Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
(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
Re: If m and n are integers, is m odd? [#permalink]
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).
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. _________________
Re: If m and n are integers, is m odd? [#permalink]
14 May 2015, 16:02
1
This post received KUDOS
Expert's post
Hi All,
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
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...
Ninety-five percent of the Full-Time Class of 2015 received an offer by three months post-graduation, as reported today by Kellogg’s Career Management Center(CMC). Kellogg also saw an increase...