# 4. CONDITION

Two two-digit numbers, written one after the other, form a four-digit number that is divisible by their product. Find these numbers.
Solution. Let $a$ and $b$ be two-digit numbers, then $100a + b$ is a four-digit number. According to the condition, $100a + b = k \cdot ab$, hence $b = a(kb - 100)$, i.e., $b$ is divisible by $a$. Therefore, $b = ma$, but $a$ and $b$ are two-digit numbers, so $m$ is a single-digit number. Since $100a + b = 100a + ma = a(100 + m)$ and $100a + b = kab$, then $a(100 + m) = kab$, i.e., $100 + m = kb$ or $100 + m = kma$, from which $100 = m(ka - 1)$. Therefore, $m$ is a divisor of the number 100, and since $m$ is a single-digit number, $m = 1, 2, 4, 5$.

Since $ka - 1 = 100 / m$, or $ka = 100 / m + 1$, and $a$ is a two-digit number, the values of $m$ of 1 and 5 are eliminated, because $100 / 1 + 1$ is not divisible by any two-digit number $a$; when $100 / 5 + 1 = 21$, we have $a = 21$ and then $b = ma = 5 \cdot 21$ is a three-digit number. When $m = 2, ka = 51, a = 17, b = 34$; when $m = 4, ka = 26, a = 13, b = 52$.

Answer. 17 and $34; 13$ and 52.