Hi richachampion,

This question is based on the same concept as the other prime-factor questions that you posted: when you prime-factor a perfect square, each of the prime factors MUST show up an EVEN number of times...

eg. 9 = (3)(3) here, there are TWO 3s.
eg. 100 = (2)(2)(5)(5) here, there are TWO 2s and TWO 5s
eg. 16 = (2)(2)(2)(2) = here, there are FOUR 2s
Etc.

We're told that P is an EVEN INTEGER and that (300)(P) is a PERFECT SQUARE. We have to be careful to incorporate ALL of the information that we're given.. Prime-factoring this number will get us...

(300)(P) =
(10)(30)(P) =
(2)(5)(2)(3)(5)(P) =
(2)(2)(3)(5)(5)(P)

We have TWO 2s and TWO 5s, but only ONE 3. We need there to be an even number of each prime factor. At first glance, you might think that P must equal 3... However, 3 is NOT an EVEN INTEGER. So we need P to contain a '3' AND and at least TWO 2s (so that P will be EVEN).

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

richachampion wrote:For any even integer P, 300 multiplied by P is square of an integer. What is the least value of P?

A) 3
B) 4
C) 10
D) 12
E) 14

OA: D

If 300P is a perfect square, then its prime factorization must contain only even exponents. Let's begin by prime factoring 300.

300 = 30 x 10 = 5 x 6 x 5 x 2 = 5 x 3 x 2 x 5 x 2 = 5^2 x 3^1 x 2^2

We can see that 300 is not a perfect square because its prime factorization contains an odd exponent (that is, 3^1). This means that P must contribute the missing factor. However, we are given that P must also be an even integer. Thus, in order to keep "an even number" of twos, P must be divisible by 2^2. Thus P must be divisible by 3^1 x 2^2 = 12. Since we want the least value of P, P must be 12.

In fact, if P = 12, then 300P = 5^2 x 3^2 x 2^4, which is a perfect square.

Answer: D