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

Revision as of 15:21, 8 January 2022

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

Since the number $11111...111$ with $n$ digits is $\frac{10^n-1}9$ , the equation $C_n - B_n = A_n^2$ can be rewritten as: $\left(a\cdot\frac{10^n-1}9\right)^2=c\cdot\frac{10^{2n}-1}9-b\cdot\frac{10^n-1}9$ or $a^2\cdot(10^{2n}-2\cdot10^n+1)=9c(10^{2n}-1)-9b(10^n-1)$ which simplifies to $a^2(10^2n)-2a^2(10^n)+a^2=9c(10^2n)-9b(10^n)+(9b-9c)$ Setting the coefficients equal (which they are unless $n = 1$ , which is only one solution, less than the intended two), we get $a^2=9c$ $2a^2=9b$ $a^2=9b-9c$ The last equation can be ignored, because it can be found from the other two equations. We get $b=2c$ and $c$ is a perfect square less than 5 (because $b$ is a digit and $c=\frac{a^2}9$ , which is a square), so $c$ is $1$ or $4$ . If $c=1$ , we find $(a,b,c)=(3,2,1)$ . If $c=4$ , then $(a,b,c)=(6,8,4)$ . Thus the maximum value of $a+b+c$ is $\boxed{\textbf{(D) } 18}$ .

Video Solution by Richard Rusczyk

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

~ dolphin7

@@ Line 64: / Line 64: @@
 ~MRENTHUSIASM (Reformatting)
+== Solution 4 ==
+Since the number <math>11111...111</math> with <math>n</math> digits is <math>\frac{10^n-1}9</math>, the equation <math>C_n - B_n = A_n^2</math> can be rewritten as:
+<cmath>\left(a\cdot\frac{10^n-1}9\right)^2=c\cdot\frac{10^{2n}-1}9-b\cdot\frac{10^n-1}9</cmath>
+or
+<cmath>a^2\cdot(10^{2n}-2\cdot10^n+1)=9c(10^{2n}-1)-9b(10^n-1)</cmath>
+which simplifies to
+<cmath>a^2(10^2n)-2a^2(10^n)+a^2=9c(10^2n)-9b(10^n)+(9b-9c)</cmath>
+Setting the coefficients equal (which they are unless <math>n = 1</math>, which is only one solution, less than the intended two), we get
+<cmath>a^2=9c</cmath>
+<cmath>2a^2=9b</cmath>
+<cmath>a^2=9b-9c</cmath>
+The last equation can be ignored, because it can be found from the other two equations. We get <math>b=2c</math> and <math>c</math> is a perfect square less than 5 (because <math>b</math> is a digit and <math>c=\frac{a^2}9</math>, which is a square), so <math>c</math> is <math>1</math> or <math>4</math>. If <math>c=1</math>, we find <math>(a,b,c)=(3,2,1)</math>. If <math>c=4</math>, then <math>(a,b,c)=(6,8,4)</math>. Thus the maximum value of <math>a+b+c</math> is <math>\boxed{\textbf{(D) } 18}</math>.
 == Video Solution by Richard Rusczyk ==

2018 AMC 10A (Problems • Answer Key • Resources)
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