Question

Remainder when 2^2003 divided by 17

Answer 1

We have to perform-

                               2^2003 / 17

Divide the last two digits of the power with 4 as "2" has a cyclicity of 4.

So , 03/4 , rem = 3

Now use this remainder  as s power in 2 = 2^3 = 8

8 /17

Rem = 8 ans

Things to remember

2 , 3 , 7 and 8 has a cyclicity of 4. So whenever you see such a problem divide the last two digit of the power with 4 and then proceed you will get your answer quickly.

Answer 2

Answer:

Remainder is 8.

Step-by-step explanation:

Given numbers are $2^{2003}$ and 17.

To find: Remainder when $2^{2003}$ divisible by 17.

We know that 2 has cyclicity of 4.

So, we can write given number as,

$(2^4)^{500}\times2^3$

$\imples(16)^{500}\times8$

We also know that

When 16 is divided by 17 it leave -1 as remainder,

So,

$\implies(-1)^{500}\times8$

$\implies1\times8$

⇒ 8

So, when 8 is divided by 17 it leaves remainder = 8.

Therefore, Remainder is 8.