Digit Sum till 999999??!?!?!?

To do this problem, I would use a technique known as generating function

Let digits be ${d}_{1}, {d}_{2}, {d}_{3}, {d}_{4}, {d}_{5}, {d}_{6}$

and ${d}_{1} + {d}_{2} + {d}_{3} + {d}_{4} + {d}_{5} + {d}_{6} = 20$ and $0 \le {d}_{i} \le 9$ .

Now, we will generate a function -

$f(x) = {(1 + x + {x}^{2} + {x}^{3} + {x}^{4} + {x}^{5} +..... {x}^{8} + {x}^{9})}^{6}$

And the coefficient of ${x}^{20}$ in $f(x)$ is our answer.

$f(x) = \frac{{(1-{x}^{10})}^{6}}{{1-x}^{6}}$

${(1-x)}^{-n} = 1 + \left( \begin{matrix} n+1-1 \\ 1 \end{matrix} \right)*x + \left( \begin{matrix} n+2-1 \\ 2 \end{matrix} \right)*{x}^{2} +..... \left( \begin{matrix} n+r-1 \\ r \end{matrix} \right)*{x}^{r}...$

Hence,

$f(x) = ({(1-{x}^{10})}^{6})*(1 + \left( \begin{matrix} 6+1-1 \\ 1 \end{matrix} \right)*x + \left( \begin{matrix} 6+2-1 \\ 2 \end{matrix} \right)*{x}^{2} +..... \left( \begin{matrix} 6+r-1 \\ r \end{matrix} \right)*{x}^{r}...)$

This is simple binomial expansion - ${(1-{x}^{10})}^{6}$

And so, the coefficient of ${x}^{20}$ is the sum of coefficients of

• ( ${x}^{20}$ in 1st part of product) * ( ${x}^{0}$ in 2nd part of product)

• ( ${x}^{0}$ in 1st part of product) * ( ${x}^{20}$ in 2nd part of product)

• ( ${x}^{10}$ in 1st part of product) * ( ${x}^{10}$ in 2nd part of product)

Hence,

coefficient of ${x}^{20}$ in the product = $\left( \begin{matrix} 6 \\ 2 \end{matrix} \right) -\left( \begin{matrix} 6+10-1 \\ 10 \end{matrix} \right) \left( \begin{matrix} 6 \\ 1 \end{matrix} \right) +\left( \begin{matrix} 6+20-1 \\ 20 \end{matrix} \right)$

= $\left( \begin{matrix} 6 \\ 2 \end{matrix} \right) -\left( \begin{matrix} 6+10-1 \\ 10 \end{matrix} \right) \left( \begin{matrix} 6 \\ 1 \end{matrix} \right) +\left( \begin{matrix} 6+20-1 \\ 20 \end{matrix} \right)$

= $\boxed{35127}$