Problems & Puzzles: Puzzles

 Puzzle 496. P*R(P)+1 = Palprime JC Rosa sent the following puzzle: Is possible that P*R(P)+1 is palprime?

Comments came from J. C. Rosa, Enoch Haga & Farideh Firoozbakht. No solution has come yet. It's expecting a new & brave world...

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Rosa wrote:

I have tested P up to 2447338141 and I have found only two palindromic numbers: 23*32+1=737 and 41*14+1=575

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Enoch wrote:

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 relatively
rare, and finding palprimes seemingly impossible. I wrote some programs that
can reverse primes, but it is slow going for me so I will give it up and
simply say that I don't see that a solution is impossible. One may be
lurking.

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 palindrome.

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Faride wrote:

I didn't find a prime p greater than 2 such that p*R(p)+1 be palprime.
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 exist some
solutions for puzzle 496. Probably the smallest solution is too large and at this
time we can not find it.

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J. K. Andersen  wrote:

I only did a short search to 10^8 with no solution above 2*2+1 = 5.
If both p and p*r(p)+1 are allowed to be composite then the only palindrome
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 shown.

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