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 13 of 14.
Jitendra said:
1 decade ago
Please help in another example similar to this.
Rajan said:
1 decade ago
Please explain briefly this sum.
Karthi said:
1 decade ago
Give some explanations to understand clearly.
Karthik said:
1 decade ago
How do u directly write (2^96+1) ?
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;
Richa said:
1 decade ago
Unable to undestand. Kindly help.
Jaiveer said:
1 decade ago
I could not understand so pls help me for full procedure
Rahul wadhwa said:
1 decade ago
Really its very confusing!!!!!!!!!!!!!!!!!!!
Anusha said:
1 decade ago
Is there any simple procedure?
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...
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