: 3581. 1789 = 4·445 + 9, 445 = 4·109 + 9, 109 = 4·25 + 9, 25 = 4·4 + 9. f(4) = 8 + 3 = 11, so f(11) = f(f(4)) = 4·4+9 = 25, f(25) = f(f(11)) = 4·11 + 9 = 53, f(53) = f(f(25)) = 4·25+9 = 109, f(109) = f(f(53)) = 4·53 + 9 = 221, f(221) = f(f(109)) = 4·109 + 9 = 445, f(445) = 893, f(893) = 1789, f(1789) = 3581. Solutions are also available in Australian Mathematical Olympiads 1979-1995 by H Lausch and P J Taylor, ISBN 0858896451, published by Australian Mathematics Trust. It can be ordered by email. 10th Aus 1989 © John Scholes jscholes@kalva.demon.co.uk 15 June 2002 Last corrected/updated 15 June 2002
10th Australian 1989 Problem B3 Let N be the positive integers. The function f : N → N satisfies f(1) = 5, f( f(n) ) = 4n + 9 and f(2 n ) = 2 n+1 + 3 for all n. Find f(1789). Solution