20080814, 19:04  #1 
May 2004
New York City
5·7·11^{2} Posts 
Triangle of Primes
A variation on a recent puzzle:
Draw 21 congruent circles in rows of 1,2,3,4,5 & 6, to form the shape of an equilateral triangle. Now fill in each circle with a different 2digit prime (there just happen to be 21 of these) such that the concatenation of primes in any row of circles in either direction is prime (that's 33 primes). Don't reverse the digits of the 2digit primes. Is this possible? 
20080827, 20:08  #2 
May 2004
New York City
5×7×11^{2} Posts 
If this problem was poorly stated or has no solution,
then the following is just an attempt to save the problem: (a) Fill in the circles so as to create as many primes as possible, instead of all rows, backward and forward, in three directions giving prime concatenations. (I think expecting to find 33 primes of average length ~7 digits was far too unlikely among the 21! different configurations.) (b) Instead of a triangle, construct a 3x7 rectangle using all the 2digit primes such that the rows and columns by concatenation are prime. 
20080905, 14:33  #3 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
I miss Mally if only for his provision of a less numbertheoretic
type of problem. Last fiddled with by davieddy on 20080905 at 14:34 
20080915, 12:52  #4  
Dec 2007
Cleves, Germany
2·5·53 Posts 
Quote:
I wrote a program to fill the triangle from the bottom up and had it dump all partial solutions with the last number (the top one) still missing. It came up with only 21 of these. However, I had previously invested some thought into pruning the search space so that no full solutions would be reported multiple times, so there may be more than just 21. I then manually added the last number and checked the four 12digit primes along the left and right edges for factors. In 16 cases, neither was prime. In four cases, one was prime. In one case, two were prime: Code:
37 89 23 31 73 71 19 67 41 13 97 53 79 61 83 47 43 59 11 17 29 So what I have I'd like to call a "31prime, 35factor" solution. A better solution in the spirit of post #2 (a) may exist, but I seriously doubt it. Cheers, Carsten 

20090702, 20:16  #5 
May 2004
New York City
4235_{10} Posts 

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