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hardest DS in 15 testset

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Re: hardest DS in 15 testset

by Andrei » Thu Oct 29, 2009 9:00 am
Hi,
If x is a multiple of 6, it can be written as:
x = 2*3*k, k - positive integer.
If y is a multiple of 14, it can be written as:
y = 2*7*m, m - - positive integer.

So, xy = 2*3*2*7*k*m.

105 can be written as: 105 = 3*5*7

So, in order that xy to be an multiple of 105, it has to be divisible by 3, 5 and 7.
As xy is already divisible by 2, 3 and 7, it has to be divisible by 5 too, thus either k or m must be divisible by 5.

a. x a multiple of 9 - NOT SUFFICIENT
b. y is multiple of 25 -> so m is divisible by 5 - SUFFICIENT

My answer is B.
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