2006 Cyprus MO/Lyceum/Problem 13

Problem

The sum of the digits of the number $10^{2006}-2006$ is

$\mathrm{(A)}\ 18006\qquad\mathrm{(B)}\ 20060\qquad\mathrm{(C)}\ 2006\qquad\mathrm{(D)}\ 18047\qquad\mathrm{(E)}\ \text{None of these}$

$10^{2006}$ is a $1$ followed by 2006 $0$ 's. When we subtract $2006$ , we will get something close to 2006 $9$ 's.

The last four digits are $10000 - 2006 = 7994$ , and so we have 2002 $9$ s followed by $7994$ .

The sum of these is $2002 \cdot 9 + 7 + 9 + 9 + 4 = 18047 \Longrightarrow \mathrm{(D)}$