Russian 1999

The digits of n strictly increase from left to right. Find the sum of the digits of 9n.

Answer

9

Solution

Suppose the digits of n are a₁a₂...a_k. Then the digits of 10n are a₁a₂...a_k0. 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 a₁ (a₂-a₁) (a₃-a₂) ... (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