## Task 22/87

We are looking for the smallest number $n \in N$, for which the cross sum is a three-digit prime number $p_{3}$. The cross sum of this number $p_{3}$ should be a two-digit prime number $p_{2}$, and finally, the cross sum of this should be a one-digit odd prime number $p_{1}$.
1. Observation: For the cross sum $Q\left(n_{3}\right)$ of any three-digit number $n_{3} \in N$, it holds that $Q\left(n_{3}\right) \leq 3 \cdot 9 = 27$.
2. Observation: For the cross sum $Q\left(p_{2}\right)$ of a two-digit prime number $p_{2}$, it holds that $Q\left(p_{2}\right)=5$ or $Q\left(p_{2}\right)=7$, since $Q\left(p_{2}\right)=3$ contradicts the prime property of $p_{2}$. From $Q\left(p_{2}\right)=5$ it follows that $p_{2}=23$ or $p_{2}=41$. From $Q\left(p_{2}\right)=7$ it follows that $p_{2}=43$ or $p_{2}=61$. Given $p_{2}=Q\left(n_{3}\right) \leq 27$, only $p_{2}=23$ remains with $p_{1}=Q\left(p_{2}\right)=5$.
3. Observation: The smallest three-digit number $n_{3} \in N$ with the cross sum $Q\left(n_{3}\right)=p_{2}=23$ is $n_{3}=599$. Indeed, this number is a prime number.
4. Observation: Since $Q(n)=n_{3}=p_{3}=599=66 \cdot 9+5$, the number

$$
n=5 \cdot 10^{66}+9 \cdot \sum_{i=0}^{65} 10^{i} \in N
$$

has the cross sum $Q(n)=p_{3}=599$. Indeed, $n$ is also the smallest number with this cross sum; for any number $m \in N$ with $m<n$, it necessarily holds that $Q(m)<Q(n)=599$.