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Given prime factorization of k: \(k = (m^2)n\), If k is a multiple of 3, we can say that either m = 3 or n = 3. So either m^2 or n^2 will be a multiple of 9 but we don't know which of them is a multiple of 9. That is why none of A, B, C and D work. (E) \((mn)^2\) includes both\(m^2\) and \(n^2\) and hence it must be a multiple of 9. _________________

Re: If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]
04 Feb 2013, 13:05

Since 'm' and 'n' are primes, In order for k to be a multiple of 3 either 'n' or 'm' must be 3. Also, In order to be a multiple of 9 the product must have two 3's as factors. A) No. 'm' could be 3 or 'n' could be 3 we don't know which one B) No. Same logic as above. C) No. 'm' and 'n' could both be 3 in which case it would work. But they could also be '5' and '7' or some other pair of primes. D) No. n^2 could be 3^2 or it could be 5^2 (or some other prime number) E) Correct Answer. Since 3 must be one of the numbers, squaring the product must yield 9 as a factor.

must be a multiple of 9 m and n are prime...lets have 2 and 3 now either of then can be a 3... a) m can be 2 and n = 3...out b) n can be 2 and m = 3...out c) m = 2 and n=3 ...out d) n can be 2 and m = 3...out e) m = 2/3 or n = 3/2 both satisfies .......must be a multiple of 9 _________________

Re: If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]
19 May 2014, 20:41

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Re: If k is a multiple of 3 and k = (m^2)n, where m and n are [#permalink]
05 Aug 2015, 22:50

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