Construction 1

Assuming E=5 ... since sum is 45, each row/column adds up to 15

Now, A + D + G = 15 ..... => D = 15 - (A+G) Also, C + F + I = 15 ...... => F = 15 - (C+I)

Now D + E + F = 15 since E = 5, D + F =10 that means 15 + 15 - (A+G+C+I) = 10 which implies A+G+C+I = 30 - 10 = 20

Hope this helps!! .....

It is required to find the value of $A+C+G+I$ .

The magic sum is $M_3=\frac{3}{2}(3^2+1)=15$ . Now, consider the following equations: $A+B+C=15$ $A+D+G=15$ $C+F+I=15$ $G+H+I=15$ $A+E+I=15$ $C+E+G=15$ Adding all these equations, we have $3A+B+3C+D+2E+F+3G+H+3I=15(6)=90$ Since $A+B+C+D+E+F+G+H+I=\text{sum of all numbers from 1 to 9}=\frac{9}{2}(1+9)=45$ rewriting the previous result gives $(A+B+C+D+E+F+G+H+I)+(2A+2C+E+2G+2I)=45+2(A+C+G+I)+E=90$ Now, $E=5$ (Why?). Thus, from the last equation, $45+2(A+C+G+I)+5=90$ and therefore $A+C+G+I=\boxed{20}$

Exercise: Prove that the central number $E=5$ .