Of a Fibonacci number,
the 'ones' digit has a cyclic period of 60.
the 'tens' digit has a cyclic period of 300 (=60*5).
the 'hundreds' digit has a cyclic period of 1500 (=60*5*5).
In the cyclic period of the first 1500, the first occurance and only occurance of the terminating digits 999 is the 1497th, thus a Fibonacci number that terminates with 999 occurs every 1497 + 1500*(n1), such that n is the nth occurance of the Fibonacci number with terminating digits 999.
Edited on November 12, 2012, 4:54 am

Posted by Dej Mar
on 20121111 23:37:40 