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Talk:Minimum positive multiple in base 10 using only 0 and 1: Difference between revisions
Talk:Minimum positive multiple in base 10 using only 0 and 1 (view source)
Revision as of 22:49, 2 March 2020
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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)
== general observations ==
If the '''n''' is only composed of ones and zeros, then the '''B10''' is equal to '''n'''.
If the '''n''' is only composed of nines, then the '''B10''' is composed of the shown digits below:
::: For '''9''', the '''B10''' digits are '''12345678''', and change the last eight to a nine.
::: For '''99''', the '''B10''' digits are '''1122334455667788''', and change the last eight to a nine.
::: For '''999''', the '''B10''' digits are '''111222333444555666777888''', and change the last eight to a nine.
::: For '''9999''', the '''B10''' digits are '''11112222333344445555666677778888''', and change the last eight to a nine.
I'm sure there is a more elegant way of expressing this, but I think showing is better than telling (the pattern seems obvious enough). -- [[User:Gerard Schildberger|Gerard Schildberger]] ([[User talk:Gerard Schildberger|talk]]) 22:48, 2 March 2020 (UTC)
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