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# Is 'n' an integer? (1) n^2 is an integer. (2) sqrt(n) is an

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Manager
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Is 'n' an integer? (1) n^2 is an integer. (2) sqrt(n) is an [#permalink]

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24 Jul 2007, 14:56
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Is 'n' an integer?
(1) n^2 is an integer.
(2) sqrt(n) is an integer.

I thought the answer was D, but it is B. I figured, if n^2 is an integer, then the sqrt(n^2) has to be an integer as well. So if sqrt(n^2)=n, then n is an integer. What am i missing?
Manager
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24 Jul 2007, 15:34
It's B.

From 1 - n can be lets's say 2 or sqrt 2. Both squares are integers. Insuff.

From 2 - sqrt of n is an integer, so n must be an integer. Suff
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24 Jul 2007, 17:57
St1:
N could be sqrt(2) and N^2 = integer 2, or N could be 2, and N^2 = integer 4. Insufficient.

St2:
If sqrt(n) = integer, then n must be an integer. Sufficient.

Ans B
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24 Jul 2007, 22:19
Clearly B

n^2 int tells you nothing
if n^2= 2 then n =1.41 which is not an integer

root (n) is integer gives that n = square of integer
which is always an integer

Hence B
24 Jul 2007, 22:19
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