Le Monde puzzle [#6]

A simple challenge in Le Monde this week: find the group of four primes such that any sum of three terms in the group is prime and the overall sum is minimised. Here is a quick exploration by simulation, using the schoolmath package (with its imperfections):


A=primes(start=1,end=53)[-1]
lengthA=length(A)

res=4*53
for (t in 1:10^4){

 B=sample(A,4,prob=1/(1:lengthA))
 sto=is.prim(sum(B[-1]))
 for (j in 2:4)
 sto=sto*is.prim(sum(B[-j]))

 if ((sto)&(sum(B)<res)){
 res=sum(B)
 sol=B}
 }
}

providing the solution 5 7 17 19.

A subsidiary question in the same puzzle is whether or not it is possible to find a group of five primes such that any sum of three terms is still prime. Running the above program with the proper substitutions of 4 by 5 does not produce any solution, even when increasing the upper boundary in A. So it is most likely that the answer is no.

One Response to “Le Monde puzzle [#6]”

  1. The solution to the five prime problem appeared yesterday in Le Monde: consider five primes

    a_1,\ldots,a_5

    satisfying the constraints. Then necessarily at most two of them takes the same value modulo 3. This implies that there exist

    a_{i_1}\equiv 0\mod 3, a_{i_2}\equiv 1\mod 3, a_{i_3}\equiv 2\mod 3.

    Hence

    a_{i_1}+a_{i_1}+a_{i_2}\equiv 0\mod 3

    cannot be a prime number.

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