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# If x is a positive integer, is Sqrt(x) an integer? (1) Sqrt

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Intern
Joined: 13 Aug 2005
Posts: 26
Location: Israel
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If x is a positive integer, is Sqrt(x) an integer? (1) Sqrt [#permalink]  14 Aug 2005, 10:19
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If x is a positive integer, is Sqrt(x) an integer?

(1) Sqrt (4x) is an integer
(2) Sqrt (3x) is not an integer

The right answer according to OG is A, but I could not understand their explanation. Can you explain?

Thanks!
Manager
Joined: 14 Jul 2005
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Location: Sofia, Bulgaria
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I don't think I can do a better job than the OG, but here goes:

(1) you know that sqrt(4x) is an integer. You can split sqrt(4x) into two factors:

sqrt(4x) = sqrt(4)*sqrt(x)

We know that sqrt(4)=2 (integer), so the only way for sqrt(4)*sqrt(x) to be an integer is for sqrt(x) to be an integer. Otherwise sqrt(4x) would be a non-integer, and we would have a contrdiction with the statement.

(2) Similar reasoning to above...

sqrt(3x) = sqrt(3)*sqrt(x) is not an integer, which means that sqrt(x) is not an odd exponent of sqrt(3), which in turn means that x does not equal 3, 27, 243 and so on.

If we assume that x=4, statement (2) holds and sqrt(x) is an integer.
If we assume that x=5, statement (2) holds, but sqrt(x) is not an integer.

=> statement (2) is not sufficient.

Vasil
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Joined: 13 Aug 2005
Posts: 26
Location: Israel
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Thanks Vasil.

I overlooked the fact that X is a positive INT and can only be 1,4,9â€¦ according to (1)
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