Question

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

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Answer 1

We have to express 2 or powers of 2 in terms of 17.We know $2^4$ = 16 = 17 - 1.

So we want write $2^{2003}$ in the form k $(17-1{)}^n$

$2^{2003}$ = 8$\left[2^{2000}\right]$

$2^{2003}$ = 8$\left[{16}^{500}\right]$

$2^{2003}$ = 8$\left[\right(17-1{)}^{500}]$

8$\left[\right(17-1{)}^{500}]$ = 8 $\left[{}^{500}{C}_0\right(17{)}^{500}-{}^{500}{C}_1(17{)}^{499}+{}^{500}{C}_2(17{)}^{498}...........-{}^{500}{C}_{4}99(17{)}^1+{}^{500}{C}_{500}]$

8$\left[\right(17-1{)}^{500}]$ = 8 $\left[{}^{500}{C}_0\right(17{)}^{500}-{}^{500}{C}_1(17{)}^{499}+{}^{500}{C}_2(17{)}^{498}...........-{}^{500}{C}_{4}99(17{)}^1+1]$

8$\left[\right(17-1{)}^{500}]$ = 8 $\left[{}^{500}{C}_0\right(17{)}^{500}-{}^{500}{C}_1(17{)}^{499}+{}^{500}{C}_2(17{)}^{498}...........-{}^{500}{C}_{4}99(17{)}^1+1-1]+8\times 1$

(We added and subtracted 8^* 1)

$\Rightarrow$ 8$\left[\right(17-1{)}^{500}]$ = 8 $\left[{}^{500}{C}_0\right(17{)}^{500}-{}^{500}{C}_1(17{)}^{499}+{}^{500}{C}_2(17{)}^{498}...........-{}^{500}{C}_{4}99\left(17{)}^1\right]+8\times 1$

8 $\left[{}^{500}{C}_0\right(17{)}^{500}-{}^{500}{C}_1(17{)}^{499}+{}^{500}{C}_2(17{)}^{498}...........-{}^{500}{C}_{4}99\left(17{)}^1\right]$ is a multiple of 17.we will write it as 17k.

$\Rightarrow$ 8$\left[\right(17-1{)}^{500}]$ = 17k + 8 $\times$ 1