Divisibility Problem

Anurag@Gurome wrote:Let us take n - 4 = 1, then n = 5 and n + 1 = 6
So, n(n + 1)(n - 4) = 5 * 6 * 1, which is clearly divisible by 3. Hence, option A.

By the same logic

Option C: n-5 =1 , then n=6 and n+1 = 7 -> This is also divisible. It was given that n>6 also.
I am really confused, need a much more basic and simpler solution.

I posted a solution here:

https://www.beatthegmat.com/problem-solv ... tml#389246

@leumas. Hope this post helps
I'm only gonna discuss the option A so lemme know if u need explanations for the rest
(A) n(n + 1)(n - 4)

Now, n-4 can be re-written as (n-1)-3 and substituting this value in the above eqn we get
n(n + 1)[(n-1)-3]. Opening only the second bracket we get

n(n+1)(n-1)-n(n+1)3
Rewriting it as

(n-1)n(n+1)-n(n+1)3

We'll break this into two parts

1st part
Now (n-1)n(n+1) are consecutive numbers. Keep in mind That the product of x consecutive numbers is always divisible by x! and therefore is divisible by x

So in the case of (n-1)n(n+1)...x=3 as there are three consecutive number and their product is obviously divisible by 3! and 3.

2nd part
n(n+1)3 . I guess you can figure that since there is a 3 in the product it will obviously be divisible by 3

So we can write the equation (n-1)n(n+1)-n(n+1)3 as Multiple of 3-Multiple of 3

Now, also keep in mind that If two multiples of a certain number are added/subtracted the resultant number will always be a multiple of that number

So if two multiples of 3 are added, obviously their sum will be divisible by 3. Hence A. However this problem is solved in a much quicker and efficient manner by just substituting numbers. Hope this clarifies ur doubts