Divisibility Problem

[leumas]

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)

Seems to be mentally draining on reading OG 12, Q82 solution. Is there any simple solution to this?

I tried with factors, but it didn't work.

[Anurag@Gurome]

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)

Seems to be mentally draining on reading OG 12, Q82 solution. Is there any simple solution to this?

I tried with factors, but it didn't work.

Note that for the product of 3 numbers to be divisible by 3, one of the three numbers should be a multiple of 3.

(A) n(n + 1)(n - 4)
Adding 3 to n - 4, we get, 3 + (n - 4) = n - 1
So, n(n + 1)(n - 1), which are 3 consecutive integers, so their product will always be divisible by 3.

So, the correct answer is A.

[leumas]

Note that for the product of 3 numbers to be divisible by 3, one of the three numbers should be a multiple of 3.

(A) n(n + 1)(n - 4)
Here n - 4 can be written as 3 + (n - 4) = n - 1
So, n(n + 1)(n - 4) = n(n + 1)(n - 1), which are 3 consecutive integers, so their product will always be divisible by 3.

So, the correct answer is A.

Thanks for the post. But can you explain the highlighted step.

[Anurag@Gurome]

Thanks for the post. But can you explain the highlighted step.

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.

[leumas]

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.

[GMATGuruNY]

I posted a solution here:

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

[knight247]

@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