Talk:Undulating numbers: Difference between revisions
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(base 7 when in base 10 remark) |
(Yes, they do long wrong at first glance.) |
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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 |
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 --[[User:Petelomax|Petelomax]] ([[User talk:Petelomax|talk]]) 21:49, 1 June 2023 (UTC) |
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: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 :) --[[User:PureFox|PureFox]] ([[User talk:PureFox|talk]]) 12:00, 2 June 2023 (UTC) |
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Latest revision as of 12:01, 2 June 2023
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)