What is the sum of the six-digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6?
I. solution: 6-digit numbers can be formed from 6 different digits - using each digit - without any repeating digits. The number of such numbers is $P_{6}=6!=720$. If we imagine these 720 numbers written one below the other in the usual way for addition, then in each column there will be 5! = 120 ones, 120 twos, ..., 120 sixes. Therefore, the sum in each column is $120(1+2+3+4+5+6)=120 \cdot 21=2520$. Taking into account the place values, the desired sum is:

$$
2520\left(1+10+10^{2}+10^{3}+10^{4}+10^{5}\right)=2520 \cdot 111111=279999720
$$

We also accepted solutions that allowed for repetition. In this case, the number of numbers that can be formed is $V_{6}^{i, 6}=6^{6}$ and the sum of all these numbers is

$$
\begin{gathered}
6^{5}(1+2+3+4+5+6)\left(1+10+10^{2}+10^{3}+10^{4}+10^{5}\right)= \\
=7776 \cdot 21 \cdot 111111=18143981856
\end{gathered}
$$

II. solution: For any permutation or repetitive variation formed number, we can create a "complementary" number such that the sum of the number and its complementary number is 777777 (e.g., $241653+536124=777777 \quad$ or $443555+$ $+334222=777777$).

The complementary number - as it is easy to see - is always necessarily a permutation, or a repetitive variation, of the given 6 elements. If we do this with all possible 6-digit numbers, then the sum of the resulting complementary numbers is obviously the same as the sum of all possible permutations, or all possible repetitive variations, of the given elements. Therefore, the sum of all possible numbers and their corresponding complementary numbers is twice the desired sum.

Thus, in the case of numbers consisting entirely of different digits, the desired sum is:

$$
777777 \cdot \frac{6!}{2}=777777 \cdot 360=279999720
$$

And in the case of numbers with repeating digits:

$$
777777 \cdot \frac{6^{6}}{2}=777777 \cdot 23328=18143981856
$$

Zoltán Deseó (Bp., X., I. László g. II. o. t.)

Note: The left sides of the equalities under (1) and (2) can also be written as:

$$
\frac{123456+654321}{2} \cdot 6!\quad \text { or } \quad \frac{111111+666666}{2} \cdot 6^{6}
$$

This shows that the sum of our number sequences can be calculated similarly to the formula $\frac{a_{1}+a_{n}}{2} \cdot n$ for the sum of an arithmetic sequence, but these number sequences are clearly not arithmetic progressions.