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Re: If x is a positive integer, is x^1/2 an integer [#permalink]
13 Jan 2014, 00:52

1

This post received KUDOS

Expert's post

SOLUTION:

If x is a positive integer, is \sqrt{x} an integer?

As given that x is a positive integer then \sqrt{x} is either an integer itself or an irrational number.

(1) \sqrt{4x} is an integer --> 2\sqrt{x}=integer --> 2\sqrt{x} to be an integer \sqrt{x} must be an integer or integer/2, but as x is an integer, then \sqrt{x} can not be integer/2, hence \sqrt{x} is an integer. Sufficient.

(2)\sqrt{3x} is not an integer --> if x=9, condition \sqrt{3x}=\sqrt{27} is not an integer satisfied and \sqrt{x}=3 IS an integer, BUT if x=2, condition \sqrt{3x}=\sqrt{6} is not an integer satisfied and \sqrt{x}=\sqrt{2} IS NOT an integer. Two different answers. Not sufficient.

If x is a positive integer, is \sqrt{x} an integer?

(1) \sqrt{4x} is an integer. (2) \sqrt{3x} is not an integer.

Statement 1) root(4x) is an integer. root(4x) = either +2root(x) or -2root(x) . As in either case, the result is an integer, then root(x) has to be an integer as (+2 or -2) is integer. Hence Sufficient. Statement 2) root(3x) is an integer. root(3) is not an integer. There can be following two cases: 1) root(x) is an integer => root(3) * root(x) is not an integer as root(3) is not an integer. 2) root(x) is not an integer => root(3) * root(x) can never be an integer except in once case where x= 3. Either case it is not sufficient.

Re: If x is a positive integer, is x^1/2 an integer [#permalink]
17 Jan 2014, 02:49

Expert's post

SOLUTION:

If x is a positive integer, is \sqrt{x} an integer?

As given that x is a positive integer then \sqrt{x} is either an integer itself or an irrational number.

(1) \sqrt{4x} is an integer --> 2\sqrt{x}=integer --> 2\sqrt{x} to be an integer \sqrt{x} must be an integer or integer/2, but as x is an integer, then \sqrt{x} can not be integer/2, hence \sqrt{x} is an integer. Sufficient.

(2)\sqrt{3x} is not an integer --> if x=9, condition \sqrt{3x}=\sqrt{27} is not an integer satisfied and \sqrt{x}=3 IS an integer, BUT if x=2, condition \sqrt{3x}=\sqrt{6} is not an integer satisfied and \sqrt{x}=\sqrt{2} IS NOT an integer. Two different answers. Not sufficient.

If x is a positive integer, is x^1/2 an integer [#permalink]
18 Jan 2014, 03:22

If x is a positive integer, is \sqrt{x} an integer?

(1)\sqrt{4x} is an integer. (2)\sqrt{3x} is not an integer.

Statement 1

\sqrt{4x} = k, k is an integer Or,\sqrt{x}= k/2

If k/2 is not an integer, then (k/2)^2 or k^2/4 is also not an integer and this implies that x is not an integer as x = k^2/4. But, it is given that x is an integer.

Therefore, k^2/4 is an integer => k/2 is an integer =>\sqrt{x} is an integer...........Sufficient....(B)(C)(E)

Statement 2

\sqrt{3x}=k, k is not an integer Or, \sqrt{x} = k/\sqrt{3}

Thus, \sqrt{x} = integer, when k is a multiple of \sqrt{3} and \sqrt{x} is not an integer when k is not a multiple of \sqrt{3}

As the value of \sqrt{x} can not be uniquely determined, statement (2) is not sufficient.................(D)

Answer: (A)

Last edited by arunspanda on 31 Jul 2014, 15:18, edited 1 time in total.