10th Australian 1989 Problem A1 Find d(1) + d(2) + ... + d(1989) mod 2, where d(n) is the number of positive divisors of n.
Even. Solution If n = p a q b ... then d(n) = (a+1)(b+1)..., so d(n) is odd iff a, b, ... are all even, in other words iff n is a square. We have 44 2 = 1936, 45 2 = 2025, so there are 44 squares in the range 1, 2, ... , 1989 and hence 44 odd d(n), and 1945 even. Hence the sum is even. Solutions are also available in Australian Mathematical Olympiads 1979-1995 by H Lausch and P J Taylor, ISBN 0858896451, published by Australian Mathematics Trust. It can be ordered by email. 10th Aus 1989 © John Scholes jscholes@kalva.demon.co.uk 6 January 2004 Last corrected/updated 6 Jan 04