Bunuel wrote:

carcass 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:

http://gmatclub.com/forum/if-x-is-an-in ... 26853.htmlHope it helps.

Hi

Bunuel,

pushpitkc its me again

you know i am trying to undeerstand the concept and have some questions:) could you please explain

so we are given 5 answer choices:

(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)

you say "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" in order for this to be true we need to TO HAVE 3 CONSECUTVE NINTEGERS , am i correct ? please say something

Next you say " 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.

Question 1) why you say " to guarantee that at least one multiple is divisible by 3, these numbers must have

different remainders upon division by 3 ?

why you mention remainders, and not that that they must be consecutive numbers ?

Question 2) what is the logic behind this "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"

i mean remainders could be any numbers, why mention 1 , and than 2 ? is it MUST BE TRUE ?

Question 3) you write "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"

How do you figure out that here

(A) n(n + 1)(n – 4) n and n+1 have different remainder upon division by 3 , i simply cant understand this, cause i dont see logic

Question 4) why do you think that for n-4, it will have the same remainder as (n-4)+3=n-1 (Also what to you mean by "the same remainder" 0 ? )

Question 5) how about other answer choices ? how you ruled them out ? i am just trying get under the skin of this concept

(B) n(n + 2)(n – 1)

(C) n(n + 3)(n – 5)

(D) n(n + 4)(n – 2)

(E) n(n + 5)(n – 6)

generis generis any comments on my questions above are apprecuated

all moderators are attending now indian premier league, but you

so i decided to tag youhere because you are not a football lover

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