Competition problem — Number theory

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Number theory Difficulty 5.3 AIME, harder Prove it Philippines

Problem:
Let $A$ be the set of all two-digit positive integers $n$ for which the number obtained by erasing its last digit is a divisor of $n$ . How many elements does $A$ have?

Solution

Solution:
Let $n = 10a + b$ . Since $a$ is a divisor of $n$ , we infer $a$ divides $b$ . Any number $n$ that ends in $0$ is therefore a solution. Assuming $b \neq 0$ , $n$ must be one of the numbers $11, 12, \ldots, 19, 22, 24, 26, 28, 33, 36, 39, 44, 48, 55, 66, 77, 88, 99$ . This gives a total of $32$ positive integers.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.