If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is? 
If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?

$729=9^3$
For any number to be divisible by $9$, the sum of the digits have to be divisible by $9$. The given number is divisible by $9$.
Then I tried dividing the given number by $9$. The quotient was like $123456789012\cdots$
So the quotients sum is $(45\times 8)+1=361$
Is this the way to proceed? Is there a shorter way?

Put $$a=10^9=729\cdot1371742+82\equiv 1+9^2\ (mod\space 9^3)$$
$$N=111111111=729\cdot 152415+576\equiv 576 \ (mod\space 9^3)$$
Let $M$ be the number so $$M=N(a^8+a^7+a^6+a^5+a^4+a^3+a^2+a+1)$$
Furthermore $$a\equiv 1+9^2\ (mod\space 9^3)$$ $$a^2\equiv 1+2\cdot9^2\ (mod\space 9^3)$$ $$a^3\equiv 1+3\cdot9^2\ (mod\space 9^3)$$ $$......$$ $$a^8\equiv 1+8\cdot9^2\ (mod\space 9^3)$$
It follows $$M\equiv N(9+9^2\cdot 9\cdot 4)\equiv 576\cdot 9\equiv \color{red} {81}\ (mod\space 9^3)$$ because $576\cdot 9=5184=7\cdot 729+81$
