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

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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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