## Task $2 / 81$

Let $n=\sum_{i=0}^{k} 10^{i} a_{i}>0$ with $0 \leq a_{i} \leq 9 ; a_{i} \in N$ be a $(k+1)$-digit natural number and $Q(n)=\prod_{i=0}^{k} a_{i}$ its "cross product". How large is the number $r$ of at most ( $\mathrm{k}+1$ )-digit natural numbers $n$, for which $Q(n)$ is a prime number?
If $Q(m)$ is to be a prime number, exactly one $a_{i}$ must be one of the four single-digit primes $2, 3, 5,$ or $7$, while for the remaining $a_{i}$, $a_{i}=1$. With each of these four single-digit primes, $s$ $s$-digit numbers $m$ can consequently be formed $(1 \leq s \leq k+1)$, since the single-digit prime can stand at any of the 3 positions. Therefore, for the desired number $r$

$$
r=4 \cdot \sum_{s=1}^{k+1} s=4 \cdot \frac{(k+1)(k+2)}{2}=2(k+1)(k+2)
$$