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Difference between revisions of "2018 AMC 10A Problems/Problem 25"
(Made the argument in Sol 1 more rigorous. Previously Sol 1 says that plugging in n = 0, but n = 0 is not a valid value. Also made some cosmetic changes to Sol 1 and made more intermediate steps.)
Line 10: Line 10:
 
By geometric series, we have
 
By geometric series, we have
 
<cmath>\begin{alignat*}{8}
 
<cmath>\begin{alignat*}{8}
−
A_n&=a\cdot\sum_{k=0}^{n-1}10^k&&=a\cdot\frac{10^n-1}{9}, \\
+
A_n&=a\sum_{k=0}^{n-1}10^k&&=a\cdot\frac{10^n-1}{9}, \\
−
B_n&=b\cdot\sum_{k=0}^{n-1}10^k&&=b\cdot\frac{10^n-1}{9}, \\
+
B_n&=b\sum_{k=0}^{n-1}10^k&&=b\cdot\frac{10^n-1}{9}, \\
−
C_n&=c\cdot\sum_{k=0}^{2n-1}10^k&&=c\cdot\frac{10^{2n}-1}{9}.
+
C_n&=c\sum_{k=0}^{2n-1}10^k&&=c\cdot\frac{10^{2n}-1}{9}.
 
\end{alignat*}</cmath>
 
\end{alignat*}</cmath>
 
By substitution, we rewrite the given equation <math>C_n - B_n = A_n^2</math> as
 
By substitution, we rewrite the given equation <math>C_n - B_n = A_n^2</math> as
Line 18: Line 18:
 
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:
 
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*}
 
<cmath>\begin{align*}
−
c\cdot\left(10^n+1\right) - b = a^2\cdot\frac{10^n-1}{9}
+
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>
 
\end{align*}</cmath>
 +
Note that <math>(\bigstar)</math> is a linear equation with <math>10^n.</math> Since it has at least two solutions of <math>n,</math> it has at least two solutions of <math>10^n.</math> We conclude that <math>(\bigstar)</math> must be an identity. So, we have 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>
  
−
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. Now we plug in <math>n=0</math> and <math>n=1</math> (or some other number), we get <math>2c - b = 0</math> and <math>11c - b= a^2</math> . Solving the equations for <math>c</math> and <math>b</math>, we get <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 <math>3</math>. If <math>a = 9</math> then <math>b</math> exceeds <math>9</math>. 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>.
+
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> If <math>a=9,</math> then <math>(b,c)=(18,9),</math> which violates the restrictions. However, if <math>a=6,</math> then <math>(b,c)=(8,4).</math> Therefore, the greatest possible value of <math>a + b + c</math> is <math>6+8+4=\boxed{\textbf{(D) } 18}.</math>
  
 
== Solution 2==
 
== Solution 2==

Revision as of 03:28, 28 August 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 $C_n - B_n = A_n^2$?

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

Solution 1

By geometric series, we have \begin{alignat*}{8} A_n&=a\sum_{k=0}^{n-1}10^k&&=a\cdot\frac{10^n-1}{9}, \\ B_n&=b\sum_{k=0}^{n-1}10^k&&=b\cdot\frac{10^n-1}{9}, \\ C_n&=c\sum_{k=0}^{2n-1}10^k&&=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*} Note that $(\bigstar)$ is a linear equation with $10^n.$ Since it has at least two solutions of $n,$ it has at least two solutions of $10^n.$ We conclude that $(\bigstar)$ must be an identity. So, we have 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 $3.$ If $a=9,$ then $(b,c)=(18,9),$ which violates the restrictions. However, if $a=6,$ then $(b,c)=(8,4).$ Therefore, the greatest possible value of $a + b + c$ is $6+8+4=\boxed{\textbf{(D) } 18}.$

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

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 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

Solution 4 (If you are running out of time)

Considering this is an AMC 10 and calculators are not allowed, the number of digits in $A_n$ and $B_n$, $n$, probably is not greater than 3. Checking cases with $n=2$ digits first, we find that if $a=9$ then there are no solutions with a two digit $B_n$. Thus we check the case with $a=8$ and find that $88^2=7744$. Thus if $c=7$ and $b=3$ then $C_n - B_n = A_n^2$ for $n=2$. If $n=1$, then $C_n - B_n = A_n^2$ if $c=7$, $a=8$, and $b=3$. If this is the case then $a+b+c=\boxed{\textbf{(D)} \text{18}}$. ~Dhillonr25

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

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