Problem 1. Determine the three-digit numbers for which, by removing the tens digit, we obtain a number 13 times smaller.

Mathematical Gazette
Problem 1. Determine the three-digit numbers for which, by removing the tens digit, we obtain a number 13 times smaller.

Mathematical Gazette

## Solution

Let $\overline{a b c}$ be a number with the property mentioned in the statement.

The given condition is equivalent to $\overline{a b c}=13 \cdot \overline{a c}$, which means $100 a+10 b+c=130 a+13 c$, thus $10 b=30 a+12 c$.

Noting that 5 divides both $10 b$ and $30 a$ (or by examining the last digit of these numbers), it follows that $12 c$ is divisible by 5, hence $c \in\{0,5\}$. . 2p If $c=0$, we arrive at $b=3 a$. Since $a$ is a non-zero digit, the valid options are: $a=1, b=3 ; a=2, b=6$ and $a=3, b=9$, which lead to the numbers 130, 260, and 390. If $c=5$, we arrive at $b=3 a+6$. Since $a$ is a non-zero digit, only the option $a=1, b=9$ is valid, which leads to the number 195. .................. 2p In conclusion, there are four numbers with the desired property: 130, 260, 390, and 195.