2018 AMC 10A Problem 25 Number Theory

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 $C_n - B_n = A_n^2$? $\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}$
Source of the problem

2018 AMC 10A/Problem 25

Number Theory
Difficulty Level
Suggested Book
Mathematical Circles (Russian Experience)

Start with hints

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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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