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(1) n + m is odd --> \(n+m=odd\). The sum of two integers is odd only if one is odd and another is even, hence \(m\) may or may not be odd. Not sufficient.

(2) n + m = n^2 + 5 --> \(m-5=n^2-n\) --> \(m-5=n(n-1)\), either \(n\) or \(n-1\) is even hence \(n(n-1)=even\) --> \(m-5=m-odd=even\) --> \(m=odd\). Sufficient.

Re: If m and n are integers, is m odd? [#permalink]
16 Aug 2013, 05:22

m and n are integers, is m odd? -Can't rephrase the question here, so you work with m = odd? It's a Yes/No DS question type.

Here's how I approached the problem.

Statement 1. Given: m + n is odd. Using even/odd number property rules: even + odd = odd. OR, odd + even = odd. Since m can be either even or odd in this case, Statement 1 is insufficient, we don't have enough information.

Statement 2. Given: \(m+n = n^2 + 5\)

Simplifying the equation: \(m - 5 = n^2 - n\) \(m - 5 = n(n - 1)\) \(m = n(n-1) +5\) m = Even + odd. Therefore, m is odd.

n(n-1) will always be even because it's a number times the number preceding it (think consecutive)...so one of those numbers has to be even (if n = 3 for example, you would have 3*(3-1) = 3*(2) = 6). You have an even*odd = even situation here.

5 is odd. So putting it all together, m = Even + Odd. You can answer with the given information that, yes, m is odd. Sufficient.

Re: If m and n are integers, is m odd? [#permalink]
23 Nov 2013, 04:04

Statement 1, m and n could be both odd or one odd, one even. Insufficient. Statement 2, when n is odd, n^2+5 is even, then m+n is even, m is odd; when n is even, n^2+5=odd, m+n is odd, then m is odd. Sufficient. So, answer is B

(1) n + m is odd --> \(n+m=odd\). The sum of two integers is odd only if one is odd and another is even, hence \(m\) may or may not be odd. Not sufficient.

(2) n + m = n^2 + 5 --> \(m-5=n^2-n\) --> \(m-5=n(n-1)\), either \(n\) or \(n-1\) is even hence \(n(n-1)=even\) --> \(m-5=m-odd=even\) --> \(m=odd\). Sufficient. Answer: B.

Another method: St1: n + m is odd, Answer is YES when n is even but answer is NO when n is odd. Insufficient!

St2: n + m = n^2 + 5 --> n^2 - n + (5-m) = 0 -- > is a quadratic equation with sum of roots = 1 and product of roots = (5-m). Clearly the the roots are consecutive integers with least among them as negative (ex. 4 & -3, 6 & -5...) That means one of the root is even. So, product of roots (5-m) is also even. For (5-m) to be even, m must be odd. Sufficient! Answer: B

Re: If m and n are integers, is m odd? [#permalink]
28 Mar 2014, 07:22

Satatment 1:Clearly insuff. Even+odd=odd.So m could be even or odd. S2:sufficient. The eq can be rearranged as N(n-1)=m-5 Now n(n-1) is a product of 2 consecutive integers so it'll definitely be even=>m-5=even So m=odd because only odd-odd=even Ans option B

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gmatclubot

Re: If m and n are integers, is m odd?
[#permalink]
28 Mar 2014, 07:22

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