M is the smallest possible sum for a set of four distinct primes such that the sum of any three is prime - (p1,p2,p3,p4}.
N is the smallest possible sum for a set of six distinct primes such that the sum of any five is prime - (q1,q2,q3,q4,q5,q6}.
Find M & N and the
corresponding sets.
For six primes, choose (5,7,11,19,29,37) giving N=108. I believe this is minimal.
That took a lot of pen and paper work. It's much easier to prove that no set of five distinct primes exists such that the sum of any four is prime.
2 can't be a member because the other four would be odd and would sum to a non-prime even number.
Say 3 is a member. If at least three of the others = 1 mod 6, then the sum of those three and 3 will be divisible by 3, an impossibility. The same argument holds if at least three = -1 mod 6. The remaining case has two primes = 1 mod 6 and two = -1 mod 6, but then their sum is an even number > 2, hence non-prime.
So 3 is out and the primes all = 1 or -1 mod 6. Four primes of the same type will give an even sum. If three = 1 and two = -1 mod 6, or vice versa, then two of the group of three added to the group of two will give an even number.
So no set of five primes satisfied the conditions of the problem.
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Posted by xdog
on 2019-02-05 18:30:47 |
solution for 6, proof 5 impossible
