Question

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 $C_n - B_n = A_n^2?$

$12$

$14$

$16$

$18$

$20$

Solution:

Using $A_n = a \cdot \tfrac{10^n - 1}{9},$ $B_n = b \cdot \tfrac{10^n - 1}{9},$ and $C_n = c \cdot \tfrac{10^{2n} - 1}{9},$ the equation $C_n - B_n = A_n^2$ becomes, after dividing by $10^n - 1$ and clearing fractions, $(9c - a^2) \cdot 10^n = 9b - 9c - a^2.$ For this to hold at two different $n,$ the coefficient of $10^n$ must be zero, so $9c = a^2$ and hence $9b - 9c - a^2 = 0.$

Then $c = \tfrac{a^2}{9}$ and $b = 2c.$ So $a \in \{3, 6, 9\}$ with $c \in \{1, 4, 9\}$ and $b \in \{2, 8, 18\};$ the case $b = 18$ is not a digit. The valid triples are $(a, b, c) = (3, 2, 1)$ and $(6, 8, 4),$ and indeed $4444 - 88 = 4356 = 66^2.$ The greater digit sum is $6 + 8 + 4 = 18.$

Thus, the correct answer is D.

2018 AMC 12A Problem 25

Solution:

Problem 25 in Other Years