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Re: If (4a)^(1/2) integer, is (a)^(1/2) integer?
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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
Hence, Answer is D _________________
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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
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?
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
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.
Re: If (4a)^(1/2) integer, is (a)^(1/2) integer?
[#permalink]
Show Tags
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
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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