11th Brazil 1989 Problem 3 Let Z be the integers. f : Z → Z is defined by f(n) = n - 10 for n > 100 and f(n) = f(f(n+11)) for n ≤ 100. Find the set of possible values of f.
{91, 92, 93, ... } Solution We show that f(n) = 91 for n ≤ 100. For n = 100, 99, ... , 90, we have f(n) = f(f(n+11)) = f(n+11-10) = f(n+1). But f(101) = 91, so f(n) = 91 for n = 100, 99, ... , 90. Now for n = 89, 88, ... , 1, 0, -1, ... we use induction. We have f(n) = f(f(n+11)) = f(91), by induction, = 91. 11th Brazil 1989 © John Scholes jscholes@kalva.demon.co.uk 4 April 2004 Last corrected/updated 4 Apr 04