Re: For 2014?
- To: mathgroup at smc.vnet.net
- Subject: [mg132146] Re: For 2014?
- From: Ulrich Arndt <ulrich.arndt at data2knowledge.de>
- Date: Thu, 26 Dec 2013 06:24:49 -0500 (EST)
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- References: <20131224071749.C3F1A69D5@smc.vnet.net> <3688DC0E-CE14-47F1-B93B-5BFA875C59E3@data2knowledge.de>
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
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> 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
>>
>
>
- References:
- For 2014?
- From: "Harvey P. Dale" <hpd@hpdale.org>
- For 2014?