The solution to the puzzle is:

23452

43214

15521

31433

12455

45214

13351

32422

25315

15433

41354

52142

A=2, B=5, C=11, D=13, E=17, F=23, G=29, H=31

J=37, K=41, L=43, M=47, N=53, P=61, Q=79, R=83

Notes from the original puzzle by John Gowland. I changed the letters of the puzzle to make it a bit easier.

1. Working backwards, XII = 2DN2 = 2.31.292 = 52142. Thus D=31
and N = 29

2. XI =2DNQ = 2.31.29.23.= 41354. Thus Q = 23

3. X = PQC = .(3,4,5). Only solution with 3,4,5 in the middle
is 61.23.11. Thus P = 61 and C = 11

4. IX = FPK = F.61.K. It must start with a 2 or 3 and not end
in 3 or 4.

67.61.13 = 53131 and 83.61.7 = 35441. It cannot
end in ...41 as there is already a 4 in the

Y-pentomino. Only solution is 83.61.5 = 25315
giving F = 83 and K = 5.

5. VIII = 2GNB = 2.G.29.B. It must start with a 3 and end with
a 2. Only satisfactory solution is

2.43.29.13 = 32422. Thus G = 43 and B = 13

6. VII ends with 1 or 5. VII = MB2 = M.132 = 79.132 = 13351. Thus M
= 79

7. VI begins with a 4 and cannot contain 3 in the last 3 digits. 2JLB
= 2JL.13 ; of the five possible

solutions, only 2.47.37.13 = 45214 begins
with a 4. Thus J = 47 and L = 37

8. V = HJK and ends with a 5 = H.47.5 = 53.47.5 or 61.47.5. But
P =61, so H =53 and HJK = 12455

9. IV = G2E = 432.E and begins with a 3. E could be 17 or 19.
However, 432.19 = 35131 and the

T-pentomino already contains a 5. Therefore,
E = 17 and 432.17 = 31433

10. III = FEC = ...2. = 83.17.11 = 15521

11. II = 2ADE = 2.A.31.17 making A = 43 or 41.

But G =43, therefore A = 41 and 2ADE
= 41.13.17 = 43214

12. Check I = 4ABC = 4.41.13.11 = 23452.