Because it's easier, I'm just using the English A->F:

The first equation is A x A = A. There are only two integers that satisfy that criteria, and they are 0 and 1. If A was 0, then the second equation would be 0 + 0 = B, and B would be 0 as well. Since A cannot equal B, A cannot be 0, and A must be 1.

then

A x A = A (1 x 1 = 1) A + A = B (1 + 1 = 2; B=2) B + A = C (1+2=3; c=3) C x B = D (2 x 3 = 6; D=6) D x C = E (6 x 3 = 18; E=18) E - B = ★ (18-2 = 16; ★ = 16)

The final answer is:

16

Because it's easier, I'm just using the English A->F:

The first equation is A x A = A. There are only two integers that satisfy that criteria, and they are 0 and 1. If A was 0, then the second equation would be 0 + 0 = B, and B would be 0 as well. Since A cannot equal B, A cannot be 0, and A must be 1.

then

1

A x A = A (1 x 1 = 1)

2

A + A = B (1 + 1 = 2; B=2)

3

B + A = C (1+2=3; c=3)

4

C x B = D (2 x 3 = 6; D=6)

5

D x C = E (6 x 3 = 18; E=18)

6

E - B = ★ (18-2 = 16; ★ = 16)

The final answer is:

16