First, we claim that [tex]\frac{10^n-1}{9}[/tex] is the [tex]n-[/tex]digit number with all digits equal to one.
Note that [tex]10^n[/tex] is a one followed by [tex]n[/tex] zeroes, so subtracting one gives [tex]n[/tex] nines. Divide that number by nine and you get [tex]n[/tex] ones, completing the proof.
Therefore, we have that [tex]A_n = a \frac{10^n - 1}{9}[/tex], [tex]B_n = b \frac{10^n-1}{9}[/tex], and [tex]C_n = c \frac{10^{2n} - 1}{9}[/tex].
Let [tex]x = 10^n[/tex]. Then, we have:
[tex]c \frac{x^2-1}{9} - b \frac{x-1}{9} = (a \frac{10^n - 1}{9})^2 = a^2 \frac{x^2 - 2x + 1}{81}[/tex].
Multiplying by [tex]81[/tex] gives:
[tex] 9c(x^2-1) - 9b(x-1) = a^2 (x^2-2x+1)[/tex]
Now, note that [tex]x=1[/tex] is not a valid input, since [tex]10^n = 1[/tex] requires [tex]n=0[/tex], so we safely divide by [tex]x-1[/tex] to get:
[tex]9c(x+1) - 9b = a^2(x-1)[/tex]
[tex]9cx + 9c - 9b = a^2x - a^2[/tex]
[tex]x(9c-a^2) = 9b-9c-a^2[/tex]
Because this is now a linear equation in [tex]x[/tex], it has either zero, one, or infinitely many solutions. Obviously, we need the latter to occur, which happens when [tex]9c=a^2[/tex] and [tex]9b-9c-a^2 = 0[/tex], since the coefficient of [tex]x[/tex] must cancel to zero and thus the RHS must equal zero as well.
Since [tex]9c = a^2[/tex], we must have [tex]a = 3 \sqrt{c}[/tex]. Since [tex]a[/tex] must be a multiple of three, we plug in values. If [tex]a = 9[/tex], we get [tex]c = 9[/tex] and thus [tex]9b - 162 = 0[/tex], which is impossible. So [tex]a = 9[/tex] doesn't work.
With [tex]a = 6[/tex], [tex]c = 4[/tex], and thus [tex]9b-72=0[/tex], so [tex]9b=72[/tex], so [tex]b=8[/tex]. This gives [tex]6+8+4=18[/tex] as one possibility.
With [tex]a = 3[/tex], [tex]c = 1[/tex], and thus [tex]9b - 18 = 0[/tex], so [tex]b = 2[/tex]. This obviously is worse.
We've gone through all the cases and the two possibilities are [tex]2[/tex] and [tex]18[/tex], so our answer is [tex]\boxed{18}[/tex].
