A wine merchant left 24 casks of wine to his three sons, and the casks were in the following conditions:
- 5 casks were full.
- 11 casks were half-full.
- 8 casks were empty.
His last will said that each son must be bequeathed with the same amount of wine and the same number of casks. To complicate it further, since each cask contained wine of a different vintage, mixing or decanting was not an option. How could the last will of the wine merchant be executed?
Note that there are three different solutions.
This can be solved through trial and error though using algebra lessen the work required for finding the solutions.
Let x, y, and z be the number of full, half-full and empty wine casks each of the sons inherits respectively. Also, let’s say that the contents of a full cask is set at 2 and a half-full cask at 1. The following simultaneous equations can be formed:
x + y + z = 8 and 2x + y = 7,
which have four solutions:
This means that there are four possible ways for each son to receive his share in the estate, three of which have to be combined so that the sum of the three x values is 5 (remember that there were 5 full casks?). There are three different ways to accomplish this.
- 0 + 2 + 3;
- 1 + 1 + 3; and
- 1 + 2 + 2.
Therefore, the estate can be distributed to the three sons A, B, and C in three different ways: