Assume a fraction k\/m, with 0\u2264k\u2264m and look at its digits written out<\/p>\n
<\/p>\n
It is surprisingly simple to prove<\/p>\n
Here a little piece of code to explore<\/p>\n
\r\nint k = 1; \/\/ for k\/m for various 1<m\r\nfor (int m = 2; m<=1111; m++) {\r\n int r= k;\r\n System.Numerics.BigInteger kTimes10to_d = k;\r\n List<int> rem = new();\r\n int ix;\r\n int d = 0; \/\/number of digits used\r\n do { \/\/ the division digit by digit, until repeated remainder (incl zero)\r\n rem.Add(r);\r\n d++;\r\n kTimes10to_d*=10;\r\n r=(int)(kTimes10to_d % m); \/\/less than m so plain int\r\n }\r\n while ((ix=rem.IndexOf(r))<0); \/\/ new remaindier?\r\n System.Numerics.BigInteger f = (kTimes10to_d-r)\/m; \/\/r was last remainders of q\/m\r\n int leadlen = ix; \/\/First time the repeated reminder reoccurs is ix steps in \r\n int replen = d-leadlen;\r\n System.Numerics.BigInteger tenToReplen = 1;\r\n for (int i = 0; i<replen; i++) tenToReplen*=10; \/\/ 10^replen\r\n System.Numerics.BigInteger rep = (tenToReplen-1)*r\/m; \/\/ we know this is integral\r\n System.Numerics.BigInteger lead = (f-rep)\/tenToReplen; \/\/clip the repeated tail.\r\n \/\/ (hack: C# D0 does not give empty string, so one extra and drop afterwards)\r\n string leads = lead.ToString("D"+(leadlen+1).ToString())[1..]; \/\/leading zeroes\r\n string reps = rep.ToString("D"+replen.ToString()); \/\/leading zeroes\r\n \/\/Strictly the conditional part is superfluous 1\/4=0.25(0)+ is not wrong.\r\n Console.WriteLine($"{k}\/{m} = {Math.Truncate((double)k\/m)}.{leads}{(r==0?"":$"({reps})+")}");\r\n}\r\n\r\n<\/pre>\n(Where one can easily output rep*m also to study the digit-structure)<\/p>\n
Obviously this in not base dependant, in 3. replace “9” with (base-1) in general, e.g “F” for hex. And “10” with base.<\/p>\n
The first can also be used the other way round. A decimal number with repeated digit-group of length j can not be an integer fraction k\/m unless j<m.
\nHere j can be upper bound as well assuming a lead with L decimals before the repetition the number is (lead+Sequence\/(10^j -1)\/10^L ), so the dividend must be a divisor in (10^j -1)*10^L. (e.g.\u00a0 12.3456456456456… three digits repeated started one decimal in, so 999*10 is a possibility :\u00a0 12.3+456\/9990 = 123333\/9990 = 41111\/3330)<\/p>\n
Note that it is NOT the case that a a prime divisor will give a long repeated sequence, take e.g. 1\/37=0.(027)+ with just three digits repeated, as 37 is a divisor in 10\u00b3-1 (=27*37)<\/p>\n
Some ‘pretty’ ones:<\/p>\n
1\/3 = 0.(3)+<\/p>\n
1\/30 = 0.0(3)+
\n1\/33 = 0.(03)+<\/p>\n
1\/300 = 0.00(3)+
\n1\/303= 0.(0033)+
\n1\/330 = 0.0(03)+
\n1\/333 = 0.(003)+<\/p>\n
1\/3000 = 0.000(3)+
\n1\/3003 = 0.(000333)+
\n1\/3030 = 0.0(0033)+
\n1\/3300 = 0.00(03)+
\n1\/3330 = 0.0(003)+
\n1\/3333 = 0.(0003)+<\/p>\n
1\/30000 = 0.0000(3)+
\n1\/30003 = 0.(00003333)+
\n1\/30030 = 0.0(000333)+
\n1\/30300 = 0.00(0033)+
\n1\/30303 = 0.(000033)+
\n1\/33000 = 0.000(03)+
\n1\/33300 = 0.00(003)+
\n1\/33330 = 0.0(0003)+
\n1\/33333 = 0.(00003)+<\/p>\n
1\/300000 = 0.00000(3)+
\n1\/300003 = 0.(0000033333)+
\n1\/300030 = 0.0(00003333)+
\n1\/300300 = 0.00(000333)+
\n1\/303000 = 0.000(0033)+
\n1\/303030 = 0.0(000033)+
\n1\/330000 = 0.0000(03)+
\n1\/330033 = 0.(00000303)+
\n1\/333000 = 0.000(003)+
\n1\/333300 = 0.00(0003)+
\n1\/333330 = 0.0(00003)+
\n1\/333333 = 0.(000003)+<\/p>\n
On the other hand numbers like 1\/3033 and 1\/3303 are not<\/em> ‘pretty’, What matter is that we by sums of 3×10^i for some i combination can get\u00a0 10\u207f-1. e.g. 330033*303 = 99999999,<\/p>\nEqually pretty if you multiply or divide by 3 obviously.
\nYou will get the same forms with 1s on one side and 9s on the other.<\/p>\n","protected":false},"excerpt":{"rendered":"
Assume a fraction k\/m, with 0\u2264k\u2264m and look at its digits written out The fraction has a finite number of digits (less m) OR is a repeated sequence with at the most m-1 digits per group A repeated sequence will have at the most n-1 zeroes next to each other, if the number m is […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[13,15],"tags":[],"class_list":["post-5243","post","type-post","status-publish","format-standard","hentry","category-math","category-pc-and-code-samples"],"_links":{"self":[{"href":"https:\/\/eskerahn.dk\/index.php?rest_route=\/wp\/v2\/posts\/5243","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/eskerahn.dk\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/eskerahn.dk\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/eskerahn.dk\/index.php?rest_route=\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/eskerahn.dk\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=5243"}],"version-history":[{"count":8,"href":"https:\/\/eskerahn.dk\/index.php?rest_route=\/wp\/v2\/posts\/5243\/revisions"}],"predecessor-version":[{"id":6545,"href":"https:\/\/eskerahn.dk\/index.php?rest_route=\/wp\/v2\/posts\/5243\/revisions\/6545"}],"wp:attachment":[{"href":"https:\/\/eskerahn.dk\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5243"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/eskerahn.dk\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5243"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/eskerahn.dk\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5243"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}