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A. n (n+1) (n-4) = 7*8*3 and if n = 8 --> 8*9*4
B. n (n+2) (n-1) = 7*9*6 and if n = 8 --> 8*10*7
C. n (n+3) (n-5) = 7*10*5; eliminate as there are no multiples of 3
D. n (n+4) (n-2) = 7*11*5; eliminate as there are no multiples of 3
E. n (n+5) (n-6) = 7*12*1 and and if n = 8 --> 8*13*2

If n is an integer greater than 6, which of the following must be divisible by 3? A. n (n+1) (n-4) B. n (n+2) (n-1) C. n (n+3) (n-5) D. n (n+4) (n-2) E. n (n+5) (n-6)

anything in the form of (n-1) (n) (n+1) is divvisible by 3. in other word, a product of any 3 consecutie intevers is divisible by 3.

A. n (n+1) (n-4) = n (n+1) ((n-1)-3) is equivalant to (n-1) (n) (n+1)
B. n (n+2) (n-1) is equivalant to (n+1) missing.
C. n (n+3) (n-5) is equivalant to (n-1) missing and n repeating.
D. n (n+4) (n-2) is equivalant to odd/even consqcutive integers
E. n (n+5) (n-6) is equivalant to (n+1) missing and n repeating.

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

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23 Mar 2012, 20:04

IMO A.

I used an arbitrary number greater than 6 and then filled each equation out. if you happen to have chosen a number that makes more than 1 answer correct, choose a different number and check the ones that were previously correct.
_________________

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

A. n (n+1) (n-4) B. n (n+2) (n-1) C. n (n+3) (n-5) D. n (n+4) (n-2) E. n (n+5) (n-6)

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 [#permalink]

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25 Mar 2012, 09:30

2

This post was BOOKMARKED

Bunuel wrote:

gregspirited wrote:

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

A. n (n+1) (n-4) B. n (n+2) (n-1) C. n (n+3) (n-5) D. n (n+4) (n-2) E. n (n+5) (n-6)

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 [#permalink]

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15 Apr 2012, 16:12

Bunuel wrote:

gregspirited wrote:

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

A. n (n+1) (n-4) B. n (n+2) (n-1) C. n (n+3) (n-5) D. n (n+4) (n-2) E. n (n+5) (n-6)

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, could you please explain how you arrive at the conclusion that "As for n-4, it will have the same remainder as (n-4)+3=n-1"? Also, is it implied that n will have a remainder of either 0 or 1, n+1 will have either 1 or 2? Thanks!

A. n (n+1) (n-4) = n (n+1) ((n-1)-3) is equivalant to (n-1) (n) (n+1)

So A is good.

Hi Bunuel, Why is n (n+1) (n-4) = n (n+1) ((n-1)-3) is equivalant to (n-1) (n) (n+1)?

Also, you said.. 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

Could you please elaborate the 2 statements above?
_________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

Can you explain why the remainder should be different upon division by 3? I atill dint understood uproach

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An integer divided by 3 can have 3 possible remainders: 0, 1, or 2.

Now, consider the product of three numbers a*b*c. If we are told that a, b, and c have different reminders upon division by three, this would mean that one of the numbers yields the remainder of zero, thus it's a multiple of 3. Thus abc is a multiple of 3.

If n is an integer greater than 6, which of the following must be divisible by 3? A. n (n+1) (n-4) B. n (n+2) (n-1) C. n (n+3) (n-5) D. n (n+4) (n-2) E. n (n+5) (n-6)

anything in the form of (n-1) (n) (n+1) is divvisible by 3. in other word, a product of any 3 consecutie intevers is divisible by 3.

A. n (n+1) (n-4) = n (n+1) ((n-1)-3) is equivalant to (n-1) (n) (n+1) B. n (n+2) (n-1) is equivalant to (n+1) missing. C. n (n+3) (n-5) is equivalant to (n-1) missing and n repeating. D. n (n+4) (n-2) is equivalant to odd/even consqcutive integers E. n (n+5) (n-6) is equivalant to (n+1) missing and n repeating.

So A is good.

I think your solution for option A is incorrect. When you write an expression like this n (n+1) ((n-1)-3) , you are multiplying -3 with the whole expression, which eventually will turn out to (-3n+3), that is not equal to what is given. Correct me if I am wrong

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

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19 Oct 2013, 02:39

Bunuel wrote:

gregspirited wrote:

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

A. n (n+1) (n-4) B. n (n+2) (n-1) C. n (n+3) (n-5) D. n (n+4) (n-2) E. n (n+5) (n-6)

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.

I took the picking numbers method but I want to understand yours as well. I understand that n (n+1), n (n+2), n (n+3)...etc will all have different remainders, and I understand the concept about the different remainders. I don't understand how you reached this statement: "As for n-4, it will have the same remainder as (n-4)+3=n-1" Can you please elaborate on that?

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

A. n (n+1) (n-4) B. n (n+2) (n-1) C. n (n+3) (n-5) D. n (n+4) (n-2) E. n (n+5) (n-6)

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.

I took the picking numbers method but I want to understand yours as well. I understand that n (n+1), n (n+2), n (n+3)...etc will all have different remainders, and I understand the concept about the different remainders. I don't understand how you reached this statement: "As for n-4, it will have the same remainder as (n-4)+3=n-1" Can you please elaborate on that?

n-1=(n-4)+3. Now, since 3 IS divisible by 3, then the remainder when (n-4)+3 is divided by by 3 is the same as n-4 is divided by 3.

For example, say n=11: The remainder when n-1=11-1=10 divided by 3 is 1. The remainder when n-4=11-4=7 divided by 3 is also 1.

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

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07 Mar 2014, 23:03

I think I solved this by luck, as per below, by choosing the only answer where the sum of the digits was divisible by 3. However, testing my method by using another possible answer where the sum of the two digits is divisible by 3, shows that it is not always correct. Best to use the test method with values of n, that are not divisible by 3, in order to eliminate incorrect answer choices.

A) 1+(-4) = -3 - only answer where the sum of the two is divisible by 3. B) 2+(-1) = 1 C) 3+(-5) = -2 D) 4+(-2) = 2 E) 5+(-6)= 1

Testing another possible answer where the sum of the two ;

n(n+2)(n-5) 2+(-5)=-3 Testing with values for n that are not divisible by 3;

n=7; 7(7+2)(7-5); 7(9)(2) - has a factor divisible by 3 n=8; 8(8+2)(8-5); 8(10)(3) - has a factor divisible by 3 n=10; 10(12)(5) - has factor divisible by 3 n=11; 11(13)(6) - has factor divisible by 3

n(n-6)(n+3) -6+3=-3 n=7 ; 7(1)(10) not divisible by 3

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

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25 May 2014, 08:27

Bunuel wrote:

gregspirited wrote:

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

A. n (n+1) (n-4) B. n (n+2) (n-1) C. n (n+3) (n-5) D. n (n+4) (n-2) E. n (n+5) (n-6)

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.

Why is the highlighted part important? Is that just another way of saying that at least one of the numbers in this sequence is divisible by 3. Correct?

Additionally, I see the correlation you made between n-4 and n-1(both leave a remainder of 1) but by that token, shouldn't (n+1) or (n) leave a remainder of 2 and 1 respectively? Meaning, (11+1) = 12/3 = no remainder and n (11) leaves a remainder of 2. So now we have remainders of 1,2,3 and therefore a consecutive set of integers. Is that the reason we want three different remainders?

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

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19 Jan 2015, 00:24

Simply look for product of 3 consecutive integers....or their equivalents....i.e if a number falls under consecutive category +/- product of 3...then it work too

A )n * (n+1)* (n-4)

equivalent of (n-4)= n-1

and n-1,n,n+1 are 3 consecutive....and this prod is divisible by 3 for any integer value of n

Ans A

gmatclubot

Re: If n is an integer greater than 6, which of the following
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19 Jan 2015, 00:24

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