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Find the remainder when $2^{89}$ is divided by $89$?









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Option(B) is correct

When we take successive powers of $2$ and find their remainders, we get the following cyclic patterns of cycle length $11$.

viz $2,4,8,16,32,64,39,78,67,45, 1$

i.e $2^{11}$ leaves a remainder $1$.

Thus, $2^{89} = (2^{11})^8 (2)$ leaves a remainder of 2.

Edit: Thank you Shireen for explaining cyclicity concept involved in the solution.

Alternate Method: As Arjunbkool  pointed out in the comments, a little trick can be used to solve this question.

Remainder of $\dfrac{N^{p-1}}{P}$ is 1 as long as $P$ is a prime number.

Now, as pointed out by Arghya Chakraborty,

$2^{89-1}=2^{88}$ leaves a remainder of 1 when divided by 89 (note that 89 is a prime number)

So, $2^{89}$ will leave remainder 2 when divided by 89.

Edit 2: For yet another alternative solution using 'modulus' method, check comment by Sravan Reddy.

(16) Comment(s)


Binomial expansion can be used as well


Only first and last numbers leave a remainders 1+1 =2


How can we find for this numbers remainder?


How to solve $2^{30}$

and any other questions like this.


You have to find the cyclic pattern as in this question and proceed using the same methodology used in the solution (or in the alternative solutions).


How can be the cyclic pattern length of powers of 2 be 11?

Sravan Reddy

Alternate solution using 'modulus' method:

$2^8 = 256\text{mod}89 = -11\text{mod}89$

By multiplying with 8 on both sides,

$2^{11} = -88\text{mod}89 = 1\text{mod}89$

Raise both sides by 8,

$2^{88} = 1\text{mod}89$

Multiply by 2 on both sides,


So, remainder is 2!!

Swatantra Singh

I am not getting it. Please help.


I recently came to know about a theorem that states reminder of N (p-1)/P is 1 as long as P is a prime number. That could be used here.


Thank you Arjunbkool for the suggestion, added your suggestion in the solution.


the quetion is good but solution is not up to satisfactory level

Mandar Dharmadhikari

Can any body suggest any other time saving trick???


The trick to solve such questions is KNOW THE CYCLICITY of that particular number.

As every number is cyclic with period 4 (see very good detailed explanation: Number System Cyclicity Explanation),

we can write the given number in terms of power of 4:

$2^89=2^{(4*22) + 1}$

using which we get the LAST DIGIT or REMAINDER as:

$2^1= \textbf{2}$

Hope this helps.

Arghya Chakraborty

Do as 'arjunbkool' has stated.

$2^88$ leaves a remainder of 1 when divided by 89(89 is prime)

So, $2^89$ will leave remainder 2 when divided by 89