5th Australian 1984 Problem B3 N is the set of positive integers. M is the set of non-negative integers. f: N → M is a function such that f(10) = 0, f(n) = 0 if the last digit of n is 3, f(mn) = f(m) + f(n). Find f(1984) and f(1985).
0, 0 Solution 0 = f(10) = f(5) + f(2) so f(5) = f(2) = 0. Hence also f(n) = 0 if no primes except 2 and 5 divide n. If n ends in the digit 1, then 3n ends in the digit 3, so 0 = f(3n) = f(3) + f(n), so f(n) = 0. If n ends in the digit 7, then 9n ends in the digit 3, so 0 = f(9n) = f(9) + f(n), so f(n) = 0. If n ends in the digit 9, then 7n ends in the digit 3, so again f(n) = 0. Given any n we may write n = bc, where no primes except 2 and 5 divide b, and c is not divisible by 2 or 5. So f(n) = f(b) + f(c) = 0. Comment: This is almost a straight copy of British MO 1977/1. Solutions are also available in Australian Mathematical Olympiads 1979-1995 by H Lausch and P J Taylor, ISBN 0858896451, published by Australian Mathematics Trust, can be ordered by email. 5th Aus 1984 © John Scholes jscholes@kalva.demon.co.uk 15 June 2002 Last corrected/updated 15 Jun 02