Too Long A Sum ? ( Sum of All The Digits of Integers )

Let the sums of unit digits, tenth digits, hundredth digits and thousand digits be $s_0$ , $s_1$ , $s_2$ and $s_3$ respectively. Then we have:

$\begin{aligned} s_0 & = \underbrace{{\color{#3D99F6}1+2+\cdots+9 + 0} + {\color{#D61F06}1+2+\cdots+9+ 0} + \cdots + {\color{#3D99F6}1+2+\cdots+9 +0}}_{\left \lfloor \frac {2008}{10}\right \rfloor \text{ times}} + 1+2+3+4+5+6+7+8 \\ & = 200 \times 45 + 36 = 9036 \end{aligned}$

Similarly,

$\begin{aligned} s_1 & = \left \lfloor \frac {2008}{100} \right \rfloor \times 45 \times 10 = 9000 \\ s_2 & = \left \lfloor \frac {2008}{1000} \right \rfloor \times 45 \times 100 = 9000 \\ s_3 & = \underbrace{1 \times 1000}_{\text{1000 to 1999}} + \underbrace{2 \times 9}_{\text{2000 to 2008}} = 1018 \end{aligned}$

Therefore, the sum of digits from 1 to 2008 is $s_0+s_1+s_2+s_3 = 9036+9000+9000+1018 = \boxed{28054}$ .