It is given that each set of 3 numbers forming a straight line gives the same sum, let's say 'K'.
∵ A, B, C, D, E, F, G, H and X represents numbers from 1 to 9.
Sum of all the numbers
A+B+X = K .... (1)
C+D+X = K .... (2)
E+F+X = K .... (3)
G+H+X = K .... (4)
Adding all four above equations, we get:
A+B+C+D+E+F+G+H+X+3X= 4K
⇒ 45+ 3X= 4K
⇒ 3(15+X) = 4K.
∴ K should be a multiple of 3 and 4K should be greater than 45.
When K= 12, we have:
45 + 3X = 48
⇒ X= 1---- Case 1 (Not Possible because X is not a perfect square)
When K= 15, we have 45+3X= 60 and X= 5---- Case 2.
When K=18, we have 45+3X= 72 and X= 9----- Case 3 (Not possible because X is not a perfect square)
We can see from equations (1), (2), (3) and (4) that A+B= C+D= E+F= G+H
CASE 2 is the only possibility.
When sum= 15 and X=5.
A+B= C+D= E+F= G+H= 15-5= 10.
And hence any pair of diametrically opposite letters can represent pairs- (1,9), (2,8), (3,7) and (4,6)
Using the extra information given in the question, we can say that the pair that contains two prime numbers are (A,B)
∵ Only (3,7) forms such a pair, A=3 and B= 7 (A takes the lower number of the two because of the 1st point mentioned in the question).
Also, the only other prime number left is 2, which is taken up by G. If G=2, H=8
We are now left with pairs (E,F) and (C,D) which can take either (1,9) or (4,6).
∵ Neither E nor F is 4, E and F must represent 1 and 9 respectively and therefore C and D takes the values 4 and 6 respectively.
So, A= 3, B=7, C=4, D=6, E=1, F=9, G=2, H=8 and X=5.