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Difference between revisions of "Mock AIME II 2012 Problems"
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==Problem 1==
 
==Problem 1==
 
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}\right)=\dfrac{m}{n},</cmath> where <math>m</math> and <math>n</math> are positive relatively prime integers, find the remainder when <math>m+n</math> is divided by <math>1000</math>.
 
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}\right)=\dfrac{m}{n},</cmath> where <math>m</math> and <math>n</math> are positive relatively prime integers, find the remainder when <math>m+n</math> is divided by <math>1000</math>.

Revision as of 23:57, 4 April 2012


Problem 1

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

Problem 2

Let $\{a_n\}$ be a recursion defined such that $a_1=1, a_2=20$, and $a_n=\sqrt{\left| a_{n-1}^2-a_{n-2}^2 \right|}$ where $n\ge 3$, and $n$ is an integer. If $a_m=k$ for $k$ being a positive integer greater than $1$ and $m$ being a positive integer greater than 2, find the smallest possible value of $m+k$.

Solution

Problem 3

The $\textit{digital root}$ of a number is defined as the result obtained by repeatedly adding the digits of the number until a single digit remains. For example, the $\textit{digital root}$ of $237$ is $3$ ($2+3+7=12, 1+2=3$). Find the $\textit{digital root}$ of $2012^{2012^{2012}}$.

Solution

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