Integers

[earth@work]

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)

Ans: A

[DanaJ]

What you have to remember is that the product of any three consecutive integers is divisible by three. This is the starting point:

n(n + 1)(n + 2) - divisible by 3

Now, another thing you should notice is that, if you have a similar series as the above, you can subtract various numbers from one or more of the terms, IF those numbers are divisible by 3, and the product will still be divisible by 3. Let me give you an example:
n(n + 1 - 3)(n + 2 - 6) = n(n - 2)(n - 4) will be divisible by 3.
n(n + 1 + 9)(n +2) = n(n + 10)(n + 2) will also be divisible by 3.
Now apply all this on your products and you get that:
a. n(n + 1)(n - 4) = n(n + 1)(n + 2 - 6) is divisible by 3.
You stop here, since this as a PS problem.

So the answer is indeed A

[earth@work]

Thanks DanaJ, this was something i never thought of!

[earth@work]

error

[pbanavara]

I took a different approach of picking numbers : Viz even and odd >6 as any number > 6 should be either odd or even.

Try with 7 - Eq 1 holds good.
Try with 8 - Still holds good.

Stop here. I'm not sure if this is the right way of doing this. DanaJ's approach is definitely better

- pradeep

[nervesofsteel]

put 11 and 13..

A will survive