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 1 of 14.
Pavi said:
2 decades ago
I cannot understand, please explain.
Devi said:
1 decade ago
I cannot understand how replace (2^96+1) = [ (2^32) ^3].
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.
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................
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
Mohsin said:
1 decade ago
Ohh Its confusing !!!!!
I cant understand y take (2^96+1)??????????????????
I cant understand y take (2^96+1)??????????????????
Mickeey said:
1 decade ago
this is some what called as hit and trial method, you have to try the options which are similar to the value (2^32+1) in the question. remaining, you can easily eliminate by seeing.
in the above, options A and D are similar to the expression given in the question.
suppose the given four options are similar, you have to try all...
this is also called as trial and error method...
in the above, options A and D are similar to the expression given in the question.
suppose the given four options are similar, you have to try all...
this is also called as trial and error method...
Anusha said:
1 decade ago
Is there any simple procedure?
Rahul wadhwa said:
1 decade ago
Really its very confusing!!!!!!!!!!!!!!!!!!!
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