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 Reversal creates average (Posted on 2018-10-10)
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 No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: computer soln | Comment 5 of 7 |
(In reply to computer soln by Steven Lord)

I'm not sure this list is quite right. For example the sum of the first 5034 primes is 116114247, giving an average of 23066.00059...

There is a list in OEIS: A050248

2, 38, 110, 3066, 60020, 740282, 2340038, 29380602, 957565746, ...

As to the second part, when n gets large, m(n) ≈ n/2, and the largest prime ≈ n. e.g, 83/38= 2.184210526. Most numbers do not come near qualifying for this test. Of those between 1 and 100, for example, only {25,52}, {37,73}, {38,83}, {49,94} are reasonably close; and there are only around 40 such pairs below 1000. 73 and 83 are prime; so are 211,613,643,673,683,743,773, giviing a total of only 9 possible candidates less than 1000.

So the chance of any of later m(n) happening to coincide with the reversal of N must be vanishingly small.

Edited on October 11, 2018, 3:54 am
 Posted by broll on 2018-10-11 03:49:10

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