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.
AnonymousType said:
1 decade ago
2^32 + 1 = 4294967297 = 641 * 6700417.
Now [A] (2^16+1) = 65537 is a prime.
[B] (2^16-1) = 3*5*17*257.
[C] (7*2^23) only is divisible by 7 and a bunch of even numbers.
[D] (2^96+1) = 79228162514264337593543950337 = 641 * 6700417 * 18446744069414584321.
So there.
Now [A] (2^16+1) = 65537 is a prime.
[B] (2^16-1) = 3*5*17*257.
[C] (7*2^23) only is divisible by 7 and a bunch of even numbers.
[D] (2^96+1) = 79228162514264337593543950337 = 641 * 6700417 * 18446744069414584321.
So there.
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.
Nitesh Jindal said:
1 decade ago
It is not comfusing. ok. try to understand the logic here.
lets assume 2^32 is x.
Now 2^96 = {2^32 }^3 .As 2^32 is x so 2^96 is x^3.
Then x^3 +1 is furter solved by formula
as a^3 + b^3 = (a+b)(a^2 -ab +b^3) .
therefore x^3 + 1 = (x+1)(x^2 - x + 1 )
lets assume 2^32 is x.
Now 2^96 = {2^32 }^3 .As 2^32 is x so 2^96 is x^3.
Then x^3 +1 is furter solved by formula
as a^3 + b^3 = (a+b)(a^2 -ab +b^3) .
therefore x^3 + 1 = (x+1)(x^2 - x + 1 )
Pranjal said:
1 decade ago
It's a simple thing;
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.
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.
Madhu said:
1 decade ago
I have one solution.
Let 232 = x. Then, (232 + 1) = (x + 1).
Let (x + 1) be completely divisible by the natural number N. Then
[(232)3 + 1] = (296 + 1) =(x3 + 1) = (x + 1)(x2 - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.
Let 232 = x. Then, (232 + 1) = (x + 1).
Let (x + 1) be completely divisible by the natural number N. Then
[(232)3 + 1] = (296 + 1) =(x3 + 1) = (x + 1)(x2 - x + 1), which is completely divisible by N, since (x + 1) is divisible by N.
Devi krishna said:
1 year ago
Let 2^32 be x.
Then (2^32+1) = (x+1).
2^96 + 1 = (2^32)^3 + 1.
(x^3+1^3) = (a^3+b^3) = (a+b) (a^2-ab+b^2)
=> (x^3+1^3) = (x+1) (x^2-x+1).
So, (x+1) (x^2-x+1)/(x+1) = (x^2-x+1).
So, which is 2^96+1 is divisible by 2^32+1.
By using the formula of(a^3+b^3).
Then (2^32+1) = (x+1).
2^96 + 1 = (2^32)^3 + 1.
(x^3+1^3) = (a^3+b^3) = (a+b) (a^2-ab+b^2)
=> (x^3+1^3) = (x+1) (x^2-x+1).
So, (x+1) (x^2-x+1)/(x+1) = (x^2-x+1).
So, which is 2^96+1 is divisible by 2^32+1.
By using the formula of(a^3+b^3).
(1)
Lupu gogoi said:
8 years ago
We are finding the answer. We don't know what is the answer above this problem. Then how to you directly put the answer 2^96+1.
How you know the answer is that? If the problem given you another then how to you solve that. Please explain.
How you know the answer is that? If the problem given you another then how to you solve that. Please explain.
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
Vilaz said:
5 years ago
Let 2^32 = x.
Lets take,
(2^96 + 1)
= [(2^32)3 + 1]
= (x^3 + 1) { we took 2^32=x}
=(x^3 + 1) { In the form of [a^3+b^3 ]=[a^2- ab+b^2 ]
= (x + 1)(x2 - x + 1).
Which is completely divisible by N, since (x + 1) is divisible by N.
Lets take,
(2^96 + 1)
= [(2^32)3 + 1]
= (x^3 + 1) { we took 2^32=x}
=(x^3 + 1) { In the form of [a^3+b^3 ]=[a^2- ab+b^2 ]
= (x + 1)(x2 - x + 1).
Which is completely divisible by N, since (x + 1) is divisible by N.
(4)
Hasan said:
1 decade ago
Suppose X = 2^32 then 2^32+1 = X+1.
2^96 + 1
= (2^32)^3 + 1
= X^3 + 1. .'. (2^32 = X)
Let X^3+1/X+1
(X+1)(X^2 - X + 1) / (X+1)
=(X^2 - X + 1).
Which is completely divisible by N, since (x + 1) is divisible by N.
2^96 + 1
= (2^32)^3 + 1
= X^3 + 1. .'. (2^32 = X)
Let X^3+1/X+1
(X+1)(X^2 - X + 1) / (X+1)
=(X^2 - X + 1).
Which is completely divisible by N, since (x + 1) is divisible by N.
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