### Putnam 1995

Problem B3

For each positive integer k with n2 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 a1a2 ... an, then the matrix has elements bij = an(i-1)+j). Find f(n) = ∑d(k), where the sum is taken over all 9·10n2-1 such integers.

Solution

Answer: f(1) = 45; f(2) = 10.452; for n ≥ 3, f(n) = 0.

Straightforward.

f(1) = 1 + 2 + ... + 9 = 45.

f(2) = ∑ (a11a22 - a12a21). Summing a11a22 over a term that is present, such as a22 gives a factor 45. Summing over a term that is not present gives a factor 10, unless the sum is over a11 in which case it gives a factor 9. Hence the sum for a11a22 is 452102. Similarly, the sum for - a12a21 is - 45290. Hence the total is 10.452.

The same approach gives zero for n ≥ 3, because there are an even number of terms involving a11, 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 a11, half with a plus sign and half with a minus sign. Again, the sum over these terms is zero.