## 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.