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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Divisibility
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Deepthi Subbu
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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.Deepthi Subbu wrote:1. If x^2 is divisible by 216 ,what is the smallest possible value for positive integer x?
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On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
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