2003 AMC 10A Problem 16

The units digit of a product depends only on the units digits of the factors, so $13^{2003}$ ends in the same digit as $3^{2003}$.

List the last digits of powers of $3$: $3^1 \to 3$, $3^2 \to 9$, $3^3 \to 7$, $3^4 \to 1$, and then $3^5 \to 3$ again. The cycle $3, 9, 7, 1$ has length $4$.

Since $2003 = 4 \cdot 500 + 3$, the exponent $2003$ sits in the same position of the cycle as exponent $3$, whose units digit is $7$.

The answer is $\boxed{\textbf{(C)}\ 7}$.

Working mod $10$, $3^4 = 81 \equiv 1$, so $3^{2003} = (3^4)^{500} \cdot 3^3 \equiv 27 \equiv 7$. Last digits of powers always cycle; find the cycle length, reduce the exponent modulo it, and be careful that remainder $0$ corresponds to the last entry of the cycle.

• Getting the remainder 2003 mod 4 = 3 right but reading off the wrong entry of the cycle, giving 9 or 1.
• Using the tens digit $1$ of $13$ as well, or assuming the cycle has length $10$.