Problems & Puzzles: Puzzles

Puzzle 917. Q=P+A*B

Vic Bold sent the following nice puzzle.

Suppose you have a prime number P with an even quantity of digits. If A & B are the two halves of P, let's calculate the quantity Q=P+A*B such that if Q is even Q=Q/2.


Are there couples of primes P1 & P2 such that Q1=Q2=prime?


Vic sent two examples.


a) (P1, P2)=(23, 37), Q1=23+2*3=29, Q2=37+3*7=58 then Q2=58/2=29

b) (P1, P2)=(29, 73), Q1=29+2*9=47, Q2=73+7*3=94 then Q2=94/2=47


In short:


a) (23,37)->29

b) (29,73)->47


Vic asked for more examples.


I (CR) made a code to search for more examples and found that there are many of such examples, and it's a kind of easy to get them.


So, I asked for a harder question: are k-tuples of primes P1, P2, ... Pk, k>2, with the same value of resulting prime Q?


Here are some of my results:


a) k=3, (102367, 237179, 261071)->139801

b) k=4, (197677, 233419, 257287, 285161)->165523


 Q. Please find examples for k>4.



Contribution came from Flavio Torasso


Flavio rote on March 18, 2018:

I found these three solutions for k=5:
a) k=5, (26083999, 37379539, 49134863, 69410521, 70130413)->36513391
b) k=5, (45217343, 46736779, 47936359, 51655181, 53574637)->39207523
c) k=5, (44459803, 46738837, 49137917, 71812259, 74611799)->44017069


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