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.

This entry was posted on February 18, 2011 at 12:05 am and is filed under R, Statistics with tags bug, Le Monde, mathematical puzzle, prime numbers, schoolmath. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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

xi'an Says:
February 19, 2011 at 10:14 am

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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