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 11 of 14.
Mahalakshmi said:
9 years ago
I can't understand please explain again in a simple way.
Shirisha said:
9 years ago
I understand this. But I want simple method.
Pranshu said:
9 years ago
Why not option A? Explain.
Sajal said:
9 years ago
I didn't understand the last step. Please explain it.
Apsana said:
9 years ago
I can't understand this please explain this in a simply way.
Sajid said:
9 years ago
I don't understand. Please, someone explain me.
Saruu said:
9 years ago
Can anyone give a simple method to explain this?
Deva said:
9 years ago
I don't understand the solution. Please, someone explain me.
Kranthi said:
9 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.
Aravinth Kumar said:
9 years ago
@Kishore Sulthana.
Good explanation. I can understand it now.
Good explanation. I can understand it now.
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