What is the remainder when \(32^{30^{32}}\) is divided by 9?
Answer: A
We want to find \(32^{30^{32}} \pmod{9}\).
Step 1: Simplify the base. \(32 \equiv 5 \pmod{9}\). The problem becomes finding \(5^{30^{32}} \pmod{9}\).
Step 2: Use Euler's totient theorem. We need to find the exponent modulo \(\phi(9)\). \(\phi(9) = 9(1 - 1/3) = 6\). So we need to find the remainder of the exponent \(30^{32}\) when divided by 6.
Step 3: Calculate the exponent's remainder. \(30^{32} \pmod{6}\). Since 30 is a multiple of 6, \(30 \equiv 0 \pmod{6}\). Therefore, \(30^{32} \equiv 0 \pmod{6}\) for any power \(>0\).
Step 4: A remainder of 0 in the exponent means we should use the totient value itself, which is 6 (as the cycle repeats). So, the problem reduces to finding \(5^6 \pmod{9}\). Since 5 and 9 are coprime, by Euler's theorem, \(5^{\phi(9)} = 5^6 \equiv 1 \pmod{9}\). The remainder is 1.