Problems & Puzzles: Puzzles

Puzzle 932. A puzzle about emirps. J. M. Bergot posed a puzzle about emirps. Here is my reformulation of it: Emirp points out to a pair of distinct primes such that any of them is the reverse of the other. Emirp = (p & p') such that p=reverse(p') and p'=reverse(p) Here are the first eight primes generating a couple of emirps: 13, 17, 31, 37, 71, 73, 79, 97, .(sequence A006567 in the OEIS) For our puzzle we will consider the couple of primes that form an emirp as a unit. (13 & 31) (17 & 71) etc... The puzzle asks to find emirps such that for both primes of the emirp, the sum of the prime and its digits produces another emirp. Schematically: For Emirp1=(p & p'), p+sod(p)=q and p'+sod(p')=r such that (q & q')=Emirp2 and (r & r')=Emirp3. This mean that Emirp1 define two Emirps: Emirp2 and Emirp 3. Example: Emirp1=(1933, 3391), sod(p)=16 -> Emirp2=(1949, 9491), Emirp3=(3407,7043) Q1. Send your largest example. Q2. Send and example where both Emirp2 and Emirp3 behave as Emirp1 (this mean that Emirp1 will define six Emirps)

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Contributions came from Jan van Delden and Simon Cavegn.

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Jan wrote, Nov 19, 2018:

For Q1 & Q2, respectively:

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Simon wrote on Nov 22, 2018:

Q1: 1000000000000000000000000000000000000000000000000005328930513

 

Q2: Did not find any. (Searched up to 188733911393.)

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On August 9, 2019 Metin Sariyar from Turkey wrote:

A larger example for Q1: 10^99+97885251=100000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000097885251 (100 digits)

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