$A=\underbrace{111\dots111}_{\text{Number of 1s = 81}}$

What is the remainder when $A$ is divided by 81?

9
36
0
54
27
18
72
45

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Let $a = \underbrace{111 111 111}_{\text{\# of 1s = 9}}$ . Then $A=a(1+10^9+10^{18}+ 10^{27}+…+10^{72})$ . Note that the sum of digit of $a$ is 9; $a$ is divisible by 9. And that the sum of digit of $1+10^9+10^{18}+ 10^{27}+…+10^{72}$ is also 9 and it is divisible by 9. Therefore $A$ is divisible by 81 and the remainder is $\boxed{0}$ .