Aptitude - Numbers - Discussion

2^32+1=2^33.
2^96+1=2^97.
So, it is completely divided.

96/3 = 32 so the divisible no is 3.

Suppose 2^32 = X.
so the first equation is X+1.
we can break 2^96 in 2 ^ 32^3 (mean (2 power 32 )power 3 )
So we can write X^3 + 1 because 2^32 =x already define.

I can't understand the last step. Can anyone help me to get it?

Here, the options a,b,c are wrong because all are the smallest value than 2^32 +1.

When you divide a number by largest number, the answer must be decimal so it not whole and not completely divisible. Then final option 2^96 is larger than 2^32. It may be divisible.


Let x= 2^32.
We can write, Power 96=32 *3.
Then 2^96+1=(2^32)^3+1,
We know that (a^3+b^3)=(a+b)(a^2-ab+b^2).
Let a=x i.e,x=2^32.
b=1.

Then
X^3+1=(x+1)(x^2-x+1).
÷by x+1i.e x+1=2^32+1.
Then we get (x^3+1)/(x+1)=x^2-x+1.
And the answer that must be the whole number.
Yes, it is completely divisible by 2^32+1.

I can't understand it, please explain me.

When we get to handle a big no then just assume for small no;

eg. 2^1(n)+1 always divides 2^3(n)+1;
This pattern followed for all the powers of 2,
For simplest one took n=1
2^1+1=3
2^3+1=9;

For n=2.
2^2+1=5
2^6+1=65;

So we can see the pattern.
Likewise, it follows the last option.

When it comes to 2^16+1, how can we find its not divisible by n?

(2^8)2+1 = x^2+1=?

Agree @Diya.

Can anyone please explain it?

I am not understanding please anyone help me to get it.