All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
2 sets of primes (Posted on 2019-02-04) Difficulty: 3 of 5
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.

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
computer soln | Comment 1 of 5
M = 5 + 7 + 17 + 19 = 48

N = 5 + 7 + 11 + 19 + 29 + 37 = 108

        program rs

c

c       determine if any 3/4 primes and any 5/6 primes add to a prime

c

        implicit none

        integer p(140),topodds,sum3,sum4,sum5,sum6,

        1 cnt,s3,s5,pp(6),i,j,ir,lim,ptest,j4,j6,i1,i2,i3,i4,i5,i6,

        1 maxindex

        topodds=740

        maxindex=30


c make the primes that are less than "top"


        p(1)=2

        p(2)=3

           cnt=2

           do 100 i=5,topodds,2

           lim=sqrt(1.*i)+1

                do j=3,lim,2

                ir=i/j

                if (j*ir.eq.i)go to 100

                enddo

           cnt=cnt+1

        if(cnt.eq.140)then

        print*,' too many primes'

        stop

        endif

           p(cnt)=i

100     enddo

         do 1 i1=1,maxindex

         pp(1)=p(i1)

          do 2 i2=i1+1,maxindex

          pp(2)=p(i2)

           do 3 i3=i2+1,maxindex

           pp(3)=p(i3)

            do 4 i4=i3+1,maxindex

            pp(4)=p(i4)

                sum4=pp(1)+pp(2)+pp(3)+pp(4)

                   do 444 s3=1,4

                   sum3=sum4-pp(s3)

                        do ptest=4,maxindex

                        if(sum3.eq.p(ptest))go to 444

                        enddo

                   go to 500

444                enddo

                if(sum4.le.48)print*,'*** ',(pp(j4),j4=1,4),sum4


500          do 5 i5=i4+1,maxindex

             pp(5)=p(i5)

              do 6 i6=i5+1,maxindex

              pp(6)=p(i6)

                sum6=pp(1)+pp(2)+pp(3)+pp(4)+pp(5)+pp(6)

                   do 666 s5=1,5

                   sum5=sum6-pp(s5)

                        do ptest=5,maxindex

                        if(sum5.eq.p(ptest))go to 666

                        enddo

                   go to 6

666                enddo

                if(sum6.le.108)print*,'*****',(pp(j6),j6=1,6),sum6

6             enddo

5            enddo

4           enddo

3          enddo

2         enddo

1        enddo

        end 

Edited on February 6, 2019, 2:37 am
  Posted by Steven Lord on 2019-02-05 00:06:52

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information