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If (4a)^(1/2) integer, is (a)^(1/2) integer?

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If (4a)^(1/2) integer, is (a)^(1/2) integer?  [#permalink]

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24 Jun 2017, 07:24
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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]

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24 Jun 2017, 07:43
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]

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01 Jul 2017, 03:33
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

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Re: If (4a)^(1/2) integer, is (a)^(1/2) integer?  [#permalink]

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25 Jul 2017, 03:56
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

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]

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25 Jul 2017, 04:41
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

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]

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25 Jul 2017, 12:13
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

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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Re: If (4a)^(1/2) integer, is (a)^(1/2) integer?   [#permalink] 25 Jul 2017, 12:13
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