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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]

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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]

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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]

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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

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

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12 May 2015, 01:17

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Even though this is an old post (and most of the original posters are probably long gone), this question serves as a nice example of some of the Number Properties that you're likely to face on Test Day. If you can spot the NPs involved, then you can move relatively quickly through the work, but even if you don't recognize the NPs, you can still TEST VALUES to prove that patterns exist. In that way, you can do some quick work and avoid "staring at the screen" and hoping that something comes to you.

We're told that M and N are INTEGERS. We're asked if M is ODD. This is a YES/NO question.

Fact 1: N + M is ODD

IF.... N = 1 M = 0 The answer to the question is NO

IF.... N = 0 M = 1 The answer to the question is YES Fact 1 is INSUFFICIENT

Fact 2: N + M = N^2 + 5

Here, the value of M depends on the value of N. Let's start with something simple and see if a pattern emerges.

IF.... N = 0 M = 5 The answer to the question is YES

IF.... N = 1 M = 5 The answer to the question is YES

IF.... N = 2 M = 7 The answer to the question is YES

IF.... N = 3 M = 11 The answer to the question is YES

It certainly looks like the answer to the question is ALWAYS YES. We can quickly TEST some negative values for N and see what happens....

IF.... N = -1 M = 7 The answer to the question is YES

IF.... N = -2 M = 11 The answer to the question is YES Fact 2 is SUFFICIENT

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

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02 Jun 2016, 04:29

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: If m and n are integers, is m odd? [#permalink]

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21 Nov 2016, 10:15

Great Question Here we are given two integers m,n And we are asked if m is odd or not Statement 1 m+n=odd hmm This Statement just tells us that m and n are of opposite even/odd nature Not sufficient Statement 2 m=n(n-1)+5 here n(n-1) is a product of two consecutive integers. hence it must be even so n=even+5 => even+odd => odd hence sufficient hence B
_________________

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

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21 Nov 2016, 10:28

zisis wrote:

If m and n are integers, is m odd?

(1) n + m is odd (2) n + m = n^2 + 5

FROM STATEMENT - I( INSUFFICIENT )

If n + m = Odd

Either n = Odd/Even and m = Odd/Even because -

1. Even + Odd = Odd 2. Odd + Even = Odd

FROM STATEMENT - II( INSUFFICIENT )

n - n^2 = 5 - m

Or, n ( n - 1 ) = 5 - m

Here, LHS must be Even, because product of 2 consecutive integers is always even ( ie, one of the 2 consecutive numbers must be even )

So, 5 - m = Evem

Or, Odd - m = Even

Hence, m = Odd Thus, Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked, answer will certainly be (B)... _________________

Thanks and Regards

Abhishek....

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