Sixteen five gallon containers form a 4x4 grid. Move only ONE container so as to make the row sums, column sums, and main diagonal sums, the same.
[1] [2] [3] [2]
[2] [2] [3] [1]
[3] [1] [2] [4]
[2] [4] [5] [3]
All containers sum 40 gallons, so each row and column must have 10 gallons. The row totals are 8, 8, 10, and 14. The column totales are 8, 9, 13, and 10.
Thus, you must take the container in the fourth row (the only row with excess) and the third column (same) and pour four of its gallons in other containers: two gallons in row 1, and two gallons in row 2; on the other hand, 2 gallons in column 1, 1 in column 2, and 1 in column 3.
The obvious solution would be to add 2 gallons at row 2, column 1, and 1 gallon at row 1, column 2, and at row 1, column 3... but can you dole out individual gallons?
Solution
