2000 9s

I observed a pattern. Let $S(n)$ denotes the sum of the digit of $n$ .

$\begin{aligned} & S(9^3) = S(729) & = 18 \\ & S(99^3) = S(970,299) & = 36 \\ & S(999^3) = S(997,002,999) & = 54 \end{aligned}$

We can conclude that $S\left( \underbrace{999...999^3}_{\small\text{number of 9s} = n }\right) = 18n$ .

For $n = 2000$ , the answer is $18 \times 2000 = 36000$ .

---

Proof

This is the generalozation of Chew-Seong Cheong's and Áron Bán-Szabó's solution.

Consider the expression below.

$\begin{aligned} \underbrace{99...99^3}_{\text{number of 9s} = n} & = (10^n - 1)^3 \\ & = 10^{3n} - 3 \cdot 10^{2n} + 3 \cdot 10^n - 1\\ & = \underbrace{100..00}_{\small \text{number of 0s} = 3n} - \underbrace{300...00}_{\small \text{number of 0s} = 2n} + \underbrace{300...00}_{\small \text{number of 0s} = n} - 1 \\ & = \overline{ \underbrace{99...99}_{\small n-1 \space \text{times}} 7 \underbrace{00...00}_{\small 2n \space \text{times}} } + \overline{ 2\underbrace{99...99}_{\small n \space \text{times}}} \end{aligned}$

Then,

$\begin{aligned} S\left(\underbrace{99...99^3}_{ \small \text{number of 9s} = n} \right) & = 9( n - 1 ) + 7 + 2 + 9n \\ & = 9( n -1 ) + 9 + 9n \\ & = 9(n-1 + 1 + n) \\ & = 9(2n) = 18n \end{aligned}$

Hence, proved.

Similar solution with Áron Bán-Szabó 's

$\begin{aligned} N & = \underbrace{999...999^3}_{\text{Number of 9s }= \ 2000} \\ & = \left(10^{2000}-1\right)^3 \\ & = \color{#3D99F6} 10^{6000} - 3\cdot 10^{4000} + \color{#D61F06} 3\cdot 10^{2000} - 1 \\ & = \color{#3D99F6}\underbrace{999...999}_{1999}7\underbrace{000...000}_{4000} + \color{#D61F06} 2\underbrace{999...999}_{2000} \\ & = \underbrace{999...999}_{1999}7\underbrace{000...000}_{1999} 2 \underbrace{999...999}_{2000} \end{aligned}$

Therefore, the sum of digits of $N$ , $S_n = 9 \cdot 2999 + 7 + 2 = \boxed{36000}$ .

$\underbrace{999\dots99}_{2000\space\text{number of}\space9}=10^{2000}-1$ $\Rightarrow (10^{2000}-1)^3=10^{6000}-3*10^{4000}+3*10^{2000}-1$ The $(3*10^{2000}-1)$ number's first digit is $2$ , the others are $9$ . From the back the first $4000$ digit are $0$ in $(10^{6000}-3*10^{4000})$ , then the $4001.$ digit from the bag is $7$ , then from the $4002.$ digit to the $6000.$ digit all of the digits are $9$ s. So the answer is: $2000*9+2+7+1999*9=\boxed{36000}$