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 14 of 14.
Mohsin said:
1 decade ago
Ohh Its confusing !!!!!
I cant understand y take (2^96+1)??????????????????
I cant understand y take (2^96+1)??????????????????
Priya said:
1 decade ago
When we make 2^32 is X,
We try to simplify to get the answer split 2^96 = 2^(32X3)
so it is = X^3
When using the formula for X^3+1, we can identify that number is divisible by 2^32
We try to simplify to get the answer split 2^96 = 2^(32X3)
so it is = X^3
When using the formula for X^3+1, we can identify that number is divisible by 2^32
Divya said:
1 decade ago
I can't to understand your explanation. Can you explain clearly and basically................
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?
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.
Devi said:
1 decade ago
I cannot understand how replace (2^96+1) = [ (2^32) ^3].
Pavi said:
2 decades ago
I cannot understand, please explain.
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