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If L <> 0 , is an integer? a. (K^2) / (L^2) is an

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If L <> 0 , is an integer? a. (K^2) / (L^2) is an [#permalink]

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New post 09 May 2010, 03:28
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If L <> 0 , is [(18 * K ) / L ] an integer?

a. (K^2) / (L^2) is an integer.
b. K - L = L
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Re: DS integer [#permalink]

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New post 09 May 2010, 04:10
Expert's post
msand wrote:
If L does not equal to 0, is (18*K)/L an integer?

(1) K^2/L^2 is an integer.
(2) K-L=L


If \(l\neq{0}\) is \(\frac{18k}{l}=integer\)?

(1) \(\frac{k^2}{l^2}=integer\). \(k=4\) and \(l=2\) - answer YES but \(k=\sqrt{6}\) and \(l=\sqrt{2}\) - answer NO. Not sufficient.

(2) \(k=2l\) --> \(\frac{18k}{l}=\frac{18*2l}{l}=36=integer\). Sufficient.

Answer: B.
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Re: DS integer [#permalink]

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New post 16 May 2010, 12:32
according to statement A K^2/L^2 is integer but that doesn't mean that K & L are integer; square of 1.732 is 3 which is an integer but 1.732 itself is not a integer. hence statement A - i nsufficient

statment B - k=2l sufficeint we get a integer from it
so B must be the answer.
Re: DS integer   [#permalink] 16 May 2010, 12:32
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