30th CanMO 1998 Problem 1 How many real x satisfy x = [x/2] + [x/3] + [x/5]? Solution
: 30. Put x = 30q + r, where q is an integer and 0 ≤ r < 30. Then x - [x/2] - [x/3] - [x/5] = -q + r - [r/2] - [r/3] - [r/5]. So x = [x/2] + [x/3] + [x/5] iff q = r - [r/2] - [r/3] - [r/5]. q, [r/2], [r/3] and [r/5] are all integers, so r must be an integer. There are 30 possible integral values of r (namely 0, 1, 2, ... , 29), so there is one solution x for each. 30th CanMO 1998 © John Scholes jscholes@kalva.demon.co.uk 15 June 2002