been confirmed that gmaas uses gmewal as his email service<br /> <br /> - Gmaas enjoys wearing gmean shorts<br /> <br /> - Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br /> <br /> - Gmaas is well known behind his stage name, Michael Stevens (also known as Vsauce XD), or his page name, Purrshanks. EDIT: Crookshanks was his brother.<br /> <br /> - Gmaas rekt sseraj at 12:54 June 4, 2016 UTC time zone. And then the Doctor chased him.<br /> <br /> - Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br /> <br /> - Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br /> <br /> - In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seen the Dark Lord's hut knows that both Gmaas and the DL (USF code name of the Dark Lord) love BBC. How Gmaas gave him a TV may be lost to history. And it has been lost.<br /> <br /> - The TV has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br /> <br /> -The reason why Dacammel left the USF is that gmaas entrusted his TV to him, and not wanting to be discovered by LF, Cobra, or Z9, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br /> <br /> - EDIT: The above fact is somewhat irrelevant.<br /> <br /> -EDIT EDIT. Dacammel gave the TV back to gmaas, and he left the dark side and their cookies alone. <br /> <br /> - Gmaas is a Super Duper Uper Cat Time Lord. He has &lt;math&gt;57843504&lt;/math&gt; regenerations and has used &lt;math&gt;3&lt;/math&gt;. &lt;cmath&gt;9\cdot12\cdot2\cdot267794=57843504&lt;/cmath&gt;. <br /> <br /> -Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part.<br /> <br /> - Gmaas loves to eat turnips. At &lt;math&gt;\frac{13}{32}&lt;/math&gt; of the sites he was spotted at, he was seen with a turnip.<br /> <br /> -Gmaas has a secret hidden garden full of turnips under sseraj's house.<br /> <br /> - Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br /> <br /> -Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br /> <br /> -Gmaas is in alliance with the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything in sight? Nobody knows (except him), not even the leader of the Cult of Skaro.<br /> <br /> -Gmaas lives in Gallifrey and in Gotham City (he has sleepovers with Batman).<br /> <br /> -Gmaas is an excellent driver. EDIT: he was to one who designed the driver's license test, although he didn't bother with the permit test.<br /> <br /> -The native location of Gmaas is the twilight zone.<br /> <br /> -Donald Trump once sang &quot;All Hail the Chief&quot; to Gmaas, 3 days after being sworn in as US President.<br /> <br /> - Gmaas likes to talk with rrusczyk from time to time.<br /> <br /> - Gmaas can shoot fire from his smelly butt.<br /> <br /> - Gmaas is the reason why the USF has the longest thread on AoPS.<br /> <br /> - Gmass is an avid watcher of the popular T.V. show &quot;Bernie Sanders and the Gauntlet of DOOM&quot;<br /> <br /> - sseraj, in 1521 Introduction to Number Theory, posted an image of Gmaas after saying &quot;Who wants to see 5space?&quot; at around 5:16 PM Mountain Time, noting Gmaas was &quot;also 5space&quot;<br /> <br /> -EDIT: he also did it in Introduction to Algebra A once.<br /> <br /> - Gmaas is now my HD background on my Mac.<br /> <br /> - In 1521 Into to Number Theory, sseraj posted an image of a 5space Gmaas fusion. (First sighting) <br /> <br /> - Also confirmed that Gmaas doesn't like ketchup because it was the only food left the photo.<br /> <br /> - In 1447 Intro to Geometry, sseraj posted a picture of Gmaas with a rubik's cube suggesting that Gmaas's has an average solve time of &lt;math&gt;-GMAAS&lt;/math&gt; seconds.<br /> <br /> -Gmaas beat Superman in a fight with ease<br /> <br /> -Gmaas was an admin of Roblox<br /> <br /> -Gmaas traveled around the world, paying so much &lt;math&gt;MONEY&lt;/math&gt; just to eat :D<br /> <br /> -Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br /> Summary:<br /> <br /> -When Gmaas subtracts &lt;math&gt;0.\overline{99}&lt;/math&gt; from &lt;math&gt;1&lt;/math&gt;, the difference is greater than &lt;math&gt;0&lt;/math&gt;.<br /> <br /> -Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br /> <br /> -Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br /> <br /> -The results of the revival are top secret, and nobody knows what happened.<br /> <br /> -sseraj, in 1496 Prealgebra 2, said that Gmaas is Santacat.<br /> <br /> -sseraj likes to post a picture of gmaas in every class he passes by.<br /> <br /> -sseraj posted a picture of gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Endor is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought gmaas is now wandering space in search for a home.<br /> EDIT: What evidence is there Endor was affected? Other Ewoks still live there.<br /> EDIT EDIT: also, glass doesn't care. He can live there no matter what the climate is.<br /> <br /> -Gmaas is the lord of the pokemans<br /> <br /> -Gmaas can communicate with, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will.<br /> <br /> -Picture of Gmaas http://i.imgur.com/PP9xi.png<br /> <br /> -Known by Mike Miller<br /> <br /> -Gmaas got mad at sseraj once, so he locked him in his own freezer<br /> <br /> -Then, sseraj decided to eat all of Gmaas's hidden turnips in the freezer as punishment<br /> <br /> -Gmaas is an obviously omnipotent cat.<br /> <br /> -ehawk11 met him<br /> <br /> -sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br /> <br /> -sseraj has posted pictures of gmaas in '&quot;intro to algebra&quot;, before class started, with the title, &quot;caption contest&quot;. anyone who posted a caption mysteriously vanished in the middle of the night. <br /> EDIT: This has happened many times, including in Introduction to Geometry 1533, among other active classes. The person writing this (Poeshuman) did participate, and did not disappear. (You could argue Gmaas is typing this through his/her account...)<br /> <br /> - gmaas has once slept in your bed and made it wet<br /> <br /> -It is rumored that rrusczyk is actually Gmaas in disguise<br /> <br /> -Gmaas is suspected to be a Mewtwo in disguise<br /> <br /> -Gmaas is a cat but has characteristics of every other animal on Earth.<br /> <br /> -Gmaas is the ruler of the universe and has been known to be the creator of the species &quot;Gmaasians&quot;.<br /> <br /> -There is a rumor that Gmaas is starting a poll<br /> <br /> -Gmaas is a rumored past ThunderClan cat who ran away, founded GmaasClan, then became a kittypet.<br /> <br /> -There is a rumored sport called &quot;Gmaas Hunting&quot; where people try to successfully capture gmaas in the wild with video/camera/eyes. Strangely, no one has been able to do this, and those that have have mysteriously disappeared into the night. Nobody knows why. The person who is writing this(g1zq) has tried Gmaas Hunting, but has never been successful.<br /> <br /> - Gmaas burped and caused an earthquake.<br /> <br /> - Gmaas once drank from your pretty teacup.<br /> <br /> -GMAAS IS HERE.... PURRRRRRRRRRRRRRRRRRRR<br /> <br /> -Gmass made, and currently owns the Matrix.<br /> <br /> -The above fact is true. Therefore, this is an illusion.<br /> <br /> -Gmaas is the reason Salah will become better than Ronaldo.<br /> <br /> -Who is Gmaas, really?<br /> -Gmass is a heavenly being<br /> <br /> -Illuminati was a manifestation of Gmaas, but Gmass decided illuminati was not great enough for his godly self.<br /> <br /> -jlikemath has met Gmaas and Gmaas is his best friend.<br /> <br /> -Gmaas loves Kpop<br /> <br /> -Gmaas read Twilight EDIT: ...and SURVIVED<br /> <br /> -there is a secret code when put into super smash, Gmass would be a playible character. Too bad he didn't say it.<br /> <br /> -Gmaas was a tribute to one of the Hunger Games and came out a Victor and now lives in District 4<br /> <br /> === Gmaas photos ===<br /> http://cdn.artofproblemsolving.com/images/f/f/8/ff8efef3a0d2eb51254634e54bec215b948a1bba.jpg<br /> <br /> === gmaas in Popular Culture ===<br /> <br /> - &lt;s&gt;Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, &quot;The Adventures of gmaas&quot;.&lt;/s&gt;Sorry, this was a rick roll troll.<br /> <br /> - BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,&quot;What are gwali?&quot; the customary answer &quot;This is gwali&quot; is returned. Scientist 5space is now looking into it.<br /> <br /> - Sullymath and themoocow are also writing a book about Gmaas<br /> <br /> -Sighting of Gmaas: https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project<br /> <br /> -Oryx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed<br /> <br /> - Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br /> <br /> - Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br /> <br /> - Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles<br /> <br /> - gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br /> <br /> -oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here<br /> <br /> -Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br /> <br /> -No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br /> <br /> -In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br /> https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br /> https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br /> https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br /> https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br /> <br /> <br /> - Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br /> <br /> -&lt;math&gt;Another&lt;/math&gt; sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br /> - Yet Another Gmaas sighting ? [https://supportforums.cisco.com/t5/user/viewprofilepage/user-id/45046]<br /> <br /> -Gmaas has been sighted several times on the Global Announcements forum<br /> <br /> -Gmaas uses the following transportation: &lt;img&gt; http://cdn.artofproblemsolving.com/images/3/6/8/368da4e615ea3476355ee3388b39f30a48b8dd48.jpg &lt;/img&gt;<br /> <br /> - When Gmaas was mad, he started world wars 1 &amp; 2. It is only because of Gmaas that we have not had World War 3.<br /> <br /> - Gmaas is the only cat to have been proved irrational and transcendental, though we suspect all cats fall in the first category.<br /> <br /> - Gmaas plays Geometry Dash and shares an account with Springhill, his username is D3m0nG4m1n9.<br /> <br /> -Gmaas likes to whiz on the wilzo<br /> <br /> -Gmaas has been spotted in AMC 8 Basics<br /> <br /> -Gmaas is cool<br /> <br /> -Gmaas hemoon card that does over 9000000 dmg<br /> <br /> -Gmaas is a skilled swordsman who should not to be mistaken for Puss in Boots. Some say he even trained the mysterious and valiant Meta Knight.<br /> <br /> -Kirby once swallowed Gmaas. Gmaas had to spit him out.<br /> <br /> -Gmaas was the creator of pokemon, and his pokemon card can OHKO anyone in one turn. He is invisible and he will always move first.<br /> <br /> -Gmass beat Dongmin in [url=https://en.wikipedia.org/wiki/The_Genius_(TV_series)]The Genius Game[/url] Season 1, 2, 3, 4, and 5.</div> Relay400 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_23&diff=89971 2017 AMC 10B Problems/Problem 23 2018-01-25T04:08:46Z <p>Relay400: /* Solution 2 */</p> <hr /> <div>==Problem 23==<br /> Let &lt;math&gt;N=123456789101112\dots4344&lt;/math&gt; be the &lt;math&gt;79&lt;/math&gt;-digit number that is formed by writing the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;44&lt;/math&gt; in order, one after the other. What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;45&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44&lt;/math&gt;<br /> <br /> ==Solution==<br /> We only need to find the remainders of N when divided by 5 and 9 to determine the answer.<br /> By inspection, &lt;math&gt;N \equiv 4 \text{ (mod 5)}&lt;/math&gt;.<br /> The remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;9&lt;/math&gt; is &lt;math&gt;1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4&lt;/math&gt;, but since &lt;math&gt;10 \equiv 1 \text{ (mod 9)}&lt;/math&gt;, we can also write this as &lt;math&gt;1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45&lt;/math&gt;, which has a remainder of 0 mod 9. Therefore, by inspection, the answer is &lt;math&gt;\boxed{\textbf{(C) } 9}&lt;/math&gt;.<br /> <br /> Note: the sum of the digits of &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;270&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Noting the solution above, we try to find the sum of the digits to figure out its remainder when divided by &lt;math&gt;9&lt;/math&gt;. From &lt;math&gt;1&lt;/math&gt; thru &lt;math&gt;9&lt;/math&gt;, the sum is &lt;math&gt;45&lt;/math&gt;. &lt;math&gt;10&lt;/math&gt; thru &lt;math&gt;19&lt;/math&gt;, the sum is &lt;math&gt;55&lt;/math&gt;, &lt;math&gt;20&lt;/math&gt; thru &lt;math&gt;29&lt;/math&gt; is &lt;math&gt;65&lt;/math&gt;, and &lt;math&gt;30&lt;/math&gt; thru &lt;math&gt;39&lt;/math&gt; is &lt;math&gt;75&lt;/math&gt;. Thus the sum of the digits is &lt;math&gt;45+55+65+75+4+5+6+7+8 = 240+30 = 270&lt;/math&gt;, and thus &lt;math&gt;N&lt;/math&gt; is divisible by &lt;math&gt;9&lt;/math&gt;. Now, refer to the above solution. &lt;math&gt;N \equiv 4 \text{ (mod 5)}&lt;/math&gt; and &lt;math&gt;N \equiv 0 \text{ (mod 9)}&lt;/math&gt;. From this information, we can conclude that &lt;math&gt;N \equiv 54 \text{ (mod 5)}&lt;/math&gt; and &lt;math&gt;N \equiv 54 \text{ (mod 9)}&lt;/math&gt;. Therefore, &lt;math&gt;N \equiv 54 \text{ (mod 45)}&lt;/math&gt; and &lt;math&gt;N \equiv 9 \text{ (mod 45)}&lt;/math&gt; so the remainder is &lt;math&gt;\boxed{\textbf{(C) }9}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=B|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Relay400 https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2010_Answer_Key&diff=78245 Mock AIME 2 2010 Answer Key 2016-04-25T23:50:49Z <p>Relay400: /* Unformatted Solutions */</p> <hr /> <div>Return to [[Mock AIME 2 2010]]<br /> <br /> # 804<br /> # 683<br /> # 401<br /> # 392<br /> # 416<br /> # 108<br /> # 756<br /> # 028<br /> # 135<br /> # 041<br /> # 164<br /> # 256<br /> # 385<br /> # 167<br /> # 860<br /> <br /> {{AIME box|year = Mock|n = 2 2010|before=[[Mock AIME 1 2010]]|after=[[Mock AIME 1 2011]]}}<br /> <br /> == Unformatted Solutions ==<br /> :''Written by the problem authors''. See the external links at [[Mock AIME 2 2010]].<br /> <br /> <br /> There are 2010 lemmings. At each step, we may separate the lemmings into groups of 5 and purge the remainder, separate them into groups of 3 and purge the remainder, or pick one lemming and purge it. Find the smallest number of steps necessary to remove all 2010 lemmings. (Alex Zhu)<br /> <br /> <br /> %%Solution 1<br /> <br /> Notice that removing as many lemmings as possible at each step (i.e., using the greedy algorithm) is optimal. This can be easily proved through strong induction. Starting from 2010, which is a multiple of 15, we must first purge 1 lemming. We can then purge 4 lemmings by using groups of 5. Then, we can use groups of 3 followed by groups of 5 to reduce it to 5 less. We can repeat this step again, and we will end up at &lt;math&gt;2010-15=1995&lt;/math&gt;. This process can be repeated every for every 15 lemmings, so since it takes 6 steps to clear 15 lemmings, it will take us &lt;math&gt;2010\cdot\frac{6}{15}=\boxed{804}&lt;/math&gt; to remove all the lemmings.<br /> <br /> <br /> %%Problem 2<br /> <br /> \setcounter{enumi}{1}<br /> Let &lt;math&gt;a_1, a_2, \ldots, a_{10}&lt;/math&gt; be nonnegative integers such that &lt;math&gt;a_1 + a_2 + \ldots + a_{10} = 2010&lt;/math&gt;, and define &lt;math&gt;f&lt;/math&gt; so that &lt;math&gt;f((a_1, a_2, \ldots, a_{10})) = (b_1, b_2, \ldots, b_{10})&lt;/math&gt;, with &lt;math&gt;0 \le b_i \le 2, 3|a_i-b_i&lt;/math&gt; for &lt;math&gt;1 \le i \le 10&lt;/math&gt;. Given that &lt;math&gt;f&lt;/math&gt; can take on &lt;math&gt;K&lt;/math&gt; distinct values, find the remainder when &lt;math&gt;K&lt;/math&gt; is divided by 1000. (Alex Zhu)<br /> <br /> <br /> %%Solution 2<br /> <br /> For any &lt;math&gt;a_1, a_2, \ldots, a_{10}&lt;/math&gt;, since &lt;math&gt;b_i\equiv a_i \pmod{3}&lt;/math&gt; for &lt;math&gt;1\le i\le 10&lt;/math&gt;, therefore, &lt;math&gt;b_1+b_2+\ldots+b_{10}\equiv a_1+a_2+\ldots+a_{10} = 2010 \equiv 0 \pmod{3}&lt;/math&gt;. Also, note that for any &lt;math&gt;(b_1, b_2, \ldots, b_{10})&lt;/math&gt; that satisfies &lt;math&gt;b_1+b_2+\ldots+b_{10}\equiv 0 \pmod{3}&lt;/math&gt;, there exists a corresponding &lt;math&gt;(a_1, a_2, \ldots, a_{10})&lt;/math&gt; such that &lt;math&gt;a_1+a_2+\ldots+a_{10}=2010&lt;/math&gt; (let &lt;math&gt;a_i = b_i&lt;/math&gt; for &lt;math&gt;1 \leq i \leq 9&lt;/math&gt;, and &lt;math&gt;a_{10} = 2010 - (b_1 + b_2 + \ldots + b_9)&lt;/math&gt;.) Therefore, we are just trying to count the number of &lt;math&gt;(b_1, b_2, \ldots, b_{10})&lt;/math&gt; that satisfy &lt;math&gt;b_1+b_2+\ldots+b_{10}\equiv 0 \pmod{3}&lt;/math&gt;. To do this, we note that for any of the &lt;math&gt;3^9&lt;/math&gt; possible ways to pick &lt;math&gt;b_1, b_2, \dots, b_9&lt;/math&gt;, there is exactly 1 possible value of &lt;math&gt;b_{10}&lt;/math&gt; that works. Therefore, the total number of possibilities is &lt;math&gt;3^9=19\boxed{683}&lt;/math&gt;.<br /> <br /> <br /> %%Problem 3<br /> <br /> \setcounter{enumi}{2}<br /> Five gunmen are shooting each other. At the same moment, each randomly chooses one of the other four to shoot. The probability that there are some two people shooting each other can be expressed in the form &lt;math&gt;\frac{a}{b}&lt;/math&gt;, where &lt;math&gt;a, b&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;a+b&lt;/math&gt;. (Tim Wu)<br /> <br /> <br /> <br /> We will count how many distinct arrangements (choices of target for each gunman) result in a pair shooting each other, and divide this by the total number of arrangements, &lt;math&gt;4^5&lt;/math&gt;. For any pair to shoot each other, the two shoot each other in exactly one way, while the other three people shoot any of their 4 targets. There are 10 possible pairs, giving a count of &lt;math&gt;4^3\cdot10=640&lt;/math&gt; this way. However, we must account for the overcount caused by 2 pairs shooting each other simultaneously. We can choose one person who is not in the pair in 5 ways, and then of the following people, say A, B, C, and D, we make A shoot one of the others in 3 possible ways (the remaining people can shoot each other in only one other way.) Finally, the lone person can shoot whomever he so desires in 4 ways, giving &lt;math&gt;5\cdot3\cdot4=60&lt;/math&gt; total possibilities. Our probability is then &lt;math&gt;\frac{640-60}{4^5}=\frac{580}{4^5}=\frac{145}{256}&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{401}&lt;/math&gt;.<br /> <br /> <br /> %%Problem 4<br /> <br /> \setcounter{enumi}{3}<br /> Anderson is bored in physics class. His favorite numbers are 1, 7, and 33. He randomly writes 0., and then writes down a long string consisting of those numbers juxtaposed against one another (e.g., 0.1337173377133...) on a sheet of paper. Since physics class is infinitely long, he writes an infinitely long decimal number. If the expected value of the number he wrote down is of the form &lt;math&gt;\frac{a}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers, find &lt;math&gt;a+b&lt;/math&gt;. (Alex Zhu)<br /> <br /> <br /> %%Solution 4<br /> <br /> Let &lt;math&gt;E&lt;/math&gt; be the expected value of the number Anderson writes. Consider the first segment of the number Anderson writes. If it is 1, then the number will begin with 1, and the remaining part will have an expected value of &lt;math&gt;\frac{E}{10}&lt;/math&gt; (the expected value of the original number only moved down a digit). Thus, in this case, his number will average &lt;math&gt;\frac{1}{10}+\frac{E}{10}&lt;/math&gt;. Similarly, if he writes 7, his number will average &lt;math&gt;\frac{7}{10}+\frac{E}{10}&lt;/math&gt;, and if he writes 33, his number will average &lt;math&gt;\frac{33}{100}+\frac{E}{100}&lt;/math&gt;. Each of these cases happens with equal probability, so we have<br /> \begin{align*}<br /> E&amp;= \dfrac{\frac{1}{10}+\frac{E}{10}+\frac{7}{10}+\frac{E}{10}+\frac{33}{100}+\frac{E}{100}}{3} \\<br /> 300E &amp;= 10+10E+70+10E+33+E \\<br /> 279E &amp;= 113 \\<br /> E &amp;= \frac{113}{279}.<br /> \end{align*}<br /> Our answer is therefore &lt;math&gt;113+279=\boxed{392}&lt;/math&gt;. <br /> <br /> <br /> %%Problem 5<br /> <br /> \setcounter{enumi}{4}<br /> Let &lt;math&gt;P(x)=(1+x)(1+2x^2)(1+3x^4)(1+4x^8)(1+5x^{16})&lt;/math&gt;. Find the three rightmost nonzero digits of the product of the coefficients of &lt;math&gt;P(x)&lt;/math&gt;. (Alex Zhu)<br /> <br /> <br /> %%Solution 5<br /> <br /> First, note that each of the &lt;math&gt;2^{5}&lt;/math&gt; possible coefficients is multiplied by a different power of &lt;math&gt;x&lt;/math&gt;, because there are also &lt;math&gt;2^5&lt;/math&gt; different powers of &lt;math&gt;x&lt;/math&gt; in total (this follows from the fact that every positive integer can be written uniquely as a sum of powers of 2; the coefficient of &lt;math&gt;x^n = x^{\sum 2^i b_i}&lt;/math&gt;, where &lt;math&gt;b_i \in \{0,1\}&lt;/math&gt;, is obtained from multiplying together all &lt;math&gt;i x^i&lt;/math&gt; with &lt;math&gt;b_i = 1&lt;/math&gt;.) Therefore, we are trying to find the product of the product of all of the elements of subsets of the set &lt;math&gt;S=\{ 1,2,3,4,5\}&lt;/math&gt; (where the product of the elements of the empty set is taken to be 1). If we pair each subset &lt;math&gt;P&lt;/math&gt; with its complement, we will have 16 pairs of subsets, each of which multiply together to &lt;math&gt;5!&lt;/math&gt;, so the product of all of the subsets is &lt;math&gt;(5!)^{16}=120^{16}&lt;/math&gt;. Since we want the last three nonzero digits, we simply wish to find &lt;math&gt;12^{16} \pmod{1000}&lt;/math&gt;. To make this easier, we will find it modulo 8 and modulo 125 and use the Chinese remainder theorem to find the answer. <br /> <br /> &lt;math&gt;12^{16}\equiv 0 \pmod{8}&lt;/math&gt;, and &lt;math&gt;12^{16}\equiv 144^8\equiv 19^8\equiv 361^4\equiv (-14)^4\equiv 196^2\equiv 71^2\equiv 41 \pmod{125}&lt;/math&gt;.<br /> <br /> We can now see that because &lt;math&gt;41\equiv 1\pmod{8}&lt;/math&gt; and &lt;math&gt;125\equiv 5\pmod{8}&lt;/math&gt;, the only number that is &lt;math&gt;0\pmod{8}&lt;/math&gt; and &lt;math&gt;41\pmod{125}&lt;/math&gt; is &lt;math&gt;41+125 \cdot 3=\boxed{416}&lt;/math&gt;.<br /> <br /> <br /> %%Problem 6<br /> <br /> \setcounter{enumi}{5}<br /> \item<br /> Let &lt;math&gt;S_n&lt;/math&gt; denote the set &lt;math&gt;\{1,2,\ldots,n\}&lt;/math&gt;, and define &lt;math&gt;f(S)&lt;/math&gt;, where &lt;math&gt;S&lt;/math&gt; is a subset of the positive integers, to output the greatest common divisor of all elements in &lt;math&gt;S&lt;/math&gt;, unless &lt;math&gt;S&lt;/math&gt; is empty, in which case it will output 0. Find the last three digits of &lt;math&gt;\sum_{S\subseteq S_{10}} f(S)&lt;/math&gt;, where &lt;math&gt;S&lt;/math&gt; ranges over all subsets of &lt;math&gt;S_{10}&lt;/math&gt;. (George Xing)<br /> <br /> <br /> %%Solution 6<br /> <br /> Let &lt;math&gt;a_i&lt;/math&gt; denote the number of subsets &lt;math&gt;S \subseteq S_{10}&lt;/math&gt; such that &lt;math&gt;f(S)=i&lt;/math&gt;. Our answer will be the sum &lt;math&gt;\sum_{i=0}^{10} ia_i&lt;/math&gt;.<br /> <br /> For any positive integer &lt;math&gt;d \leq 10&lt;/math&gt;, we note that the non-empty sets satisfying the property that &lt;math&gt;f(S)&lt;/math&gt; is a multiple of &lt;math&gt;d&lt;/math&gt; are exactly the non-empty sets whose elements are multiples of &lt;math&gt;d&lt;/math&gt;. Hence, the number of such sets is &lt;math&gt;2^k - 1&lt;/math&gt;, where &lt;math&gt;k=\lfloor \frac{10}{d} \rfloor&lt;/math&gt; (as there are &lt;math&gt;k&lt;/math&gt; multiples of &lt;math&gt;d&lt;/math&gt; between 1 and 10 inclusive.) On the other hand, the number of such sets is &lt;math&gt;a_d + a_{2d} + \ldots + a_{kd}&lt;/math&gt;, that is, the sum of the number of sets &lt;math&gt;S&lt;/math&gt; such that &lt;math&gt;f(S)&lt;/math&gt; is &lt;math&gt;d&lt;/math&gt;, &lt;math&gt;f(S)&lt;/math&gt; is &lt;math&gt;2d&lt;/math&gt;, &lt;math&gt;f(S)&lt;/math&gt; is &lt;math&gt;3d&lt;/math&gt;, etc., so we have that &lt;math&gt;a_d + a_{2d} + \ldots + a_{kd} = 2^k - 1&lt;/math&gt;. By setting &lt;math&gt;d = 1, 2, 3, \ldots, 10&lt;/math&gt;, we see that <br /> <br /> \begin{align*}<br /> a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 + a_{10} &amp;= 2^{10} - 1 = 1023 \\<br /> a_2 + a_4 + a_6 + a_8 + a_{10} &amp;= 2^5 - 1 = 31 \\<br /> a_3 + a_6 + a_9 &amp;= 2^3 - 1 = 7 \\<br /> a_4 + a_8 &amp;= 2^2 - 1 = 3 \\<br /> a_5 + a_{10} &amp;= 2^2 - 1 = 3 \\<br /> a_6 &amp;= 2^1 - 1 = 1 \\<br /> a_7 &amp;= 2^1 - 1 = 1 \\<br /> a_8 &amp;= 2^1 - 1 = 1 \\<br /> a_9 &amp;= 2^1 - 1 = 1 \\<br /> a_{10} &amp;= 2^1 - 1 = 1<br /> \end{align*}<br /> We can easily solve this system of equations to find &lt;math&gt;a_1 = 983&lt;/math&gt;, &lt;math&gt;a_2 = 26&lt;/math&gt;, &lt;math&gt;a_3 = 5&lt;/math&gt;, &lt;math&gt;a_4 = 2&lt;/math&gt;, &lt;math&gt;a_5 = 2&lt;/math&gt;, and &lt;math&gt;a_6 = a_7 = a_8 = a_9 = a_{10} = 1&lt;/math&gt;. Our answer is therefore &lt;math&gt;1 \cdot 983 + 2 \cdot 26 + 3 \cdot 5 + 4 \cdot 8 + 5 \cdot 2 + 6\cdot 1 + 7 \cdot 1 + 8 \cdot 1 + 9 \cdot 1 + 10 \cdot 1 = 1\boxed{108}&lt;/math&gt;.<br /> <br /> <br /> %%Problem 7<br /> <br /> <br /> <br /> \setcounter{enumi}{6}<br /> Find the number of functions &lt;math&gt;f&lt;/math&gt; from &lt;math&gt;\{1,2,3,4,5\}&lt;/math&gt; to itself such that &lt;math&gt;f(f(f(x))) = f(f(x))&lt;/math&gt; for all &lt;math&gt;x \in \{1,2,3,4,5\}&lt;/math&gt;. (Alex Zhu)<br /> <br /> <br /> %%Solution 7<br /> <br /> For a function &lt;math&gt;f&lt;/math&gt;, let &lt;math&gt;A= \{i : f(i)=i\}&lt;/math&gt; be the set of values that are fixed, let &lt;math&gt;B = \{i : f(i) \in A \text{ and } i \not\in A\}&lt;/math&gt; be the set of values that evaluate to an element in &lt;math&gt;A&lt;/math&gt; but are not elements of &lt;math&gt;A&lt;/math&gt; themselves, and let &lt;math&gt;C=\{1,2,3,4,5\} - (A\cup B)&lt;/math&gt; be the set of the remaining values. Notice that all possible values of &lt;math&gt;f(f(x))&lt;/math&gt; are in &lt;math&gt;A&lt;/math&gt;, so &lt;math&gt;A&lt;/math&gt; must be nonempty. We will now use casework on the size of &lt;math&gt;A&lt;/math&gt;.<br /> <br /> \textbf{Case 1: } &lt;math&gt;|A|=1&lt;/math&gt;.\\<br /> WLOG, let &lt;math&gt;f(1)=1&lt;/math&gt;, so we will multiply our result by 5 at the end. Now, let &lt;math&gt;|B|=x&lt;/math&gt;, so &lt;math&gt;|C|=4-x&lt;/math&gt;. We must have &lt;math&gt;f(f(x))=1&lt;/math&gt; for all &lt;math&gt;x&lt;/math&gt;, so each element &lt;math&gt;c&lt;/math&gt; in &lt;math&gt;C&lt;/math&gt; must satisfy &lt;math&gt;f(c)=b&lt;/math&gt; for some &lt;math&gt;b&lt;/math&gt; in &lt;math&gt;B&lt;/math&gt;, because &lt;math&gt;f(c)\neq 1&lt;/math&gt; and the only other numbers for which &lt;math&gt;f(x)=1&lt;/math&gt; are the elements of b. This also implies &lt;math&gt;|B|&gt;0&lt;/math&gt;. Conversely, it is easy to check any function satisfying &lt;math&gt;f(c)=b&lt;/math&gt; works, so the total number of functions in this case is &lt;math&gt;{5\sum_{x=1}^4 \binom{4}{x} x^{4-x}}&lt;/math&gt; because there are &lt;math&gt;\binom{4}{x}&lt;/math&gt; ways to choose the elements in &lt;math&gt;B&lt;/math&gt;, and each of the &lt;math&gt;4-x&lt;/math&gt; elements in C can map to any of the &lt;math&gt;x&lt;/math&gt; elements in B. This sum evaluates to &lt;math&gt;5(4+6\cdot 4+4\cdot 3+1)=205&lt;/math&gt;, so there are 205 functions in this case.<br /> <br /> \textbf{Case 2: } &lt;math&gt;|A|=2&lt;/math&gt;.\\<br /> This is very similar to case 1, except for the fact that each element in &lt;math&gt;B&lt;/math&gt; can now correspond to one of two possible elements in &lt;math&gt;A&lt;/math&gt;, so this adds a factor of &lt;math&gt;2^x&lt;/math&gt;. The sum is &lt;math&gt;{\binom{5}{2}\sum_{x=1}^3\binom{3}{x} 2^x x^{3-x}}=10(3\cdot 2+3\cdot 4\cdot 2+8)=380&lt;/math&gt;, so there are 380 functions in this case.<br /> <br /> \textbf{Case 3: } &lt;math&gt;|A|=3&lt;/math&gt;.\\<br /> The sum is &lt;math&gt;{\binom{5}{3}\sum_{x=1}^2\binom{2}{x} 3^x x^{2-x}}=10(2\cdot 3+9)=150&lt;/math&gt;, so there are 150 functions in this case.<br /> <br /> \textbf{Case 4: } &lt;math&gt;|A|=4&lt;/math&gt;.\\<br /> There are clearly &lt;math&gt;5(4)=20&lt;/math&gt; functions in this case. <br /> <br /> \textbf{Case 5: } &lt;math&gt;|A|=5&lt;/math&gt;.\\<br /> There is a total of 1 function (the identity) in this case. <br /> <br /> Adding everything up, we see that our final answer is &lt;math&gt;205+380+150+20+1=\boxed{756}&lt;/math&gt;.<br /> <br /> <br /> %%Problem 8<br /> <br /> \setcounter{enumi}{7}<br /> In triangle &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;BA = 15&lt;/math&gt;, &lt;math&gt;AC = 20&lt;/math&gt;, and &lt;math&gt;BC = 25&lt;/math&gt;. In addition, there is a point &lt;math&gt;D&lt;/math&gt; lying on segment &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;BD = 16&lt;/math&gt;. Given that the length of the radius of the circle through &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; that is tangent to side &lt;math&gt;AC&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime integers, find &lt;math&gt;p + q&lt;/math&gt;. (Alex Zhu)<br /> <br /> <br /> %%Solution 8<br /> \begin{sol1}<br /> Let &lt;math&gt;E&lt;/math&gt; be the point of tangency of the circle with &lt;math&gt;AC&lt;/math&gt;. By power of a point, &lt;math&gt;CD \cdot CB = CE^2&lt;/math&gt;, that is, &lt;math&gt;9 \cdot 25 = C^2&lt;/math&gt;, giving &lt;math&gt;CE = 15&lt;/math&gt; and &lt;math&gt;AE = 20 - 15 = 5&lt;/math&gt;. Since &lt;math&gt;15^2 + 20^2 = 25^2&lt;/math&gt;, &lt;math&gt;m \angle A = 90^{\circ}&lt;/math&gt;, so &lt;math&gt;BE = 5 \sqrt{10}&lt;/math&gt; and &lt;math&gt;\sin m \angle AEB = \frac{15}{5 \sqrt{10}}&lt;/math&gt;. Since the circle is tangent to &lt;math&gt;AC&lt;/math&gt;, we have that &lt;math&gt;\angle AEB \cong \angle EDB&lt;/math&gt;. By the extended law of sines, the circumdiameter of &lt;math&gt;\triangle BED&lt;/math&gt; (that is, the radius of our circle) is &lt;math&gt;\frac{BE}{\sin m \angle EDB} = \frac{BE}{\sin m \angle AEB} = \frac{50}{3}&lt;/math&gt;. Therefore, the radius is &lt;math&gt;\frac{25}{3}&lt;/math&gt;, giving an answer of &lt;math&gt;25 + 3 = \boxed{028}&lt;/math&gt;. <br /> \end{sol1}<br /> <br /> \begin{sol2}<br /> Let O be the center of the circle, and let points E and F to be the perpendiculars from O to AC and BC, respectively. First, by power of a point, we have &lt;math&gt;CD \cdot CB=CE^2&lt;/math&gt;, so &lt;math&gt;9\cdot 25=CE^2&lt;/math&gt;, giving &lt;math&gt;CE=15&lt;/math&gt; and &lt;math&gt;EA=20-15=5&lt;/math&gt;. Now, let &lt;math&gt;r&lt;/math&gt; be the radius of the circle. We have &lt;math&gt;AF=EO=r&lt;/math&gt;, so &lt;math&gt;FB=AB-AF=15-r&lt;/math&gt;. By the Pythagorean theorem on triangle FOB, we have &lt;math&gt;FB^2+FO^2=BO^2\iff 225-30r+r^2+FO^2=r^2 \iff FO=\sqrt{30r-225}&lt;/math&gt;. Finally, we have &lt;math&gt;5=EA=FO=\sqrt{30r-225}&lt;/math&gt;, so &lt;math&gt;25=30r-225&lt;/math&gt;. Hence, &lt;math&gt;r=\frac{25}{3}&lt;/math&gt;, giving us an answer of &lt;math&gt;25+3=\boxed{028}&lt;/math&gt;.<br /> \end{sol2}<br /> <br /> %%Problem 9<br /> <br /> \setcounter{enumi}{8}<br /> Given that &lt;math&gt;x,y,z&lt;/math&gt; are reals such that &lt;math&gt;16 \sin^4(x+y) + 49 \cos^4(x+y) = \sin^2(2x + 2y)(8\sin^2(x+z) + 6 \cos^2(y+z))&lt;/math&gt;, the largest possible value of &lt;math&gt;3\cos^2(x+y+z)&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac{a+b\sqrt{c}}{d}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers, &lt;math&gt;c&lt;/math&gt; is a positive integer not divisible by the square of any prime, and &lt;math&gt;d&lt;/math&gt; is a positive integer such that &lt;math&gt;\gcd(a,b,d)=1&lt;/math&gt;. Find &lt;math&gt;a+b+c+d&lt;/math&gt;. (Alex Zhu)<br /> <br /> <br /> %%Solution 9<br /> <br /> \begin{align*}<br /> 16\sin^4(x+y) + 49\cos^4(x+y) <br /> &amp;= \sin^2(2x + 2y)\left(8\sin^2(x+z) + 6\cos^2(y+z)\right) \\<br /> &amp;\leq 14 \sin^2(2x + 2y) \\<br /> &amp;= 56(\sin(x+y) \cos(x+y))<br /> \end{align*}<br /> Hence, &lt;math&gt;(4 \sin^2(x+y) - 7 \cos^2(x+y)) \leq 0&lt;/math&gt;, yielding &lt;math&gt;\tan^2(x+y) = \frac{7}{4}&lt;/math&gt;, or &lt;math&gt;x + y = \pm \arctan{\frac{\sqrt{7}}{2}} + a\pi&lt;/math&gt;, for some integer &lt;math&gt;a&lt;/math&gt;. Furthermore, we must have &lt;math&gt;\sin^4(x+z) = \cos^4(y+z) = 1&lt;/math&gt;, so &lt;math&gt;x + z = \frac{\pi}{2} + b\pi&lt;/math&gt; and &lt;math&gt;y + z = c \pi&lt;/math&gt; for some integers &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;. Hence, &lt;math&gt;2x + 2y + 2z = \pm \arctan\frac{\sqrt{7}}{2} + \frac{\pi}{2} + \pi(a+b+c)&lt;/math&gt;, so &lt;math&gt;|\cos(2x + 2y + 2z)| = |\sin \arctan\frac{\sqrt{7}}{2}| = \frac{\sqrt{77}}{11}&lt;/math&gt;. Hence, &lt;math&gt;2\cos(x+y+z)^2 - 1 \leq |2\cos(x+y+z)^2 - 1| = |\cos(2x + 2y + 2z)| = \frac{\sqrt{77}}{11}&lt;/math&gt;, giving us &lt;math&gt;3 \cos(x+y+z)^2 \leq \frac{33 + 3 \sqrt{77}}{22}&lt;/math&gt;. This is attained when &lt;math&gt;\cos(2x + 2y + 2z) &gt; 0&lt;/math&gt;, which indeed occurs if we pick &lt;math&gt;x + y = \arctan \frac{\sqrt{7}}{2}&lt;/math&gt;, &lt;math&gt;x + z = \frac{\pi}{2} + \pi&lt;/math&gt;, and &lt;math&gt;z + y = 0&lt;/math&gt;, so our answer is &lt;math&gt;33 + 3 + 77 + 22 = \boxed{135}&lt;/math&gt;. <br /> <br /> <br /> %%Problem 10<br /> <br /> \setcounter{enumi}{9}<br /> How many positive integers &lt;math&gt;n \le 2010 &lt;/math&gt; satisfy &lt;math&gt;\phi(n) | n&lt;/math&gt;, where &lt;math&gt;\phi(n)&lt;/math&gt; is the number of positive integers less than or equal to &lt;math&gt;n&lt;/math&gt; relatively prime to &lt;math&gt;n&lt;/math&gt;? (Alex Zhu)<br /> <br /> <br /> %%Solution 10<br /> <br /> If &lt;math&gt;n = 2^k&lt;/math&gt;, then &lt;math&gt;\varphi(2^k) = 2^{k-1} | 2^k = n&lt;/math&gt;. Now, suppose &lt;math&gt;n&lt;/math&gt; is not a power of 2. Let &lt;math&gt;n = 2^a b&lt;/math&gt;, where &lt;math&gt;b&lt;/math&gt; is odd. &lt;math&gt;\phi(n) = 2^{a-1} \phi(b)&lt;/math&gt;. If &lt;math&gt;b = pqm&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are distinct odd prime numbers and &lt;math&gt;m&lt;/math&gt; is some integer, then &lt;math&gt;\phi(b) = \phi(p)\phi(q) \phi(m) = (p-1)(q-1) \phi(m)&lt;/math&gt;. Since &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are odd, &lt;math&gt;p-1&lt;/math&gt; and &lt;math&gt;q-1&lt;/math&gt; are even, so &lt;math&gt;4 | \phi(b)&lt;/math&gt;. Hence, &lt;math&gt;2^{a+1} | \phi(n)&lt;/math&gt;, so &lt;math&gt;\phi(n) \not | n&lt;/math&gt;. <br /> <br /> It follows that &lt;math&gt;n&lt;/math&gt; has at most one other prime factor, so &lt;math&gt;n&lt;/math&gt; is of the form &lt;math&gt;2^k p^j&lt;/math&gt;. &lt;math&gt;\varphi(2^k p^j) = 2^{k-1} p^{j-1}(p-1) | 2^k p^j&lt;/math&gt;, so &lt;math&gt;p - 1 | 2p&lt;/math&gt;. Hence, &lt;math&gt;p - 1 | 2p - 2(p - 2) = 2&lt;/math&gt;, so &lt;math&gt;p = 2, 3&lt;/math&gt;. But &lt;math&gt;p \neq 2&lt;/math&gt;, so we must have that &lt;math&gt;p = 3&lt;/math&gt;. <br /> <br /> We now note that &lt;math&gt;\varphi(3^b)&lt;/math&gt;, &lt;math&gt;b &gt; 0&lt;/math&gt;, is even, and hence does not divide &lt;math&gt;3^k&lt;/math&gt;. But &lt;math&gt;\varphi(2^a 3^b) = 2^a 3^{b-1} | 2^a 3^b&lt;/math&gt;, so we wish to find the number of integers less than or equal to 2010 of the form &lt;math&gt;2^a 3^b&lt;/math&gt; such that &lt;math&gt;b \neq 0&lt;/math&gt; implies &lt;math&gt;a \neq 0&lt;/math&gt;. <br /> <br /> When &lt;math&gt;a=0&lt;/math&gt;, &lt;math&gt;b=0&lt;/math&gt; giving one possible value of &lt;math&gt;b&lt;/math&gt;. For any larger &lt;math&gt;a &lt; 10&lt;/math&gt; (as &lt;math&gt;2^{11} &gt; 2010&lt;/math&gt;), we have &lt;math&gt;0 \leq b \leq \lfloor \log_2 \frac{2010}{2^a} \rfloor&lt;/math&gt;, yieding &lt;math&gt;\lfloor \log_2 \frac{2010}{2^a} \rfloor + 1&lt;/math&gt; possible values of &lt;math&gt;b&lt;/math&gt;. Summing these values for &lt;math&gt;1 \leq a \leq 9&lt;/math&gt; gives us the answer of &lt;math&gt;1 + 7 + 6 + 6 + 5 + 4 + 4 + 3 + 2 + 2 = \boxed{041}&lt;/math&gt;. <br /> <br /> <br /> %%Problem 11<br /> <br /> \setcounter{enumi}{10}<br /> Let &lt;math&gt;f:\mathbb{N}\to\mathbb{N}&lt;/math&gt; be a function such that &lt;cmath&gt;f(n) =\sum_{abc = n | a,b,c \in \mathbb{N}} ab + bc + ca.&lt;/cmath&gt; For example, &lt;math&gt;f(5) = (1 \cdot 1 + 1 \cdot 5 + 5 \cdot 1) + (1 \cdot 5 + 5 \cdot 1 + 1 \cdot 1) + (5 \cdot 1 + 1 \cdot 1 + 1 \cdot 5) = 33&lt;/math&gt;, where we are summing over the triples &lt;math&gt;(a,b,c) = (1,1,5), (1,5,1)&lt;/math&gt;, and &lt;math&gt;(5,1,1)&lt;/math&gt;. Find the last three digits of &lt;math&gt;f(30^{3})&lt;/math&gt;. (Mitchell Lee)<br /> <br /> <br /> %%Solution 11<br /> <br /> &lt;math&gt;\sum_{abc = n | a,b,c \in \mathbb{N}} ab + bc + ca = abc \sum_{abc = n | a,b,c \in \mathbb{N}} \frac{1}{a} + \frac{1}{b} + \frac{1}{c}&lt;/math&gt;. For any &lt;math&gt;a&lt;/math&gt; that divides &lt;math&gt;n&lt;/math&gt;, the number of sums that &lt;math&gt;\frac{1}{a}&lt;/math&gt; will be in is exactly three the number of ordered pairs &lt;math&gt;(b,c)&lt;/math&gt; we can find such that &lt;math&gt;abc = n&lt;/math&gt;, which in turn is just &lt;math&gt;\tau(\frac{n}{a})&lt;/math&gt;, where &lt;math&gt;\tau(n)&lt;/math&gt; is the number of divisors of &lt;math&gt;n&lt;/math&gt; (&lt;math&gt;b&lt;/math&gt; can be any divisor of &lt;math&gt;\frac{n}{a}&lt;/math&gt;, and &lt;math&gt;c = \frac{n}{ab}&lt;/math&gt;.) (The factor of three arises from the fact that &lt;math&gt;\frac{1}{b} + \frac{1}{a} + \frac{1}{c}&lt;/math&gt; and &lt;math&gt;\frac{1}{b} + \frac{1}{c} + \frac{1}{a}&lt;/math&gt; are to be taken account as well.) Hence, &lt;math&gt;abc \sum_{abc = n | a,b,c \in \mathbb{N}} \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{n} \sum_{a | n} \frac{\tau(\frac{n}{a})}{a}&lt;/math&gt;. Since the set &lt;math&gt;\{d : d &gt; 0, d | n\}&lt;/math&gt; is the same as the set &lt;math&gt;\{\frac{n}{d} : d &gt; 0, d | n\}&lt;/math&gt;, &lt;math&gt;n \sum_{a | n} \frac{\tau(\frac{n}{a})}{a} = n \sum_{a | n} \frac{\tau(a)}{\frac{n}{a}} = \sum_{a | n} a \tau(a)&lt;/math&gt;, that is, &lt;math&gt;f(n) = 3 \sum_{a | n} a \tau(a)&lt;/math&gt; for all &lt;math&gt;n&lt;/math&gt;. <br /> <br /> Note that if &lt;math&gt;r&lt;/math&gt; and &lt;math&gt;s&lt;/math&gt; are relatively prime integers, then &lt;math&gt;\tau(rs) = \tau(r)\tau(s)&lt;/math&gt; and &lt;math&gt;rs = (r)(s)&lt;/math&gt;. Hence, <br /> &lt;cmath&gt;\sum_{a | rs} a \tau(a) = \left(\sum_{a | r} a \tau(a)\right) \left(\sum_{a | s} a \tau(a) \right),&lt;/cmath&gt; since every divisor of &lt;math&gt;rs&lt;/math&gt; can be expressed uniquely as a product of a divisor of &lt;math&gt;r&lt;/math&gt; and as a divisor of &lt;math&gt;s&lt;/math&gt; (upon expanding the product using the distributive property, it follows that we are summing over the set of all numbers of the form &lt;math&gt;r_1 s_1&lt;/math&gt;, where &lt;math&gt;r_1 | r&lt;/math&gt; and &lt;math&gt;s_1 | s&lt;/math&gt;, which is exactly the set divisors of &lt;math&gt;rs&lt;/math&gt;.) <br /> <br /> It follows that <br /> \begin{align*}<br /> 3 f(30^3) <br /> &amp;= 3 \sum_{d | 30^3} d \tau(d) \\<br /> &amp;= 3 \left(\sum_{d | 2^3} d \tau(d) \right) \left(\sum_{d | 3^3} d \tau(d) \right) \left(\sum_{d | 5^3} d \tau(d) \right) \\<br /> &amp;= 3 \left(1 \cdot 1 + 2 \cdot 2 + 3 \cdot 4 + 4 \cdot 8 \right) \left(1 \cdot 1 + 2 \cdot 3 + 3 \cdot 9 + 4 \cdot 27\right) \left( 1 \cdot 1 + 2 \cdot 5 + 3 \cdot 25 + 4 \cdot 125 \right) \\<br /> &amp;= 3 \cdot 49 \cdot 142 \cdot 586<br /> \end{align*}<br /> The last three digits of this product can easily be computed to be &lt;math&gt;\boxed{164}&lt;/math&gt;. <br /> <br /> <br /> %%Problem 12<br /> <br /> \setcounter{enumi}{11}<br /> Let &lt;cmath&gt;K = \sum_{a_1=0}^8\sum_{a_2=0}^{a_1}...\sum_{a_{100}=0}^{a_{99}}\binom{8}{a_1}\binom{a_1}{a_2}...\binom{a_{99}}{a_{100}}.&lt;/cmath&gt; Find the sum of digits of &lt;math&gt;K&lt;/math&gt; in base-100. (Tim Wu)<br /> <br /> <br /> %%Solution 12<br /> \begin{sol1}<br /> Let &lt;cmath&gt;f(n,k)=\sum_{a_1=0}^k \sum_{a_2=0}^{a_1}...\sum_{a_{n}=0}^{a_{n-1}}\binom{k}{a_1}\binom{a_1}{a_2}...\binom{a_{n-1}}{a_{n}}.&lt;/cmath&gt; We seek to find &lt;math&gt;f(100,8)&lt;/math&gt;.<br /> <br /> We shall prove by induction that &lt;math&gt;f(n,k) = (n+1)^k&lt;/math&gt;. Our base case, &lt;math&gt;n=1&lt;/math&gt;, follows from the fact that &lt;math&gt;f(1,k)= \sum_{a_1=0}^k \binom{k}{a_1}=2^k&lt;/math&gt; by the binomial theorem. Now, we note that &lt;math&gt;f(n,k)=\sum_{i=0}^{k} \binom{k}{i} f(n-1,i)&lt;/math&gt;. Assuming that &lt;math&gt;f(m,k) = (m+1)^k&lt;/math&gt; for all integers &lt;math&gt;m&lt;/math&gt; with &lt;math&gt;1 \leq m \leq n-1&lt;/math&gt;, we have that <br /> \begin{align*}<br /> f(n,k)&amp;=\sum_{i=0}^{k} \binom{k}{i} f(n-1,i) \\<br /> &amp;= \sum_{i=0}^k \binom{k}{i} n^i \\<br /> &amp;= (n+1)^k,<br /> \end{align*}<br /> where the last step follows from the binomial theorem. <br /> <br /> Thus, we have &lt;math&gt;f(100,8)=101^8&lt;/math&gt;. To express this in base-100, we notice that &lt;math&gt;101^8=(100+1)^8 = 100^8+\binom{8}{1}100^7+\binom{8}{2}100^6+\cdots+100^0&lt;/math&gt;. Since &lt;math&gt;\binom{8}{4} = 70 &lt; 100&lt;/math&gt;, there are no carry-overs,'' so the sum of digits in base-100 will simply be &lt;math&gt;\binom 80+\binom{8}1+\binom 82 + \cdots +\binom{8}{8} = 2^8=\boxed{256}&lt;/math&gt;.<br /> \end{sol1}<br /> <br /> \begin{sol2}<br /> Let &lt;math&gt;A_0 = \{1,2,3,4,5,6,7,8\}&lt;/math&gt;. The number of 101-tuples of sets &lt;math&gt;(A_{100}, A_{99}, \ldots, A_1, A_0)&lt;/math&gt;, such that &lt;math&gt;A_{100} \subseteq A_{99} \subseteq \ldots \subseteq A_1 \subseteq A_0&lt;/math&gt; is equal to &lt;math&gt;K&lt;/math&gt;; &lt;math&gt;\binom{8}{a_1}&lt;/math&gt; tells the number of ways to pick &lt;math&gt;a_1&lt;/math&gt; elements from &lt;math&gt;A_0&lt;/math&gt;, &lt;math&gt;a_2&lt;/math&gt; elements from &lt;math&gt;A_1&lt;/math&gt;, etc. <br /> <br /> We shall now count &lt;math&gt;K&lt;/math&gt; another way. Consider all 8-digit (leading digits can be 0) base 101 integers; there are clearly &lt;math&gt;101^8&lt;/math&gt; such integers. Let &lt;math&gt;A_i&lt;/math&gt; be the set of all &lt;math&gt;i&lt;/math&gt; such that the &lt;math&gt;i&lt;/math&gt;-th digit is greater than or equal to &lt;math&gt;i&lt;/math&gt;; then we have that &lt;math&gt;A_{100} \subseteq A_{99} \subseteq \ldots \subseteq A_1&lt;/math&gt;. Conversely, for any 101-tuple of sets &lt;math&gt;(A_{100}, A_{99}, \ldots, A_1, A_0)&lt;/math&gt; such that &lt;math&gt;A_{100} \subseteq A_{99} \subseteq \ldots \subseteq A_1 \subseteq A_0&lt;/math&gt;, we can construct an 8-digit base 101 integer from this set, by having the digit &lt;math&gt;i&lt;/math&gt; be the largest integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;i \in A_k&lt;/math&gt;. <br /> <br /> Hence, &lt;math&gt;K = 101^8&lt;/math&gt;. We proceed as we did before to find the sum of the digits of &lt;math&gt;K&lt;/math&gt; when it is written in base 100 to arrive at our answer of &lt;math&gt;\boxed{256}&lt;/math&gt;. <br /> \end{sol2}<br /> <br /> %%Problem 13<br /> <br /> \setcounter{enumi}{12}<br /> &lt;math&gt;\triangle ABC&lt;/math&gt; is inscribed in circle &lt;math&gt;\mathcal{C}&lt;/math&gt;. The radius of &lt;math&gt;\mathcal{C}&lt;/math&gt; is 285, and &lt;math&gt;BC = 567&lt;/math&gt;. When the incircle of &lt;math&gt;\triangle ABC&lt;/math&gt; is reflected across segment &lt;math&gt;BC&lt;/math&gt;, it is tangent to &lt;math&gt;\mathcal{C}&lt;/math&gt;. Given that the inradius of &lt;math&gt;\triangle ABC&lt;/math&gt; can be expressed in the form &lt;math&gt;a\sqrt{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers and &lt;math&gt;b&lt;/math&gt; is not divisible by the square of any prime, find &lt;math&gt;a+b&lt;/math&gt;. (Alex Zhu)<br /> <br /> <br /> %%Solution 13<br /> \begin{sol1}<br /> We shall prove that in a circle &lt;math&gt;\mathcal{C}&lt;/math&gt; with center &lt;math&gt;O&lt;/math&gt;, radius &lt;math&gt;R&lt;/math&gt;, a chord &lt;math&gt;BC&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt; such that &lt;math&gt;OM = m&lt;/math&gt;, and a point &lt;math&gt;A&lt;/math&gt; on &lt;math&gt;\mathcal{C}&lt;/math&gt; such that the incircle &lt;math&gt;\omega&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt; with radius &lt;math&gt;r&lt;/math&gt;, when reflected across &lt;math&gt;BC&lt;/math&gt;, is tangent to &lt;math&gt;\mathcal{C}&lt;/math&gt;, we have that &lt;math&gt;r = 4m&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, let &lt;math&gt;I'&lt;/math&gt; be the reflection of &lt;math&gt;I&lt;/math&gt; across &lt;math&gt;BC&lt;/math&gt; (hence, it is the center of the reflection of the incircle of &lt;math&gt;\triangle ABC&lt;/math&gt; across &lt;math&gt;BC&lt;/math&gt;), and let &lt;math&gt;O'&lt;/math&gt; be the reflection of &lt;math&gt;O&lt;/math&gt; across &lt;math&gt;BC&lt;/math&gt;. &lt;math&gt;IOO'I'&lt;/math&gt; is an isosceles trapezoid, so it is cyclic. Hence, by Ptolemy's theorem, &lt;math&gt;II' \cdot OO' + IO \cdot I'O' = O'I \cdot OI'&lt;/math&gt;. In other words, &lt;math&gt;(2r)(2m) + IO^2 = OI'^2&lt;/math&gt;, since &lt;math&gt;I'O' = IO&lt;/math&gt; and &lt;math&gt;O'I = OI'&lt;/math&gt;. But &lt;math&gt;OI' = R - r&lt;/math&gt;, since the reflection of &lt;math&gt;\omega&lt;/math&gt; across &lt;math&gt;BC&lt;/math&gt; is tangent to &lt;math&gt;\mathcal{C}&lt;/math&gt;, and &lt;math&gt;IO^2 = R^2 - 2Rr&lt;/math&gt; by Euler's distance formula, so &lt;math&gt;4rm + R^2 - 2Rr = R^2 - 2Rr + r^2&lt;/math&gt;, yielding &lt;math&gt;r = 4m&lt;/math&gt;. <br /> <br /> It therefore suffices to find &lt;math&gt;m&lt;/math&gt;. By the Pythagorean theorem, since &lt;math&gt;OMB&lt;/math&gt; is right, we have that &lt;math&gt;m^2 + R^2 = \frac{BC^2}{4}&lt;/math&gt;, so <br /> \begin{align*}<br /> m<br /> &amp;= \sqrt{285^2 - \frac{567^2}{4}} \\<br /> &amp;= \sqrt{\frac{570^2 - 567^2}{4}} \\<br /> &amp;= \frac{\sqrt{(570 - 567)(570 + 567)}}{2} \\<br /> &amp;= \frac{\sqrt{3 \cdot 1137}}{2} \\<br /> &amp;= \frac{3 \sqrt{379}}{2}.<br /> \end{align*}<br /> Hence, &lt;math&gt;r = 4m = 6 \sqrt{379}&lt;/math&gt;, giving us an answer of &lt;math&gt;6 + 379 = \boxed{385}&lt;/math&gt;. <br /> \end{sol1}<br /> <br /> \begin{sol2}<br /> We shall prove that in another way that a circle &lt;math&gt;\mathcal{C}&lt;/math&gt; with center &lt;math&gt;O&lt;/math&gt;, radius &lt;math&gt;R&lt;/math&gt;, a chord &lt;math&gt;BC&lt;/math&gt; with midpoint &lt;math&gt;M&lt;/math&gt; such that &lt;math&gt;OM = m&lt;/math&gt;, and a point &lt;math&gt;A&lt;/math&gt; on &lt;math&gt;\mathcal{C}&lt;/math&gt; such that the incircle &lt;math&gt;\omega&lt;/math&gt; of &lt;math&gt;\triangle ABC&lt;/math&gt; with radius &lt;math&gt;r&lt;/math&gt;, when reflected across &lt;math&gt;BC&lt;/math&gt;, is tangent to &lt;math&gt;\mathcal{C}&lt;/math&gt;, we have that &lt;math&gt;r = 4m&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;I&lt;/math&gt; be the incenter of &lt;math&gt;\triangle ABC&lt;/math&gt;, and let &lt;math&gt;I'&lt;/math&gt; be the reflection of &lt;math&gt;I&lt;/math&gt; across &lt;math&gt;BC&lt;/math&gt; (hence, it is the center of the reflection of the incircle of &lt;math&gt;\triangle ABC&lt;/math&gt; across &lt;math&gt;BC&lt;/math&gt;.) Let &lt;math&gt;E&lt;/math&gt; be the foot of the perpendicular from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;II'&lt;/math&gt;. We have that &lt;math&gt;IE = r - m&lt;/math&gt; and &lt;math&gt;OI^2 = R^2 - 2Rr&lt;/math&gt; by Euler's distance formula, so &lt;math&gt;OE^2 = R^2 - 2Rr - (r-m)^2 = R^2 - 2Rr + 2rm - r^2 - m^2&lt;/math&gt;. In addition, &lt;math&gt;EI' = m + r&lt;/math&gt; and &lt;math&gt;OI' = R - r&lt;/math&gt;, since the reflection of &lt;math&gt;\omega&lt;/math&gt; across &lt;math&gt;BC&lt;/math&gt; is tangent to &lt;math&gt;\mathcal{C}&lt;/math&gt;. Sine &lt;math&gt;\triangle OEI'&lt;/math&gt; is right, we have that &lt;math&gt;OE^2 + EI'^2 = OI'^2&lt;/math&gt;, that is, &lt;math&gt;R^2 - 2Rr + 2rm - r^2 - m^2 + m^2 + 2rm + r^2 = R^2 - 2Rr + r^2&lt;/math&gt;, again yielding &lt;math&gt;r = 4m&lt;/math&gt;. <br /> <br /> We simply compute the answer the way we did before to arrive at an answer of &lt;math&gt;\boxed{385}&lt;/math&gt;. <br /> \end{sol2}<br /> <br /> %%Problem 14<br /> <br /> \setcounter{enumi}{13}<br /> Alex and Mitchell decide to play a game. In this game, there are 2010 pieces of candy on a table, and starting with Alex, the two take turns eating some positive integer number of pieces of candy. Since it is bad manners to eat the last candy, whoever eats the last candy loses. The two decide that the amount of candy a person can pick will be a set equal to the positive divisors of a number less than 2010 that each person picks (individually) from the beginning. For example, if Alex picks 19 and Mitchell picks 20, then on each turn, Alex must eat either 1 or 19 pieces, and Mitchell must eat 1, 2, 4, 5, 10, or 20 pieces. Mitchell knows Alex well enough to determine with certainty that Alex will either be immature and pick 69, or be clich\'ed and pick 42. How many integers can Mitchell pick to guarantee that he will not \emph{lose the game}? (George Xing)<br /> <br /> <br /> %%Solution 14<br /> <br /> We claim that Mitchell has a winning strategy if and only if the number he picks is a multiple of 12. To show that he has a winning strategy whenever the number he picks is a multiple of 5, it is clearly sufficient to show that Mitchell has a winning strategy when Mitchell's number is 12. <br /> <br /> Note that Mitchell has 1, 2, 3, and 4 among his divisors, and that no matter which number Alex chooses, Alex will not have a multiple of 4. Therefore, after Alex's first turn, let Mitchell choose one of his divisors so that the remaining number of pieces of candy on the table is congruent to 1 modulo 4. As Alex cannot decrease the amount of candy by a multiple of 4, he cannot force Mitchell to lose (as Mitchell can only be forced to lose if he is left with 1 piece of candy on the table.) Because Mitchell cannot be forced to lose, Mitchell has a winning strategy. <br /> <br /> We shall now show that if Mitchell does not pick a multiple of 3, then Mitchell cannot guarantee a victory. Alex can pick 42, which have 1, 2, and 3 among its divsiors. Therefore, Alex can play so that he always forces Mitchell to carry out his move when the number of pieces remaining on the table is congruent to 1 modulo 3. As the number of pieces of candy that Mitchell removes cannot be a multiple of 3, Alex will never have to move on a table with the number of pieces on the table congruent to 1 modulo 3. Hence, Alex cannot be forced to lose, so Alex has a winning strategy, and Mitchell does not. <br /> <br /> We shall now show that if Mitchell does not pick a multiple of 4, then he cannot guarentee a victory. Alex can pick 42, which have 1, 2, and 3 among its divisors. On Alex's first turn, he can pick a 1. Thereon, as Mitchell cannot pick a number that is a multiple of 4, Alex can play so that after his move, the number of pieces of candy remaining on the board is congruent to 1 modulo 4. Again, as Mitchell cannot pick a number that is a multiple of 4, Alex will never have to make his move when the number of pieces of candy remaining is congruent to 1 modulo 4, so he can never be forced to lose. Hence, he has a winning stratgecy, and Mitchell does not. <br /> <br /> First, let us consider what happens if Alex picks 69. If Mitchell picks an odd number, then each person can only take odd numbers of candies. This means that on Alex's turn, he will always have an even amount left, so he will always be able to move, whereas Mitchell will always have an odd number left, and eventually be forced to take the last candy. However, if Mitchell picks an even number, then he can take an even amount of candies on his first turn and force Alex to always have an odd amount of candies, forcing him to lose. Therefore, Mitchell must pick an even number.<br /> <br /> With this information, our answer simply becomes the number of multiples of 12 below 2010, which is &lt;math&gt;\left\lfloor \frac{2010}{12} \right\rfloor = \boxed{167}&lt;/math&gt;.<br /> <br /> \emph{Remark: }I lost. <br /> <br /> <br /> %%Problem 15<br /> <br /> \setcounter{enumi}{14}<br /> Given that &lt;math&gt;\sum_{a = 1}^{491}\sum_{b = 1}^{491} \frac{1}{(a+bi)^4 - 1}&lt;/math&gt; can be expressed in the form &lt;math&gt;\frac{p + qi}{r}&lt;/math&gt;, where &lt;math&gt;\gcd(p, q, r) = 1&lt;/math&gt; and &lt;math&gt;r &gt; 0&lt;/math&gt;, find the unique integer &lt;math&gt;k&lt;/math&gt; between 0 and 982 inclusive such that &lt;math&gt;983&lt;/math&gt; divides &lt;math&gt;(p + q) - kr&lt;/math&gt;. [Note: 983 is a prime.] (Alex Zhu)<br /> <br /> <br /> %%Solution 15<br /> <br /> Let &lt;math&gt;S_1 = S&lt;/math&gt;, &lt;math&gt;S_2 = \{iz : z \in S\}&lt;/math&gt;, &lt;math&gt;S_3 = \{i^2 z : z \in S\}&lt;/math&gt;, &lt;math&gt;S_4 = \{i^3 z : z \in S\}&lt;/math&gt;, and let &lt;math&gt;S' = S_1 \cup S_2 \cup S_3 \cup S_4&lt;/math&gt;. Note that the following: <br /> <br /> The sets &lt;math&gt;\{z^4 : z \in S_i\}&lt;/math&gt;, for &lt;math&gt;1 \leq i \leq 4&lt;/math&gt;, are equal (since &lt;math&gt;i^4 = 1&lt;/math&gt;.)<br /> &lt;math&gt;\{iz : z \in S'\} = \{z : z \in S'\}&lt;/math&gt;, since &lt;math&gt;z \in S'&lt;/math&gt; if and only if &lt;math&gt;iz \in S'&lt;/math&gt;. <br /> The sets &lt;math&gt;\{z^2 : z \in S_1 \mbox{ or } z \in S_4\}&lt;/math&gt; is equal to the set &lt;math&gt;\{z^2 : z \in S_2 \mbox{ or } z \in S_3 \}&lt;/math&gt;, since &lt;math&gt;z \in S_1&lt;/math&gt; or &lt;math&gt;z \in S_4&lt;/math&gt; if and only if &lt;math&gt;-z \in S_2&lt;/math&gt; or &lt;math&gt;-z \in S_3&lt;/math&gt;. <br /> &lt;math&gt;S_4 = \{\bar{z} : z \in S_1\}&lt;/math&gt;. This is because &lt;math&gt;S_4&lt;/math&gt; is the set of all complex numbers of the form &lt;math&gt;i(a + bi) = -b + ai&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers between 1 and 491 inclusive, which is the same as the set of all complex numbers of the form &lt;math&gt;a - bi&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers between 1 and 491 inclusive. The latter set is clearly comprised of exactly the conjugates of the elements of &lt;math&gt;S_1&lt;/math&gt;. <br /> <br /> It follows that<br /> <br /> \begin{align*}<br /> \sum_{z \in S} \frac{1}{z^4 - 1} <br /> &amp;= \frac{1}{4} \sum_{z \in S'} \frac{1}{z^4 - 1} \\ <br /> &amp;= \frac{1}{8} \sum_{z \in S'} \left(\frac{1}{z^2 - 1} - \frac{1}{z^2 + 1} \right) \\<br /> &amp;= \frac{1}{8} \left(\sum_{z \in S'} \frac{1}{z^2 - 1} - \sum_{z \in S'} \frac{1}{z^2 + 1} \right) \\<br /> &amp;= \frac{1}{8} \left(\sum_{z \in S'} \frac{1}{z^2 - 1} - \sum_{z \in S'} \frac{1}{(iz)^2 + 1} \right) \\<br /> &amp;= \frac{1}{8} \left(\sum_{z \in S'} \frac{1}{z^2 - 1} - \sum_{z \in S'} \frac{1}{-z^2 + 1}\right) \\<br /> &amp;= \frac{1}{4} \left(\sum_{z \in S'} \frac{1}{z^2 - 1}\right) \\<br /> &amp;= \frac{1}{2} \left(\sum_{z \in S_1 \cup S_4} \frac{1}{z^2 - 1}\right) \\ <br /> &amp;= \sum_{z \in S_1} \frac{\frac{1}{z^2 - 1} + \frac{1}{\bar{z}^2 - 1}}{2} \\<br /> &amp;= \sum_{z \in S_1} \frac{\frac{1}{z^2 - 1} + \overline{\left(\frac{1}{z^2 - 1}\right)}}{2} \\<br /> &amp;= \sum_{z \in S} \mbox{Re } \frac{1}{z^2 - 1} \\<br /> &amp;= \frac{1}{2}\mbox{Re} \left(\sum_{z \in S} \frac{1}{z-1} - \frac{1}{z+1} \right) \\<br /> &amp;= \frac{1}{2} \mbox{Re} \left(\sum_{b=1}^{491} \sum_{a=1}^{491} \frac{1}{(a-1) + bi} - \frac{1}{(a+1) + bi}\right) \\<br /> &amp;= \frac{1}{2} \mbox{Re} \left(\sum_{b=1}^{491} \left(\frac{1}{bi} + \frac{1}{1 + bi} - \frac{1}{491 + bi} - \frac{1}{492 + bi} \right)\right) \\<br /> &amp;= \frac{1}{2} \sum_{b=1}^{491} \left(\mbox{Re} \left(\frac{1}{bi} + \frac{1}{1 + bi} - \frac{1}{491 + bi} - \frac{1}{492 + bi} \right)\right) \\<br /> &amp;= \frac{1}{2} \left(\sum_{b=1}^{491} \left(\frac{1}{1 + b^2} - \frac{491}{491^2 + b^2} - \frac{492}{492^2 + b^2}\right) \right)<br /> \end{align*}<br /> We seek the integer &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;983 | p - qk&lt;/math&gt;, that is, we seek the value of &lt;math&gt;\frac{p}{q}&lt;/math&gt; modulo 983, that is, the value of &lt;math&gt;pq^{-1}&lt;/math&gt; modulo 983. Since the product of a quadratic residue and a quadratic nonresidue is a quadratic nonresidue, and -1 is a quadratic residue mod 983 (as 983 is congruent to 3 modulo 4), it follows that &lt;math&gt;b^2 + 1&lt;/math&gt;, &lt;math&gt;b^2 + 492^2&lt;/math&gt;, and &lt;math&gt;b^2 + 491^2&lt;/math&gt; are never divisible by &lt;math&gt;983&lt;/math&gt;, so the denominators of the above sum is never divisible by &lt;math&gt;983&lt;/math&gt;. (Alternatively, upon noticing that &lt;math&gt;r \neq 0&lt;/math&gt;, the problem statement implies that none of the denominators are divisible by &lt;math&gt;983&lt;/math&gt;.) As the above sum that we have arrived at is real, and none of its denominators are divisible of &lt;math&gt;983&lt;/math&gt;, we may manipulate the sum modulo 983. <br /> <br /> We note that &lt;math&gt;491 \equiv -492 \pmod{983}&lt;/math&gt;, so &lt;math&gt;491^2 \equiv 492^2 \pmod{983}&lt;/math&gt;, yielding &lt;math&gt;\frac{491}{491^2 + b^2} + \frac{492}{492^2 + b^2} \equiv \frac{491 + 492}{491^2 + b^2} = \frac{983}{491^2 + b^2} \equiv 0 \pmod{983}&lt;/math&gt;. Hence, our sum taken modulo &lt;math&gt;p&lt;/math&gt; is just &lt;math&gt;\frac{1}{2} \sum_{a=1}^{491} \frac{1}{1 + b^2}&lt;/math&gt; mod &lt;math&gt;p&lt;/math&gt;. Let this sum equal &lt;math&gt;T&lt;/math&gt;. We have that <br /> \begin{align*}<br /> 8T <br /> &amp;\equiv 4\sum_{b=1}^{491} \frac{1}{1 + b^2} \\<br /> &amp;\equiv 2\left(\sum_{b=1}^{491} \frac{1}{1 + b^2} + \sum_{b=1}^{491} \frac{1}{1 + (983 - b)^2}\right) \\<br /> &amp;\equiv 2\left(\sum_{b=1}^{491} \frac{1}{1 + b^2} + \sum_{b=492}^{982} \frac{1}{1+b^2}\right) \\<br /> &amp;\equiv 2 \sum_{b=1}^{982} \frac{1}{1+b^2} \pmod{983}.<br /> \end{align*}<br /> Since &lt;math&gt;\{\frac{1}{1}, \frac{1}{2}, \ldots, \frac{1}{p-1}\}&lt;/math&gt; is a permutation of &lt;math&gt;\{1, 2, \ldots, p-1\}&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt;, we have that <br /> \begin{align*}<br /> 2 \sum_{b=1}^{982} \frac{1}{1+b^2} <br /> &amp;\equiv \sum_{b=1}^{982} + \sum_{b=1}^{982} \frac{1}{1 + \left(\frac{1}{b}\right)^2} \\<br /> &amp;\equiv \sum_{b=1}^{982} \frac{1}{b^2+1} + \sum_{b=1}^{982} \frac{b^2}{b^2 + 1} \\<br /> &amp;\equiv \sum_{b=1}^{982} 1 \\<br /> &amp;= 982 \pmod{983}. <br /> \end{align*}&lt;math&gt;<br /> Hence, <br /> \begin{align*}<br /> T <br /> &amp;\equiv \frac{982}{8} \\<br /> &amp;\equiv \frac{491}{4} \\ <br /> &amp;\equiv \frac{491 + 983}{4} \\<br /> &amp;\equiv \frac{737}{2} \\<br /> &amp;\equiv \frac{737 + 983}{2} \\<br /> &amp;\equiv 860 \pmod{983}<br /> \end{align*}<br /> giving us an answer of &lt;/math&gt;\boxed{860}\$.</div> Relay400