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2024/25 AHSMC Part 1 Problem 2

Problem 2 of 16

nn is a positive integer with at least two digits. When we erase the last digit of nn, the number we obtain is exactly n13\dfrac{n}{13}. How many such nn's are there?

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Solution

Let aa be the last digit of nn and AA the number obtained after erasing it. Then 13A=n=10A+a  ⟹  3A=a13A = n = 10A + a \implies 3A = a. Since aa is a nonzero digit, a∈{3,6,9}a \in \{3,6,9\}. There are three such numbers: 1313, 2626, 3939. The answer is (D).

University of Alberta