3rd Mexico 1989 Problem B1 Find a positive integer n with decimal expansion a m a m-1 ...a 0 such that a 1 a 0 a m a m-1 ...a 2 0 = 2n.
10 526 315 789 473 684 210 or 789 473 684 210 526 315 Solution Clearly a 0 = 0 or 5 and a 1 = 1. Take a 0 = 0. Now we can calculate the digits successively from the right. Doubling ...10 gives ...20, so a 2 = 2. Doubling ...210 gives ...420, so a 3 = 4. Doubling ...4210 gives ...8420 so a 4 = 8. Doubling ...84210 gives ...68420 so a 5 = 6, and so on. We stop when the result of the doubling starts with 10. That gives the first answer above. Similarly, starting with ...15 gives the second answer above. 3rd Mexican 1989 © John Scholes jscholes@kalva.demon.co.uk 25 February 2004 Last corrected/updated 25 Feb 04