If three discs the 1st disc has a six on it the 2nd a 7 and the 3rd an eight what are the numbers on the back when the discs are flipped the numbers facing up must be 15 16 17 18 20 21 22 and 23.?

Answer 1

The numbers on the back are 1, 6 and 10, in that order; the possible sums

are then

* 6 + 7 + 8 = 21, * 6 + 7 + 10 = 23, * 6 + 6 + 8 = 20, * 6 + 6 + 10 = 22, * 1 + 7 + 8 = 16, * 1 + 7 + 10 = 18, * 1 + 6 + 8 = 15, and * 1 + 6 + 10 = 17.