Discussion
What is the remainder when \(32^{32^{32}}\) is divided by 9?
Answer: A
Correct answer: 4
Reduce the base first: \(32 \equiv 5 \pmod 9\), so the problem becomes \(5^{32^{32}} \bmod 9\).
The powers of 5 modulo 9 run in a cycle of length 6:
\[5,\ 7,\ 8,\ 4,\ 2,\ 1,\ 5,\ 7,\ \ldots\]
So only the exponent modulo 6 matters. Since \(32 \equiv 2 \pmod 6\), we need \(2^{32} \bmod 6\). Powers of 2 alternate between 2 and 4 modulo 6, and an even exponent gives 4, so \(32^{32} \equiv 4 \pmod 6\).
The fourth entry of the cycle is 4, so \(32^{32^{32}} \equiv 5^4 = 625 \equiv 4 \pmod 9\).
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