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.
Saad said:
6 years ago
I can't understand. Please, anyone, explain it briefly.
Sonu said:
6 years ago
Your explanation is very complicated.
Deva.Harshitha Patel said:
5 years ago
WKT,(x^n+a^n) is divisible by (x+a),if n is odd.
Here,if we take x=2^32 and a=1,x+a=2^32+1 divides (x^n+a^n),if n is odd i.e,i may be 1 or 3 or 5 or 7..
If we take n=1,(2^32)^1+1^1=2^32+1 is not in the option,
if n=3,(2^32)^3+1^3=2^96+1 is the correct answer.
Here,if we take x=2^32 and a=1,x+a=2^32+1 divides (x^n+a^n),if n is odd i.e,i may be 1 or 3 or 5 or 7..
If we take n=1,(2^32)^1+1^1=2^32+1 is not in the option,
if n=3,(2^32)^3+1^3=2^96+1 is the correct answer.
Octavia said:
5 years ago
I don't understand. Please explain in detail.
Anuj yadav said:
5 years ago
I think the formula will be 2^(n.3)+1.
Yaduvanshi said:
5 years ago
@Deva Harshitha Patel.
You are absolutely right, thanks.
You are absolutely right, thanks.
Shyam Ingalagi said:
2 months ago
What is the exponentiation rule? Please explain to me.
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