Question

The remainder when $2^{2003}$ is divided by 17 is

• A. 7
• B. 8
• C. 2
• D. 1
• E. 0

Answer 1

Answer: (B)

We have,
$2^{2000}=\left(2^{4}\right)^{500}=(17-1)^{500} $
$={ }^{500} C_{0} \,17^{500}-{ }^{500} C_{1}\, 17^{499}+\ldots-{ }^{500} C_{499}\,17+(-1)^{500}$
$17 m+1$, where $m$ is some positive integer
$=2^{2003}=(8)\left(2^{2000}\right)=8(17 m+1)$
$=17(8 m)+8$
Hence, the remainder will be $8$ .