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Re: Is x the square of an integer [#permalink]
24 Apr 2010, 23:30

5

This post received KUDOS

From Stmt 1 , x= 6(2k+1) where k = positive integer. 2k + 1 is always odd. To be a square of an integer, 2k+1 has to be a multiple of 6 and that is not possible. So, x can't be square of an integer.

Stmt 1 sufficient.

From Stmt 2, x= 3(q+3) putting different values of q can have 2 different results. Example put q=1, x =12 ... not a square of an integer put q=24 , x = 81 ... square of an integer 9.

Re: Is x the square of an integer? [#permalink]
13 Nov 2013, 01:13

2

This post received KUDOS

Expert's post

Is x the square of an integer?

(1) x = 12k + 6, where k is a positive integer --> \(x=6(2k+1)=2*3(2k+1)\). Now, \(x\) to be a perfect square it should have an even power of its primes, but \(2k+1\) is an odd number and can no way produce 2 for \(x\). Thus \(x\) is not a perfect square. Sufficient.

(2) x = 3q + 9, where q is a positive integer --> \(x=3(q+3)\). If \(q=1\) then \(x=12\) and the answer is NO but if \(q=9\) then \(x=36\) and the answer is YES (basically if (q+3)=3*any perfect square then x will be a perfect square and if (q+3) is some other type of number then x won't be a perfect square). Not sufficient.

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