46th Putnam 1985 Problem A4 Let a n be the sequence defined by a 1 = 3, a n+1 = 3 k , where k = a n . Let b n be the remainder when a n is divided by 100. Which values b n occur for infinitely many n? Solution
: 87. We find fairly quickly that 3 20 = 1 mod 100. So a 3 = 3 27 = 3 7 = 87 mod 100. Hence a 4 = 3 87 = 3 7 = 87 mod 100. So b n = 87 for n > 2. 46th Putnam 1985 © John Scholes jscholes@kalva.demon.co.uk 7 Jan 2001