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.
Lupu gogoi said:
9 years ago
We are finding the answer. We don't know what is the answer above this problem. Then how to you directly put the answer 2^96+1.
How you know the answer is that? If the problem given you another then how to you solve that. Please explain.
How you know the answer is that? If the problem given you another then how to you solve that. Please explain.
Shyam said:
9 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:
8 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:
8 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.
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