jmc

The base-$b$ representation of $2013$ ends in $3$ if and only if $2013$ leaves a remainder of $3$ when divided by $b$: that is, if $2010$ is a multiple of $b.$ Since $2010=2^1\cdot 3^1\cdot 5^1\cdot 67^1,$ it has $(1+1)(1+1)(1+1)(1+1)=16$ positive divisors. However, since $3$ is a valid digit in base $b,$ we must have $b>3,$ so we must subtract $3$ from our count (since $1,$ $2,$ and $3$ are all divisors of $2010$). Therefore, the answer is $16-3=\boxed{13}.$