Triangle of Primes

[davar55]

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 2-digit 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 2-digit primes.

Is this possible?

[davar55]

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
2-digit primes such that the rows and columns by concatenation
are prime.

[davieddy]

I miss Mally if only for his provision of a less number-theoretic
type of problem.

[ckdo]

Originally Posted by davar55

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 2-digit 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 2-digit primes.

Is this possible?

At the risk of being proven wrong: no.

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 12-digit 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:

          37
        89  23
      31  73  71
    19  67  41  13
  97  53  79  61  83
47  43  59  11  17  29

378931199747 and 479719318937 are both prime, however 372371138329 = 409 * 910442881 and 298313712337 = 116707 * 2556091.

So what I have I'd like to call a "31-prime, 35-factor" solution. A better solution in the spirit of post #2 (a) may exist, but I seriously doubt it.

Cheers,
Carsten

[davar55]

Originally Posted by davieddy

I miss Mally if only for his provision of a less number-theoretic
type of problem.

Yes, his interest in old math did help some of us
learn from the past.