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If m and n are non-negative integers, mn=?

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If m and n are non-negative integers, mn=?  [#permalink]

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New post 01 Aug 2017, 00:59
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Question Stats:

63% (01:15) correct 37% (01:40) wrong based on 58 sessions

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If m and n are non-negative integers, mn=?

1) \(9^n=3^m\)
2) \(2^n=5^m\)

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Re: If m and n are non-negative integers, mn=?  [#permalink]

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New post 01 Aug 2017, 01:31
1
1. \(9^n=3^m\)
This is possible when the values of n and m are as follows:
a) n=1,m=2 nm=2
b) n=2,m=4, nm=8

We are not able to come up with an unique value for the expression mn(Insufficient)

2. \(2^n=5^m\)
This is possible if and only if n=m=0.

The value of nm=0 (Sufficient)(Option B)
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Re: If m and n are non-negative integers, mn=?  [#permalink]

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New post 03 Aug 2017, 01:25
==> In the original condition, there are 2 variables (m,n) and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), from con 1), you get \(9^n=(3^2)^n=3^{2n}=3^m\), which becomes \(2n=m\). In order for con 2) to satisfy as well, you only get m=n=0, hence it is unique and sufficient. The answer is C. However, this is an integer question, one of the key questions, so you apply CMT 4 (A: if you get C too easily, consider A or B). For con 1), the way to satisfy \(9^n=(3^2)^n=3^{2n}=3^m\) to \(2n=m\) is not unique and not sufficient. For con 2), from \(2^n=5^m\), you get \(2^n=even\) and \(5^m=odd\), so even≠odd. Only m=n=0 satisfies this, hence it is unique and sufficient.

Therefore, the answer is B, not C.
Answer: B
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Re: If m and n are non-negative integers, mn=?   [#permalink] 03 Aug 2017, 01:25
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If m and n are non-negative integers, mn=?

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