How do I solve this?

1. Find all integers n such that n! is divisible by 716 but not by 717.
   

Hey evthim.

First you need to know what the factors are for 716 and also for 717.

We know that if you have something as a product of primes, then something can only be divided into something else (factored) if they have the same primes (that have a power of 1 or more) and you can only divide one thing by another if the denominator has the prime powers which are less than or equal to those in the numerator.

Now given that you need n! and 716/717 to have the same prime factors to be divisible, what can you conclude about n and what it should be?