Question

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

For a positive integer n and nonzero digits a, b and c, let
`A_n` be the n-digit integer each of whose digits is equal to a; let
`B_n` be the n-digit integer whose digits is equal to b; let
`C_n` be the 2n digit integer (not n digit) each of whose digits is equal to c. What is the greatest possible value of a + b + c for which there are at least two values of n such that
`C_n - B_n = A^2_n`?

*This was #25, and the questions have been posted to the Art of Problem Solving website, so I'm guessing discussion is allowed.

Answer 1

First, we claim that
`(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 (10^n - 1)/(9)`,
`B_n = b (10^n-1)/(9)`, and
`C_n = c (10^(2n) - 1)/(9)`.

Let
`x = 10^n`. Then, we have:


`c (x^2-1)/(9) - b (x-1)/(9) = (a (10^n - 1)/(9))^2 = a^2 (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 √(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}`.