, 2012

Solution:

For a positive integer $n$, let $s(n)$ be the sum of digits when $n$ is expressed in base 12. We claim that $s(n) \equiv n \pmod{11}$ for all positive integers $n$. Indeed, if $n = d_{k} 12^{k} + d_{k-1} 12^{k-1} + \cdots + d_{0}$ with each $d_{i}$ an integer between 0 and 11, inclusive, because $12 \equiv 1 \pmod{11}$, reducing modulo 11 gives exactly $s(n)$. Thus, our answer is congruent to $N = 7^{6^{5^{4^{3^{2^{1}}}}}}$ modulo 11, and furthermore must be a one-digit integer in base 12; these two conditions uniquely determine the answer.

By Fermat's Little Theorem, $7^{10} \equiv 1 \pmod{11}$, and also observe that $6^{5^{4^{3^{2^{1}}}}} \equiv 6 \pmod{10}$ because $6 \cong 0 \pmod{2}$ and $6 \cong 1 \pmod{5}$. Thus, $N \equiv 7^{6} \equiv 343^{2} \equiv 2^{2} \equiv 4 \pmod{11}$, which is our answer. (Additionally, we note that this process of writing the number in base twelve and summing the digits must eventually terminate because the value decreases after each step.)