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

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{{duplicate|[[2018 AMC 10A Problems/Problem 25|2018 AMC 10A #25]] and [[2018 AMC 12A Problems/Problem 25|2018 AMC 12A #25]]}}
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== Problem ==
 
== 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) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math>
  
 
== Solution 1==
 
== Solution 1==
 +
By geometric series, we have
 +
<cmath>\begin{alignat*}{8}
 +
A_n&=a\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=a\left(1+10+10^2+\cdots+10^{n-1}\right)&&=a\cdot\frac{10^n-1}{9}, \\
 +
B_n&=b\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=b\left(1+10+10^2+\cdots+10^{n-1}\right)&&=b\cdot\frac{10^n-1}{9}, \\
 +
C_n&=c\bigl(\phantom{ }\underbrace{111\cdots1}_{2n\text{ digits}}\phantom{ }\bigr)&&=c\left(1+10+10^2+\cdots+10^{2n-1}\right)&&=c\cdot\frac{10^{2n}-1}{9}.
 +
\end{alignat*}</cmath>
 +
By substitution, we rewrite the given equation <math>C_n - B_n = A_n^2</math> 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> it follows that <math>10^n > 1.</math> We divide both sides by <math>\frac{10^n-1}{9}</math> and then rearrange:
 +
<cmath>\begin{align*}
 +
c\left(10^n+1\right) - b &= a^2\cdot\frac{10^n-1}{9} \\
 +
9c\left(10^n+1\right) - 9b &= a^2\left(10^n-1\right) \\
 +
\left(9c-a^2\right)10^n &= 9b-9c-a^2. &&(\bigstar)
 +
\end{align*}</cmath>
 +
Let <math>y=10^n.</math> Note that <math>(\bigstar)</math> is a linear equation with <math>y,</math> and <math>y</math> is a one-to-one function of <math>n.</math> Since <math>(\bigstar)</math> has at least two solutions of <math>n,</math> it has at least two solutions of <math>y.</math> We conclude that <math>(\bigstar)</math> must be an identity, so we get the following system of equations:
 +
<cmath>\begin{align*}
 +
9c-a^2&=0, \\
 +
9b-9c-a^2&=0.
 +
\end{align*}</cmath>
 +
The first equation implies that <math>c=\frac{a^2}{9}.</math> Substituting this into the second equation gives <math>b=\frac{2a^2}{9}.</math>
  
 +
To maximize <math>a + b + c = a + \frac{a^2}{3},</math> we need to maximize <math>a.</math> Clearly, <math>a</math> must be divisible by <math>3.</math> The possibilities for <math>(a,b,c)</math> are <math>(9,18,9),(6,8,4),</math> or <math>(3,2,1),</math> but <math>(9,18,9)</math> is invalid. Therefore, the greatest possible value of <math>a + b + c</math> is <math>6+8+4=\boxed{\textbf{(D) } 18}.</math>
  
−
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
+
~CantonMathGuy (Solution)
−
<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)
 
  
−
== Solution 2 (quicker?) ==
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~MRENTHUSIASM (Revision)
 +
 
 +
== Solution 2==
  
 
Immediately start trying <math>n = 1</math> and <math>n = 2</math>. These give the system of equations <math>11c - b = a^2</math> and <math>1111c - 11b = (11a)^2</math> (which simplifies to <math>101c - b = 11a^2</math>). These imply that <math>a^2 = 9c</math>, so the possible <math>(a, c)</math> pairs are <math>(9, 9)</math>, <math>(6, 4)</math>, and <math>(3, 1)</math>. The first puts <math>b</math> out of range but the second makes <math>b = 8</math>. We now know the answer is at least <math>6 + 8 + 4 = 18</math>.
 
Immediately start trying <math>n = 1</math> and <math>n = 2</math>. These give the system of equations <math>11c - b = a^2</math> and <math>1111c - 11b = (11a)^2</math> (which simplifies to <math>101c - b = 11a^2</math>). These imply that <math>a^2 = 9c</math>, so the possible <math>(a, c)</math> pairs are <math>(9, 9)</math>, <math>(6, 4)</math>, and <math>(3, 1)</math>. The first puts <math>b</math> out of range but the second makes <math>b = 8</math>. We now know the answer is at least <math>6 + 8 + 4 = 18</math>.
  
−
We now only need to know whether <math>a + b + c = 20</math> might work for any larger <math>n</math>. We will always get equations like <math>100001c - b = 11111a^2</math> where the <math>c</math> coefficient is very close to being nine times the <math>a</math> coefficient. Since the <math>b</math> term will be quite insignificant, we know that once again <math>a^2</math> must equal <math>9c</math>, and thus <math>a = 9, c = 9</math> is our only hope to reach <math>20</math>. Substituting and dividing through by <math>9</math>, we will have something like <math>100001 - b/9 = 99999</math>. No matter what <math>n</math> really was, <math>b</math> is out of range (and certainly isn't <math>2</math> as we would have needed).
+
We now only need to know whether <math>a + b + c = 20</math> might work for any larger <math>n</math>. We will always get equations like <math>100001c - b = 11111a^2</math> where the <math>c</math> coefficient is very close to being nine times the <math>a</math> coefficient. Since the <math>b</math> term will be quite insignificant, we know that once again <math>a^2</math> must equal <math>9c</math>, and thus <math>a = 9, c = 9</math> is our only hope to reach <math>20</math>. Substituting and dividing through by <math>9</math>, we will have something like <math>100001 - \frac{b}{9} = 99999</math>. No matter what <math>n</math> really was, <math>b</math> is out of range (and certainly isn't <math>2</math> as we would have needed).
  
−
The answer then is <math>\boxed{\textbf{(D)} \text{ 18}}</math>.
+
The answer then is <math>\boxed{\textbf{(D) } 18}</math>.
  
 
== Solution 3 ==
 
== Solution 3 ==
−
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>.
+
 
 +
The given equation can be written as
 +
<cmath>c \cdot (\phantom{ } \overbrace{1111 \ldots 1111}^{2n\text{ digits}}\phantom{ }) - b \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ })^2.</cmath>
 +
Divide by <math>\overbrace{11 \ldots 11}^{n\text{ digits}}</math> on both sides:
 +
<cmath>c \cdot (\phantom{ } \overbrace{1000 \ldots 0001}^{n+1\text{ digits}}\phantom{ }) - b = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }).</cmath>
 +
Next, split the first term to make it easier to deal with:
 +
<cmath>\begin{align*}
 +
2c + c \cdot (\phantom{ }\overbrace{99 \ldots 99}^{n\text{ digits}}\phantom{ }) - b &= a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) \\
 +
2c - b &= (a^2 - 9c) \cdot (\phantom{ }\overbrace{11 \ldots 11}^{n\text{ digits}}\phantom{ }).
 +
\end{align*}</cmath>
 +
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
 +
<cmath>\begin{align*}
 +
2c&=b, \\
 +
a^2&=9c.
 +
\end{align*}</cmath>
 +
Knowing that <math>a,b,</math> and <math>c</math> are single digit positive integers and that <math>9c</math> must be a perfect square, the 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) } 18}.</math>
 +
 
 +
~LegionOfAvatars (Solution)
 +
 
 +
~MRENTHUSIASM (Reformatting)
 +
 
 +
==Solution 4==
 +
By [[PaperMath’s sum]], the answer is <math>6+8+4=\boxed{\textbf{(D) } 18}</math>
 +
 
 +
== Video Solution by Richard Rusczyk ==
 +
 
 +
https://artofproblemsolving.com/videos/amc/2018amc10a/470
 +
 
 +
~ dolphin7
  
 
==See Also==
 
==See Also==
Line 29: Line 79:
 
{{AMC12 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}}
 
{{MAA Notice}}
 +
 +
[[Category:Intermediate Number Theory Problems]]

Revision as of 15:50, 21 January 2024

The following problem is from both the 2018 AMC 10A #25 and 2018 AMC 12A #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 $C_n - B_n = A_n^2$?

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

Solution 1

By geometric series, we have \begin{alignat*}{8} A_n&=a\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=a\left(1+10+10^2+\cdots+10^{n-1}\right)&&=a\cdot\frac{10^n-1}{9}, \\ B_n&=b\bigl(\phantom{ }\underbrace{111\cdots1}_{n\text{ digits}}\phantom{ }\bigr)&&=b\left(1+10+10^2+\cdots+10^{n-1}\right)&&=b\cdot\frac{10^n-1}{9}, \\ C_n&=c\bigl(\phantom{ }\underbrace{111\cdots1}_{2n\text{ digits}}\phantom{ }\bigr)&&=c\left(1+10+10^2+\cdots+10^{2n-1}\right)&&=c\cdot\frac{10^{2n}-1}{9}. \end{alignat*} By substitution, we rewrite the given equation $C_n - B_n = A_n^2$ 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 $n > 0,$ it follows that $10^n > 1.$ We divide both sides by $\frac{10^n-1}{9}$ and then rearrange: \begin{align*} c\left(10^n+1\right) - b &= a^2\cdot\frac{10^n-1}{9} \\ 9c\left(10^n+1\right) - 9b &= a^2\left(10^n-1\right) \\ \left(9c-a^2\right)10^n &= 9b-9c-a^2. &&(\bigstar) \end{align*} Let $y=10^n.$ Note that $(\bigstar)$ is a linear equation with $y,$ and $y$ is a one-to-one function of $n.$ Since $(\bigstar)$ has at least two solutions of $n,$ it has at least two solutions of $y.$ We conclude that $(\bigstar)$ must be an identity, so we get the following system of equations: \begin{align*} 9c-a^2&=0, \\ 9b-9c-a^2&=0. \end{align*} The first equation implies that $c=\frac{a^2}{9}.$ Substituting this into the second equation gives $b=\frac{2a^2}{9}.$

To maximize $a + b + c = a + \frac{a^2}{3},$ we need to maximize $a.$ Clearly, $a$ must be divisible by $3.$ The possibilities for $(a,b,c)$ are $(9,18,9),(6,8,4),$ or $(3,2,1),$ but $(9,18,9)$ is invalid. Therefore, the greatest possible value of $a + b + c$ is $6+8+4=\boxed{\textbf{(D) } 18}.$

~CantonMathGuy (Solution)

~MRENTHUSIASM (Revision)

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 $\boxed{\textbf{(D) } 18}$.

Solution 3

The given equation can be written as \[c \cdot (\phantom{ } \overbrace{1111 \ldots 1111}^{2n\text{ digits}}\phantom{ }) - b \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ })^2.\] Divide by $\overbrace{11 \ldots 11}^{n\text{ digits}}$ on both sides: \[c \cdot (\phantom{ } \overbrace{1000 \ldots 0001}^{n+1\text{ digits}}\phantom{ }) - b = a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }).\] Next, split the first term to make it easier to deal with: \begin{align*} 2c + c \cdot (\phantom{ }\overbrace{99 \ldots 99}^{n\text{ digits}}\phantom{ }) - b &= a^2 \cdot (\phantom{ } \overbrace{11 \ldots 11}^{n\text{ digits}} \phantom{ }) \\ 2c - b &= (a^2 - 9c) \cdot (\phantom{ }\overbrace{11 \ldots 11}^{n\text{ digits}}\phantom{ }). \end{align*} 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 \begin{align*} 2c&=b, \\ a^2&=9c. \end{align*} Knowing that $a,b,$ and $c$ are single digit positive integers and that $9c$ must be a perfect square, the values of $(a,b,c)$ that satisfy both equations are $(3,2,1)$ and $(6,8,4).$ Finally, $6 + 8 + 4 = \boxed{\textbf{(D) } 18}.$

~LegionOfAvatars (Solution)

~MRENTHUSIASM (Reformatting)

Solution 4

By PaperMath’s sum, the answer is $6+8+4=\boxed{\textbf{(D) } 18}$

Video Solution by Richard Rusczyk

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

~ dolphin7

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

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