Remainder when $64^{32^{32}}$ is divided by 9 is equal to ______.
Step-by-Step Solution
Key Concept: Write $64=8^2$ so $64^{32^{32}}=8^{2\cdot32^{32}}$. Then note $64=(9-1)$, so $64^t=(9-1)^t\equiv(-1)^t\pmod{9}$. Here $t=32^{32}$ is even, so remainder is 1.
Let $32^{32}=t$. $64^t = (9-1)^{2t}$.
By binomial theorem $(9-1)^{2t}=9k+1$ for some integer $k$.
Hence remainder $= 1$.
Correct Answer: 1