The product of k consecutive integers is divisible by k

Hi aces021,

To help you with this concept, you should think about how a number line "works":

Let's start with a consecutive series of numbers.....

1 2 3 4 5 6 7 8 9 10 11 12 etc.

When we're told that (N+2) is a multiple of 3, then we have to start with 3 or 6 or 9 etc.

Now, let's say that (N+2) = 9....what OTHER values do we know would also be multiples of 3?

If we go "up", then the next multiple is 12.....that would be 3 more than (N+2), so (N+5) is a multiple of 3.

If we go "down", then the next multiple is 6....the would be 3 less than (N+2), so (N-1) is a multiple of 3.

If we go "down" again, then the next multiple is 3...the would be 3 less than (N-1), so (N-4) is a multiple of 3.
Etc.

GMAT assassins aren't born, they're made,
Rich

aces021 wrote:If we know that n+2 is a multiple of 3, how would we know that n-4, n-1, and n+5 are also multiples of 3? What math are you doing to get those answers? What is your thinking? Where do you start?

We can also use some algebra.

GIVEN: n+2 is a multiple of 3
This means that n+2 = 3k for some integer k

What about n-4?
We can rewrite n-4 as follows:
n-4 = (n+2) - 6
= 3k - 6
= 3(k - 2)
Aha! If n-4 = 3(k - 2), then n-4 is a multiple of 3

What about n+5?
We can rewrite n+5 as follows:
n+5 = (n+2) + 3
= 3k + 3
= 3(k + 1)
Aha! If n+5 = 3(k + 1), then n+5 is a multiple of 3

And so on.

Cheers,
Brent