Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AMC 10A Problems/Problem 25"

(Created page with "For a positive integer <math>n</math> and nonzero digits <math>a</math>, <math>b</math>, and <math>c</math>, let <math>A_n</math> be the <math>n</math>-digit integer each of w...")
 
Line 1: Line 1:
 +
== Problem ==
 +
 
For a positive integer <math>n</math> and nonzero digits <math>a</math>, <math>b</math>, and <math>c</math>, let <math>A_n</math> be the <math>n</math>-digit integer each of whose digits is equal to <math>a</math>; let <math>B_n</math> be the <math>n</math>-digit integer each of whose digits is equal to <math>b</math>, and let <math>C_n</math> be the <math>2n</math>-digit (not <math>n</math>-digit) integer each of whose digits is equal to <math>c</math>. What is the greatest possible value of <math>a + b + c</math> for which there are at least two values of <math>n</math> such that <math>C_n - B_n = A_n^2</math>?
 
For a positive integer <math>n</math> and nonzero digits <math>a</math>, <math>b</math>, and <math>c</math>, let <math>A_n</math> be the <math>n</math>-digit integer each of whose digits is equal to <math>a</math>; let <math>B_n</math> be the <math>n</math>-digit integer each of whose digits is equal to <math>b</math>, and let <math>C_n</math> be the <math>2n</math>-digit (not <math>n</math>-digit) integer each of whose digits is equal to <math>c</math>. What is the greatest possible value of <math>a + b + c</math> for which there are at least two values of <math>n</math> such that <math>C_n - B_n = A_n^2</math>?
  
 
<math>\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}</math>
 
<math>\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}</math>
 +
 +
== Solution ==
 +
 +
 +
Observe <math>A_n = a(1 + 10 + \dots + 10^{n - 1}) = a \cdot \tfrac{10^n - 1}{9}</math>; similarly <math>B_n = b \cdot \tfrac{10^n - 1}{9}</math> and <math>C_n = c \cdot \tfrac{10^{2n} - 1}{9}</math>. The relation <math>C_n - B_n = A_n^2</math> rewrites as
 +
<cmath>c \cdot \frac{10^{2n} - 1}{9} - b \cdot \frac{10^n - 1}{9} = a^2 \cdot \left(\frac{10^n - 1}{9}\right)^2.</cmath>Since <math>n > 0</math>, <math>10^n > 1</math> and we may cancel out a factor of <math>\tfrac{10^n - 1}{9}</math> to obtain
 +
<cmath>c \cdot (10^n + 1) - b = a^2 \cdot \frac{10^n - 1}{9}.</cmath>This is a linear equation in <math>10^n</math>. Thus, if two distinct values of <math>n</math> satisfy it, then all values of <math>n</math> will. Matching coefficients, we need
 +
<cmath>c = \frac{a^2}{9} \quad \text{and} \quad c - b = -\frac{a^2}{9} \implies b = \frac{2a^2}{9}.</cmath>To maximize <math>a + b + c = a + \tfrac{a^2}{3}</math>, we need to maximize <math>a</math>. Since <math>b</math> and <math>c</math> must be integers, <math>a</math> must be a multiple of 3. If <math>a = 9</math> then <math>b</math> exceeds 9. However, if <math>a = 6</math> then <math>b = 8</math> and <math>c = 4</math> for an answer of <math>\boxed{\textbf{(D)} \text{ 18}}</math>. (CantonMathGuy)
 +
 +
==See Also==
 +
 +
{{AMC10 box|year=2018|ab=A|num-b=24|after=Last Problem}}
 +
{{AMC12 box|year=2018|ab=A|num-b=24|after=Last Problem}}
 +
{{MAA Notice}}

Revision as of 15:47, 8 February 2018

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 $C_n - B_n = A_n^2$?

$\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}$

Solution

Observe $A_n = a(1 + 10 + \dots + 10^{n - 1}) = a \cdot \tfrac{10^n - 1}{9}$; similarly $B_n = b \cdot \tfrac{10^n - 1}{9}$ and $C_n = c \cdot \tfrac{10^{2n} - 1}{9}$. The relation $C_n - B_n = A_n^2$ rewrites as \[c \cdot \frac{10^{2n} - 1}{9} - b \cdot \frac{10^n - 1}{9} = a^2 \cdot \left(\frac{10^n - 1}{9}\right)^2.\]Since $n > 0$, $10^n > 1$ and we may cancel out a factor of $\tfrac{10^n - 1}{9}$ to obtain \[c \cdot (10^n + 1) - b = a^2 \cdot \frac{10^n - 1}{9}.\]This is a linear equation in $10^n$. Thus, if two distinct values of $n$ satisfy it, then all values of $n$ will. Matching coefficients, we need \[c = \frac{a^2}{9} \quad \text{and} \quad c - b = -\frac{a^2}{9} \implies b = \frac{2a^2}{9}.\]To maximize $a + b + c = a + \tfrac{a^2}{3}$, we need to maximize $a$. Since $b$ and $c$ must be integers, $a$ must be a multiple of 3. If $a = 9$ then $b$ exceeds 9. However, if $a = 6$ then $b = 8$ and $c = 4$ for an answer of $\boxed{\textbf{(D)} \text{ 18}}$. (CantonMathGuy)

See Also

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_25&oldid=90471"