2006 Cyprus MO/Lyceum/Problem 13

Problem

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

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

Solution

$10^{2006}$ 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 $2002 \cdot 9 + 7 + 9 + 9 + 4 = 18047 \Longrightarrow \mathrm{(D)}$

See also

2006 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30