Take any three different digits.
1. Create every possible two-digit combination.
2. Find their sum.
3. Sum the original three digits.
4. Divide your answer from part 2 by your answer to part 3.
What number do you always get, and why?
Let the three digits be x, y, and z.
1. There are six possible 2-digit combinations of these three digits and, they are:
10x+y, 10y+x, 10x+z, 10z+x, 10y+z and 10z+y
2. Adding them, we have:
10(x+y+x+z+y+z) + (y+y+z+x+z+y)
=(20+2)(x+y+z)
=22(x+y+z)
3. Summing the original three digits, we have: x+y+z
4. Dividing 22(x+y+z) by (x+y+z) we get 22
Consequently, we will always get 22 as the end result at the conclusion of the 4-step procedure.
Edited on April 14, 2022, 7:06 am
Puzzle Solution
