Russian 1999 Problem 1 The digits of n strictly increase from left to right. Find the sum of the digits of 9n.
9 Solution Suppose the digits of n are a 1 a 2 ...a k . Then the digits of 10n are a 1 a 2 ...a k 0. Now subtract n. The units digit is 10-ak and we have carried one from the tens place. But because ai > ai-1, no further carries are needed and the digits of 9n are a 1 (a 2 -a 1 ) (a 3 -a 2 ) ... (a k -a k-1 -1) (10-a k ). Thus the sum of the digits is 9. Russian 1999 © John Scholes jscholes@kalva.demon.co.uk 4 March 2004 Last corrected/updated 4 Mar 04