Talk:Permuted multiples: Difference between revisions
→Ranges to be checked
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At the above zhilongji notes "since x and 3x have the same digits,x%3 == 0 will always be true, so we can search with the start as 10^i+2,and the step as 3." which is correct--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 15:24, 17 August 2021 (UTC)
Extending this for bases other than 10 then for an n digit number the range is (base<sup>n-1</sup>+step-1)<sub>base</sub>..step<sub>base</sub>..(base<sup>(n)</sup>/max multiplier)<sub>base</sub>, step is the lcm of the multipliers which meet the condition (base-1)%multiplier=0. So for 10 digit numbers with multipliers 1..7 and in base 13: 100000000B<sub>13</sub>..C<sub>13</sub>..1B1B1B1B1B<sub>13</sub>. Note that the digital root of the candidates is step--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 11:40, 20 August 2021 (UTC)
== Incorrect extended output for Pascal ==
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{`101111010`, 378}}
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::Our collective "1st digit must be 1"
:::"Our collective "1st digit must be 1" assumption is starting to look decidedly wrong."<BR>But I don't think so.How will you determine the count of digits with trillions of "0" in front ;-)<BR>But it would be nice, not to fix to "1", if the base is 13 then "2" at start can succeed in 1x..6x -[[User:Horst.h|Horst.h]] ([[User talk:Horst.h|talk]]) 17:56, 18 August 2021 (UTC)
==Step counts in Phix==
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