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 2 of 14.
Jaiveer said:
1 decade ago
I could not understand so pls help me for full procedure
Richa said:
1 decade ago
Unable to undestand. Kindly help.
Nagaraj said:
1 decade ago
It's a simple thing;
((2^96)+1)/((2^32)+1)
=((x^3)+1)/(x+1)
=(x+1)(x^2-x+1)/(x+1); here easy to delete the (x+1);
So we get only a (x^2-x+1)as anatural number;
((2^96)+1)/((2^32)+1)
=((x^3)+1)/(x+1)
=(x+1)(x^2-x+1)/(x+1); here easy to delete the (x+1);
So we get only a (x^2-x+1)as anatural number;
Karthik said:
1 decade ago
How do u directly write (2^96+1) ?
Karthi said:
1 decade ago
Give some explanations to understand clearly.
Rajan said:
1 decade ago
Please explain briefly this sum.
Jitendra said:
1 decade ago
Please help in another example similar to this.
Arun said:
1 decade ago
I understand but I want another example.
Manohar said:
1 decade ago
I cannot understand please explain again.
Mala said:
1 decade ago
Its confusing. Is there any simple methods?
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