# Aptitude - Numbers - Discussion

Discussion Forum : Numbers - General Questions (Q.No. 3)

3.

It is being given that (2

^{32}+ 1) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?Answer: Option

Explanation:

Let 2^{32} = *x*. Then, (2^{32} + 1) = (*x* + 1).

Let (*x* + 1) be completely divisible by the natural number N. Then,

(2^{96} + 1) = [(2^{32})^{3} + 1] = (*x*^{3} + 1) = (*x* + 1)(*x*^{2} - *x* + 1), which is completely divisible by N, since (*x* + 1) is divisible by N.

Discussion:

129 comments Page 1 of 13.
VENKY said:
6 months ago

I can't understand, anyone can explain this?

(4)

Rrai said:
8 months ago

1&2 option will be eliminated as they are less than the denominator, and in 3rd option even * odd = even.

And the denominator is even + odd = odd, and even no. Can't divide the odd so it will also be eliminated.

And the denominator is even + odd = odd, and even no. Can't divide the odd so it will also be eliminated.

(5)

Ronny said:
11 months ago

@Tanvi.

Here (a^3+b^3)identity is used. Please check it.

Here (a^3+b^3)identity is used. Please check it.

(1)

Nasreen.A said:
1 year ago

(X^3+1)=(x+1)(x^2-x+1).

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

= 2^32+1-2^64+2^32-1

= 2^32-2^32 = 0.

((2^32)^3+1) = 0.

((2^96+1)) = 0.

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

= 2^32+1-2^64+2^32-1

= 2^32-2^32 = 0.

((2^32)^3+1) = 0.

((2^96+1)) = 0.

(2)

Mile said:
1 year ago

From where 3 came?

There 2^32 * 3 instead of 3 we can use any other numbers or not. Anyone explain, please.

There 2^32 * 3 instead of 3 we can use any other numbers or not. Anyone explain, please.

(2)

Tanvi said:
2 years ago

Can anyone help me. What identity is used to determine whether a number is divisible or not completely?

RITIK GUPTA said:
2 years ago

@All.

You have to remember a concept that (x+1) is always a factor of x^n + 1 if n is an odd integer.

You have to remember a concept that (x+1) is always a factor of x^n + 1 if n is an odd integer.

(1)

Anil said:
2 years ago

It's too hard to understand. Anyone, please help me to get it.

Vivek Singh said:
2 years ago

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.

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.

(2)

Vilaz said:
2 years ago

Let 2^32 = x.

Lets take,

(2^96 + 1)

= [(2^32)3 + 1]

= (x^3 + 1) { we took 2^32=x}

=(x^3 + 1) { In the form of [a^3+b^3 ]=[a^2- ab+b^2 ]

= (x + 1)(x2 - x + 1).

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

Lets take,

(2^96 + 1)

= [(2^32)3 + 1]

= (x^3 + 1) { we took 2^32=x}

=(x^3 + 1) { In the form of [a^3+b^3 ]=[a^2- ab+b^2 ]

= (x + 1)(x2 - x + 1).

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

(1)

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