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 $C_n - B_n = A_n^2$ ?

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

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

Video Solution by Richard Rusczyk

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

~ dolphin7

2018 AMC 10A (Problems • Answer Key • Resources)
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2018 AMC 12A (Problems • Answer Key • Resources)
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