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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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