2021 AMC 12A Fall 第 10 题

$9 \equiv -1 \pmod 5,$ so $9^k \equiv (-1)^k \pmod 5$

Take the alternating sum of the base-nine digits (from the right)

Since $9 \equiv -1 \pmod 5,$ each power $9^k \equiv (-1)^k,$ so $N$ is congruent to the alternating sum of its base-nine digits.

The nonzero digits, with their positions from the right, are $2$ (position $0$), $5$ (position $1$), $6$ (position $6$), $7$ (position $9$), and $2$ (position $10$). The alternating sum is $2 - 5 + 6 - 7 + 2 = -2$ $\equiv 3 \pmod 5.$

Thus, the correct answer is D.