2018 AMC 10A Problems/Problem 25

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 $6+8+4=\boxed{\textbf{(D) } 18}.$

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

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

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

Page

Toolbox

Search