Division by 6

[kbikov]

What is the proof that the multiple of three consequtive integers is divisible by 6 i.e. k(k-1)(k+1) mod 6 = 0?

Thank you,
Kal

[kbikov]

Actually, I figured it out.

To prove division by 6 we need to prove division by 2 and by 3.

Obviuosly, there will be at least one intiger among the three consequtive intigers which will be even and so their product will also be an even number (divisible by 2).

Since every third integer starting from 1 is divisible by 3, we know that exactly one of the three consequtive intigers will be divisible by 3 and therefore their product will also be divisible by 3.

[gmatplayer]

Multiple of any 3 consicative number is divisible by 6.

[Ian Stewart]

Yes, exactly. You can extend your proof to any number of consecutive integers:

The product of n consecutive integers is always divisible by n!

So, if you multiply any five consecutive integers, you can be sure that the result will be divisible by 5! = 120, for example.