# Mathmatical Induction Problem (Divisibility)

1. May 10, 2007

### Kb1jij

1. The problem statement, all variables and given/known data
Use Mathematical Induction to prove that $$12^n + 2(5^{n-1})$$ is divisible by 7 for all $$n \in Z^+$$

2. Relevant equations

3. The attempt at a solution

First, show that it works for n = 1:
$$12^1 + 2 \cdot 5^0 = 14$$ , 14/7 = 2

Next assume:
$$12^k + 2(5^{k-1}) = 7A$$

Then, prove for k + 1:
$$12^{k+1} + 2(5^k)$$

I can't figure out how to prove this. I know that this can be changed to:
$$12 \cdot 12^{k} + 2 \cdot 5 (5^{k-1})$$
But that doesn't seem to help me much.

I also tried substituting values for 12^k and 5^(k-1) from above:
$$12^k = 7A - 2(5^{k-1})$$
$$2(5^{k-1}) = 7A - 12^k$$

This doesn't seem too help either, I can reduce it to:
$$189A - (12 \cdot 2(5^{k-1})+5(12^k))$$

Any suggestions?
Thanks,
Tom

2. May 10, 2007

### neutrino

Actually, having come till

$$12.12^k + 2.5(5^{k-1})$$,

the next step should have been

$$7.12^k + 5.12^k + 2.5(5^{k-1})$$.

3. May 10, 2007

### Kb1jij

Ah, got it now. Thank you. I don't like these induction problems...