2023 AMC 10A Problem 12

Reversing the digits of $N$ makes its hundreds digit the new units digit. A number is divisible by $5$ exactly when its last digit is $0$ or $5$. The hundreds digit of a three-digit number cannot be $0$, so it must be $5$: $N$ lies between $500$ and $599$.

Now count multiples of $7$ in that range. The smallest is $504 = 7 \cdot 72$ and the largest is $595 = 7 \cdot 85$, so there are
$$85 - 72 + 1 = 14$$
of them.

(If $N$ ends in $0$, such as $560$, the reversal $065 = 65$ is still divisible by $5$, so no cases are lost.)

The answer is $\boxed{\textbf{(B)}\ 14}$.

Divisibility by $5$ or $2$ is a statement about one digit only, so a condition on the reversed number is really a condition on the original's leading digit. Once that digit is pinned down, the problem collapses to counting multiples of $7$ in a block of $100$ consecutive integers, which is $14$ or $15$ depending on the endpoints.

• Allowing the hundreds digit to be 0, or miscounting the multiples of 7 in the 500s (504 through 595 is 14, not 13).
• Testing divisibility of the reversal by $7$ as well, or requiring $N$ itself to end in $0$ or $5$.