Aptitude - Numbers - Discussion

How you take 2^96 +1 exactly?

Confused with this problem. Let me know the shortcut method?

I can't understand please explain me easily.

I can't understand. Please explain it.

I don't understand this problem. Please explain it in a simple way.

a^n + b^n is divisible by a + b only when n is odd. Suppose 2^32 is a then (2^32) ^3 = 2^96 i.e. a^3 which is odd means a^3 + b^3 here b=1. i.e, a^3 + b^3 is divisible by a + b.

Was it possible to use power cycle of 2? Can anyone please explain how to use this?

I could not understand the problem please explain me easily.

It is written in question no should be completely divisible.

For that number should be greater than given number i.e. 2^32+1.

Ex 1/2 = 0.5 (not completely divisible).

4/2 = 2 (completely divisible).

So option A, B, C all three numbers are less than given number 2^32 +1.

Now come on option D = 2^96+1. Clearly shows number is greater than 2^32+1.

This is what up to which some logic can b made.

Rest how it is possible to refer to given explanation of the question.

Thanks.

I can't understand please explain again in a simple way.