We are given three distinct non-zero digits. On the board, we will write all three-digit numbers that can be formed from these digits, using all three digits for each number. The sum of the written numbers is 1776.

Which three digits did we work with? Determine all solutions.
Let's denote the given digits as $a, b, c$. The calculation on the board then corresponds to this sum:

$$
\begin{gathered}
a b c \\
a c b \\
b a c \\
b c a \\
c a b \\
c b a \\
\hline 1776
\end{gathered}
$$

In each column, we are adding the same six digits, namely $a+a+b+b+c+c$. From the total result, it is clear that the sum of these six digits is 16, i.e., $a+b+c=8$. The only ways to write 8 in this form are $8=1+1+6=1+2+5=1+3+4=2+2+4=2+3+3$. The two triplets of digits that satisfy the conditions in the problem are $1,2,5$ and $1,3,4$.

Note. The previous calculation on the board can be written as $(100a + 10b + c) + (100a + 10c + b) + (100b + 10a + c) + (100b + 10c + a) + (100c + 10a + b) + (100c + 10b + a) = 1776$, which simplifies to $222a + 222b + 222c = 1776$. After simplification, we get $a + b + c = 8$, and the conclusion is the same.