
Number Theory
Let n and a be positive integers and let p be a prime number. The p^a is said to exactly divide n. If p^a exactly divides n and p^a+1 does not exactly divide n. Assume p^a exactly divides m and p^b exactly divides n.
What power of p exactly divides m+n. Prove.

Lemma: If $\displaystyle p^a$ exactly divides n, then no greater power of p divides n, and conversely, if $\displaystyle p^a  n$ and no greater power of p divides n, then $\displaystyle p^a$ exactly divides n.
The answer to your question is obviously a + b, but why? Can you use the lemma to show how?

yes you can use the lemma.

It is clear that if $\displaystyle p^a  m$ and $\displaystyle p^b  n$, then $\displaystyle p^{(a+b)}  mn$. Suppose $\displaystyle p^{(a+b+k)}  mn$ for some positive integer k and $\displaystyle p^a, p^b$ both exactly divide m and n, respectively. Then since a is the greatest power of p that divides m, it must be true that $\displaystyle p^{(b+k)}  n$, which is impossible by the lemma. Therefore, $\displaystyle p^{(a+b)}$ divides mn exactly.
Edit: never mind, I had the question wrong.

I don't think you have enough information.
Consider the following series of integers m, n:
$\displaystyle m = p(p  1), n = p, m + n = p^2$
$\displaystyle m = p(p^2  1), n = p, m + n = p^3$
$\displaystyle m = p(p^3  1), n = p, m + n = p^4$
etc.
In each case it is clear that p exactly divides both m and n, but the power of p that exactly divides m + n is not determined by this fact.