GmatPrep Test arithmetic question- Need Help

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I stumbled on this question while taking the GmatPrep test and I couldn't get to understand the reasoning behind the answer. I'd really appreciate if any of you could post the reasoning behind the answer.

"If n is a multiple of 5 and n=psquare x q , where p and q are prime numbers. Which of the following must be a multiple of 25?

(a)psquare (b)qsquare (c) pq (d)psquare x qsquare (e) pcube x q

p.s: psquare is as in "p to the power of 2" ; didnt know how to type the exponent.

The answer is option (d).
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ildude02 wrote:I stumbled on this question while taking the GmatPrep test and I couldn't get to understand the reasoning behind the answer. I'd really appreciate if any of you could post the reasoning behind the answer.

"If n is a multiple of 5 and n=psquare x q , where p and q are prime numbers. Which of the following must be a multiple of 25?

(a)psquare (b)qsquare (c) pq (d)psquare x qsquare (e) pcube x q

p.s: psquare is as in "p to the power of 2" ; didnt know how to type the exponent.

The answer is option (d).
If n is a multiple of 5, then (P^2)(Q) must also be a multiple of 5 (since n=(P^2)(Q)).

Since 5 is prime, if (P^2)(Q) is a multiple of 5, then either P or Q must be a multiple of 5. However, we have no way to determine which of them is the multiple of 5, so any answer choice with just P or Q in it doesn't have to be a multiple of 25.

Choice (c), PQ, has to be a multiple of 5, but not necessarily a multiple of 25. Same with choice (e), (P^3)(Q) (i.e. if Q is a multiple of 5 but P isn't, choice (e) doesn't have to be a multiple of 25).

Now, since P or Q is a multiple of 5, we know for sure that either P^2 or Q^2 will be a multiple of 5^2, or 25. But again, we have no way of knowing which of the terms is a multiple of 25, so any choice with just P^2 or Q^2 in it doesn't have to be a multiple of 25.

So, the only way to guarantee that a number will be a multiple of 25 is if it contains BOTH P^2 and Q^2. Therefore, (d) is the right choice.
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by ildude02 » Mon Jan 14, 2008 10:16 am
Thanks for the explanation, it makes sense.