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# Understand the 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 each of whose digits is equal to $b$, and let $C_n$ be the $2n$-digit (not $n$-digit) integer 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 $C_n - B_n = A_n^2$? $\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}$

# 2018 AMC 10A/Problem 25

Number Theory
Medium
##### Suggested Book
Mathematical Circles (Russian Experience)

Do you really need a hint? Try it first!

Use $$A_{n}= a(1+10+10^2….+10^{n-1}) = a \times \frac {10^{n -1}}{9}$$ , similarly $$B_{n} = b \times \frac {10^{n -1}}{9}$$ and $$C_{n} = c \times \frac {10^{2n -1}}{9}$$ . Then proceed .
Using $$n > 0 \Rightarrow 10^n > 1$$ as $$n \in Z^+$$ , $$\$$ arrive at $$c \times (10^n + 1) – b = a^2 \times \frac {10^n -1}{9}$$ .
Observe that this expression $$c \times (10^n + 1) – b = a^2 \times \frac {10^n -1}{9}$$ is a linear combination of $$10^n$$ . As per the question , $$c \times (10^n + 1) – b = a^2 \times \frac {10^n -1}{9}$$ is true for at least two distinct values of $$n$$ . $$\Rightarrow c \times (10^n + 1) – b = a^2 \times \frac {10^n -1}{9}$$ is an identity . i.e. $$c \times (10^n + 1) – b = a^2 \times \frac {10^n -1}{9}$$ is true $$\forall n \in Z^+$$ . Then compare the coeffiecients .

Comparing coefficients we have $$c= \frac {a^2} {9}$$ and $$c-b= – \frac {a^2} {9} \Rightarrow b= \frac {2a^2} {9}$$ . So , $$a+b+c = a + \frac {a^2}{3}$$ . As $$a, b \ and \ c$$ are nonzero digits $$\Rightarrow a + \frac {a^2}{3}$$ is an integer $$\Rightarrow 3|a^2$$ $$\Rightarrow 3|a$$ . Now to maximize $$a+b+c = a + \frac {a^2}{3}$$ , put highest value of $$a$$ i.e. $$9$$ but it will give $$b= 18$$ which is not possible . [ as $$a, b \ and \ c$$ are nonzero digits ] . Then putting $$a= 6$$ , $$max (a+b+c) =18$$ .

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