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^2 - 1 is an odd integer --> \(n^2-1=odd\) --> \(n^2=odd+1=even\). Now, since \(n\) is an integer, then in order \(n^2\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some irrational number (square root of an even number), for example \(\sqrt{2}\), so not an even integer.

(2) 3n + 4 is an even integer --> \(3n + 4=even\) --> \(3n=even-4=even\). The same here, since \(n\) is an integer, then in order \(3n\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some fraction, for example \(\frac{2}{3}\), so not an even integer.

(1) n^2 - 1 is odd. The next consecutive integer is n^2 and is therefore even. This means that n must be even too, because squaring a number does NOT change this. --> sufficient

(2) 3n + 4 is even. So 3n is even, too. This means that the prime factorization of 3n includes at least one 2. Dividing by 3 (to get from 3n to n) does NOT eliminate recude the number of twos in the prime factorization, so n is even. --> sufficient

The correct answer is D. Both statements are individually sufficient.

i will go with d both options individually can get the required info -as its an integer no more fractions integer when squared gives the same type value i.e odd gives odd and even gives even same way multiplying odd with even gives even and odd with odd gives odd

hence both are individually sufficient _________________

(1) n^2 - 1 is an odd integer --> \(n^2-1=odd\) --> \(n^2=odd+1=even\). Now, since \(n\) is an integer, then in order \(n^2\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some irrational number (square root of an even number), for example \(\sqrt{2}\), so not an even integer.

(2) 3n + 4 is an even integer --> \(3n + 4=even\) --> \(3n=even-4=even\). The same here, since \(n\) is an integer, then in order \(3n\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some fraction, for example \(\frac{2}{3}\), so not an even integer.

(1) n^2 - 1 is an odd integer --> \(n^2-1=odd\) --> \(n^2=odd+1=even\). Now, since \(n\) is an integer, then in order \(n^2\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some irrational number (square root of an even number), for example \(\sqrt{2}\), so not an even integer.

(2) 3n + 4 is an even integer --> \(3n + 4=even\) --> \(3n=even-4=even\). The same here, since \(n\) is an integer, then in order \(3n\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some fraction, for example \(\frac{2}{3}\), so not an even integer.

(1) n^2 - 1 is an odd integer --> \(n^2-1=odd\) --> \(n^2=odd+1=even\). Now, since \(n\) is an integer, then in order \(n^2\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some irrational number (square root of an even number), for example \(\sqrt{2}\), so not an even integer.

(2) 3n + 4 is an even integer --> \(3n + 4=even\) --> \(3n=even-4=even\). The same here, since \(n\) is an integer, then in order \(3n\) to be even \(n\) must be even. Sufficient. Notice that if we were not told that \(n\) is an integer, then \(n\) could be some fraction, for example \(\frac{2}{3}\), so not an even integer.

Answer: D.

Since n is a integer, can we not try with n as 0?

Yes, n can be 0 but 0 is even too. _________________

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. _________________

(1) n^2 - 1 is an odd integer. (2) 3n + 4 is an even integer.

Given: n is an integer Required: is n even?

Statement 1: \(n^2\) - 1 is an odd integer \(n^2\) - 1 = (n-1)(n+1) = odd. This means both n-1 and n+1 are odd Odd*Odd = Odd Odd*Even = Even Even*Even = Even

n-1, n, n+1 are three consecutive integers. Since we know that both n-1 and n+1 are odd Hence n has to be even. SUFFICIENT

Statement 2: 3n + 4 is an even integer Even + Even = Even Even + Odd = Odd Odd + Odd + Odd

Since 3n+4 = even and 4 is an even integer. Hence 3n = even. Therefore n = even SUFFICIENT

(1) n^2 - 1 is an odd integer. (2) 3n + 4 is an even integer.

We need to determine whether integer n is even. Let's review four facts about even and odd integers: 1) An integer and its square are either both even or both odd. 2) The sum (or difference) between an even integer and an odd integer is always odd. 3) The sum of two even integers (or two odd integers) is always even. 4) If the product of two integers is even, at least one of them must be even.

Statement One Alone:

(n^2) - 1 is an odd integer.

Since (n^2) - 1 is an odd integer, we know that n^2 must be even and thus n must be even. Statement one is sufficient to answer the question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

3n + 4 is an even integer.

Since 3n + even = even integer, we know that 3n must be even, and since 3 is odd, n must be even. Statement two is sufficient to answer the question.

The answer is D. _________________

Jeffrey Miller Scott Woodbury-Stewart Founder and CEO

We need to find if x is even or odd given => x is an integer statement 1 =>n^2-1=odd => n^2=> even => n must be even Rule used => POWER Does not effect the even / odd nature of any number

statement 2 => 3n+4=even=> 3n=even => n must even Rulw => if XY=even => at-least one of them must be even Hence suff

Smash that D _________________

Give me a hell yeah ...!!!!!

gmatclubot

If n is an integer, is n even?
[#permalink]
22 Aug 2016, 02:18

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...