Let $N=10^{3}+10^{4}+10^{5}+10^{6}+10^{7}+10^{8}+10^{9}$. The sum of the digits of $N$ is
(A) 12
(B) 1
(C) 6
(D) 9
(E) 7
$\quad$ Since $N$ is written out as a sum of powers of 10 , then $N$ can be written as 1111111000 , and so the sum of the digits is 7 .

ANSWER: (E)