### Putnam 1998

**Problem B4**

For what m, n > 0 is ∑_{0}^{mn-1} (-1)^{[i/m] + [i/n]} = 0?

**Solution**

Answer: The highest powers of 2 dividing m and n are different.

*Moderately hard.*

If m and n are both odd, then there are an odd number of terms, so the sum must be non-zero. If just one of m and n is odd, then m+n is odd and hence [i/m] + [i/n] has opposite parity to [(mn-1-i)/m] + [(mn-1-i)/n] = n - [i/m] -1 + m - [i/n] - 1= (m + n) + ([i/m] + [i/n]) - 2([i/m] + [i/n] + 1). So in this case the sum is zero.

Finally, suppose m=2m' and n=2n'. Then for any i satisfying 0 <= i < m'n', we have [i/m'] + [i/n'] = [2i/m] + [2i/n] = [(2i+1)/m] + [(2i+1)/n]. Moreover, [(j+m'n')/m] + [(j+m'n')m] = m'+n' + [j/m] + [j/n]. So for m'+n' both even or both odd, [j/m] + [j/n] = [i/m'] + [i/n'] for j = 2i, 2i+1, m'n'+2i, m'n'+2i+1 and hence the sum for m, n has 4 times the value for m', n'. For one of m', n' even and the other odd, m'+n' is odd and hence the two terms j = 2i, 2i+1 cancel with the two terms j = m'n'+2i, m'n'+2i+1, so the sum is zero.

Putnam 1998

© John Scholes

jscholes@kalva.demon.co.uk

12 Dec 1998