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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.

User Dprado
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1 Answer

2 votes
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}.


User Dishan TD
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8.5k points