Solutions to the 63 rd

A3 Note that each of the sets {1}, {2}, . . . , {n} has the desired property. Moreover, for each set S with integer average m that does not contain m, S ∪ {m} also has average m, while for each set T of more than one element with integer average m that contains m, T \{m} also has average m. Thus the subsets other than {1}, {2}, . . . , {n} can be grouped in pairs, so Tn −n is even. A4 (partly due to David Savitt) Player 0 wins with optimal play. In fact, we prove that Player 1 cannot prevent Player 0 from creating a row of all zeroes, a column of all zeroes, or a 2 × 2 submatrix of all zeroes. Each of these forces the determinant of the matrix to be zero.