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 8 of 14.
Ram said:
1 decade ago
I can't understand.
Priya said:
1 decade ago
It is difficult. But if you we observe carefully we can get it. Think of it more and more.
Varsha Pathak said:
1 decade ago
(296 + 1) = [(232)3 + 1] = (x3 + 1) = (x + 1)(x2 - x + 1) please explain this part.
Ashwini said:
1 decade ago
I can't understand please explain briefly?
Dibya said:
1 decade ago
Why it will not be divisible by 2^16+1?
As if 2^32 then why not 2^16?
As if 2^32 then why not 2^16?
Pari said:
10 years ago
Please give a simple method with which one can get answer in like just 40 secs. Please do help.
Radhika said:
10 years ago
I can't understand please again reply.
Manoj said:
10 years ago
I can understand the problem. But is there any simplest method to solve this?
Manoj said:
10 years ago
I can understand the problem. But is there any simplest method to solve this?
Kapis said:
10 years ago
Please give me a basic and simple methods.
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