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:

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

Show Tags

09 Jul 2013, 01:16

Expert's post

VeritasPrepKarishma wrote:

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.

Responding to a pm:

As said here, I prefer to use logic rather than number plugging. Number plugging is usually the last resort. Given above is the logical solution to this problem. For more on this topic, check:

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

Show Tags

09 Jul 2013, 06:42

VeritasPrepKarishma wrote:

VeritasPrepKarishma wrote:

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.

Responding to a pm:

As said here, I prefer to use logic rather than number plugging. Number plugging is usually the last resort. Given above is the logical solution to this problem. For more on this topic, check:

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

Show Tags

09 Jul 2013, 21:34

Expert's post

JaJaIrie wrote:

I wish someone would solve this question step by step.

That's the point - there are no steps here. There is no process you have to follow. You just need to understand the logic and you will have your answer.

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

Show Tags

08 Oct 2013, 21:25

1

This post received KUDOS

Expert's post

khairilthegreat wrote:

Dear Bunuel,

"Now, to guarantee that at least one multiple is divisible by 3, these numbers must have different remainders upon division by 3"

Why is this? Why the numbers must have different reminders?

Any positive number will take one of three forms: 3m, (3m+1) or (3m+2) i.e. either it will be divisible by 3, will leave remainder 1 or will leave remainder 2 when divided by 3. If the number takes the form 3m, the number after it is of the form 3m+1 and the one after it is of the form 3m+2.

If we have 3 consecutive numbers such as a, (a+1), (a+2), we know for sure that at least one of them is divisible by 3 since one of them will be of the form 3m. We don't know which one but one of them will be divisible by 3.

So given numbers such as (n-1)*n*(n+1), we know that the product is divisible by 3. In the given options, we don't know whether n is divisible by 3 or not. We need to look for the option which has 3 consecutive numbers i.e. in which the terms leave a remainder of 0, 1 and 2 to be able to say that the product will be divisible by 3.

Note a product such as n(n+3)(n+6). When this is divided by 3, we cannot say whether it is divisible or not because all three factors will leave the same remainder, 1. Say n = 4. Product 4*7*10. All these factors are of the form 3m+1. We don't have any 3m factor here. So we need the factors to have 3 different remainders so that one of them is of the form 3m. _________________

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

Show Tags

09 Oct 2013, 21:43

Expert's post

ygdrasil24 wrote:

(n-4)+3=n-1

How to conclude on this ?

When talking about 'divisibility by 3', (n - 4) and (n - 1) are of the same form. This means that if (n-1) is divisible by 3, so is (n - 4). If (n-1) leaves a remainder of 1 when divided by 3, so does (n-4). If (n-1) leaves a remainder of 2 when divided by 3, so does (n-4). the reason for this is that (n-1) and (n-4) have a difference of 3 between them.

If (n-4) = 5, (n-1) = 8. Both leave remainder 2 when divided by 3 If (n-4) = 6, (n-1) = 9. Both leave remainder 0 when divided by 3 If (n-4) = 7, (n-1) = 10. Both leave remainder 1 when divided by 3 _________________

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

Show Tags

10 Oct 2013, 01:44

VeritasPrepKarishma wrote:

ygdrasil24 wrote:

(n-4)+3=n-1

How to conclude on this ?

When talking about 'divisibility by 3', (n - 4) and (n - 1) are of the same form. This means that if (n-1) is divisible by 3, so is (n - 4). If (n-1) leaves a remainder of 1 when divided by 3, so does (n-4). If (n-1) leaves a remainder of 2 when divided by 3, so does (n-4). the reason for this is that (n-1) and (n-4) have a difference of 3 between them.

If (n-4) = 5, (n-1) = 8. Both leave remainder 2 when divided by 3 If (n-4) = 6, (n-1) = 9. Both leave remainder 0 when divided by 3 If (n-4) = 7, (n-1) = 10. Both leave remainder 1 when divided by 3

Okay. Thanks. Its a bit awkward to get the logic initially. usually when asked for division by 3, i would look for a 3N type number. If the question had , say, to check for 4, then in that case we should be having 4 multipliers of n something like N(N-2)(N+3)(N-4) and so on ? So that we can have remainders as 0,1,2,3

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

Show Tags

10 Oct 2013, 04:35

Expert's post

ygdrasil24 wrote:

VeritasPrepKarishma wrote:

ygdrasil24 wrote:

(n-4)+3=n-1

How to conclude on this ?

When talking about 'divisibility by 3', (n - 4) and (n - 1) are of the same form. This means that if (n-1) is divisible by 3, so is (n - 4). If (n-1) leaves a remainder of 1 when divided by 3, so does (n-4). If (n-1) leaves a remainder of 2 when divided by 3, so does (n-4). the reason for this is that (n-1) and (n-4) have a difference of 3 between them.

If (n-4) = 5, (n-1) = 8. Both leave remainder 2 when divided by 3 If (n-4) = 6, (n-1) = 9. Both leave remainder 0 when divided by 3 If (n-4) = 7, (n-1) = 10. Both leave remainder 1 when divided by 3

Okay. Thanks. Its a bit awkward to get the logic initially. usually when asked for division by 3, i would look for a 3N type number. If the question had , say, to check for 4, then in that case we should be having 4 multipliers of n something like N(N-2)(N+3)(N-4) and so on ? So that we can have remainders as 0,1,2,3

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

Show Tags

29 Oct 2013, 16:50

Try with 1, and that will limit us to option A & D. Alternatively, looking at all the options, only A will result to 3. This is how I did 1. N+1 + N -4 = 2n-3 (Any no. will give a multiple of 3). 2. N-1 + N+2 = 2n+1 3. N-5 + N+3 = 2n-2 4. N-2 + N+4 = 2n +2 5. N-6 + N+5 = 2n -1

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

Show Tags

30 Oct 2013, 18:50

multiplication of three consecutive numbers is always divisible by 3, i.e (n-1)*n*(n+1) or n*(n+1)*(n+2) or (n-2)*(n-1)*n check the option that represents any of the above types.

a) satisfies (n-1)*n*(n+1) condition (n-4)= (n-1) because difference of both is 3 for verification substitute n=13, 14, 19, 10

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

Show Tags

09 Mar 2014, 11:00

Multiple brilliant answers in this wonderful forum... as always. However... I found a way that I -personally- understood better, and a way I think I can use in a time-constrained manner in the actual exams.

According to GMAT materials, the ways where an integer can be divided by 3 are 1) when the sum of the integer DIGITS is divisible by 3 2) the integer have multiple of 3's 3) a consecutive set of 3 integers can be divisible by both 2 and 3 (factor foundation rule)

There are two ways to solve this 1) Plugging in: Undoubtedly, plugging in numbers to the 5 options below is perhaps the simplest and easiest way to answer this question. To save time on computation, using the sum of integer DIGITS may accelerate your process, or looking for multiples of 3s. For example: Option A: n(n + 1)(n - 4) = 7(7 +1)(7 - 4) = (7)(8)(3)

2) Look for patterns Given the answer options, which are adjusting the position of n on the number lines, another way is to scan if any of these answer choices provide clues to consecutive numbers.

In Option (A): we have a partial consecutive set of N, N+1. It looks like it is missing a N+2 or N-1. OR is it? Now look closer at (N-4). From a consecutive set perspective, (N-4) is also the same "position" as (N-1) Therefore this answer choice has N, N+1 and N-1. This is a consecutive set that is both divisible by 2 and 3.

All other options will not have such a relationship: Option (B) n(n + 2)(n – 1): We have N, N+2, and (N-1 or N+3....). Therefore missing N+1 to make a consecutive set

Option (C) n(n + 3)(n – 5) We have N, N+3, and (N-5 or N-2 or N+1....) Therefore missing N+2 to make a consecutive set

Option (D) n(n + 4)(n – 2) We have N, N+4 and (N-2, N+1, N+4...). Therefore missing N+2 to make a consecutive set

Option (E) n(n + 5)(n – 6) We have N, N+5 and (N-6, N-3, N, N+3....). Therefore missing N+2 to make a consecutive set

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

Show Tags

09 Mar 2014, 11:02

Multiple brilliant answers in this wonderful forum... as always. However... I found a way that I -personally- understood better, and a way I think I can use in a time-constrained manner in the actual exams.

Concept According to GMAT materials, the ways where an integer can be divided by 3 are 1) when the sum of the integer DIGITS is divisible by 3 2) the integer have multiple of 3's 3) a consecutive set of 3 integers can be divisible by both 2 and 3 (factor foundation rule)

There are two ways to solve this 1) Plugging in: Undoubtedly, plugging in numbers to the 5 options below is perhaps the simplest and easiest way to answer this question. To save time on computation, using the sum of integer DIGITS may accelerate your process, or looking for multiples of 3s. For example: Option A: n(n + 1)(n - 4) = 7(7 +1)(7 - 4) = (7)(8)(3)

2) Look for patterns Given the answer options, which are adjusting the position of n on the number lines, another way is to scan if any of these answer choices provide clues to consecutive numbers.

In Option (A): we have a partial consecutive set of N, N+1. It looks like it is missing a N+2 or N-1. OR is it? Now look closer at (N-4). From a consecutive set perspective, (N-4) is also the same "position" as (N-1) Therefore this answer choice has N, N+1 and N-1. This is a consecutive set that is both divisible by 2 and 3.

All other options will not have such a relationship: Option (B): n(n + 2)(n – 1): We have N, N+2, and (N-1 or N+3....). Therefore missing N+1 to make a consecutive set

Option (C): n(n + 3)(n – 5) We have N, N+3, and (N-5 or N-2 or N+1....) Therefore missing N+2 to make a consecutive set

Option (D): n(n + 4)(n – 2) We have N, N+4 and (N-2, N+1, N+4...). Therefore missing N+2 to make a consecutive set

Option (E): n(n + 5)(n – 6) We have N, N+5 and (N-6, N-3, N, N+3....). Therefore missing N+2 to make a consecutive set

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

Show Tags

20 Mar 2015, 11:22

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

Show Tags

01 Dec 2015, 03:28

EMPOWERgmatRichC: can you also provide some insights into this question? What principle is GMAT testing here? What's the most efficient way to tackle it etc. Thanks _________________

Please consider giving 'kudos' if you like my post and want to thank

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

Show Tags

01 Dec 2015, 21:43

1

This post received KUDOS

Expert's post

Hi MensaNumber,

This question can be solved rather easily by TESTing VALUES, although the work itself will take a bit longer than average and it would help a great deal if you could spot the subtle Number Properties involved.

From the question stem, you can see that we're dealing with division by 3 (or the 'rule of 3', if you learned the concept that way). You don't actually have to multiply out any of the answer choices though - you just need to find the one answer that will ALWAYS have a '3' in one of its 'pieces.' The subtle Number Property I referred to at the beginning is the 'spacing out' of the terms.

(1)(2)(3) is a multiple of 3, since it's 3 times some other integers.

(5)(6)(7) is also a multiple of 3, since we can find a 3 'inside' the 6, so we have 3x2 times some other integers.

Looking at the answer choices to this question, we're clearly NOT dealing with consecutive integers, but the 'cycle' of integers is something that we can still take advantage of.

For example, we know that... When n is an integer, (n+1)(n+2)(n+3) will include a multiple of 3, since it's 3 consecutive integers (one of those 3 terms MUST be a multiple of 3, even if you don't know exactly which one it is).

You can take this same concept and 'move around' any (or all) of the pieces:

(n+1)(n+2)(n+6) will also include a multiple of 3 (that third term is 3 'more' than 'n+3').

Instead of adding a multiple of 3 to a term, you could also subtract a multiple of 3 from a term.

eg. (n-2)(n+2)(n+3) will also include a multiple of 3 (that first timer is 3 'less' than 'n+1').

The correct answer to this question subtracts a multiple of 3 from one of the terms.

All things being equal, I'd still stick to TESTing VALUES (and not approaching the prompt with math theory) - the math is easy and you can put it 'on the pad' with very little effort.

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...

This highly influential bestseller was first published over 25 years ago. I had wanted to read this book for a long time and I finally got around to it...