## Task 32/79

It is given that $9 \cdot 45=405$. For which products of a one-digit and a two-digit number does it hold that the product can be obtained by inserting a zero between the first and the second digit of the two-digit factor?
Let $a$ be the single-digit factor, $10 b+c$ the two-digit one, where $a ; b ; c \in N, 1 \leq b ; c \leq 9,2 \leq a \leq 9$ holds. Then (according to the conditions of the problem) the following Diophantine equation must be solved:

$$
a(10 b+c)=100 b+c
$$

If we solve this equation for $b$, we get

$$
b=\frac{c(a-1)}{10(10-a)}
$$

Since the denominator of the fraction on the right side of this equation contains the factor 10, either (1) $c$ is even and $a-1=5$, so $a=6$, or (2) $c=5, a-1$ is even. From (1) it immediately follows that $b=\frac{c}{8}$, so $c=8, b=1$. Indeed, $6 \cdot 18=108$.

From (2) we get

$$
b=\frac{a-1}{2(10-a)}
$$

If we substitute these values for $a$, we only get integer values for $b$ for $a=7$ and $a=9$:

$$
a=7: \quad b=\frac{6}{2 \cdot 3}=1 \quad ; \quad a=9: \quad b=\frac{8}{2 \cdot 1}=4
$$

Here too, the check confirms: $7 \cdot 15=105,9 \cdot 45=405$. Thus, there are exactly three products of the required type.