Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 3)
3.
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?
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 12 of 14.
Shyam said:
8 years ago
Let 2^32 be x.
2^32+1 = x+1----------->(1)
2^96+1 = (2^32)^3+1
= x^3+1
= (x+1)(x^2-x+1)--------->(2)
By (2)÷(1)
2^96+1/2^32+1 = (x+1)(x^2-x+1)/(x+1).
= x^2-x+1.
So, 2^96+1 Is completely divisible by 2^32+1.
2^32+1 = x+1----------->(1)
2^96+1 = (2^32)^3+1
= x^3+1
= (x+1)(x^2-x+1)--------->(2)
By (2)÷(1)
2^96+1/2^32+1 = (x+1)(x^2-x+1)/(x+1).
= x^2-x+1.
So, 2^96+1 Is completely divisible by 2^32+1.
Karna said:
8 years ago
Here, We know that (x^n+y^n) is divisible by (x+a) for natural numbers.
Yuvraj said:
8 years ago
I can't understand to which way solve this question please explain in simple way.
Yuvraj said:
8 years ago
I Can't understand, how to solve this? Someone help me.
Priya DP said:
8 years ago
By trial and error method only option A&D are applicable
Akriti said:
7 years ago
I can't understand last step how it will completely divisible?
Please explain again in simple step.
Please explain again in simple step.
Shreya said:
7 years ago
Hi;
Actually, if any no is of the form a^n +b^n if n is odd.
So, here if in 2^96 + 1 we take n=3 it will become (2^32)^3 + (1^32)^3..to tell you we took n=3 so that we get 32 after dividing...now from the above-mentioned formula (2^32)^3 + (1^32)^3 will always be divisible by 2^32 + 1^32 as n is odd.
Therefore 2^32 + 1 completely divides 2^96 + 1.
Actually, if any no is of the form a^n +b^n if n is odd.
So, here if in 2^96 + 1 we take n=3 it will become (2^32)^3 + (1^32)^3..to tell you we took n=3 so that we get 32 after dividing...now from the above-mentioned formula (2^32)^3 + (1^32)^3 will always be divisible by 2^32 + 1^32 as n is odd.
Therefore 2^32 + 1 completely divides 2^96 + 1.
Habib said:
7 years ago
I can't understand properly, please anyone explain me easily.
Siri said:
7 years ago
I can't understand properly, please anyone explain me easily.
Kannan said:
7 years ago
Here, 2^96 / 2^32 = 0 is completely divided.
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