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

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== Solution 3 (Cheating) ==
 
== Solution 3 (Cheating) ==
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Notice that <math>(0.\overline{3})^2 = 0.\overline{1}</math> and <math>(0.\overline{6})^2 = 0.\overline{4}</math>. Setting <math>a = 3</math> and <math>c = 1</math>, we see <math>b = 2</math> works for all possible values of <math>n</math>. Similarly, if <math>a = 6</math> and <math>c = 4</math>, then <math>b = 8</math> works for all possible values of <math>n</math>. The second solution yields a greater sum of <math>\boxed{\textbf{(D)} \text{ 18}}</math>.
 
Notice that <math>(0.\overline{3})^2 = 0.\overline{1}</math> and <math>(0.\overline{6})^2 = 0.\overline{4}</math>. Setting <math>a = 3</math> and <math>c = 1</math>, we see <math>b = 2</math> works for all possible values of <math>n</math>. Similarly, if <math>a = 6</math> and <math>c = 4</math>, then <math>b = 8</math> works for all possible values of <math>n</math>. The second solution yields a greater sum of <math>\boxed{\textbf{(D)} \text{ 18}}</math>.
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== Solution 4 ==
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The given equation can be written as:
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<cmath>c \cdot ( \overbrace{1111 \ldots 1111}^\text{2n}) - b \cdot ( \overbrace{11 \ldots 11}^\text{n} ) = a^2 \cdot ( \overbrace{11 \ldots 11}^\text{n} )^2</cmath>
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Divide by <math>\overbrace{11 \ldots 11}^\text{n}</math> on both sides:
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<cmath>c \cdot ( \overbrace{1000 \ldots 0001}^\text{n+1}) - b = a^2 \cdot ( \overbrace{11 \ldots 11}^\text{n} )</cmath>
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Next, split the first term to make it easier to deal with.
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<cmath>2c + c \cdot (\overbrace{99 \ldots 99}^\text{n}) - b = a^2 \cdot ( \overbrace{11 \ldots 11}^\text{n} )</cmath>
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<cmath>2c - b = (a^2 - 9c) \cdot (\overbrace{11 \ldots 11}^\text{n})</cmath>
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Because <math>2c - b</math> and <math>a^2 - 9c</math> are constants and because there must be at least two distinct values of <math>n</math> that satisfy, <math>2c - b = a^2 - 9c = 0</math>. Thus, we have:
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<cmath>2c=b</cmath>
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<cmath>a^2=9c</cmath>
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Knowing that <math>a</math>, <math>b</math>, and <math>c</math> are single digit positive integers and that <math>9c</math> must be a perfect square, the possible values of <math>(a,b,c)</math> that satisfy both equations are <math>(3,2,1)</math> and <math>(6,8,4).</math> Finally, <math>6 + 8 + 4 = \boxed{\textbf{(D)} \text{18}}</math>.
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~LegionOfAvatars
  
 
== Video Solution by Richard Rusczyk ==
 
== Video Solution by Richard Rusczyk ==

Revision as of 22:55, 11 January 2021

The following problem is from both the 2018 AMC 12A #25 and 2018 AMC 10A #25, so both problems redirect to this page.

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 1

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. Now we plug in $n=0$ and $n=1$ (or some other number), we get $2c - b = 0$ and $11c - b= a^2$ . Solving the equations for $c$ and $b$, we get \[c = \frac{a^2}{9} \quad \text{and} \quad c - b = -\frac{a^2}{9} \implies b = \frac{2a^2}{9}.\]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 $\boxed{\textbf{(D)} \text{ 18}}$.

Solution 2

Immediately start trying $n = 1$ and $n = 2$. These give the system of equations $11c - b = a^2$ and $1111c - 11b = (11a)^2$ (which simplifies to $101c - b = 11a^2$). These imply that $a^2 = 9c$, so the possible $(a, c)$ pairs are $(9, 9)$, $(6, 4)$, and $(3, 1)$. The first puts $b$ out of range but the second makes $b = 8$. We now know the answer is at least $6 + 8 + 4 = 18$.

We now only need to know whether $a + b + c = 20$ might work for any larger $n$. We will always get equations like $100001c - b = 11111a^2$ where the $c$ coefficient is very close to being nine times the $a$ coefficient. Since the $b$ term will be quite insignificant, we know that once again $a^2$ must equal $9c$, and thus $a = 9, c = 9$ is our only hope to reach $20$. Substituting and dividing through by $9$, we will have something like $100001 - \frac{b}{9} = 99999$. No matter what $n$ really was, $b$ is out of range (and certainly isn't $2$ as we would have needed).

The answer then is $\boxed{\textbf{(D)} \text{ 18}}$.

Solution 3 (Cheating)

Notice that $(0.\overline{3})^2 = 0.\overline{1}$ and $(0.\overline{6})^2 = 0.\overline{4}$. Setting $a = 3$ and $c = 1$, we see $b = 2$ works for all possible values of $n$. Similarly, if $a = 6$ and $c = 4$, then $b = 8$ works for all possible values of $n$. The second solution yields a greater sum of $\boxed{\textbf{(D)} \text{ 18}}$.

Solution 4

The given equation can be written as: \[c \cdot ( \overbrace{1111 \ldots 1111}^\text{2n}) - b \cdot ( \overbrace{11 \ldots 11}^\text{n} ) = a^2 \cdot ( \overbrace{11 \ldots 11}^\text{n} )^2\] Divide by $\overbrace{11 \ldots 11}^\text{n}$ on both sides: \[c \cdot ( \overbrace{1000 \ldots 0001}^\text{n+1}) - b = a^2 \cdot ( \overbrace{11 \ldots 11}^\text{n} )\] Next, split the first term to make it easier to deal with. \[2c + c \cdot (\overbrace{99 \ldots 99}^\text{n}) - b = a^2 \cdot ( \overbrace{11 \ldots 11}^\text{n} )\] \[2c - b = (a^2 - 9c) \cdot (\overbrace{11 \ldots 11}^\text{n})\] Because $2c - b$ and $a^2 - 9c$ are constants and because there must be at least two distinct values of $n$ that satisfy, $2c - b = a^2 - 9c = 0$. Thus, we have: \[2c=b\] \[a^2=9c\] Knowing that $a$, $b$, and $c$ are single digit positive integers and that $9c$ must be a perfect square, the possible values of $(a,b,c)$ that satisfy both equations are $(3,2,1)$ and $(6,8,4).$ Finally, $6 + 8 + 4 = \boxed{\textbf{(D)} \text{18}}$.

~LegionOfAvatars

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2018amc10a/470

~ dolphin7

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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