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If (4a)^(1/2) integer, is (a)^(1/2) integer?
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Bunuel
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If (4a)^(1/2) integer, is (a)^(1/2) integer? [#permalink] New post  24 Jun 2017, 06:24
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If \(\sqrt{4a}\) integer, is \(\sqrt{a}\) integer?

(1) \(a\) is a positive integer

(2) \(a = n^6\), where n is an integer
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Re: If (4a)^(1/2) integer, is (a)^(1/2) integer? [#permalink] New post  24 Jun 2017, 06:43
Bunuel wrote:
If is \(\sqrt{4a}\) integer, is \(\sqrt{a}\) integer?

(1) \(a\) is a positive integer

(2) \(a = n^6\), where n is an integer


\sqrt{4a} is an integer means either a = a perfect square or 1 or 4^3 or perfect square*4. If a is either then \sqrt{a} will be an integer

(1) no new information....but since stem was sufficient ..Sufficient

(2) ok...Sufficient (if n^6 is a then \sqrt{a} is n^3 which will be an integer)

I would say D

Please correct me if i am wrong
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Re: If (4a)^(1/2) integer, is (a)^(1/2) integer? [#permalink] New post  01 Jul 2017, 02:33
If is \(\sqrt{4a}\) integer, is \(\sqrt{a}\) integer?

\(\sqrt{4a}\) = Integer

This is only possible when \(\sqrt{a} = Integer\)

(1) \(a\) is a positive integer

This is sufficient as per the given information - \(\sqrt{4a}\) = Integer

Hence, (1) =====> is SUFFICIENT

(2) \(a = n^6\), where n is an integer

As n is an integer and \(a = n^6\)

\(\sqrt{a} = \sqrt{n^6} = n^3\)

As n is integer, \(n^3\) will also be an integer.

Hence, (2) =====> is SUFFICIENT

Hence, Answer is D
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Re: If (4a)^(1/2) integer, is (a)^(1/2) integer? [#permalink] New post  25 Jul 2017, 02:56
ydmuley wrote:
If is \(\sqrt{4a}\) integer, is \(\sqrt{a}\) integer?

\(\sqrt{4a}\) = Integer

This is only possible when \(\sqrt{a} = Integer\)

(1) \(a\) is a positive integer

This is sufficient as per the given information - \(\sqrt{4a}\) = Integer

Hence, (1) =====> is SUFFICIENT

(2) \(a = n^6\), where n is an integer

As n is an integer and \(a = n^6\)

\(\sqrt{a} = \sqrt{n^6} = n^3\)

As n is integer, \(n^3\) will also be an integer.

Hence, (2) =====> is SUFFICIENT

Hence, Answer is D


Hi,
Nice explanation, just one question though-are we trying to say that number 2 multiplied by a sq. root can shall give an integer only when the sq root itself is an integer or to generalize, a sq root can when doubled always gives an integer? If Yes, can u please provide few examples for the same, If No, please explain why statement 1 is suff?

Thanx
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Re: If (4a)^(1/2) integer, is (a)^(1/2) integer? [#permalink] New post  25 Jul 2017, 03:41
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saurabhsavant wrote:
ydmuley wrote:
If is \(\sqrt{4a}\) integer, is \(\sqrt{a}\) integer?

\(\sqrt{4a}\) = Integer

This is only possible when \(\sqrt{a} = Integer\)

(1) \(a\) is a positive integer

This is sufficient as per the given information - \(\sqrt{4a}\) = Integer

Hence, (1) =====> is SUFFICIENT

(2) \(a = n^6\), where n is an integer

As n is an integer and \(a = n^6\)

\(\sqrt{a} = \sqrt{n^6} = n^3\)

As n is integer, \(n^3\) will also be an integer.

Hence, (2) =====> is SUFFICIENT

Hence, Answer is D


Hi,
Nice explanation, just one question though-are we trying to say that number 2 multiplied by a sq. root can shall give an integer only when the sq root itself is an integer or to generalize, a sq root can when doubled always gives an integer? If Yes, can u please provide few examples for the same, If No, please explain why statement 1 is suff?

Thanx


Generally \(\sqrt{integer}\) is either an integer itself or an irrational number, it cannot be some reduced fraction like 1/2 or 2/3.

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Re: If (4a)^(1/2) integer, is (a)^(1/2) integer? [#permalink] New post  25 Jul 2017, 11:13
saurabhsavant wrote:
ydmuley wrote:
If is \(\sqrt{4a}\) integer, is \(\sqrt{a}\) integer?

\(\sqrt{4a}\) = Integer

This is only possible when \(\sqrt{a} = Integer\)

(1) \(a\) is a positive integer

This is sufficient as per the given information - \(\sqrt{4a}\) = Integer

Hence, (1) =====> is SUFFICIENT

(2) \(a = n^6\), where n is an integer

As n is an integer and \(a = n^6\)

\(\sqrt{a} = \sqrt{n^6} = n^3\)

As n is integer, \(n^3\) will also be an integer.

Hence, (2) =====> is SUFFICIENT

Hence, Answer is D


Hi,
Nice explanation, just one question though-are we trying to say that number 2 multiplied by a sq. root can shall give an integer only when the sq root itself is an integer or to generalize, a sq root can when doubled always gives an integer? If Yes, can u please provide few examples for the same, If No, please explain why statement 1 is suff?

Thanx



it is sufficient because \sqrt{4a} = \sqrt{4 * 1/4} = \sqrt{1} = 1(an integer)
however a = 1/4 so \sqrt{1/4} = 1/2 = not an integer, therefore statement 1 is sufficient,otherwise \sqrt{a} could be 1/2.
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