13th VMO 1975 Problem B1 Find all terms of the arithmetic progression -1, 18, 37, 56, ... whose only digit is 5.
5...5 with 18k+5 digits for k = 0, 1, 2, 3, ... Solution The general term is 19k - 1 and this must equal 5(10 n - 1)/9 for some n. Hence 5·10 n = 171k - 4 = -4 mod 19. So we certainly require 10 n = 3 mod 19 and hence n = 5 mod 18. Conversely if n = 5 mod 18, then 10 n = 3 mod 19, so 5·10 n = -4 mod 19, so 5·10 n = 19h - 4 for some h. So 5(10 n - 1) = 19h - 9. Now lhs and 9 are divisible by 9, so 19h must be divisible by 9, so h must be divisible by 9. Put h = 9k and we get 19k - 1 = 5(10 n - 1)/9. Thanks to Suat Namli 13th VMO 1975 © John Scholes jscholes@kalva.demon.co.uk 6 March 2004 Last corrected/updated 6 Mar 04