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Question on Perfect Square 4

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Question on Perfect Square 4

by richachampion » Mon Oct 17, 2016 2:48 am
If X is a perfect square, is X also a perfect cube?

(1) X = K^6, where K is a positive integer.
(2) X^(1/6) =N , where N is a positive integer.
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by richachampion » Mon Oct 17, 2016 2:50 am
OA: D
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by richachampion » Mon Oct 17, 2016 2:53 am
I think both statement 1 and statement 2 are same thing.
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by fiza gupta » Mon Oct 17, 2016 4:47 am
Yes, both the statements should be same
both represents positive integers

1) X = K^6, where K is a positive integer.
X^(1/3) = K^2 (k is integer)
X^(1/3) will always result in an integer
SUFFICIENT

(2) X^(1/6) =N , where N is a positive integer.
X^(1/3) = N^2 (N is integer)
SUFFICIENT
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by Jay@ManhattanReview » Tue Dec 06, 2016 2:42 am
Let us take another perspective of the question.

For a number X to be a perfect square as well as a perfect cube, it should be in the form of K^(6n), where K and n are positive integers. We took 6 as it is the LCM of 2 and 3.

Both the statements are in fact one and the same.

Hope this helps!

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by Matt@VeritasPrep » Thu Dec 08, 2016 9:27 pm
Easier way:

S1:

x = k� = (k²)³

So x is a cube.

S2:

Raise both sides to the sixth power:

x = n� = (n²)³

So x is a cube.
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by Matt@VeritasPrep » Thu Dec 08, 2016 9:28 pm
richachampion wrote:I think both statement 1 and statement 2 are same thing.
More or less!

A good follow up question for you, though: taken together, does this mean that k = n?
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