Aptitude - Numbers - Discussion

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...

Above mentioned are totally absurd. Few of them having tried good but not upto the mark.

Now see this carefully.

We are trying to get our given no. 2^32+1 from the choices given and I tried all d choices. And atlast 2^96+1 is easily broken to get 2^32+1.

Now I take an example to explain it in detail. Since 2^32+1 is completely div by N therefore N<=2^32+1 (less than equal to).

Ex: If 4 is completely div by x then x is always less than or equal to 4.

Now take first option 2^16+1 here it will be completely div N when N<=2^16+1 but wt if N>2^16+1 since value of N is ranging from 0 to 2^32+1 so wt if N>2^16+1 then 2^16+1 is not completely div.

Since we exactly don't know value of N we are checking its range to determine.

Therefore option A is not answer. Now similarly B is not answer not.

Option D is answer because all the range of N completely divides it which is explained how by indiabix.

If you don't understand this answer just try to focus on range of N. And again see my example of 4 completely divided.

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.

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.

This question is to solve using the formula:- a^3+b^3 = (a+b)(a^2-ab+b^2).

Only option D satisfies the above formula.

We can write (2^96+1) as ((2^32)^3 + 1). So if we solve ((2^32)^3 + 1) using the above formula then we get:

((2^32)^3 + 1) = (2^32+1)(2^64-2^32+1).

Now, divide (2^32+1)(2^64-2^32+1) by (2^32+1). If we divide these two then (2^32+1) gets cancelled and we get the result as (2^64-2^32+1). Hence option D is correct.

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.

this is some what called as hit and trial method, you have to try the options which are similar to the value (2^32+1) in the question. remaining, you can easily eliminate by seeing.
in the above, options A and D are similar to the expression given in the question.
suppose the given four options are similar, you have to try all...
this is also called as trial and error method...

@All

Here's my explanation.

Given: (2^32 + 1) is completely divisible by a whole number.
Let ( 2^32 + 1 ) = (2^n + 1 ), and since ( 2^n + 1 ) is divisible by a whole number,
we can use the exponentiation rule:
(a^m )^n = a^(m * n).

Similarly,
(( 2^n )^ m + 1 ) is also divisible by the same whole number.
2^96 = ( ( 2^32)^3) + 1.

Therefore, Option D: (2^96 + 1) is correct!

Take 2^32 (2 power 32 as x).

Given 2^32 + 1.

So x + 1.

Formula a^3 + b^3 (a cube + b cube) = (a+b) (a^2 - a.b + b^2).

So take x as a, 1 as b. So (x+1) (x^2 - x.1 + 1^2).

1^2 (1 square = 1) and x.1 = x......

So they asked divisible by x+1.

So (x + 1)(x^2 - x + 1)/(x + 1).

So we get (x^2 - x + 1). Because (x + 1) in numerator and denominator gets cancelled.

Given: (2^32 + 1) is completely divisible by a whole number.
Let ( 2^32 + 1 ) = (2^n + 1 ), and since ( 2^n + 1 ) is divisible by a whole number,
we can use the exponentiation rule:

Formula:
( a^m )^n = a^(m * n).

Similarly,
(( 2^n )^ m + 1) is also divisible by the same whole number.
2^96 = ( ( 2^32)^3)+1.

Therefore, Option D: (2^96 + 1) is correct!