2013 AMC 10A Problem 19

1. What is the last digit in base $b$? In base $10$, the last digit of $2013$ is the remainder when you divide by $10$ (it is $3$). In the same way, the last digit in base $b$ is the remainder when you divide by $b$. Check with the example: $2013 = 9 \times 223 + 6$, and indeed base $9$ ends in $6$.
2. Turn it into divisibility. We want $2013$ to leave remainder $3$ when divided by $b$. That means $2013 - 3 = 2010$ divides evenly by $b$. So $b$ must be a divisor of $2010$.
3. The digit 3 must exist. Base $b$ uses the digits $0$ to $b - 1$. To have a digit $3$, we need $b \ge 4$.
4. Count the divisors of 2010. $2010 = 2 \times 3 \times 5 \times 67$. A divisor is made by deciding, for each of these $4$ primes, whether to use it or not. That is $2 \times 2 \times 2 \times 2 = 16$ divisors.
5. Remove the small ones. The divisors less than $4$ are $1, 2, 3$. (The next one is $5$.) So $16 - 3 = 13$ bases work.

The answer is $\boxed{\textbf{(C)}\ 13}$.

"Ends in the digit $d$ in base $b$" means "leaves remainder $d$ when divided by $b$", and also $d$ must be smaller than $b$. The remainder condition turns into "$b$ divides $N - d$", so the problem becomes counting divisors, plus a quick check at the small end. In remainder notation, $2013 \equiv 3 \pmod{b}$ means exactly "$2013$ leaves remainder $3$ when divided by $b$."