Problem: John needs to pay 2010 dollars for his dinner. He has an unlimited supply of 2,5, and 10-dollar notes. In how many ways can he pay? Solution: We are trying to compute the number of non-negative integral solutions to $2a + 5b + 10c = 2010$. Taking mod $5$, we see that we must have $2a \equiv 0 \pmod 5$ meaning $5 \mid a$. We can write $a = 5a_{0}$ for some non-negative $a_{0}$ and get that $(a,b,c)$ is an integral solution to $2a + 5b + 10c = 2010$ if and only if $(a_{0},b,c)$ is an integral solution to $10a_{0} + 5b + 10c = 2010$. Dividing by $5$, we get $2a_{0} + b + 2c = 402$. From here, we notice taking mod $2$ that $b$ must be even, so write $b = 2b_{0}$ for some non-negative $b_{0}$. Now, we see that $(a,b,c)$ is a solution to the original equation if and only if $(a_{0},b_{0}c)$ is a solution to $2a_{0}+2b_{0}+2c = 402 \iff a_{0}+b_{0}+c = 201$. From here, the number of integral solutions is $\binom{201+3-1}{3-1} = \binom{203}{2}$.