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.
Are there any additional i.e. "numbers over 100" solutions?
(In reply to computer soln
by Steven Lord)
Actually it is Sp(n)/n, aka m(n), that equals 38, as required, rather than Sp(n), which is the sum, rather than the average. Likewise on the heading of the column of averages.
The answer of course that capital N is 83 is correct. All that definitional complexity was in fact confusing, and took some unraveling to get straight.
Posted by Charlie
on 2018-10-10 11:08:03