Hey Greg, can you help me with this? I used the pattern method and got remainders 2, 0, 0, 0, so I said the answer was 2, but it’s actually 0. Can you help me figure out what went wrong? Sorry to bother.
6^{29} = 36 \cdot 6^{27} = 4 \cdot \underbrace{9(6^{27})}_k = 4k
6 divided by 4 leaves a remainder of 2
6^29 behaves like 2^29 when divided by 4
2^1 divided by 4 → remainder 2
2^2 divided by 4 → remainder 0
2^3 divided by 4 → remainder 0
2^4 divided by 4 → remainder 0
From 2^2 onward, every power of 2 is divisible by 4
So 2^29 is divisible by 4 → remainder 0
→ 6^29 is divisible by 4 → remainder 0