Talk:Minimum multiple of m where digital sum equals m: Difference between revisions

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Petelomax

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Revision as of 06:11, 9 February 2022 (view source) Petelomax (talk \| contribs) m (→‎next number divisible by 11) ← Older edit		Latest revision as of 17:20, 19 February 2022 (view source) Petelomax (talk \| contribs) (→‎I'm out)
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Line 169: n<sub>20</sub>+n<sub>19</sub> must be at least 7. Each time n<sub>20</sub>+n<sub>19</sub> is incremented 1 must be subtracted from either No or Ne. The table above depicts the possibilities and is in fact a perfectly balanced binary tree. When a candidate is identified No and Ne must be expanded to the set of 8 digits summing to No and the set of 9 digits summing to Ne, and all combinations considered. In this case it is easy as 72 can only be 8 nines and 81 can only be 9 nines. Thus it only remains to prove that 70999999999999999995 is divisible by 3.--[[User:Nigel Galloway\|Nigel Galloway]] ([[User talk:Nigel Galloway\|talk]]) 13:37, 8 February 2022 (UTC) ==I'm out== ~~==next number divisible by 11==~~ Gave up on 370. 275 still a little tardy/not done 25. Pretty pleased with 200 in 0.7s though. --[[User:Petelomax\|Pete Lomax]] ([[User talk:Petelomax\|talk]]) 12:21, 14 February 2022 (UTC) ~~Hit a slight snag in my [new and not yet posted] counting-divisible-by-11s-with-same-digit-sum algorithm. <br>~~ ~~From 59998906 I'm getting 66098989 but the brute force is (correctly) telling me the next is 60949999.<br>~~ :I'm trying to decide on an adjective to describe this. I might have to toss a coin to choose between insane and impressive.--[[User:Nigel Galloway\|Nigel Galloway]] ([[User talk:Nigel Galloway\|talk]]) 13:41, 14 February 2022 (UTC) ~~I think this is one (the first I've hit) of those tricky transitions where 11 is transferred between the odd and even digits.<br>~~ ::I'm insanely impressed. Chapeau! Kudos to Pete :-) --[[User:Horst.h\|Horst.h]] ([[User talk:Horst.h\|talk]]) 16:14, 14 February 2022 (UTC) ~~I have (actually) got some (unposted & untested) code that is supposed to be doing this which isn't even triggering, but...~~ ~~<pre>~~ ::: FWIW, you may like to consider submitting this to OEIS for inclusion on the [[oeis:A131382\|A131382]] page. They only have a list of the first 90 terms at the moment. If you are so inclined. --[[User:Thundergnat\|Thundergnat]] ([[User talk:Thundergnat\|talk]]) 16:56, 14 February 2022 (UTC) ~~59998906~~ ::: Good idea, done. Update: that page links to [[oeis:A002998\|A002998]] which aleady had '''1000''' terms (gulp) and a c++ program... --[[User:Petelomax\|Pete Lomax]] ([[User talk:Petelomax\|talk]]) 15:59, 15 February 2022 (UTC) ~~+ 4 0 1 6~~ ::: OK, now I really, really, really am out ''(please, pretty please!)'' --[[User:Petelomax\|Pete Lomax]] ([[User talk:Petelomax\|talk]]) 17:19, 19 February 2022 (UTC) ~~- 5 6~~ ~~= 99999460~~ ~~> 60949999 <<-- pairwise swap all 8 [which feels right compared to other cases].~~ ~~</pre>~~ ~~as in add max from left [which feels wrong], or is it add max from right [which feels right and matches the subtraction]:~~ ~~<pre>~~ ~~59998906~~ ~~+ 1 0 1 9~~ ~~- 5 6~~ ~~= 69999490~~ ~~> 60949999 <<-- reverse the last 7 [which feels wrong and would break other cases].~~ ~~</pre>~~ I guess it could be a completely different method altogether. I guess what I am asking is that given 59998906 how would you as a human using either maths or intuition determine the next number, that still sums to 55 and is still divisible by 11? --[[User:Petelomax\|Pete Lomax]] ([[User talk:Petelomax\|talk]]) 05:47, 9 February 2022 (UTC)