2023 AMC 10A Problem 12

1. Use the second condition first, because it pins down a digit. Take an example: $N = 532$ reversed is $235$. The first digit of $N$ becomes the last digit of the reversed number.
2. A number is divisible by $5$ only when its last digit is $0$ or $5$. So the first digit of $N$ is $0$ or $5$.
3. A three-digit number cannot start with $0$. So $N$ starts with $5$: it is somewhere from $500$ to $599$.
4. Any number in the $500$s works for the second condition. Even $N = 560$ reverses to $065 = 65$, which is still a multiple of $5$.
5. Now count the multiples of $7$ from $500$ to $599$. Since $7 \times 71 = 497$ and $7 \times 72 = 504$, the first one is $504$. Since $7 \times 85 = 595$ and $7 \times 86 = 602$, the last one is $595$.
6. The multiples are $7 \times 72, 7 \times 73, \ldots, 7 \times 85$. That is $85 - 72 + 1 = 14$ numbers.

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

Divisibility by $5$ (or by $2$ or $10$) depends only on the last digit. So a rule about the reversed number is really a rule about just one digit of $N$, its first digit. Once that is fixed, the problem becomes counting multiples of $7$ in a block of $100$ numbers. When counting from the $72$nd multiple to the $85$th, remember to add $1$.

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 $\boxed{\textbf{(B)}\ 14}$.