dave13 wrote:

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.

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!

Hi dave,

it basically depends on the divisor..

here you are dividing a term by 3 to find the remainder that is why whatever remainder n-4 leaves, n-4+3 or n-4+6 will leave..

say n is 28, so n-4 = 28-4 = 24, remainder will be 0, when div by 3..

now if you

add/subtract 3 or a multiple of 3 to 24, remainder will always be 0..

say n was 30, n-4 would be 26 and would leave a remainder of 2 when div by 3..

so 26-3 = 23 or 26-6=20 will also leave remainder of 2 when div by 3

_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372

Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-