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 6 of 14.
A.Athiban Sakthivel said:
1 decade ago
But its too complicated procedure.
Is not there more simple procedure to solve this problem?
Is not there more simple procedure to solve this problem?
Divya said:
1 decade ago
I can't to understand your explanation. Can you explain clearly and basically................
D.raja said:
1 decade ago
I don't understand this answer please explain briefly. How to replace this answer (2^96+1)?
Arjun said:
1 decade ago
[(x^n) + 1] will be divisible by (x + 1) only when n is odd.
Hope this will help you guys.
Hope this will help you guys.
Priya said:
1 decade ago
It is difficult. But if you we observe carefully we can get it. Think of it more and more.
Divya said:
10 years ago
Multiplication of 36*2 gives correct solution. As friends said use hit and trial method.
DIYA said:
6 years ago
When it comes to 2^16+1, how can we find its not divisible by n?
(2^8)2+1 = x^2+1=?
(2^8)2+1 = x^2+1=?
Varsha Pathak said:
1 decade ago
(296 + 1) = [(232)3 + 1] = (x3 + 1) = (x + 1)(x2 - x + 1) please explain this part.
Ambika said:
9 years ago
Was it possible to use power cycle of 2? Can anyone please explain how to use this?
Rahul said:
1 decade ago
Hi Dear,
2^32^3 here multiplication of power will take place. So 32x3 = 96 Ok.
2^32^3 here multiplication of power will take place. So 32x3 = 96 Ok.
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