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

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

Start with hints

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 \( a= 6 \) , \( max (a+b+c) =18 \) .

Connected Program at Cheenta

Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

Similar Problems

2016 AMC 8 Problem 24 Number Theory

This beautiful application is from 2016 AMC 8 Problem 24 based on Number Theory . Sequential hints are given to understand and solve the problem .

2017 AMC 8 Problem 21 Number Theory

This beautiful application from 2017 AMC 8 Problem 21 is based on Number Theory . Sequential hints are provided to study and solve the problem .

SMO(senior)-2014 Problem 2 Number Theory

This beautiful application from SMO(senior)-2014 is based on the concepts of Number Theory . Sequential hints are provided to understand and solve the problem .

SMO (senior) -2014/problem-4 Number Theory

This beautiful application from SMO(senior)-2014/Problem 4 is based Number Theory . Sequential hints are provided to understand and solve the problem .

The best exponent for an inequality

Understand the problemLet be positive real numbers such that .Find with proof that is the minimal value for which the following inequality holds:Albania IMO TST 2013 Inequalities Medium Inequalities by BJ Venkatachala Start with hintsDo you really need a hint?...

2018 AMC 10A Problem 25 Number Theory

This beautiful application from AMC 2018 is based on Number Theory. Sequential hints are given to understand and solve the problem .

A functional inequation

Understand the problemFind all functions such thatholds for all . Benelux MO 2013 Functional Equations Easy Functional Equations by BJ Venkatachala Start with hintsDo you really need a hint? Try it first!Note that the RHS does not contain $latex y$. Thus it should...

Mathematical Circles Inequality Problem

A beautiful inequality problem from Mathematical Circles Russian Experience . we provide sequential hints . key idea is to use arithmetic mean , geometric mean inequality.

RMO 2019

Regional Math Olympiad (RMO) 2019 is the second level Math Olympiad Program in India involving Number Theory, Geometry, Algebra and Combinatorics.

AMC 2019 12A Problem 15 Diophantine Equation

Beautiful application of Logarithm and Diophantine Equation in American Mathematics Competition (2019) 12A