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