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 divisible 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 $5$ will always end in $0$ or $5$ , and since the numbers have to be a three-digit numbers (otherwise it would be a two-digit number), it cannot start with 0, narrowing our choices to 3-digit numbers starting with $5$ . Since the numbers must be divisible by 7, all possibilities have to be in the range from $7 \cdot 72$ to $7 \cdot 85$ inclusive.

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

~walmartbrian ~Shontai ~andliu766 ~andyluo ~ESAOPS

Solution 2 (solution 1 but more thorough)

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

~ Technodoggo

Solution 3 (modular arithmetic)

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 $1+2+1+2+1+1+2+1+2+1=\boxed{\textbf{(B) } 14.}$

~ Technodoggo

Solution 4

The key point is that when reversed, the number must start with a $0$ or a $5$ based on the second restriction. But numbers can't start with a $0$ .

So the problem is simply counting the number of multiples of $7$ in the $500$ s.

$7 \times 72 = 504$ , so the first multiple is $7 \times 72$ .

$7 \times 85 = 595$ , so the last multiple is $7 \times 85$ .

Now, we just have to count $7\times 72, 7\times 73, 7\times 74,\cdots, 7\times 85$ .

We have a set that numbers $85-71=\boxed{\textbf{(B) 14}}$

~Dilip ~boppitybop ~ESAOPS (LaTeX)

Video Solution

https://youtu.be/UYHCNlRDZBo

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://www.youtube.com/watch?v=4uKo5NR2o9Y

-paixiao

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/N2lyYRMuZuk?si=6B-mTB070UP2yuDF&t=435

~Math-X

Note

According to the official answer key, choice (B) is correct. However, some have argued that it is ambiguous whether the number $560$ should be included in the count, since its reversal, $065$ , has a leading zero. It is assumed that $065$ denotes the two-digit number $65$ , which is divisible by $5$ , but MAA should have clarified what happens when a number with trailing zeros is reversed.

~A_MatheMagician ~ESAOPS ~sdpandit

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
All AMC 10 Problems and Solutions

Page

Toolbox

Search

2023 AMC 10A Problems/Problem 12

Contents

Problem

Solution 1

Solution 2 (solution 1 but more thorough)

Solution 3 (modular arithmetic)

Solution 4

Video Solution

Video Solution

Video Solution by Math-X (First understand the problem!!!)

Note

See Also