Hi, wenonah42, just back from a day out. I see that others have confirmed your calculation of HNPPXX as 14652.
In fact, I followed a slightly different path from you at this stage. I'd already got the 12 letters A, B, D, H, J, P, Q, R, T, V, X, Y, and I then worked out G and M from 13a and N from 12a. I actually missed the fact that I could then have calculated C from 6a, and instead used brute force computation on all the remaining combinations of primes to determine C and K from 26d. So I didn't check 6d until I had all the letters solved. However, if you have worked out the 12 letters plus G, M and N, there are only 11 primes left, and calculating the 11 different possible values of YY(A+C+G) gives only one prime number possibility for C that yields a Pythagorean triple.
This shows that there are several different routes to identifying the prime numbers - if I'd solved for C using 6d, I'd probably have next solved for I and S simultaneously using 21d.
The endgame was, I thought, almost a substantive puzzle in its own right. Only six of the 42 answers could be entered directly, and this phase of the puzzle alone took me around an hour. I started in the bottom right hand corner, where several of the clues lead to three or four-digit values for c in the Pythagorean triples, hence limiting the possible answers to be entered in the grid, and then just used trial and error to work out which possibilities would fit.