Ten Problems

This is a simple one. Take the coin at the very bottom and place it on top of the coin at the corner. Voila!

First, we let the ages of the three sons be x, y, and z. From the problem, we know that

xyz=36, and
x+y+z = some integer

factor 36 to find all possible combinations of x, y, and z, then calculation x+y+z for these combinations, we find

x | y | z | sum
—————
1 | 1 | 36| 38
1 | 2 | 18| 21
1 | 3 | 12| 16
1 | 4 | 9| 14
1 | 6 | 6| 13
2 | 2 | 9| 13
2 | 3 | 6| 11
4 | 3 | 3| 10

Because the first mathematician could not determine x, y, and z given this chart and the sum, we know that there must be some ambiguity, namely that two sets of data gives the same sum: 1, 6, 6 and 2, 2, 9 both add up to 13.

Since the king claims that there exist an ‘oldest’ son, the ages must be 2, 2, 9 because the other set of numbers doesn’t provide an unique oldest age.

The prisoners realized that the pile of dirt they digged up was high enough for them to climb on and escape through the window.