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 3 of 14.
Octavia said:
5 years ago
I don't understand. Please explain in detail.
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.
Dilip said:
5 years ago
I'm not understanding this problem.
(1)
Sonu said:
6 years ago
Your explanation is very complicated.
Sasikumar dpi said:
6 years ago
Easy to understand this Answer.
So,
2^32 = 32x32 = 1024 + 1(this question value)
2^16 = 16x16 = 256 + 1 (not covered value)
2^16 = 16x16 = 256 - 1 (not covered value)
2^23 = 23x23 = 529 x 7 (not covered value)
2^96 = 96x96 = 9216 + 1(so this option is fully covered for question value so we can choose D option)
So,
2^32 = 32x32 = 1024 + 1(this question value)
2^16 = 16x16 = 256 + 1 (not covered value)
2^16 = 16x16 = 256 - 1 (not covered value)
2^23 = 23x23 = 529 x 7 (not covered value)
2^96 = 96x96 = 9216 + 1(so this option is fully covered for question value so we can choose D option)
(7)
Saad said:
6 years ago
I can't understand. Please, anyone, explain it briefly.
Jyoshna said:
6 years ago
I am not understanding please anyone help me to get it.
Nayudu said:
6 years ago
Agree @Diya.
Can anyone please explain it?
Can anyone please explain it?
DIYA said:
6 years ago
When it comes to 2^16+1, how can we find its not divisible by n?
(2^8)2+1 = x^2+1=?
(2^8)2+1 = x^2+1=?
Tej said:
6 years ago
When we get to handle a big no then just assume for small no;
eg. 2^1(n)+1 always divides 2^3(n)+1;
This pattern followed for all the powers of 2,
For simplest one took n=1
2^1+1=3
2^3+1=9;
For n=2.
2^2+1=5
2^6+1=65;
So we can see the pattern.
Likewise, it follows the last option.
eg. 2^1(n)+1 always divides 2^3(n)+1;
This pattern followed for all the powers of 2,
For simplest one took n=1
2^1+1=3
2^3+1=9;
For n=2.
2^2+1=5
2^6+1=65;
So we can see the pattern.
Likewise, it follows the last option.
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