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Re: For 2014?

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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@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
>>>
>>
>>
>
>
>

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Re: For 2014? carlson@wolfram.com - 2014-01-03 09:34 +0000

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