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Re: DS: x multiple of y? [#permalink]
04 May 2008, 19:55

chineseburned wrote:

If x and y are both integers greater than 1, is x a multiple of y?

(1) 3y^2 + 7y = x (2) x^2 -x is a multiple of y

A.

Given: x > 1 y > 1 n, x, y = integer Asking: Is x = ny?

(1) x = y*(3y + 7) Because y is integer, (3y + 7) must be integer; therefore, x must equal integer * y SUFFICIENT

(2) x^2 - x = ny Plug in numbers to satisfy above condition... Say x=3, n=1, then y=6. In this case, x is not a multiple of y. Say x=6, n=15, then y=2. In this case, x is a multiple of y. The solution actually depends on what n is, and the only condition we have is n is integer. Therefore, it is INSUFFICIENT

Re: DS: x multiple of y? [#permalink]
17 Jul 2010, 07:12

1

This post received KUDOS

Expert's post

dauntingmcgee wrote:

So from x(x-1)=yk, can we not derive:

x=y*k/(x-1)=yk?

In which case the answer is C, not A.

We don't know whether \(\frac{k}{x-1}\) is an integer, hence we can not write \(x=yn\) (where n is an integer) from \(x(x-1)=yk\).

If x and y are integers great than 1, is x a multiple of y?

Is \(x=ny\), where \(n=integer\geq{1}\)?

(1) \(3y^2+7y=x\) --> \(y(3y+7)=x\) --> as \(3y+7=integer\), then \(y*integer=x\) --> \(x\) is a multiple of \(y\). Sufficient.

(2) \(x^2-x\) is a multiple of \(y\) --> \(x^2-x=my\) --> \(x(x-1)=my\) --> \(x\) can be multiple of \(y\) (\(x=2\) and \(y=2\)) OR \(x-1\) can be multiple of \(y\) (\(x=3\) and \(y=2\)) or their product can be multiple of \(y\) (\(x=3\) and \(y=6\)). Not sufficient.

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