why is math so painful

[lamchopz]

I'm unsure if he'd accept that but it's something used liberally in arithmetics.

It follows from this more general statement: For any n consecutive integers, at least one of them is divisible by n.

The proof is relatively painless:

Each number can be represented as nq + r where r < n (due to Euler's integer theorem, or whatever it's called. can't remember now xD)

r is therefore any integer from 0 to n -1

since u have n numbers and they're consecutive, they are essentially:

nq + 0
nq +1
...
nq + (n - 1)

or any valid combination (we don't illustrate this in the formal answer, however).

Normally, you just say since the remainders include all values from 0 to n -1, and there are n consecutive numbers, one of them has to have a remainder of 0, which means the same as one of them is divisible by n.

Again check with your professor if he'll accept this without the burden of proof.

and LOL @ your random rage, Des.