Question

remainder of 2^2005/17​

Answer 1

Answer:

15

Step-by-step explanation:

To find the remainder when $2^{2005}$/17 we have to find out its remainder cycle :

2 / 17 = 2 (remainder)

$2^{2}$/17 = 4

$2^{3}$/17 = 8

$2^{4}$/17 = 16 or -1

So the remainder came -1 which means when $2^{8}$ is divided by 17 the remainder will be 1 , so the remainders follow a cycle of 8 .

Let the remainder be n , n will also be a term in this cycle of 8.

To find out n is which term in the series we have to find out the remainder when 2005 is divided by 8 (remainder cycle)

so the remainder when 2005 is divided by 8 is 5

so n is 5th term in the series , so the 5th term will be when $2^{5}$ is divided by 17 , so here we know the cycle till 4 so we will split  $2^{5}$ into $2^{4}$ * $2^{1}$ so the remainder will be :

[16 (remainder when $2^{4}$ is divided by 17)*2( remainder when $2^{1}$ is divided by 17)] / 17

= 32/17

= 15 (remainder)