Talk:Undulating numbers

From Rosetta Code

It surprised me that all 4 digit undulating numbers in base 7 are a multiple of 10/end in 0 when printed in decimal, but I figured it out: The powers of 7 (0..3) are 1,7,49,343, so they will always be some a*(1+49)+b*(7+343), aka 50a+350b, and yep, all k*50 too --Petelomax (talk) 21:49, 1 June 2023 (UTC)

Yeah, I thought at first that I'd made a mistake until I realized that those numbers must be multiples of 50 though I printed them all out in base 7 just to make sure I wasn't doing anything silly :) --PureFox (talk) 12:00, 2 June 2023 (UTC)