Question

The units' digit of the sum $1+9+9^2+....{.9}^{1006}$ is:

• A. $2$
• B. $1$
• C. $9$
• D. $0$

Answer 1

Answer: (D)

$0$

The units' digit of each term successively is $1,9,1,9$,..........as the power of $9$ is odd in the second term and even in the third term and so on.

First term = $1$
The units' digit of first two term $=0$ as $1+9=10$
The units' digit of first three terms $=1$ as $1+9+1=11$
The units' digit of first four terms $=0$ as $1+9+1+9=20$
Units' digit of the sum of $1006$ terms i.e. even number of terms = $0$