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Reversal creates average (Posted on 2018-10-10) Difficulty: 3 of 5
Let Sp(n) be a sum of all primes from p(1) to p(n) inclusive.
Let m(n) be the average value of all those primes, i.e. m(n)= Sp(n)/n.

Find the n-th prime N such that m(n) equals the reversal of N.

Example for n=6: Sp(6)=2+3+5+7+11+13= 41; m(6)=41/6; not the reversal of 13, that is 31 .
So 13 is not our prime.
Another 2-digit number is.

Find it.

Are there any additional i.e. "numbers over 100" solutions?

No Solution Yet Submitted by Ady TZIDON    
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computer soln | Comment 1 of 7
For n=23, p(n)=83, Sp(n)/n=38

The code lists all integer Sp(n)/n for p(n) < 100,000 and this is the only one that complies. (I do not see how to prove there are no more.)


           n         P(n)     Sp(n)/n
----------------------------------------------------
          23          83   38.0000000    

          53         241   110.000000    

         853        6599   3066.00000    

        5034       48989   23066.0000    

        5233       51137   24091.0000    

        6162       61223   28921.0000    

        6658       66809   31535.0000    

        6752       67783   32033.0000    

        6974       70379   33213.0000    

        7098       71707   33874.0000    

        7177       72559   34295.0000    

        7228       73091   34567.0000    

        7231       73133   34583.0000    

        7396       75017   35465.0000    

        7698       78439   37084.0000    

        7706       78539   37127.0000    

        7738       78889   37299.0000    

        7840       80051   37848.0000    

        7975       81547   38575.0000    

        8006       81883   38742.0000    

        8127       83233   39394.0000    

        8793       90901   43007.0000    

        8970       92849   43972.0000    

        9050       93787   44408.0000    

        9085       94169   44599.0000  


         program pp

        implicit none

        integer i,n,cnt,sum,iave

        real ave

        sum=2

        cnt=1

           do i=3,100000,2

                call isprime(i,n)

                if(n.eq.1)then

                cnt=cnt+1

                sum=sum+i

                ave=(1.*sum)/(1.*cnt)

                iave=ave

                if(abs(iave-ave).lt.0.001)print*,cnt,i,ave

                endif

           enddo

        end


        subroutine isprime(i,n)

        implicit none

        integer i,j,k,l,m,n

        n=0

        k=sqrt(1.*i)

                do j=3,k

                m=(1.*i)/(1.*j)

                l=m*j

                if(l.eq.i)go to 1

                enddo

            n=1  

1       return

        end       

Edited on October 10, 2018, 8:32 pm
  Posted by Steven Lord on 2018-10-10 08:57:25

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