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

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Manager
Joined: 27 Dec 2009
Posts: 170
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Kudos [?]: 66 [0], given: 3

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

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09 May 2010, 03:28
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Question Stats:

100% (02:59) correct 0% (00:00) wrong based on 10 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 32644
Followers: 5655

Kudos [?]: 68707 [0], given: 9817

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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.

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Manager
Joined: 16 Feb 2010
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Kudos [?]: 11 [0], given: 11

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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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