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 nth 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 2digit number is.
Find it.
Are there any additional i.e. "numbers over 100" solutions?
(In reply to
computer soln by Steven Lord)
Looking over your logic on there being no further solutions, I notice a couple of discrepancies:
Due to the sparseness of larger primes, eventually it is the average, Sp(n)/n that will have fewer digits than p(n).
Also n=9085 is not the point at which the same number of digits becomes impossible. Just to take the point at which my own program is now at, at n=530522, p(n)=6276343 and the average, Sp(n)/n, is 3021610.23.... If an integral average were to occur in this region, it would still have 7 digits just like the primes in this neighborhood.

Posted by Charlie
on 20181010 11:37:03 