Talk:Permuted multiples: Difference between revisions
→Ranges to be checked
(Step counts in Phix) |
|||
| (7 intermediate revisions by 2 users not shown) | |||
Line 24:
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 ==
Line 133 ⟶ 135:
rekCount 26734671088</pre>
-[[User:Horst.h|Horst.h]] ([[User talk:Horst.h|talk]]) 10:06, 18 August 2021 (UTC)
::That is curious. You may be interested to know there ''are'' in fact zillions of base 2 solutions, the first 11 with decimal equivalents in the right hand column are shown below, and it appears that k-5 aspect is at least initially holding:
<pre>
{{`000111`, 7},
{`001110`, 14},
{`010101`, 21},
{`011100`, 28},
{`100011`, 35},
{`101010`, 42}}
{{`0001110`, 14},
{`0011100`, 28},
{`0101010`, 42},
{`0111000`, 56},
{`1000110`, 70},
{`1010100`, 84}}
{{`0001111`, 15},
{`0011110`, 30},
{`0101101`, 45},
{`0111100`, 60},
{`1001011`, 75},
{`1011010`, 90}}
{{`00011100`, 28},
{`00111000`, 56},
{`01010100`, 84},
{`01110000`, 112},
{`10001100`, 140},
{`10101000`, 168}}
{{`00011110`, 30},
{`00111100`, 60},
{`01011010`, 90},
{`01111000`, 120},
{`10010110`, 150},
{`10110100`, 180}}
{{`00011111`, 31},
{`00111110`, 62},
{`01011101`, 93},
{`01111100`, 124},
{`10011011`, 155},
{`10111010`, 186}}
{{`000101101`, 45},
{`001011010`, 90},
{`010000111`, 135},
{`010110100`, 180},
{`011100001`, 225},
{`100001110`, 270}}
{{`000111000`, 56},
{`001110000`, 112},
{`010101000`, 168},
{`011100000`, 224},
{`100011000`, 280},
{`101010000`, 336}}
{{`000111100`, 60},
{`001111000`, 120},
{`010110100`, 180},
{`011110000`, 240},
{`100101100`, 300},
{`101101000`, 360}}
{{`000111110`, 62},
{`001111100`, 124},
{`010111010`, 186},
{`011111000`, 248},
{`100110110`, 310},
{`101110100`, 372}}
{{`000111111`, 63},
{`001111110`, 126},
{`010111101`, 189},
{`011111100`, 252},
{`100111011`, 315},
{`101111010`, 378}}
</pre>
::Our collective "1st digit must be 1" assumption is starting to look decidedly wrong. --[[User:Petelomax|Pete Lomax]] ([[User talk:Petelomax|talk]]) 17:03, 18 August 2021 (UTC)
:::"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==
Hi Pete. I don't know if it's deliberate, but your step counts don't include the steps which actually find the numbers. --[[User:Nig|Nig]] ([[User talk:Nig|talk]]) 10:57, 18 August 2021 (UTC)
:Semi-deliberate - for me the first place I look is zero steps away. However the "bump" ''is'' arguably a step. So I'm left with two choices: either increment steps before the i=7 check, or reset steps to 1 on bump. Since I can't quite decide, and it's not really important, I'll do neither. --[[User:Petelomax|Pete Lomax]] ([[User talk:Petelomax|talk]]) 13:08, 18 August 2021 (UTC)
| |||