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Divisibility

Expert replies
by Deepthi Subbu » Wed Nov 17, 2010 7:20 pm
1. If x^2 is divisible by 216 ,what is the smallest possible value for positive integer x?

I just have the explanation and the OA and no answer choices.

I will post the OA after some discussion.
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Source: — Problem Solving |

by beat_gmat_09 » Wed Nov 17, 2010 7:49 pm
36.
Hope is the dream of a man awake
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by Deepthi Subbu » Wed Nov 17, 2010 8:43 pm
Correct , thanks
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by Rahul@gurome » Wed Nov 17, 2010 9:51 pm
Deepthi Subbu wrote:1. If x^2 is divisible by 216 ,what is the smallest possible value for positive integer x?
For smallest possible positive integer value of x, x² will be smallest too. Thus, we have to find smallest possible x² such that it is divisible by 216 (= 6³). Say, x² = 6³n. In other words, we have to find minimum possible positive integer value of n for which 6³n becomes a perfect square. Clearly n = 6 => x² = 6^4 => x = 6² = 36.
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