2023 AMC 10B Problem 8

The units digit of a power depends only on the units digit of the base, so we need the last digits of $2^{2023}$ and $3^{2022}$.

Powers of $2$ end in $2, 4, 8, 6, 2, \dots$ with period $4$. Since $2023 = 4 \cdot 505 + 3$, the exponent has remainder $3$, and $2^{2023}$ ends in the third entry, $8$.

Powers of $3$ end in $3, 9, 7, 1, 3, \dots$, also period $4$. Since $2022 = 4 \cdot 505 + 2$, $3^{2022}$ ends in the second entry, $9$.

The sum ends in the units digit of $8 + 9 = 17$, namely $7$.

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

Arithmetic mod $10$ respects addition and multiplication, so replace each base by its last digit and each exponent by its position in the four-step cycle. Remainder $0$ means the fourth entry of the cycle, not the first, which is the one detail to watch.

• Swapping the exponents (2^2022 ends in 4, 3^2023 ends in 7) and answering 1.
• Using the remainder as an index into the cycle without care, e.g. reading remainder $3$ as $2^3$'s successor ($6$) instead of $8$.