Mock AIME II 2012 Problems/Problem 3

Problem

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

The sum of the digits when we get this down to a one digit number is the same thing as the number modulo 9. This is because let’s say we start with a number $n$ . When we take the sum of the digits modulo 9, this is the same thing as taking the number modulo 9. However, the sum of the digits modulo 9 is the same thing as the sum of the sum of the digits modulo 9, and this pattern repeats.

We are left with finding $2012^{2012^{2012}}\pmod{9}$ . Note that $2012=5\pmod{9}$ , therefore we get $5^{2012^{2012}}\pmod{9}$ . By Euler’s Totient Theorem, $5^{\phi(9)}\equiv 1\pmod{9}$ or $5^6\equiv 1\pmod{9}$ . We now need to find $2012^{2012}\pmod{6}$ , which is equal to $2^{2012}\pmod{6}$ . Note that $2^1\equiv 2\pmod{6}, 2^2\equiv 4\pmod 6$ , and $2^3\equiv 2\pmod 6$ , so this pattern repeats $\pmod 2$ in the exponents. Since $2012\equiv 0\pmod2$ , we get $2^{2012}\pmod 6\equiv 4\pmod 6$ . From this, we get $2012^{2012}\pmod{6}\equiv 4\pmod{6}$ , so we need to find $5^{4}\pmod{9}$ . Since $5^4=625$ , we get an answer of $6+2+5\pmod 9=\boxed{004}\pmod 9$ .

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Mock AIME II 2012 Problems/Problem 3

Problem

Solution