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Number System Doubt

Harsh Bavishi
New Member

For the Euler's Theorem example given during class session (Sahil Sir), 17^81/100, why can't we use Cyclicity to find the remainder in this question? 81/4 is 20 and the remainder comes as 1. So, 17^1 is 17, and 17/100 gives 17 as the remainder.

In the second example of the same class, where using both Remainder and Euler Theorem, we could have used Cyclicity to derive the answer. The question was 21^865/17. Here, 865 divided by 4, gives 1 as the remainder. So, 21^1 is 21, and 21/17 gives 4 as the remainder. 

So, are there any exceptions or reasons for not using the Cyclicity method?

Topic starter Posted : 05/07/2021 4:46 pm
Rahul Singh

We cannot use Cyclicity here, it is used to find the last term of any power. It was a coincidence that using cyclicity of 4 we are getting the same answer.

Let’s look at the below example-

The remainder of 39/17, as per your logic, if we use cyclicity here, 9 divided by 4 gives 1, so 31/17 will give remainder 3.

But if we see, 39 = 19683 if we divide this number by 17 it will five Remainder as 14.

Hence, we can’t use cyclicity to find our remainders.

It can be used to find out Remainder when a number is divided by 10, because basically when we divide any number by 10 then the remainder will be nothing but the last digit of that number.

I hope this helps!

Posted : 30/08/2021 11:07 am