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 $C_n - B_n = A_n^2?$

$12$

$14$

$16$

$18$

$20$

Answer: D

Concepts:digits geometric sequence algebraic manipulation

Difficulty rating: 2390

Solution:

The repeated-digit numbers can be written as $\begin{aligned} A_n&=a\dfrac{10^n-1}{9},\\ B_n&=b\dfrac{10^n-1}{9},\\ C_n&=c\dfrac{10^{2n}-1}{9}. \end{aligned}$

Substituting these expressions into $C_n-B_n=A_n^2$ and dividing by $(10^n-1)/9$ gives $c(10^n+1)-b=a^2\dfrac{10^n-1}{9}.$

Rearranging yields $(9c-a^2)10^n=9b-9c-a^2.$ If this holds for two different values of $n,$ subtracting the two equations shows that $9c-a^2=0.$ The displayed equation then also forces $9b-9c-a^2=0.$

Hence $c=a^2/9$ and $b=2a^2/9.$ Because $a,b,c$ are nonzero digits, the candidates are $(a,b,c)=(3,2,1),$ $(6,8,4),$ and $(9,18,9).$ The last triple is invalid because $18$ is not a digit. The greatest valid sum is $6+8+4=18.$ Thus, D is the correct answer.

2018 AMC 10A Problem 25

Answer: D

Difficulty rating: 2390

Solution:

Problem 25 in Other Years