31st IMO 1990 shortlist Problem 1 Is there a positive integer which can be written as the sum of 1990 consecutive positive integers and which can be written as a sum of two or more consecutive positive integers in just 1990 ways?
Yes Solution We require N = m + (m+1) + ... + (m+1989) = 1990m + 1989·1990/2 = 5·199(2m+1989). So N must be odd and divisible by 5 and 199. We also require that there are just 1990 pairs m, n such that N = m + (m+1) + ... + (m+n) = (n+1) (n+2m)/2, or equivalently, such that 2N = (n+1)(n+2m). Note that n+1  1, we can take n+1 to be A and m to be (B+1-A)/2. Since N is odd, just one of A, B will be even, so (B+1-A) is even and m is integral. Thus we require 2N to have 2·1991 factors (the extra two are the pair 1 and N which do not give a solution) = 2·11·181. We know that 2N has factors 2, 5, 199, so it cannot have any others. The factor 2 1 gives rise to the factor (1+1) = 2, the other two must give rise to the 11 and the 181. So N = 5 10 199 180 or 5 180 199 10 . 31st IMO shortlist 1990 © John Scholes jscholes@kalva.demon.co.uk 7 Aug 2003 Last updated/corrected 7 Aug 2003