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 350,000 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:

Re: a, b, and c are consecutive even integers... [#permalink]
13 Dec 2011, 13:54

4

This post received KUDOS

Expert's post

metallicafan wrote:

If a, b, and c are consecutive even integers and a < b < c, all of the following must be divisible by 4 EXCEPT

(A) a + c (B) b + c (C) ac (D) (bc)/2 (E) (abc)/4

Is there a way to solve it quickly without picking numbers? Algebraicaly, requires more than 2 min.

You can rule out C, D and E quite quickly if you know one thing: if you have a list of consecutive even numbers, every second number in that list will be a multiple of 4. So if a, b and c are consecutive even numbers, in order, then either b is divisible by 4, or a and c are both divisible by 4. Either way, ac will be divisible by at least 4, bc will be divisible by at least 8, and abc will divisible by at least 16, from which C, D and E must all be divisible by 4.

To decide between A and B, you can pick numbers, or you can write the three consecutive even integers as 2k, 2k + 2 and 2k + 4. Then a + c = (2k) + (2k + 4) = 4k + 4 = 4(k + 1), so a+c must be divisible by 4. That leaves B as the only possible answer.

I think picking numbers is perfectly fine here, since if you use 2, 4 and 6, say, it doesn't take too long. In questions with setups different from the one here, however, you may want to use a more conceptual approach. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: a, b, and c are consecutive even integers... [#permalink]
13 Dec 2011, 15:47

IanStewart wrote:

In questions with setups different from the one here, however, you may want to use a more conceptual approach.

Thank you! Do you have some rules or principles that could help us in deciding whether we should picki numbers or solve it algebraically?

IMO, I think that always we have to try to solve it algebraicaly because it is safer. We have to pick numbers when we are sure that one case or few cases represent every possible situation. _________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

Re: a, b, and c are consecutive even integers... [#permalink]
13 Dec 2011, 21:26

1

This post was BOOKMARKED

I did get the correct answer within 80 seconds and it was not by luck either. I did not pick numbers but just used the concept stated by Ian.

The 3 numbers can be written as a, (a + 2) & (a + 4). If 'a' is divisible by 4, then even 'c' or 'a + 4' is divisible by 4. However, is 'b' is divisible by 4, then both 'a' and 'a + 4' are still divisible by 2.

A - (a + c) = a + (a + 4) = 2a + 4 = 2(a + 2) = 2b. 2b will always be divisible by 4 even if 'b' is not divisible by 4. Reason: 'b' already has a prime factorization of at least a '2'. Hence '2b' has two 2s. C - ac = a(a+4). If, as stated above, one of them is divisible by 4, then the product is divisible. If both of them are not divisible by 4, then the product is still divisible by 4 because of the presence of two 2s again in the prime factorization. D - bc/2 = (a + 2)(a + 4)/2. Either b or c is divisible by 2. Hence, if we assume that b is divisible by 2 and not divisible by 4, then it leaves us just one possibility. Is c divisible by 4? It has to be because c is the next consecutive even integer. E - abc/4 = a(a + 2)(a + 4)/4. One of these integers is divisible by 4 already. If we again assume 'b' to be that integer divisible by 4, then we are left with the question - Is a(a + 4) divisible by 4? This is the same as option C.

B - b + c = (a + 2) + (a + 4) = 2a + 6 = 2(a + 3). (a + 3) will never be divisible by 2 because it is an odd integer. Hence, 2(a + 3), although divisible by 2, will not be divisible by 4 because it has just one 2 in its prime factorization.

As a whole, whether you choose numbers (2, 4 & 6 being the easiest) or solve conceptually, the answer is still easily obtainable within 2 minutes.

Re: a, b, and c are consecutive even integers... [#permalink]
15 Dec 2011, 04:34

Expert's post

kdas wrote:

Great reply Ian. How much do these books cost. How will I get these books. What does "advanced" mean - 50, 51 level

I'll make a more detailed announcement about my materials soon, but I have problem sets designed for test takers in about the 45-51 range, and topic-specific books which are aimed at above average test takers. If you'd like more information, feel free to send me an email at the address in my signature. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: a, b, and c are consecutive even integers... [#permalink]
01 Nov 2013, 11:34

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: a, b, and c are consecutive even integers... [#permalink]
06 Dec 2014, 07:23

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. _________________

How the growth of emerging markets will strain global finance : Emerging economies need access to capital (i.e., finance) in order to fund the projects necessary for...

One question I get a lot from prospective students is what to do in the summer before the MBA program. Like a lot of folks from non traditional backgrounds...