2013 AMC 10A Problem 19

The last digit of a number written in base $b$ is its remainder upon division by $b$. So we need $2013 \equiv 3 \pmod b$, which means $b$ divides $2013 - 3 = 2010$. We also need $3$ to be a legitimate digit in base $b$, which requires $b > 3$, i.e. $b \ge 4$.

Factor: $2010 = 2 \cdot 3 \cdot 5 \cdot 67$. With four distinct primes, the number of divisors is $2^4 = 16$.

The divisors that are too small are $1, 2, 3$ (the next divisor is $5$). Removing them leaves $16 - 3 = 13$ valid bases.

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

"Ends in digit $d$ in base $b$" is the congruence $N \equiv d \pmod b$ together with the requirement $d < b$. The congruence is a divisibility statement about $N - d$, so the count reduces to counting divisors, with a small boundary check for bases where the digit does not exist. The base-$9$ example in the statement is just a hint about how last digits work.

• Counting all 16 divisors of 2010, forgetting that the digit 3 only exists in bases larger than 3 (b = 1, 2, 3 must be excluded).
• Looking for divisors of $2013 = 3 \cdot 11 \cdot 61$ instead of $2010$.