Aptitude - Numbers - Discussion

I have one solution.

Let 232 = x. Then, (232 + 1) = (x + 1).

Let (x + 1) be completely divisible by the natural number N. Then
[(232)3 + 1] = (296 + 1) =(x3 + 1) = (x + 1)(x2 - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.

Given :(2)^32+1 is a whole number
we find that: which number from the following option is divisible by this whole number.

I have one best solution
let 2^32=X, now,given: (2^32)+1=X+1,
THEN CHECK any option,
first OPTION A, (2^16)+1,

IT IS TOO small number than given number,so it is not divisible by the given number.
therefore,it is wrong.

SECOND OPTION B, (2^16)-1,

IT IS also a TOO small number than given number,so it is not divisible by the given number.
therefore,it is wrong.

THIRD OPTION C,7*(2^23)

Here (2^23) is too small number than the given number.
so it is not divisible by given number

then Final FOURTH OPTION D, (2^96)+1,

here 2^96 is biggest number than (2^32)
so (2^96)+1 is divisible by (2^32)+1,
EXPLANATION: let (2^32)=X,
here let write(2^32)^3=2^96(for our convenience)
option4:(2^96)+1, therefore, [(X)^3]+1

we know the formula:(a^3+b^3)=(a+b)(a^2-ab+b^2)
from the above :write [(x^3)+(1)^3)=(x+1)(x^2-x+1)
here (x+1)(x^2-x+1)is divisible by (X+1) that is given number..
we know that (X+1)=(2^32)+1=given number..

Therefore the option4 is divisible by given number.
so,it is a correct answer friends...

I am confusing. (2^32 + 1) is completely divisible by (2^96 + 1) OR (2^96 + 1) is completely divisible by (2^32 + 1). Sorry, perhaps I can't understood the question.

Suppose X = 2^32 then 2^32+1 = X+1.

2^96 + 1
= (2^32)^3 + 1
= X^3 + 1. .'. (2^32 = X)

Let X^3+1/X+1

(X+1)(X^2 - X + 1) / (X+1)
=(X^2 - X + 1).

Which is completely divisible by N, since (x + 1) is divisible by N.

In simple words.

Guy's just look at the power of each no. It is lesser than the (2^32+1) they ask us to completely divisible by (2^32+1).

So Only (2^96+1) Has The Greater power Than the other given No. So Answer is (2^96+1).

Because to divide NR. completely power of the NR. Or no in NR. it has to be greater than Dr.

Put X =3 in place of X in Expression (X+1)(X^2-x-1),

Then 4 * (9-3-1).
= 4 * 5.
= 20.

Which is divided by (x+1) means 4.

Hence BY if "a" is divided by "b"
b is divided by C.
Then a is divided by C.


Hence x^3+1 is divided by X+1 , & X^3+1 is 2^96+1.

I can't understand how to replace (2^96+1) please explain?

I don't understand this answer please explain briefly. How to replace this answer (2^96+1)?

Its very simple try to understand friend.

Here,
Let suppose x+1 divisible by n where n is natural number as like 0. 1.2.3....

If put x=1 then,
x+1=1+1=2.
Which is divisible by n.

As like,
2/2=0.

Thus,
2^32=x suppose.
Then we can write,
x+1= 2^23+1.

Now (2^23)^3+1) = 2^96+1.

Now we can write,
Formula of (x^3+1) = (x+1)(x^2-x+1) which is divisible by n nis natural no As like 0.1.2.....n
since x+1 divisible by n.

I cannot understand how this (2^23)^3+1 is created.

Please explain sir.