Data Sufficiency

If n = p/q and p, q are integers, is n an integer?

1) n^2 is an integer
2) n^3 is an integer

[spoiler]OA is D, but can someone please explain how/why? Thanks[/spoiler]

n = integer / integer -> n is rational, hence if n^a (a is positive integer) is integer, then n is integer also.

capnx wrote:If n = p/q and p, q are integers, is n an integer?

1) n^2 is an integer
2) n^3 is an integer

[spoiler]OA is D, but can someone please explain how/why? Thanks[/spoiler]

Hi capnx,

Try rephrasing each part, and hopefully it will make sense. This question is essentially asking you if p is divisable by q. In other words, does p have all the prime factors of q?

1) Since the original statement tells us that n = p/q, we can extrapolate that n^2 = (p^2)/(q^2) = (p*p)/(q*q). If n^2 is an integer, we know that (p*p)/(q*q) is an integer. In other words, p*p holds all the prime factors of q*q. If we know this, then we know that p holds all the prime factors of q.

Testing this out with numbers should help you see it if it's not already clear. If we let p^2=36 and q^2=9, n^2=4 (36/9=4). This is easy enough to do in our head, but to get the theory, let's look at the prime factorization:

p^2 = 36 = 2*2*3*3
q^2 = 9 = 3*3
(2*2*3*3)/(3*3) = 2*2 = 4

Now, if we take the square root of this equation (to tie this back to the original question, is p/q an integer?) we get:

p = 6 = 2*3
q = 3
(2*3)/(3) = 2

If you try this out with any set of perfect squares that meet the criteria of statement 1, you'll always end up with the same results. Once you know you'll always end up with the same results, no need to try it out . This statement is sufficient.

2) This follows essentially the same logic as statement 1, except this time we have a perfect cube rather than a perfect square. Essentiall, since (p*p*p)/(q*q*q) is an integer, we can quickly extrapolate that all prime factors of q are contained in p, and therefore p is divisible by q. This statement is sufficient.

I hope I wasn't rambling, and I hope that helps.