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Number properties question

by Troika » Tue Apr 10, 2012 5:49 pm
If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 105?

1. x is a multiple of 9
2. y is a multiple of 25

OA: B

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by Bill@VeritasPrep » Tue Apr 10, 2012 6:17 pm
To be a multiple of 105, xy must contain the factors 3, 5, and 7. Since x is a multiple of 6, it has factors of 3 and 2. Since y is a multiple of 14, it has factors of 7 and 2. We can see that the 3 and the 7 are already covered; we just need to find a 5 in either x or y.

1. X being a multiple of 9 gives us another 3, but we still don't know for sure if we have a 5; if x=54, then no, but if x=270, then yes. Insufficient.

2. If y is a multiple of 25, it has two 5's as factors. Thus, we have the missing piece that we needed to guarantee that xy is a multiple of 105. Sufficient.
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by Anurag@Gurome » Tue Apr 10, 2012 7:11 pm
HG10 wrote:If positive integer x is a multiple of 6 and positive integer y is a multiple of 14, is xy a multiple of 105?

1. x is a multiple of 9
2. y is a multiple of 25

OA: B

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x = 2 * 3
y = 2 * 7
105 = 3 * 5 * 7
x * y = 2² * 3 * 7
For xy to be a multiple of 105, we need at least one 3, 5, and 7 among the factors. So, the missing number is 5, as 3 and 7 are already there.

(1) x is a multiple of 9 but this does not imply anything about the missing factor, 5; NOT sufficient.

(2) y is a multiple of 25 and 25 = 5 * 5, which answers the question; SUFFICIENT.

The correct answer is B.
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by GMAT Kolaveri » Tue Apr 10, 2012 7:35 pm
x = 6,12,18 .... 3 x 2
y = 14,28....... 7 x 2

105 = 5 x 7 x 3 ( we need atleast one 5, one 7 AND one 3 to show xy is a multiple of 105)

xy= 3 x 7 x 2 x 2 ( we need a 5) in analyzing we see that 5 is present in B. hence ans could be B or D.

To prove AO is NOT D, we see that A gives 3 which we ve already know. hence D is wrong.
AO: B

Key to this is knowing PRIME FACTORIZATION
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by Troika » Wed Apr 11, 2012 6:53 am
Guys, thank you for the solutions. Those were very helpful!
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