Aptitude - Numbers - Discussion

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

Numbers

It is being given that (2³² + 1) is completely divisible by a whole number. Which of the following numbers is completely divisible by this number?

(2¹⁶ + 1)

(2¹⁶ - 1)

(7 x 2²³)

(2⁹⁶ + 1)

Answer: Option

Explanation:

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

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

(2⁹⁶ + 1) = [(2³²)³ + 1] = (x³ + 1) = (x + 1)(x² - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.

Discussion:

137 comments Page 9 of 14.

Priya DP said: 8 years ago

By trial and error method only option A&D are applicable

Srujana said: 1 decade ago

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

Yaduvanshi said: 5 years ago

@Deva Harshitha Patel.

You are absolutely right, thanks.

Ronny said: 3 years ago

@Tanvi.

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

(7)

Devi said: 1 decade ago

I cannot understand how replace (2^96+1) = [ (2^32) ^3].

Jaiveer said: 1 decade ago

I could not understand so pls help me for full procedure

Mahalakshmi said: 9 years ago

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

Binod Barai said: 7 years ago

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

Yuvraj said: 8 years ago

I Can't understand, how to solve this? Someone help me.

Jyoshna said: 6 years ago

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

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