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

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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]

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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]

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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]

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08 Oct 2013, 21:25

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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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]

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20 Mar 2015, 11:22

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

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01 Dec 2015, 03:28

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

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01 Dec 2015, 21:43

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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.

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