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If n is an integer greater than 6, which of the following

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

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New post 19 Sep 2016, 20:11
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.

Answer: A.

Hope it helps.


"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. "

Can anyone please explain -
1. why the remainders must be different?
2. how do you know the remainders are different ?
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If n is an integer greater than 6, which of the following  [#permalink]

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New post 13 Oct 2016, 03:08
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.

Answer: A.

Similar question to practice: if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html

Hope it helps.


Hi Bunuel ,

how general can we make this concept ( you mentioned " since 3 is prime" so does this mean it works for all primes but not for all non primes??)
also to me the remainders idea is the same as saying ( 3 consecutive integers) am i right or am i missing something ?

would appreciate your feedback.
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Re: If n is an integer greater than 6, which of the following  [#permalink]

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New post 13 Oct 2016, 03:48
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)


for any to be divisible by 3 it must be the product of 3 consecutive numbers or their equivalent.

E re written in terms of equivalent is n*n*(n-1) no consecutive
d re written in terms of equivalent is n ( n+1) (n+1) non
c ...................................................... is n*n*(n+1)
b ..........................................................n * (n-1) (n-1)
a........................................................... n (n+1) (n-1) ..... answer
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Re: If n is an integer greater than 6, which of the following  [#permalink]

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New post 25 Jan 2017, 07:46
1) Every third integer is divisible by 3, consequently the option that contains three consecutive integers or integers that are +/-3n from the three consecutive integers, then their product will be divisible by 3.
2) The desireable combination is n(n+1)(n+2)
3) A) n(n+1)(n-4) - if we add 2*3 to the last term, we get the desired outcome - n(n+1)(n+2)

A must be divisible by 3, and it must be the correct answer.

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

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New post 27 Feb 2017, 03:06
We can say that any integer can be expressed as 3a, 3a+1, 3a+2.if the expression holds true for the divisibility of 3 for all integer expression, then that will be the answer.
Option A:N = 3a, it will be 3a(3a+1)(3a-4). DIVISIBLE
N=3a+1, it will be 3a+1(3a+2)(3a-3). DIVISIBLE
N = 3a+2, it will be (3a+2)(3a+3)(3a-2). DIVISIBLE

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

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New post 31 Jul 2018, 00:27
GMAT TIGER 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)


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.


Would you please explain why ((n-1)-3) is equivalent to (n-1)
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If n is an integer greater than 6, which of the following  [#permalink]

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New post 02 Oct 2018, 09:36
GMAT TIGER 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)


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.



can you show how A is equal to (n-1)(n)(n+1)?

edit: i see that you have n (n+1) ((n-1)-3) but I can't get from there to (n-1)(n)(n+1)
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If n is an integer greater than 6, which of the following  [#permalink]

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New post 03 Nov 2018, 07:07
Bunuel would it be right to assume as follows:
Scenario 1: assume n = multiple of 3 hence all options are divisible by 3
Scenario 2: assume n not = multiple of 3 hence we look at the individual numbers being added

simply take the sum of the digits being added to n among all options and A is the only one that leaves us with a sum divisible by 3 hence A
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If n is an integer greater than 6, which of the following &nbs [#permalink] 03 Nov 2018, 07:07

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