Question

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

• A. $1$
• B. $2$
• C. $8$
• D. $11$

Answer 1

The correct option is C $8$

$2^{2003}=2^3\times 2^{2000}=8(2^4{)}^{500}=8(16{)}^{500}=8(17-1{)}^{500}=8[{}^{500}{C}_0\cdot {17}^{500}-{}^{500}{C}_1\cdot {17}^{499}+\dots +{}^{500}{C}_{500}]=8[17k+1],k\in N=17\times 8k+8$

Hence, the remainder is $8$.