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 5 of 14.
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.
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.
Priya DP said:
8 years ago
By trial and error method only option A&D are applicable
Yuvraj said:
8 years ago
I Can't understand, how to solve this? Someone help me.
Yuvraj said:
8 years ago
I can't understand to which way solve this question please explain in simple way.
Karna said:
8 years ago
Here, We know that (x^n+y^n) is divisible by (x+a) for natural numbers.
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.
Lupu gogoi said:
8 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.
Aravinth Kumar said:
8 years ago
@Kishore Sulthana.
Good explanation. I can understand it now.
Good explanation. I can understand it now.
Kranthi said:
8 years ago
Remember the condition that (x^n+1^n) is divisible by (x+1). Given number (2^32+1) is divisor. He said which of the following number is divided by this number i.e. (2^32+1). The dividend should be in the form of (x^n+1^n). Let x=2^32(from divisor).
Substituae it in dividend. [(2^32)^n+1^n].
Substitube n= 1,3,5,... 4 th option suits when n=3.
Substituae it in dividend. [(2^32)^n+1^n].
Substitube n= 1,3,5,... 4 th option suits when n=3.
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