Bunuel wrote:

For integers x and y, 1 < x < y. Is x^y a factor of 11!?

(1) x > 3

(2) x is prime

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:For this problem, it's helpful to factor out 11! to see what factors it will contain. 11! = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1.

With that, you should see relatively quickly that certain values of x cannot be taken to an exponent greater than 1 and still remain a factor of 11!. 11 and 7, for example, are prime numbers that only appear once. So because the base of the exponent (x) has to be taken to an exponent greater than itself (y), those numbers are certainly eliminated.

So with that in mind, you'll need a relatively low value of x so that you can maintain a relatively low exponent, too. 2 would work (in 11!, 10 (which is 5 * 2), 8 (which is 2 * 2 * 2), 6 (which is 2 * 3), 4 (which is 2 * 2), and 2 all give you factors of 2, so you can take 2 all the way to the 8th power and it will still work). And 3 would work (3, 6, and 9 - which is 3 * 3 and therefore gives you two 3s - are part of 11!, so you can take 3 to the 4th power, too). But that's it.

So when statement 1 tells you that the lowest possible value of x is 4, then you should consider 4 to the 5th. 4 can be expressed as 2 * 2, so that leaves you with 2 to the 10th power - and since there are only eight 2s contained within 11!, you cannot have x = 4 and y = 5 as options. So

statement 1 is sufficient.

Statement 2 is not sufficient, as it allows for x = 2 and y = "anything between 2 and 9" (all of which would give you "yes") but also allows for a prime number of x like 11 (which would mean "no").