2023 AMC 10A Problem 12

Below is the professionally curated solution for Problem 12 of the 2023 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 10A solutions, or check the answer key.

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Concepts:divisibility digits counting integers in a range

Difficulty rating: 1440

12.

How many three-digit positive integers $N$ satisfy both of the following properties: $N$ is divisible by $7,$ and the number formed by reversing the digits of $N$ is divisible by $5?$

$13$

$14$

$15$

$16$

$17$

Solution:

When we reverse $N,$ its last digit is the first digit of $N.$ For the reversal to be divisible by $5,$ that digit is $0$ or $5.$ A three-digit number can't start with $0,$ so $N$ starts with $5,$ meaning $500 \le N \le 599$ (and the reversal ends in $5,$ always fine). Now just count multiples of $7$ here: from $7 \cdot 72 = 504$ to $7 \cdot 85 = 595,$ that's $14$ numbers. Therefore, the answer is B.

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