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# m and n are both integers, and m ≠ n, does n = 0?

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Re: m and n are both integers, and m ≠ n, does n = 0? [#permalink]
Bunuel wrote:
DelSingh wrote:
m and n are both integers, and m#n, does n = 0?

(1) m = 7
(2) mn = n^2

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Got this question from a Kaplan advanced quant library. I don't understand why statement 2 is sufficient. I understand that m cannot be equal to n, however, how is n forced to be equaled to be 0? I haven't found a question similar to this in the OG yet. If you have similar questions, please post! Thanks

m and n are both integers, and m#n, does n = 0?

(1) m = 7. Clearly insufficient the value of n.

(2) mn = n^2 --> $$mn-n^2=0$$ --> $$n(m-n)=0$$ --> $$n=0$$ or $$m-n=0$$ ($$m=n$$). Since given that $$m\neq{0}$$, then only the first case is left, thus $$n=0$$. Sufficient.

Similar question to practice: if-xy-0-does-xy-y-3-1-xy-3-2-y-105766.html

Hope it helps.

Wow, I didn't even catch that. Careless algebra mistake. Thank you, both! Also, thank you for the similar questions!
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Re: m and n are both integers, and m ≠ n, does n = 0? [#permalink]
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Re: m and n are both integers, and m ≠ n, does n = 0? [#permalink]
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