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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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Comments: ( Back to comment list | You must be logged in to post comments.)
re: computer soln | Comment 3 of 7 |
(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 2018-10-10 11:37:03
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