2nd Irish 1989 Problem A3 The sequence a 1 , a 2 , a 3 , ... is defined by a 1 = 1, a 2n = a n , a 2n+1 = a 2n + 1. Find the largest value in a 1 , a 2 , ... , a 1989 and the number of times it occurs.
10, 5 Solution a n = no. of binary digits in n (easily proved by induction). We have 1989 = 1024 + 512 + 256 + 128 + 64 + 4 + 1 = 11111000101 2 . So the maximum value is obviously 10, achieved by 1111111111, 10111111111, 11011111111, 11101111111, 11110111111, so 5 times. 2nd Irish 1989 (C) John Scholes jscholes@kalva.demon.co.uk 11 July 2003 Last updated/corrected 11 Jul 03