Putnam 1995 Problem B3 For each positive integer k with n 2 decimal digits (and leading digit non-zero), let d(k) be the determinant of the matrix formed by writing the digits in order across the rows (so if k has decimal form a 1 a 2 ... a n , then the matrix has elements b ij = a n(i-1)+j ). Find f(n) = ∑d(k), where the sum is taken over all 9·10 n 2 -1 such integers. Solution
: f(1) = 45; f(2) = 10.45 2 ; for n ≥ 3, f(n) = 0. Straightforward. f(1) = 1 + 2 + ... + 9 = 45. f(2) = ∑ (a 11 a 22 - a 12 a 21 ). Summing a 11 a 22 over a term that is present, such as a 22 gives a factor 45. Summing over a term that is not present gives a factor 10, unless the sum is over a 11 in which case it gives a factor 9. Hence the sum for a 11 a 22 is 45 2 10 2 . Similarly, the sum for - a 12 a 21 is - 45 2 90. Hence the total is 10.45 2 . The same approach gives zero for n ≥ 3, because there are an even number of terms involving a 11 , half with a plus sign and half with a minus sign. Thus the sum over these terms is zero. Similarly, there are an even number of terms not involving a 11 , half with a plus sign and half with a minus sign. Again, the sum over these terms is zero. Putnam 1995 © John Scholes jscholes@kalva.demon.co.uk 12 Dec 1998