From reader Chris Smith:

Pick a three-digit number in which all the digits are different. Example: 314.

Now list every possible combination of two digits from the chosen number. In our example, these are 13, 14, 31, 34, 41, and 43.

Divide the sum of these two-digit numbers by the sum of the three digits in the original number, and you’ll always get 22. In our example, (13 + 14 + 31 + 34 + 41 + 43) / (3 + 1 + 4) = 176/8 = 22.

This works because 10*a* + *b*, 10*a* + *c*, 10*b* + *a*, 10*b* + *c*, 10*c* + *a*, and 10*c* + *b* sum to 22*a* + 22*b* + 22*c* = 22(*a* + *b* + *c*), so dividing by *a* + *b* + *c* will always give 22.

(Thanks, Chris.)

06/08/2024 Reader Tom Race points out that essentially the same trick can be performed using the entire number: If you add all six permutations of the original 3 digits, then divide that total by the sum of the 3 digits, the answer is always 222.

For example, using 561:

561 + 516 + 156 + 165 + 651 + 615 = 2664

5 + 6 + 1 = 12

2664 / 12 = 222

“This works because in the first sum each of the three digits (*a*, *b* and *c*) occurs twice in each of the three columns, so the sum is 222*a* + 222*b* + 222*c* = 222(*a* + *b* + *c*).” (Thanks, Tom.)