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# If L ≠ 0, is 18K/L an integer?

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Manager
Joined: 27 Dec 2009
Posts: 164

Kudos [?]: 135 [0], given: 3

If L ≠ 0, is 18K/L an integer? [#permalink]

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09 May 2010, 02:28
00:00

Difficulty:

55% (hard)

Question Stats:

45% (00:52) correct 55% (00:45) wrong based on 38 sessions

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If L ≠ 0, is 18K/L an integer?

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

(2) K - L = L
[Reveal] Spoiler: OA

Kudos [?]: 135 [0], given: 3

Math Expert
Joined: 02 Sep 2009
Posts: 43335

Kudos [?]: 139549 [0], given: 12794

Re: If L ≠ 0, is 18K/L an integer? [#permalink]

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09 May 2010, 03:10
Expert's post
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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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Kudos [?]: 139549 [0], given: 12794

Manager
Joined: 16 Feb 2010
Posts: 178

Kudos [?]: 34 [0], given: 17

Re: If L ≠ 0, is 18K/L an integer? [#permalink]

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16 May 2010, 11:32
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.

Kudos [?]: 34 [0], given: 17

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Re: If L ≠ 0, is 18K/L an integer? [#permalink]

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16 Sep 2017, 02:17
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Re: If L ≠ 0, is 18K/L an integer?   [#permalink] 16 Sep 2017, 02:17
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