## Task 29/68

Determine the smallest six natural numbers with the following properties:

1. Each number consists of at least the digits 3, 5, and 9, placed at any position and in any order.
2. Each number is divisible by at least 3, 5, and 8.
The last digit of the smallest number $n_{1}$ is $a$, the second last is $b$ and so on: $n_{1}=$...edcba.

Because of 2, $n$ must be divisible by $5 \cdot 8=40$, meaning $a=0$ and $b$ is even. If $n_{1}$ were a 5-digit number, then $n_{1}=359 b 0$ would have to be true.

Even with any permutation of the digits 3, 5, and 9, $b$ could only be 2 or 6 due to the divisibility of $n_{1}$ by 8 (all three numbers 3, 5, and 9 are odd). However, the cross sum would not be divisible by 3 in that case, so $n_{1}$ must be at least 6 digits long. The smallest 6th digit would be 1, provided that a corresponding $b$ ($b=2$ or $b=6$) exists. In fact, the cross sum of $n_{1}=1359 b 0$ is divisible by 3 for $b=6$. This leads to $n_{1}=135960$.

The remaining sought digits result from permutations of the digits 3, 5, and 9, as the given conditions are not violated:

$$
n_{2}=139560 ; n_{3}=153960 ; n_{4}=159360 ; n_{5}=193560 ; n_{6}=195360
$$