Let an be the sequence defined by a1 = 3, an+1 = 3k, where k = an. Let bn be the remainder when an is divided by 100. Which values bn occur for infinitely many n?
Solution
Answer: 87.
We find fairly quickly that 320 = 1 mod 100. So a3 = 327 = 37 = 87 mod 100. Hence a4 = 387 = 37 = 87 mod 100. So bn = 87 for n > 2.
© John Scholes
jscholes@kalva.demon.co.uk
7 Jan 2001