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There are four rows and four columns, so we need eight different sums. The smallest eight sums (if possible) would be $0, 1, 2, 3, ... , 7$. Since each draughts is counted towards the sum of a row and the sum of a column, we would need $\frac{1}{2}(0+1+2+ ...+7) = 14$ draughts. The diagram shows it is possible to place $14$ draughts on the board to create the eight smallest sums (the numbers in the cells represent how many draughts there are in each cell, and the column and row totals are shown).
Example 1:
Example 2:

This problem is taken from the UKMT Mathematical Challenges.