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

[Gmat_mission]

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. \(n(n+5)(n−6)\)

[spoiler]OA=A[/spoiler]

Source: Official Guide

[anonylfc]

The basic concept here is to make three numbers, which have different remainders when dividing by three, 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 that atleast one of them is divisible by 3.

In A option n-4 will have the same remainder as n-1, we already have n and n+1, so basically we have 3 consecutive numbers, so atleast one of them has to be divisible by 3.

Just to explain in detail, in option B-> n (n+2) (n-1),
n-1 and n+2 will have the same remainder when divisible by 3 ( For eg - 6 and 9, 5 and 8 -> all have the same remainder when divided by 3)

So hence the equation basically boils down to (n+2)(n) but what if (n+1) is divisible by 3 ? Hence not sufficient.

[Scott@TargetTestPrep]

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. \(n(n+5)(n−6)\)

[spoiler]OA=A[/spoiler]

Source: Official Guide

Solution:

Since all the expressions have the factor n, if n is a multiple of 3, then any of the expressions is divisible by 3. So let’s assume that n is not a multiple of 3. That is, n is either 3k + 1 or 3k + 2 for some integer k.

If n = 3k + 1, then n - 4 = 3k - 3 = 3(k - 1), n - 1 = 3k, and n + 5 = 3k + 6 = 3(k + 2) are all multiples of 3. This eliminates choice C and D.

If n = 3k + 2, then only n + 1 = 3k + 3 = 3(k + 1) is a multiple of 3. Therefore, choice A is the correct answer since no matter what n is, one of its factors will be a multiple of 3.

Answer: A