Trailing zeroes!

$trailing.zeroes = (\frac{n}{5})+(\frac{n}{5^{2}})+(\frac{n}{5^{3}})+(\frac{n}{5^{4}})+...$

This is the trailing zeroes equation. Now to get each addend, divide $2002$ by $5$ continuously until you get a quotient less than 5 . Add all the quotients (which are the addends) and you'll get the number of trailing zeroes, in which you will get $\boxed{499}$