2018 AMC 10A Problems/Problem 25

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

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

2018 AMC 10A (Problems • Answer Key • Resources)
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2018 AMC 10A Problems/Problem 25

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