Question

I recently (yesterday) took the AMC 10A, can anyone help me with this problem?

Answer 1

First, we claim that $\frac{10^n-1}{9}$ is the $n-$digit number with all digits equal to one.

Note that $10^n$ is a one followed by $n$ zeroes, so subtracting one gives $n$ nines.  Divide that number by nine and you get $n$ ones, completing the proof.

Therefore, we have that $A_n = a \frac{10^n - 1}{9}$, $B_n = b \frac{10^n-1}{9}$, and $C_n = c \frac{10^{2n} - 1}{9}$.

Let $x = 10^n$.  Then, we have:

$c \frac{x^2-1}{9} - b \frac{x-1}{9} = (a \frac{10^n - 1}{9})^2 = a^2 \frac{x^2 - 2x + 1}{81}$.

Multiplying by $81$ gives:

$9c(x^2-1) - 9b(x-1) = a^2 (x^2-2x+1)$

Now, note that $x=1$ is not a valid input, since $10^n = 1$ requires $n=0$, so we safely divide by $x-1$ to get:

$9c(x+1) - 9b = a^2(x-1)$

$9cx + 9c - 9b = a^2x - a^2$

$x(9c-a^2) = 9b-9c-a^2$

Because this is now a linear equation in $x$, it has either zero, one, or infinitely many solutions.  Obviously, we need the latter to occur, which happens when $9c=a^2$ and $9b-9c-a^2 = 0$, since the coefficient of $x$ must cancel to zero and thus the RHS must equal zero as well.

Since $9c = a^2$, we must have $a = 3 \sqrt{c}$.  Since $a$ must be a multiple of three, we plug in values.  If $a = 9$, we get $c = 9$ and thus $9b - 162 = 0$, which is impossible.  So $a = 9$ doesn't work.

With $a = 6$, $c = 4$, and thus $9b-72=0$, so $9b=72$, so $b=8$.  This gives $6+8+4=18$ as one possibility.

With $a = 3$, $c = 1$, and thus $9b - 18 = 0$, so $b = 2$.  This obviously is worse.

We've gone through all the cases and the two possibilities are $2$ and $18$, so our answer is $\boxed{18}$.