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:
138 comments Page 4 of 14.
Mickeey said:
2 decades 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:
2 decades ago
Is there any simple procedure?
Rahul wadhwa said:
2 decades ago
Really its very confusing!!!!!!!!!!!!!!!!!!!
Jaiveer said:
2 decades ago
I could not understand so pls help me for full procedure
Richa said:
2 decades ago
Unable to undestand. Kindly help.
Nagaraj said:
2 decades 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:
2 decades ago
How do u directly write (2^96+1) ?
Karthi said:
2 decades 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.
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