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.
February 19, 2011 at 10:14 am
The solution to the five prime problem appeared yesterday in Le Monde: consider five primes
satisfying the constraints. Then necessarily at most two of them takes the same value modulo 3. This implies that there exist
Hence
cannot be a prime number.