## Task 17/88

Determine all three-digit natural numbers $n$ that are equal to five times their cross product!
Let $n=100a+10b+c=5abc$ with $a, b, c \in \mathbb{N}, a, b, c \leq 9$. Since $n \neq 0$, it follows that $a, b, c \neq 0$. Clearly, $n$ is divisible by 5. Therefore, since $c \neq 0$, it follows that $c=5$:

$$
n=100a+10b+5=25abc
$$

Dividing by 5 gives $20a+2b+1=5ab$. Consequently, $2b+1$ must be divisible by 5. Thus, the possible values for $b$ are $b_1=2$ and $b_2=7$. Since with $b_1=2$, the fivefold product of the digits, and thus $n$, would be divisible by 10, which contradicts $c=5$, only $b_2=b=7$ remains. Therefore, $20a+15=35a$ and $a=1$.

Thus, there is exactly one number $n$ with the required property: $n=175$ (the verification confirms the correctness).