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If (4a)^(1/2) integer, is (a)^(1/2) integer? 24 Jun 2017, 07:24
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D
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58% (01:28) correct 42% (01:15) wrong based on 118 sessions

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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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D
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If (4a)^(1/2) integer, is (a)^(1/2) integer? 24 Jun 2017, 07:43
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Bunuel
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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If (4a)^(1/2) integer, is (a)^(1/2) integer? 01 Jul 2017, 03:33
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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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If (4a)^(1/2) integer, is (a)^(1/2) integer? 25 Jul 2017, 03:56
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ydmuley
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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If (4a)^(1/2) integer, is (a)^(1/2) integer? 25 Jul 2017, 04:41
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ydmuley
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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If (4a)^(1/2) integer, is (a)^(1/2) integer? 25 Jul 2017, 12:13
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ydmuley
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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If (4a)^(1/2) integer, is (a)^(1/2) integer? 28 Jan 2024, 04:27
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ydmuley
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


for statement 1 lets assume a=4 then the output will be interger but if a = 6 [m][square_root]4a will be integer but [square_root]a will be not
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If (4a)^(1/2) integer, is (a)^(1/2) integer? 28 Jan 2024, 04:42
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karanbishti

for statement 1 lets assume a=4 then the output will be interger but if a = 6 \(\sqrt{4a}\) will be integer but \(\sqrt{a}\) will be not

If a = 6, then \(\sqrt{4a}=\sqrt{24}=2\sqrt{6}\), which is not an integer.
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If (4a)^(1/2) integer, is (a)^(1/2) integer? 29 Jan 2024, 08:10
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karanbishti

for statement 1 lets assume a=4 then the output will be interger but if a = 6 \(\sqrt{4a}\) will be integer but \(\sqrt{a}\) will be not

If a = 6, then \(\sqrt{4a}=\sqrt{24}=2\sqrt{6}\), which is not an integer.

But if a = 4 then \sqrt{4}[/m], is an integer.. doesn't it make the statement insufficient ??
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If (4a)^(1/2) integer, is (a)^(1/2) integer? 29 Jan 2024, 09:27
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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


for statement 1 lets assume a=4 then the output will be interger but if a = 6 \(\sqrt{4a}\) will be integer but \(\sqrt{a}\) will be not

If a = 6, then \(\sqrt{4a}=\sqrt{24}=2\sqrt{6}\), which is not an integer.

But if a = 4 then \(\sqrt{4}\), is an integer.. doesn't it make the statement insufficient ??

Not sure what you are trying to say there. We are given that \(\sqrt{4a}=2\sqrt{a}\) is an integer. And we want to determine whether \(\sqrt{a}\) is an integer.

(1) says that \(a\) is a positive integer.

    Crucial to know for the GMAT: if \(x\) is a positive integer, then \(\sqrt{x}\) is either a positive integer or an irrational number. (It cannot be a reduced fraction such as \(\frac{7}{3}\) or \(\frac{1}{2}\)).

Hence, for \(2\sqrt{a}\) to be an integer, \(\sqrt{a}\) must be an integer because if it's not, then \(2\sqrt{a}\) will be irrational.

Can you give an example where both \(2\sqrt{a}\) and \(a\) are integers, and \(\sqrt{a}\) is NOT an integer?
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If (4a)^(1/2) integer, is (a)^(1/2) integer? 29 Jan 2024, 09:35
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(1) says that \(a\) is a positive integer.

    Crucial to know for the GMAT: if \(x\) is a positive integer, then \(\sqrt{x}\) is either a positive integer or an irrational number. (It cannot be a reduced fraction such as \(\frac{7}{3}\) or \(\frac{1}{2}\)).

Hence, for \(2\sqrt{a}\) to be an integer, \(\sqrt{a}\) must be an integer because if it's not, then \(2\sqrt{a}\) will be irrational.

Can you give an example where both \(2\sqrt{a}\) and \(a\) are integers, and \(\sqrt{a}\) is NOT an integer?[/quote]

ohh my god !! i see my mistake I made a blunder I got it now , thanks . I was making a very basic mistake
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