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Difference between revisions of "Mock AIME II 2012 Problems/Problem 1"

(Created page with "== Problem== Given that <cmath>\left(\dfrac{6^2-1}{6^2+11}\right)\left(\dfrac{7^2-2}{7^2+12}\right)\left(\dfrac{8^2-3}{8^2+13}\right)\cdots\left(\dfrac{2012^2-2007}{2012^2+2017}\...")
 
(→‎Solution)
 
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<cmath>\frac{6^2-1}{2012^2+2017}</cmath>
 
<cmath>\frac{6^2-1}{2012^2+2017}</cmath>
  
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Note that <math>5|(6^2-1)</math> however <math>2012^2+2017\equiv 4+2\equiv 1\pmod{5}</math>.  Also, note that <math>7|(6^2-1)</math> however <math>2012^2+2017\equiv 25+4\equiv 1\pmod{7}</math>.  Since <math>6^2-1=5*7</math>, we know that <math>\gcd(6^2-1, 2012^2+2017)=1</math>.  Now note that we want <math>35+2012^2+2017 \pmod{1000}</math>, therefore we use <math>2012^2\equiv 12\pmod{1000}</math> and <math>2017\equiv 17\pmod{1000}</math> to give us <math>m+n=35+144+17=\boxed{196}\pmod{1000}</math>.
+
Note that <math>5|(6^2-1)</math> however <math>2012^2+2017\equiv 4+2\equiv 1\pmod{5}</math>.  Also, note that <math>7|(6^2-1)</math> however <math>2012^2+2017\equiv 25+4\equiv 1\pmod{7}</math>.  Since <math>6^2-1=5*7</math>, we know that <math>\gcd(6^2-1, 2012^2+2017)=1</math>.  Now note that we want <math>35+2012^2+2017 \pmod{1000}</math>, therefore we use <math>2012^2\equiv 12^2\pmod{1000}</math> and <math>2017\equiv 17\pmod{1000}</math> to give us <math>m+n=35+144+17=\boxed{196}\pmod{1000}</math>.

Latest revision as of 09:53, 13 March 2013

Problem

Given that \[\left(\dfrac{6^2-1}{6^2+11}\right)\left(\dfrac{7^2-2}{7^2+12}\right)\left(\dfrac{8^2-3}{8^2+13}\right)\cdots\left(\dfrac{2012^2-2007}{2012^2+2017}\right)=\dfrac{m}{n},\] where $m$ and $n$ are positive relatively prime integers, find the remainder when $m+n$ is divided by $1000$.

Solution

Consider $\frac{k^2-(k-5)}{k^2+(k+5)}$. We note that $(k+1)^2-(k+1-5)=k^2+(k+5)$, thus we have a telescoping sequence and we need only consider the first numerator and last denominator.

\[\frac{6^2-1}{2012^2+2017}\]

Note that $5|(6^2-1)$ however $2012^2+2017\equiv 4+2\equiv 1\pmod{5}$. Also, note that $7|(6^2-1)$ however $2012^2+2017\equiv 25+4\equiv 1\pmod{7}$. Since $6^2-1=5*7$, we know that $\gcd(6^2-1, 2012^2+2017)=1$. Now note that we want $35+2012^2+2017 \pmod{1000}$, therefore we use $2012^2\equiv 12^2\pmod{1000}$ and $2017\equiv 17\pmod{1000}$ to give us $m+n=35+144+17=\boxed{196}\pmod{1000}$.

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