2023 AMC 10A Problems/Problem 12

Problem

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 $5$ .

$\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17$

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

Alternate solution:

We first proceed as in the above solution, up to $N=500+10a+b$ . We then use modular arithmetic:

$\begin{align*} 0&\equiv N \:(\text{mod }7)\\ &\equiv500+10a+b\:(\text{mod }7)\\ &\equiv3+3a+b\:(\text{mod }7)\\ 3a+b&\equiv-3\:(\text{mod }7)\\ &\equiv4\:(\text{mod }7)\\ \end{align*}$

We know that $0\le a,b<10$ . We then look at each possible value of $a$ :

If $a=0$ , then $b$ must be $4$ .

If $a=1$ , then $b$ must be $1$ or $8$ .

If $a=2$ , then $b$ must be $5$ .

If $a=3$ , then $b$ must be $2$ or $9$ .

If $a=4$ , then $b$ must be $6$ .

If $a=5$ , then $b$ must be $3$ .

If $a=6$ , then $b$ must be $0$ or $7$ .

If $a=7$ , then $b$ must be $4$ .

If $a=8$ , then $b$ must be $1$ or $8$ .

If $a=9$ , then $b$ must be $5$ .

Each of these cases are unique, so there are a total of $1+2+1+2+1+1+2+1+2+1=\boxed{\text{(B) }14.}$

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2023 AMC 10A Problems/Problem 12

Contents

Problem

Solution 1

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

See Also