Getting $22$ as the answer always
Solution 1:
If your original digits are $x, y, $ and $z$, then among your two-digit numbers are:
two with $x$ in the ones place,
two with $y$ in the ones place,
two with $z$ in the ones place,
two with $x$ in the tens place,
two with $y$ in the tens place, and
two with $z$ in the tens place.
Therefore they sum to
$2\cdot 10(x+y+z)+2\cdot(x+y+z)=22(x+y+z)$.
So when you divide you are left with $22$.
Solution 2:
You can trace it backwards via $$ 22(a+b+c) = (10a+ b) + (10a + c) + (10b +a) + (10b + c) + (10c + a) + (10c + b) $$