3rd Junior Balkan 1999 Problem 2 Let a n = 2 3n + 3 6n+2 + 5 6n+2 . Find gcd(a 0 , a 1 , a 2 , ... , a 1999 ).
7 Solution a 0 = 35, so we need only consider whether 5 and 7 must be divisors. Consider first 5. Consider a 1 . 2 3n = 8 = 3 mod 5. 3 6n+2 = 3 8 = 1 mod 5, and 5 6n+2 = 0 mod 5, so a 1 = 4 mod 5, and is not divisible by 5. Now consider 7. We have 2 3n = 1 n = 1 mod 7. Since 3 6 = 5 6 = 1 mod 7 (Fermat's little theorem), we have 3 6n+2 = 2 mod 7, and 5 6n+2 = 4 mod 7, so a n = 0 mod 7. Thanks to Suat Namli 3rd Junior Balkan 1999 © John Scholes jscholes@kalva.demon.co.uk 29 July 2003 Last corrected/updated 29 Jul 03