The "x modulo 2 is zero if and only if x is even" idea is the simplest way to do it, if you want other programmers who you haven't met yet to understand it. So that's how I would always do it.
However if you are comfortable with bit-fiddling you can start with the fact that numbers whose least significant bit is zero are always even. So a modified version of your code might look like:
This might even be faster, since integer division takes longer than a simple bit calculation. But only by a tiny amount, and it's not impossible that the run-time might optimize division by 2 into a bit calculation. I wouldn't even consider that code unless I expected my application to be spending most of its time deciding whether numbers were even or not, because I find it a lot less readable.
That will not test for odd numbers. It will only find positive odd numbers. Use Paul C's suggestion; it will definitely run faster than the % operator. But it only works if you are trying to divide by exact powers of 2.
Can't .... do .... plaid .... So I did this tiny ad instead: