[ March 1999 ]
An original puzzle involving primes
Enoch Haga

Find more palindromes that are the concatenation of the nth prime with  the sum of the primes smaller or equal to this nth prime.

Enoch Haga himself discovered two nice solutions for this hard problem.

At 73, the 21^st prime, the sum of the primes <= 73 is 712
from which the palindrome 21712 is formed.
At 4177, the 574^th prime, the sum of the primes <= 4177 is 1111475
from which the palindrome 5741111475 is formed.

Apparently not easy to find, Enoch dares to challenge you to find more solutions !
"I have now checked to 199909, the 17978 ^th prime, and found nothing else to con-cat-enate!
Perhaps there are no more, but then I shall offer a prize of $5.95
(the sum of 21 and 574 divided by 100 -- just because it forms a palindrome)
to anyone discovering the next one in sequence (or who proves that it is impossible)."

[ August 15, 2002 ]
Jean Claude Rosa distinguished more cases that could be examined.

Let P be the prime number, N its rank number,
S the sum of the prime numbers < = P,
& the concatenation operation and
PP the result that must be palindromic.

JCR proposes the following six 'equations' to solve !

N & S = PP (Enoch's puzzle) S & N = PP N & P = PP P & N = PP P & S = PP S & P = PP

By varying P from 2 up to 1175497783 JCR obtained the following results :

	P	N	S	PP
N & S	73 4177	21 574	712 1111475	21712 5741111475
S & N	?	?	?	?
N & P	17 183661 61241363	7 16638 3631421		717 16638183661 363142161241363
P & N	491 1823 6883 757063 9642461 329147719	94 281 886 60757 642469 17741923		49194 1823281 (prime curios!) 6883886 75706360757 9642461642469 32914771917741923
P & S	2 7	1 4	2 17	22 717
S & P	2	1	2	22