2013 AMC 10A Problem 19

Note that the units digit represents the remainder when the number is divided by the base. The question then boils down to finding all numbers, $b,$ such that $2013$ leaves a remainder of $3$ when divided by $b.$ This means that $b$ must divide $2010.$ Also note that $b \geq 4,$ since otherwise the remainder cannot be $3.$ The prime factorization of $2010$ is $$2010 = 2 \cdot 3 \cdot 5 \cdot 67.$$ Then, $2010$ has $$(1 + 1)^4 = 2^4 = 16$$ factors. It has $3$ factors less than $4,$ namely $1, 2,$ and $3.$ This means there are $16 - 3 = 13$ valid values for $b.$ Thus, C is the correct answer.