Re: For 2014?
- To: mathgroup at smc.vnet.net
- Subject: [mg132153] Re: For 2014?
- From: carlson at wolfram.com
- Date: Fri, 3 Jan 2014 04:44:07 -0500 (EST)
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- References: <20131224071749.C3F1A69D5@smc.vnet.net> <3688DC0E-CE14-47F1-B93B-5BFA875C59E3@data2knowledge.de> <20131226112449.4994F6A09@smc.vnet.net>
I wrote a blog post about this a while ago which has a downloadable notebook with code. There's also a reference to a post by Hans Havermann on the same topic. http://blog.wolfram.com/2012/02/02/happy-109876-54321/ Chris On Dec 26, 2013, at 5:24 AM, Ulrich Arndt <ulrich.arndt at data2knowledge.de> wrote: > Actually it has to be SeedRandom[1992] - was late ;-). > > But also an complete check is possible - and therefore much better... > > char = CharacterRange["1", "9"] > type = {"+", "-", "*", "/", ""} > > t = Tuples[type, 8]; > r = StringJoin[Riffle[char, #]] & /@ t; > e = ToExpression[#] & /@ r; > p = Position[e, 2014] > Extract[r, p] > > Works also for reversed number list. > > Ulrich > > Am 24.12.2013 um 22:05 schrieb Ulrich Arndt: > >> 123 + 45*6*7 - 8 + 9 >> >> Generate is maybe a bit wrong - search ;-) >> >> char = CharacterRange["1", "9"] >> type = {"+", "-", "*", "/", ""} >> RandomSeed[1992] >> r = Table[StringJoin[Riffle[char, RandomChoice[type, 8]]], {1000000}]; >> e = ToExpression[#] & /@ r; >> p = Position[e, 2014] >> Union[Extract[r, p]] >> >> 2015 >> 1*2-3+4*567*8/9, 12*3+45*6*7+89, 12*34*5-6*7+8+9 >> >> >> >> >> >> >> Am 24.12.2013 um 08:17 schrieb Harvey P. Dale: >> >>> There are some nice (very simple) math puzzles using consecutive integers that produce years. For example, 10+(9 x 8 x (7/6) x 5 x 4)+321 and 0-12+(34 x 56)+7 x (8+9) both yield 2011, and (10 x 9 x 8) + 7 + 6 - 5 + (4 x 321) yields 2012. >>> >>> Two questions: (1) can anyone generate a similar puzzle yielding 2014 and (2) is there a general Mathematica program that can generate these? >>> >>> Best, >>> >>> Harvey >>> >> >> > > >