Talk:Undulating numbers

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Revision as of 21:53, 1 June 2023 by Petelomax (talk | contribs)

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)

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