2023 AMC 10A Problems/Problem 12

How many three-digit positive integers $N$ satisfy the following properties?

The number $N$ is divisible by $7$ .

The number formed by reversing the digits of $N$ is divisble by $5$ .

Solution 1

Multiples of $5$ always end in $0$ or $5$ and since it is a three-digit number (otherwise it would be a two-digit number), it cannot start with 0. All possibilities have to be in the range from $7 \cdot 72$ to $7 \cdot 85$ inclusive.

$85 - 72 + 1 = 14$ . $\boxed{(B)}$ .

~walmartbrian ~Shontai ~andliu766 ~andyluo

Solution 2 (solution 1 but more thorough + alternate way)

Let $N=\overline{cab}=100c+10a+b.$ We know that $\overline{bac}$ is divisible by $5$ , so $c$ is either $0$ or $5$ . However, since $c$ is the first digit of the three-digit number $N$ , it can not be $0$ , so therefore, $c=5$ . Thus, $N=\overline{5ab}=500+10a+b.$ There are no further restrictions on digits $a$ and $b$ aside from $N$ being divisible by $7$ .

The smallest possible $N$ is $504$ . The next smallest $N$ is $511$ , then $518$ , and so on, all the way up to $595$ . Thus, our set of possible $N$ is $\{504,511,518,\dots,595\}$ . Dividing by $7$ for each of the terms will not affect the cardinality of this set, so we do so and get $\{72,73,74,\dots,85\}$ . We subtract $71$ from each of the terms, again leaving the cardinality unchanged. We end up with $\{1,2,3,\cdots,14\}$ , which has a cardinality of $14$ . Therefore, our answer is $\boxed{\text{(B) }14.}$