\section*{Problem 4 - 171024}

If a natural number \(Z \neq 0\) is represented in the decimal system by the sequence of digits \(a_{n} a_{n-1} a_{n-2} \ldots a_{1} a_{0}\) (with \(0 \leq a_{i} \leq 9\) for \(i=0, \ldots, n\) and with \(a_{n} \neq 0\)), then we denote the sum of the digits of \(Z\) as the cross sum \(Q(Z)\) of the number \(Z\):

\[
Q(Z)=a_{n}+a_{n-1}+a_{n-2}+\ldots+a_{1}+a_{0}
\]

and the product of the digits of \(Z\) as the cross product \(P(Z)\) of the number \(Z\):

\[
P(Z)=a_{n} \cdot a_{n-1} \cdot a_{n-2} \cdot \ldots \cdot a_{1} \cdot a_{0}
\]

Determine all natural numbers \(Z\) with \(0<Z<1000\) for which \(Q(Z)+P(Z)=Z\) holds!