Talk:Minimum positive multiple in base 10 using only 0 and 1: Difference between revisions

(→‎general observations: exchange of factors to reduce runtime)
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:No idea, however I had four (unrelated) minor insights: 1) B10(n) where n is all-nines is all-ones, in fact 9*length(n) ones. 2) No smallest multiplier ends with 0. 3) B10(2n) is either B10(n) [If n ends in 5, I think] or B10(n)*10, and the multiplier must end in 5. 4) for B10(k*10+n) where n is odd {1,3,5,7,9}, the multiplier must end in {1,7,(2/4/6/8),3,9}. Note that between them, rules 3 and 4 cover a trailing {1..9}, once each. None of those really help explain anything though. --[[User:Petelomax|Pete Lomax]] ([[User talk:Petelomax|talk]]) 15:36, 2 March 2020 (UTC)
 
:I've added a reference "How to find Minimum Positive Multiple in base 10 using only 0 and 1" which I think helps.--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 13:49, 3 March 2020 (UTC)
 
== general observations ==
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