2018 AMC 10A Problem 25

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 $\frac{10^n-1}{9}$ 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 $c=\frac{a^2}{9}$ and $b=\frac{2a^2}{9}.$ 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.