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:

I like this one.I do not find it discussed earlier on the board.

OA later

Pick odd and even numbers, e.g. 7 and 8:

A) (7)(8)(3) Y ; (8)(9)(5) Y B) (7)(9)(6) Y ; (8)(10)(7) N C) (7)(10)(2) N; No need to check for 8 D) (7)(11)(5) N; No need to check for 8 E) (7)(12)(1) Y; (8)(13)(2) N

Since 3 is a prime number then in order the product to be divisible by 3 either of the multiples must be divisible by 3. Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3, meaning that one of them should have the remainder of 1, another the reminder of 2 and the third one the remainder of 0, so be divisible by 3.

For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers.

Re: If n is an integer greater than 6, which of the following mu [#permalink]

Show Tags

22 Sep 2012, 10:06

Bunuel wrote:

carcass wrote:

If n is an integer greater than 6, which of the following must be divisible by 3 ?

Since 3 is a prime number then in order the product to be divisible by 3 either of the multiples must be divisible by 3. Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3, meaning that one of them should have the remainder of 1, another the reminder of 2 and the third one the remainder of 0, so be divisible by 3.

For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers.

Hi Bunuel - I didn't quite follow this logic. Could you please elaborate. Thanks.

If n is an integer greater than 6, which of the following must be divisible by 3 ?

Since 3 is a prime number then in order the product to be divisible by 3 either of the multiples must be divisible by 3. Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3, meaning that one of them should have the remainder of 1, another the reminder of 2 and the third one the remainder of 0, so be divisible by 3.

For option A: n and n+1 have different remainder upon division by 3. As for n-4, it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers.

Hi Bunuel - I didn't quite follow this logic. Could you please elaborate. Thanks.

Cheers

Can you please tell me which part didn't you understand?

Shortcut: In every set of 3 consecutive numbers, ONE of them must be divisible by 3 when we are multiplying each of the digits

1*2*3 2*3*4 3*4*5 4*5*6

(A) n(n + 1)(n – 4)

The easiest is if we have something like n(n+1)(n+2) We know in this case we DEFINITELY have an expression that is divisible by 3. n=1 => 1*2*3 n=2 => 2*3*4 n=3 => 3*4*5

All are divisible by 3.

Any expression must pass our 3 consecutive integer test.

Even if n = 4 we have: 4*5*0 = 0 which is divisible by 3.

If n = 16 16*17*12 is divisible by 3.

So (A) passes all the tests and one of numbers in the expression will be divisible by 3 so the whole expression when multiplied together will be divisible by 3.
_________________

Re: If n is an integer greater than 6, which of the following mu [#permalink]

Show Tags

25 Sep 2012, 20:36

gmatpill wrote:

Shortcut: In every set of 3 consecutive numbers, ONE of them must be divisible by 3 when we are multiplying each of the digits

1*2*3 2*3*4 3*4*5 4*5*6

(A) n(n + 1)(n – 4)

The easiest is if we have something like n(n+1)(n+2) We know in this case we DEFINITELY have an expression that is divisible by 3. n=1 => 1*2*3 n=2 => 2*3*4 n=3 => 3*4*5

All are divisible by 3.

Any expression must pass our 3 consecutive integer test.

Even if n = 4 we have: 4*5*0 = 0 which is divisible by 3.

If n = 16 16*17*12 is divisible by 3.

So (A) passes all the tests and one of numbers in the expression will be divisible by 3 so the whole expression when multiplied together will be divisible by 3.

11 has a reminder 4 on division by 7 12 has a reminder 5 on division by 7 What is the reminder x * y divide by 7? - I can multiply reminders as long as I correct the excess amount... so 4 *5 = 20 20 - 7*2 = 6, which is the reminder when x * y / 7 - SO FAR CORRECT?

Reminder when 99 / 15 give me a reminder of 9 But when i factorize the den into primes = 9 * 11 / 3 * 5 => 9/3 leaves reminder 0 and 11/5 reminder leaves 1 On multiplying the reminders, we get zero. So I should not factorize the denominator unless the denominator factor into the same primes

Re: If n is an integer greater than 6, which of the following mu [#permalink]

Show Tags

01 Oct 2012, 08:11

I tried plugging in odd and even values for n and got the answer A ...

If we assume n as being either 7 , 8 or 9 we can easily see why A is the correct answer .. Because in every instance , either of the 3 numbers multiplied will be divisible by 3 and the entire product will meet the requirements..

if n =7 , then n-4 is divisible by 3 ... If n= 8 then n+1 is divisible by 3 If n= 9 , then n is divisible by 3 ...
_________________

"When you want to succeed as bad as you want to breathe, then you’ll be successful.” - Eric Thomas

Re: If n is an integer greater than 6, which of the following mu [#permalink]

Show Tags

08 Feb 2013, 06:14

just draw a table for each ans option plug in starting 7,8,9 etc. It should not take more than 75 s.
_________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

The best way to tackle these kind of questions is by assuming values.. Since its says, n has to be greater than 6...assume the value of n to be 7....by this u will be able to eliminate options 3 and 4... Now take the value of n as 8....by this u will be able to eliminate options 2 and 5... Now we are left with only option 1...and that is the answer..

Re: If n is an integer greater than 6, which of the following mu [#permalink]

Show Tags

25 Feb 2013, 20:28

the most easy way is to pick numbers and plug in each choice. in the test room, we have no time for thinking hard way. og explantion explicitely declare this point.

Re: If n is an integer greater than 6, which of the following mu [#permalink]

Show Tags

31 Mar 2013, 04:55

4

This post received KUDOS

2

This post was BOOKMARKED

Since this is a must be true question it must be true for any n and as we know any number is divisible by 3 if the sum of its digits is divisible by 3, so if we add up the digits or terms in the options we get the answer

(A) n + (n + 1) + (n – 4) = 3n - 3 ---- divisible by 3 for any n (B) n + (n + 2) + (n – 1) = 3n + 1 (C) n + (n + 3) + (n – 5) = 3n - 2 (D) n + (n + 4) + (n – 2) = 3n + 2 (E) n + (n + 5) + (n – 6) = 3n - 1
_________________

Life is a highway I wanna ride it all night long

Last edited by prasun9 on 08 Oct 2013, 04:38, edited 2 times in total.

Re: If n is an integer greater than 6, which of the following mu [#permalink]

Show Tags

03 Jul 2013, 19:58

1

This post received KUDOS

We need to check only for 3 values of n because the answer will be the same for every 3 consecutive integers. Actually we need not consider one of those values as it is always a multiple of 3.

So we can take the values of n as 7 and 8 or 10 and 11 or 13 and 14 etc

If we plug in the values 7 and 8 we get the answer once we check A as it satisfies both the values. We need not actually check other choices as they would fail for either 7 or 8.
_________________

Which method is better in questions like these to apply on the test? to pick the numbers or to apply concept of divisibility by 3 ( Remainder 0,1,2)

In my opinion, using logic is almost always better than plugging in. Plugging in is full of possibilities of making mistakes - incorrect calculation, not considering all possibilities, getting lost in the options etc.

Logic is far cleaner.

I know that talking about positive integers, in any set of 3 consecutive positive integers, one integer will be divisible by 3 and the other 2 will not be.

So I am looking for 3 consecutive positive integers e.g. (n-1)n(n+1)

Note that (n-4) is equivalent to (n-1) since if (n-4) is divisible by 3, so is (n-1). If (n-4) is not divisible by 3, neither is (n-1) (because the difference between these two integers is 3) Hence (A) is equivalent to 3 consecutive integers. Answer (A)

On same lines, note that n is equivalent to (n-3), (n + 3), (n + 6) etc.
_________________

Re: If n is an integer greater than 6, which of the following mu [#permalink]

Show Tags

04 Jul 2013, 01:04

VeritasPrepKarishma wrote:

pavan2185 wrote:

Which method is better in questions like these to apply on the test? to pick the numbers or to apply concept of divisibility by 3 ( Remainder 0,1,2)

In my opinion, using logic is almost always better than plugging in. Plugging in is full of possibilities of making mistakes - incorrect calculation, not considering all possibilities, getting lost in the options etc.

Logic is far cleaner.

I know that talking about positive integers, in any set of 3 consecutive positive integers, one integer will be divisible by 3 and the other 2 will not be.

So I am looking for 3 consecutive positive integers e.g. (n-1)n(n+1)

Note that (n-4) is equivalent to (n-1) since if (n-4) is divisible by 3, so is (n-1). If (n-4) is not divisible by 3, neither is (n-1) (because the difference between these two integers is 3) Hence (A) is equivalent to 3 consecutive integers. Answer (A)

On same lines, note that n is equivalent to (n-3), (n + 3), (n + 6) etc.

Your explanation completely cleared my doubt. Thank you

gmatclubot

Re: If n is an integer greater than 6, which of the following mu
[#permalink]
04 Jul 2013, 01:04

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...