\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!
}

1) Suppose for a single-digit number \(Z\) condition (1) were satisfied. Then it would follow that \(a_{0}+a_{0}=a_{0}\) and thus \(a_{0}=0\), which contradicts the assumption.

2) If a two-digit number has property (1), it follows that

\[
a_{1}+a_{0}+a_{1} a_{0}=10 a_{1}+a_{0} \Rightarrow a_{1} a_{0}=9 a_{1}
\]

since \(a_{1} \neq 0\), it follows that \(a_{0}=9\).

Therefore, a two-digit number \(Z\) can only satisfy conditions (1) if it ends with the digit 9. For every such number, indeed,

\[
a_{1}+a_{0}+a_{1} a_{0}=a_{1}+9 a_{1}=10 a_{1}+9=10 a_{1}+a_{0}
\]

so condition (1) is satisfied.

3) Suppose for a three-digit number \(Z\) condition (1) were satisfied. Then it would follow that

\[
a_{2}+a_{1}+a_{0}+a_{2} a_{1} a_{0}=100 a_{2}+10 a_{1}+a_{0} \Rightarrow a_{2} a_{1} a_{0}=99 a_{2}+9 a_{1}
\]

since \(9 \geq a_{0}\), it follows that \(9 a_{1} a_{2} \geq 99 a_{2}+9 a_{1} \geq 99 a_{2}\). This would result in a contradiction \(a_{1} \geq 11\) because \(a_{2}>0\).

Thus, for \(0<Z<1000\), exactly the numbers 19, 29, 39, 49, 59, 69, 79, 89, and 99 satisfy condition (1).

Solutions of the II. Round 1977 taken from [5]

\subsection*{7.19.3 III. Round 1977, Class 10}