loadedx wrote:Need help with this concept:

When n is odd
n^3-n or (n-1)n(n+1) is divisible by 24

Let's take n equal to 1
1^3-1 = 0
1 is odd
0 is even
2 is even

Thank you for your help!

I'm not sure what you're asking with the part in blue, but I can address the statement that (n-1)n(n+1) is divisible by 24

First notice that n-1, n, and n+1 represent 3 consecutive integers (for any integer n)
Next, there's a nice rule that says "among any n integers, one the numbers will be divisible by n"
So, one of the three integers (n-1, n, and n+1) will be divisible by 3
Next notice that, if n is odd, then n-1 and n+1 are even (i.e., divisible by 2)
Also notice that every second even number is divisible by 4.
So, of the 3 consecutive integers, we know that one is divisible by 2, one is divisible by 3, and one is divisible by 4
This means that their product must be divisible by (2)(3)(4) or 24

Cheers,
Brent

loadedx wrote:Brent, thank you very much! Now the concept is clearer.

About the part in blue. The point I was trying to make is the following. If a problem is asking:

"n is odd, is (n-1)n(n+1) is divisible by 24?"

I can not answer this question, unless I know that n is positive.

Do you agree with me?

No, (n-1)n(n+1) is divisible by 24 for any integer n

First of all, 0 is divisible by 24. Some students don't like this, but the formal definition of divisibility says that integer N is divisible by integer D if N = kD for some integer k.
So, 0 is divisible by 24 since 0 = 0(24)

Using the formal definition, we can also show that the rule applies to negative numbers as well.
For example, -24 is divisible by 24 since -24 = (-1)(24)

Having said all of that, in most GMAT questions, the numbers are typically restricted to positive values (to make things easier for us).

Cheers,
Brent