Sum Mystery. Please explain why after.

Choose three different digits from 1-9.
Make all the possible two-digit numbers using these three digits. (6 combos)
Add all the two digit numbers together.
Divide by the sum of the three original digits.

Example:
A. 1, 3, 8
B. 13, 18, 31, 38, 81, 83
C. 13+18+31+38+81+83 = 264
D. 1+3+8 = 12 ; 264/12 = 22

Will the answer always equal 22 no matter what 3 numbers you choose? Why?
(Please explain WHY!!! i need that part most.)

Thanks. (:

Well, we take digits a,b,c

so if we sum ab+ac+bc+ba+cb+ca
which is
(10*a+b)+(10*a+c)+(10*b+c)+(10*b+a)+(1…
= 10 * (2*a+2*b+2*c) + (2*a+2*b+2*c)
= 20*(a+b+c) + 2*(a+b+c)
= 22*(a+b+c)

so if you divide now by a+b+c you are left with 22.

Example:
A. 1, 3, 8
B. 13, 18, 31, 38, 81, 83
C. 13+18+31+38+81+83 = 264
D. 1+3+8 = 12 ; 264/12 = 22

Will the answer always equal 22 no matter what 3 numbers you choose? Why?
Let 0 < a, b, c < 10.
Now a, b, c are three natural numbers which are such that 1 - or < a, b, c < or = 9.
Sum of a, b, c = (a + b + c)
Two digit numbers formed with a, b, c are 10a + b, 10a + c, 10b + a, 10b + c, 10c + a, 10c + b
Sum of such two digit numbers = 20(a + b + c) + 2(a + b + c) = 22(a + b + c)
Dividing thew Sum of Two digit numbers by the Sum of Digits we get
22(a + b + c) / (a + b + c) = 22 , No matter what a, b, c are.

A. a=1, b=3, c=8
B. 13=(10a+b), 18=(10a+c), 31=(10b+a),
38=(10b+c), 81=(10c+a), 83=(10c+b)
C. (10a+b)+(10a+c)+(10b+a)+(10b+c)
+(10c+a)+(10c+b) = 22a + 22b +22c
D. (22a + 22b + 22c)/(a + b + c)
= 22(a + b + c)/1(a +b + c)
= 22/1
= 22