2003 AMC 10A Problem 16

Only the last digit of $13$ affects the last digit of its powers, so this is the same question as the last digit of $3^{2003}$.

1. Why only the last digit matters: when you multiply two numbers, the last digit of the answer comes only from the last digits of the two numbers. For example $13 \times 13 = 169$ ends in $9$, just like $3 \times 3 = 9$.
2. List the last digits of the first few powers of $3$:

1. The last digits repeat in a block of four: $3, 9, 7, 1$. Each time the power goes up by $4$, you land on the same digit again.
2. Divide $2003$ by $4$: $2003 = 4 \times 500 + 3$. So $3^{2003}$ is $500$ full blocks past $3^3$, and it has the same last digit as $3^3 = 27$.
3. That digit is $7$.

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

The last digits of powers always fall into a repeating pattern, because each new last digit depends only on the previous one. Find how long the pattern is, divide the exponent by that length, and use the remainder to pick the spot. Be careful with a remainder of $0$: it means the last spot in the block, not the first.

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 $\boxed{\textbf{(C)}\ 7}$.