# Thread: 2 Induction Problems

1. ## 2 Induction Problems

1. Prove, by induction, that for every positive integer n, $4^n +14 \equiv 0 (mod 6)$

Base case, n=1, $6 \mid 18-0$

Assume: $4^{n+1} +14 \equiv 0 (mod 6)$

Then I could write is as $4^n * 4 +14 \equiv 0 (mod 6)$

I have no idea how to proceed.

2. If we assume $4^n + 14 \equiv 0 (mod 6)$ that means $4^n \equiv 4 mod 6$

So times it by 4 for $4^{n+1} \equiv 4 \times 4 \equiv 4 (mod 6)$ and there you go.

3. Originally Posted by Matt Westwood
If we assume $4^n + 14 \equiv 0 (mod 6)$ that means $4^n \equiv 4 mod 6$

So times it by 4 for $4^{n+1} \equiv 4 \times 4 \equiv 4 (mod 6)$ and there you go.
so all we cares about is the 4^n, it is always divisible by 6, the constant doesn't matter?

4. No, 4^n always leaves a remainder 4 when divided by 6. As 14 leaves a remainder 2 when divided by 6, so add them together and the sum is then divisible by 6.