p^a*q^b*r^c*s^d = x, where x is a perfect square. If p, q, : DS Archive
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# p^a*q^b*r^c*s^d = x, where x is a perfect square. If p, q,

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p^a*q^b*r^c*s^d = x, where x is a perfect square. If p, q, [#permalink]

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24 Jun 2007, 18:25
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p^a*q^b*r^c*s^d = x, where x is a perfect square. If p, q, r, and s are prime integers, are they distinct?

(1) 18 is a factor of ab and cd

(2) 4 is not a factor of ab and cd

(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.

i have NO clue
CIO
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24 Jun 2007, 18:49
Here are my assumptions:

1. Statement one means that 18 is a factor of ab alone, and also cd alone.
2. Statement two means that 4 is neither a factor of ab, nor a factor of cd.

Under those conditions, the answer is B. Here's my logic:

1. It's not hard to make x be perfect square - as long as all the expondents (a,b,c, and d) are even, x will be a perfect square.

2. If they are all even, then there is no way to answer this question, because the statements don't tell us anything about p, q, r, or s. If there are some odd ones, however, that would mean that the prime numbers ARE NOT distinct, because we'd have to have two identical primes with odd exponents to make an even exponent.

For example, if we had 3^3, we would also need something like 3^5 in order to combine and get 3^8, which is even, and thus a perfect square.

3. Statement 1 tells us that both ab and cd are multiples of 18. If they were actually 18, then that would mean only one of each pair could be even. But they could also be 36 or higher, and thus you'd be able to make each one of the exponents even. So statement 1 is not enough.

4. Statement 2 tells us that neither of the pairs is divisible by 4. We need them to be AT LEAST multiples of four to begin hoping that both variables in each pair are even. Because they are not multiples of 4, we know they cannot all be even.

So with statement 2, we must have some odd exponents, so we must have some repeated prime numbers, so the answer is distinctly, NO.

B
24 Jun 2007, 18:49
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