From the 1st condition, it follows that $a ; b ; c \in N, a ; b ; c \leq 9$, from the 3rd $a ; b ; c>0$. The three conditions can be summarized in the following equation:

$$
a c+b c+c(a-b)=a^{c} b^{c} c^{a-b} \rightarrow 2=a^{c-1} b^{c} c^{a-b-1}
$$

From the uniqueness of the prime factorization, it follows that one of the factors on the right side of the last equation must be equal to 2 and the other two must be equal to 1.

Case distinction:

1. $a^{c-1}=2$ yields $a=2, c=2$ in contradiction to the 1st condition.
2. $b^{c}=2$ results in $b=2, c=1$. Then, due to the 1st condition, $a \geq 3$. This leads to $n_{1}=321, n_{2}=$ $4211, n_{3}=52111, n_{4}=621111, n_{5}=7211111, n_{6}=82111111, n_{7}=921111111$.
3. $c^{a-b-1}=2$ with $a^{c-1}=1$ and $b^{c}=1$. Since $c>1$, $a=b=1$ would have to be true, which contradicts the 1st condition.

Thus, the numbers found under 2. and all numbers that arise from permutations of the digits from them are the only ones that satisfy the conditions of the problem (the case distinction is complete). There are 238 numbers.
## Task 10/88

We are looking for all numbers $n \in N$ that have the following properties:

1. The decimal representation requires exactly three pairwise different digits $a$, $b$, and $c$.
2. Here, $a$ and $b$ each appear exactly $c$ times; $c$, on the other hand, appears $(a-b)$ times.
3. The sum of the digits and the product of the digits of $n$ are equal.