2023 AMC 10B Problem 8

The last digit of a product depends only on the last digits of the numbers being multiplied. So $2022^{2023}$ ends in the same digit as $2^{2023}$, and $2023^{2022}$ ends in the same digit as $3^{2022}$.

1. Powers of $2$. The last digits are
$$2^1 = 2,\quad 2^2 = 4,\quad 2^3 = 8,\quad 2^4 = 16,\quad 2^5 = 32,\ \ldots$$
so the last digit goes $2, 4, 8, 6$ and then repeats every $4$ steps.
2. Where is $2023$ in that cycle? $2023 = 4 \times 505 + 3$, so the remainder is $3$. Step $3$ of the cycle is $8$. So $2^{2023}$ ends in $8$.
3. Powers of $3$. The last digits are $3, 9, 7, 1$ ($3, 9, 27, 81$) and then repeat every $4$ steps.
4. Where is $2022$ in that cycle? $2022 = 4 \times 505 + 2$, so the remainder is $2$. Step $2$ is $9$. So $3^{2022}$ ends in $9$.
5. Add. $8 + 9 = 17$, which ends in $7$.

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

When you add or multiply, the last digit of the answer depends only on the last digits you started with. So a huge power becomes a small power of a single digit, and those repeat in short cycles. Divide the exponent by the cycle length and use the remainder. Watch one detail: a remainder of $0$ means the last step of the cycle, not the first.

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