TITLE: Getting $22$ as the answer always
QUESTION [20 upvotes]: I am puzzled by the below exercise:
Step 1: Select any number having 3 digits (all different from one another). Ex. $125$.
Step 2: Now, write all possible combination of two digit number forming from selected digits. Here it is $12$,$21$,$15$,$51$,$25$,$52$. Add all of them. 
here, $ 12+21+15+51+25+52=176$
Step 3: Divide the addition, (here $176$) by sum of all 3 digits selected. i.e.
$ \dfrac{176}{1+2+5} = 22$.
Always. Why, so? 
I have tried many combination, it works. Can anyone give proof and explain the reason behind this?

REPLY [31 votes]: 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 $22$.

REPLY [14 votes]: You can trace it backwards via 
$$
22(a+b+c) = (10a+ b) + (10a + c) + (10b +a) + (10b + c) + (10c + a) + (10c + b)
$$