We have 0 ≤ xi < b for i = 0, 1, ... , n and xn > 0, xn-1 > 0. If a > b, and xnxn-1...x0 represents the number A base a and B base b, whilst xn-1xn-2...x0 represents the number A' base a and B' base b, prove that A'B < AB'.
Solution
We have anbm > bnam for n > m. Hence anB' > bnA'. Adding anbn to both sides gives anB > bnA. Hence xnanB > xnbnA. But xnan = A - A' and xnbn = B - B', so (A - A')B > (B - B')A. Hence result.
Note that the only purpose of requiring xn-1 > 0 is to prevent A' and B' being zero.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
6 Oct 1998
Last corrected/updated 14 Mar 03