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)

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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