Putnam 1995, Problem B-3

To each positive integer with \( n^2 \) decimal digits, we associate the determinant of the matrix obtained by writing the digits in order across the rows. For example, for \( n = 2 \), to the integer 8617 we associate \( \det\begin{pmatrix} 8 & 6 \\ 1 & 7 \end{pmatrix} = 50 \). Find, as a function of \( n \), the sum of all the determinants associated with \( n^2 \)-digit integers. (Leading digits are assumed to be nonzero; for example, for \( n = 2 \), there are 9000 determinants.)

Solution:
For \( n = 1 \) we obviously get 45, while for \( n = 3 \) the answer is 0 because it both changes sign (because determinants are alternating) and remains unchanged (by symmetry) when you switch any two rows other than the first one. So only \( n = 2 \) is left. By the multilinearity of the determinant, the answer is the determinant of the matrix whose first (resp. second) row is the sum of all possible first (resp. second) rows. There are 90 first rows whose sum is the vector \( (450, 405) \), and 100 second rows whose sum is \( (450, 450) \). Thus the answer is \( 450 \times 450 - 450 \times 405 = 45 \times 450 = 20250 \).