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"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.
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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
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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

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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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
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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.

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

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01 Dec 2015, 23:09

Thanks for your reply EMPOWERgmatRichC! I have read every single expert and non expert reply on this question on this and other forums. This by far the best and most lucid explanation I have seen. I have learnt a very powerful technique here- to test for the divisibility of 3, you should see if you're able to create three consecutive numbers by arithmetic manipulation using 3. I guess similarly for divisibility by 4 we can create 4 consec numbers by arithmetic manupulation using 4 etc. To appreciate beauty of your approach I needed the theory explained so brilliantly by VeritasPrepKarishma in her blog posts on this topics. Thanks guys!

I also agree that number testing could be far straightforward approach. but I felt it was time consuming so was looking for a more intuitive conceptual approach. Thanks again
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My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

I stumbled upon your explanation while working on this OG 16 question. I found it to be quite intuitive and helpful. Sorry, for asking question on something posted so long back

I have a question on this approach (relevant part highlighted). Is it necessary that if a number is divisible 3, the sum of individual digits of its factors should be divisible by 3?

Lets take the case of 30 which is divisible by 30. If I write 30 = 2*5*3 - the sum of digits = 2+5+3 = 10 is not divisible by 3. What am I missing here?

Thanking you in advance!

carcass wrote:

I attacked the problem in this was, tell me if pron of errors

n is an integers, so we can choose 7, 8, 9 and so on. Now, a number divisible by 3 the sum of number MUST be divisible by 3.

1) n (n+1)(n-4) ---> 8 * 9 * 4 ---> without perform multiplication the SUM of 8 + 9 + 4 = 21 and is divisible by 3 without reminder

To solve semi logically, take n in each. In choice A, either adding 1 to n or subtracting 4 from n will give you a multiple of 3. If neither n will be a multiple of 3.So it has to be divisible by 3.

To illustrate take the extreme values. In A, (n-4) and (n+1) are the extreme values. There is a difference of 5. (3,8), (4,9), (5,10) are examples . (n-4), (n+1) and n resp. are the multiples of 3 in these examples, making the expression divisible by 3.

Seeing 2 values at a time reduces the clutter.
_________________

I stumbled upon your explanation while working on this OG 16 question. I found it to be quite intuitive and helpful. Sorry, for asking question on something posted so long back

I have a question on this approach (relevant part highlighted). Is it necessary that if a number is divisible 3, the sum of individual digits of its factors should be divisible by 3?

Lets take the case of 30 which is divisible by 30. If I write 30 = 2*5*3 - the sum of digits = 2+5+3 = 10 is not divisible by 3. What am I missing here?

Thanking you in advance!

carcass wrote:

I attacked the problem in this was, tell me if pron of errors

n is an integers, so we can choose 7, 8, 9 and so on. Now, a number divisible by 3 the sum of number MUST be divisible by 3.

1) n (n+1)(n-4) ---> 8 * 9 * 4 ---> without perform multiplication the SUM of 8 + 9 + 4 = 21 and is divisible by 3 without reminder

The rest of choices do not work if you try.

thanks

This is not true. 8*9*4 is divisible by 3 because one of the factors is 9. So of course the product has 3 as a factor.

I stumbled upon your explanation while working on this OG 16 question. I found it to be quite intuitive and helpful. Sorry, for asking question on something posted so long back

I have a question on this approach (relevant part highlighted). Is it necessary that if a number is divisible 3, the sum of individual digits of its factors should be divisible by 3?

Lets take the case of 30 which is divisible by 30. If I write 30 = 2*5*3 - the sum of digits = 2+5+3 = 10 is not divisible by 3. What am I missing here?

Thanking you in advance!

carcass wrote:

I attacked the problem in this was, tell me if pron of errors

n is an integers, so we can choose 7, 8, 9 and so on. Now, a number divisible by 3 the sum of number MUST be divisible by 3.

1) n (n+1)(n-4) ---> 8 * 9 * 4 ---> without perform multiplication the SUM of 8 + 9 + 4 = 21 and is divisible by 3 without reminder

The rest of choices do not work if you try.

thanks

This is not true. 8*9*4 is divisible by 3 because one of the factors is 9. So of course the product has 3 as a factor.

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.

Bunuel good day! can you please explain this part "it will have the same remainder as (n-4)+3=n-1, so also different than the remainders of the previous two numbers" I dont understand the logic behind this (n-4)+3=n-1 have a great day thanks!