P*R(P)+1 = Palprime
JC Rosa sent the following puzzle:
Is possible that P*R(P)+1
Comments came from J. C. Rosa, Enoch Haga & Farideh Firoozbakht. No
solution has come yet. It's expecting a new & brave world...
I have tested P up to 2447338141 and I have found only two
palindromic numbers: 23*32+1=737 and 41*14+1=575
At any rate, I have studied the puzzle at some length and have
come to the
conclusion that a solution is likely, but that finding primes is
rare, and finding palprimes seemingly impossible. I wrote some
can reverse primes, but it is slow going for me so I will give it up
simply say that I don't see that a solution is impossible. One may
All said and done it seems an excellent puzzle.
One interesting prime is 677*776+1 where we have two concatenated
palindromes, 525353 (353 is a palprime whereas 525 of course is a
I didn't find a prime p greater than 2 such that p*R(p)+1 be
But all palprimes of the form m*R(m)+1 that I found are: 2, 5, 11,
101 & 1008001.
It's interesting that the palprime 1008001 has two representations
of the form
m*R(m)+1 where m>R(m) (m1=24000, m2=42000).
There is no a reason that puzzle has no solution and maybe there
solutions for puzzle 496. Probably the smallest solution is too
large and at this
time we can not find it.
J. K. Andersen wrote:
I only did a short search to 10^8 with no solution above 2*2+1 =
If both p and p*r(p)+1 are allowed to be composite then the only
results below 10^8 are 10^n * 1 + 1 = 10^n+1 for every n, and:
2 * 2 + 1 = 5
13 * 31 + 1 = 404
14 * 41 + 1 = 575
23 * 32 + 1 = 737
40 * 4 + 1 = 161
24000 * 42 + 1 = 1008001
42000 * 24 + 1 = 1008001
318317 * 713813 + 1 = 227218812722
43029128 * 82192034 + 1 = 3536651551566353
When p and r(p) have the same length, only the case with p < r(p) is