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Re: Prime Puzzle with Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg92720] Re: [mg92680] Prime Puzzle with Mathematica
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sat, 11 Oct 2008 06:45:26 -0400 (EDT)
  • References: <200810100835.EAA07106@smc.vnet.net> <CA8420BF-8E49-4DC8-885E-73A0A0D0E24C@mimuw.edu.pl> <10b625070810100804n6e75a193t47f29ee4c086c099@mail.gmail.com>

One can get this with just one more line:

Intersection[AverageIsPrime,AverageOfEachPairIsPrime]

{{11, 47, 71}}

Best regards

Andrzej Kozlowski


On 11 Oct 2008, at 00:04, W F wrote:

> The third vector is the correct and unique answer.
>
> I think the following requirement must not be in the formulas:
>
>          (The average of all three is also a prime number.)
>
> Thanks for the clean code!
>
> ~A
>
> On Fri, Oct 10, 2008 at 7:26 AM, Andrzej Kozlowski  
> <akoz at mimuw.edu.pl> wrote:
>
> On 10 Oct 2008, at 17:35, amzoti wrote:
>
> Hi All,
>
> trying to find an efficient way to this in Mathematica.
>
> I found the answer - but it was a manual list manipulation - and it
> was ugly!
>
> Any suggestions?
>
> 3 Nice Primes:
>
> Find three 2-digit prime numbers such that:
>
>   * The average of any two of the three is a prime number, and
>   * The average of all three is also a prime number
>
> Thanks!
>
> ~A
>
>
>
> I assume that you do not want any repetitions in your lists of three  
> numbers (i.e. {11,11,11} does not count). So
>
> ls1 = Select[Range[11, 99], PrimeQ];
> ls2 = Tuples[ls1, {3}];
> ls3 = DeleteCases[Union[Sort /@ ls2], {___, x_, ___, x_, ___}];
>
> Now
>
> AverageIsPrime = Select[ls3, PrimeQ[Mean[#]] &];
>
> Length[AverageIsPrime]
> 144
>
>
>
> The other one is shorter:
>
>  AverageOfEachPairIsPrime = Select[ls3, And @@ PrimeQ /@ Mean /@  
> Partition[#1, 2, 1, {1, 1}] & ]
>
> {{11, 23, 71}, {11, 23, 83}, {11, 47, 71}, {13, 61, 73}, {17, 29,  
> 89}, {23, 59, 83}, {29, 53, 89}}
>
>
> Andrzej Kozlowski
>
>
>
>



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