       Re: Re: Prime Puzzle with Mathematica

• To: mathgroup at smc.vnet.net
• Subject: [mg92745] Re: [mg92707] Re: [mg92680] Prime Puzzle with Mathematica
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Sun, 12 Oct 2008 04:31:49 -0400 (EDT)

```AverageOfEachPairIsPrime should be based on AverageIsPrime rather than ls3

ls1 = Select[Range[11, 99], PrimeQ];
ls2 = Tuples[ls1, {3}];
ls3 = DeleteCases[Union[Sort /@ ls2], {___, x_, ___, x_, ___}];

AverageIsPrime = Select[ls3, PrimeQ[Mean[#]] &];

AverageOfEachPairIsPrime =
Select[AverageIsPrime,
And @@ PrimeQ /@ Mean /@ Partition[#1, 2, 1, {1, 1}] &]

{{11, 47, 71}}

Bob Hanlon

---- 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

--

Bob Hanlon

```

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