## Prime number dividing fractions.

Let p be a prime number.
Let A be an integer divisible by p but B be an integer not be divisible by p.
Let A/B be an integer.

How do I prove that A/B is divisible by p?

This sounds like a simple question but I just can't get it. I'm doing it in relation to proving Fermat's little theorem. (a^p = a mod p for all integers a) I'm trying to understand why the binomial coefficients in the binomial expansion of (1+a)^n are all divisible by p (=0 mod p) for all the terms with powers [1, p-1].

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 Recognitions: Science Advisor Decompose A and B into prime factors. Since A/B is an integer all the B factors cancel A factors. However B did not contain p, so p remains a factor in A/B.
 Thank you, mathman! That instantly resolved my question (and I was banging my head against it for like an hour)! Damn that fundamental theorem of arithmetic!

 Tags binomial, number theory