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 > > > >

**References**:**Prime Puzzle with Mathematica***From:*amzoti <amzoti@gmail.com>