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

We are looking for

$ x = 10^{80} + 10^{79} + \dots + 10^2 + 10 + 1 \mod{9^3}$

$ 10^k = (9 + 1)^k = \sum\limits_{i=0}^{k}{k \choose i}9^i $ (Binomial theorem)

Considering the second equality $\mod{9^3} $, we only need first three summands of the sum. So $ 10^k \equiv 1 + k\cdot 9 + \frac{k\cdot(k-1)}{2}\cdot 9^2\mod{9^3}$. So what we're looking for is

$$ x\mod{9^3} \equiv\sum\limits_{k=0}^{80}1 + 9k + 81\frac{k\cdot(k-1)}{2} \mod{9^3} = \sum\limits_{k = 0}^{80}1 - \frac{63}{2}k + \frac{81}{2}k^2$$

Using formulas $ \sum_{k=0}^n k = \frac{n(n+1)}{2}, \sum\limits_{k=0}^n k^2 = \frac{n(n+1)(2n+1)}{6} $ we get that

$$ x \mod{9^3} = 81 + \frac{-63}{2}\cdot80\cdot81\cdot\frac{1}{2} + \frac{81}{2}\cdot80\cdot81\cdot161\cdot\frac{1}{6} \mod{9^3}$$

$$ x \mod{9^3} = 81 + 81((-63)\cdot20 + 20\cdot27\cdot161) \mod{9^3}$$

Notice the term in brackets is divisible by $ 9 $, hence the answer is $ 81 $.

I'm not sure that's the best way to approach this. Please correct me if any of the calculations were wrong

According to Wolfy, or by repeated use of $x^3-1 =(x-1)(x^2+x+1) $, $\frac{x^{81}-1}{x-1} = (x^2+x+1) (x^6+x^3+1) (x^{18}+x^9+1) (x^{54}+x^{27}+1) $.

Since $x^{3n} \equiv 1 \bmod (x^3-1)$, $x^{6n}+x^{3n}+1 \equiv 3 \bmod (x^3-1) $.

Therefore the right 3 factors are all $\equiv 3 \bmod (x^3-1) $.

Therefore the whole product $\equiv 27(x^2+x+1) \bmod (x^3-1) $.

Since

$\begin{array}\\ x^2+x+1 &=((x-1)+1)^2+(x-1)+2\\ &=(x-1)^2+2(x-1)+1+(x-1)+2\\ &=(x-1)^2+3(x-1)+3\\ \end{array} $

we have

$\begin{array}\\ 27(x^2+x+1) &=27((x-1)^2+3(x-1)+3)\\ &=27(x-1)^2+81(x-1)+81\\ &\equiv 81 \bmod 729 \qquad\text{setting }x=10 \text{ since }27\cdot 9^2, 81\cdot 9 \equiv 0 \bmod 729\\ \end{array} $

Therefore the answer is $81$.