What is the largest number with exactly 3 factors that divides n! ? 1)

Not sure how the tags are generated for these questions, but this certainly isn't a sub-600 level problem. It should be a 700+ level problem, and it's also too awkward (and time-consuming) for the GMAT.

If a number has exactly 3 divisors, it must be the square of a prime, p^2 (which has the three divisors 1, p and p^2). We might first look at some random factorial, like 50!, say. The largest square of a prime that divides 50! will be 23^2, because if we write out the product 50! in full, we'll find both 23 and 46 = 2(23) in there, so we'll have two 23's. We won't be able to divide 50! by any larger square of a prime, because, considering the next larger prime, we'll only have one 29 (since we'd need to have 58 in our product to get another 29), and the same will be true for larger primes. So n! will only be divisible by p^2, where p is prime, if p < n/2.

If 47 divides n!, then n must be at least 47. If 47 is the largest prime factor of n!, then n cannot be 53 or greater, because then n! would be divisible by 53 too. So Statement 1 tells us 46 < n < 53. In each case, the largest prime less than or equal to n/2 is 23, so Statement 1 is sufficient.

If n! ends in 12 zeros, then n! is divisible by 10^12 = (2^12)(5^12). For any value of n > 1, n! will be divisible by a lot more 2s than 5s, so if n! is divisible by 5^12, it will always be divisible by 10^12. If you consider 50!, we'll have ten multiples of 5 if we write out our product in full, so 50! is divisible by at least 5^10. But we also have a 25 and a 50 in that product, which give us two extra 5's. So 50! is divisible by 5^12, and so are the factorials from 51! through 54!. Once we get to 55!, we'll have a number divisible by 10^13 though, so Statement 2 tells us 49 < n < 55. Again, in each case the largest prime less than or equal to n/2 is 23, so Statement 2 is also sufficient.