## Task $6 / 88$

Determine all three-digit natural numbers $n$ whose cross product $Q_{p}(n)=120$ and whose cross sum $Q_{s}(n)=2^{m}$ is a power of two with $m \in N$.
It is $Q_{p}(n)=120$ to be decomposed into three factors $f_{1}, f_{2}$ and $f_{3}$ with $f_{i} \in N, 0<f_{i} \leq 9$ for $i=1 ; 2 ; 3$.

From the prime factorization of $120=2^{3} \cdot 3 \cdot 5$ it follows immediately that for one $i$ (w.l.o.g. let this be $i=3$) it holds that $f_{i}=5$. Then for $f_{1}$ and $f_{2}$ there remain two possibilities:
$f_{11}=8, f_{21}=3$ and $f_{12}=4, f_{22}=6$.

Only for the first possibility it holds that

$$
Q_{s}(n)=f_{11}+f_{21}+f_{3}=8+3+5=16=2^{m} \text { with } m=4 \in N
$$

the second possibility, however, yields

$$
Q_{s}(n)=f_{12}+f_{22}+f_{3}=4+6+5=15 \neq 2^{m} \text { with } m=4 \in N
$$

so it is ruled out. Thus, the digits of $n$ are fixed with 3, 5, and 8. Their $3!=6$ permutations result in 6 values for $n: 358 ; 385 ; 538 ; 583 ; 835 ; 853$.