5th Mexico 1991 Problem A1 Find the sum of all positive irreducible fractions less than 1 whose denominator is 1991.
900 Solution 1991 = 11·181 (and 181 is prime). So the 180 fractions 11/1991, 22/1991, ... , 11·180/1991 are reducible, and the 10 fractions 181/1991, 363/1991, ... , 1810/1991 are reducible. All the others are irreducible. Thus their sum is (1/1991)(1 + 2 + ... + 1990 - 11(1 + 2 + ... + 180) - 181(1 + 2 + ... + 10)) = (1990·1991/2 - 11·180·181/2 - 181·10·11/2)/1991 = 995 - 90 - 5 = 900. 5th Mexican 1991 © John Scholes jscholes@kalva.demon.co.uk 28 February 2004 Last corrected/updated 28 Feb 04