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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 16:58, 3 March 2020
, 4 years ago→general observations: added a blank line.
m (→How does it work "C code for Ed Pegg Jr's 'Binary' Puzzle algorithm" ?: corrected ten =( 10^x-1)%num to //ten == 10^(x-1)%num) |
m (→general observations: added a blank line.) |
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:: Yuppers, I wrongedlyness bidirectionally confuseducated '''B10''' with the multiplier. -- [[User:Gerard Schildberger|Gerard Schildberger]] ([[User talk:Gerard Schildberger|talk]]) 23:43, 2 March 2020 (UTC)
::: I think about prime decomposition of the integer to test, because of the observation that 3 is equivilant to 37 both are factors of 111, so when a number has a factor of 37 I can substitute it by 3, like 2 and 5 both are the factors of 10.<BR>But this is only possible, when both numbers are prime.<BR>7 and 143= 11*13 with 7*(11*13) =1001 so 143*37 = 5291 can't be transformed to 3*7 = 21 with the same result,but into 143*3.
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