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 10 of 14.
B.nandu said:
1 decade ago
2^32+1 is divisible by all whole numbers.
So here we multiply the power of 32 with whole numbers. i.e,
32 x 1=32.
32 x 2=64.
32 x 3=96.
We have this option 96. So 2696+1 is correct one.
So here we multiply the power of 32 with whole numbers. i.e,
32 x 1=32.
32 x 2=64.
32 x 3=96.
We have this option 96. So 2696+1 is correct one.
Soumya said:
1 decade ago
(2^32+1).
Take x = 2^32, Then (2^32+1) = (x+1) ------> Eq1.
2^96 = (2^32) ^3, ie x^3.
Then (2^96+1) = (x^3+1).
We know that.
(x^3+1) = (x+1) (x^2-x+1) ------> Eq2.
Then dividing eq2 by eq1.
(x+1) (x^2-x+1) / (x+1) = (x^2-x+1).
If we give any value to x we get only natural numbers.
Take x = 2^32, Then (2^32+1) = (x+1) ------> Eq1.
2^96 = (2^32) ^3, ie x^3.
Then (2^96+1) = (x^3+1).
We know that.
(x^3+1) = (x+1) (x^2-x+1) ------> Eq2.
Then dividing eq2 by eq1.
(x+1) (x^2-x+1) / (x+1) = (x^2-x+1).
If we give any value to x we get only natural numbers.
Satyajit chakraborty said:
1 decade ago
Simply remember this:
(a^n +b^n) is always divisible by same value of 'a' and 'b' if 'n' is odd in number. Here 'n' denotes power of a number.
(a^n +b^n) is always divisible by same value of 'a' and 'b' if 'n' is odd in number. Here 'n' denotes power of a number.
Deependra said:
1 decade ago
It is very easy question guys. Try again to understand the logic of mathematics which applied in this problem.
Deepa said:
1 decade ago
How can we know (x^3+1) is divisible by n?
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.
Sandeep kumar jangir said:
1 decade ago
2^32 = x.
So (2^32+1) = x+1.
And option A, B n C have less power compared to given value.
So 2^96 = 2^(32*3). Which is completely divisible by 2^32.
So (2^32+1) = x+1.
And option A, B n C have less power compared to given value.
So 2^96 = 2^(32*3). Which is completely divisible by 2^32.
Senthil kumar said:
1 decade ago
I cannot understand how this (2^23)^3+1 is created.
Please explain sir.
Please explain sir.
Navin kumar kamti said:
1 decade ago
Its very simple try to understand friend.
Here,
Let suppose x+1 divisible by n where n is natural number as like 0. 1.2.3....
If put x=1 then,
x+1=1+1=2.
Which is divisible by n.
As like,
2/2=0.
Thus,
2^32=x suppose.
Then we can write,
x+1= 2^23+1.
Now (2^23)^3+1) = 2^96+1.
Now we can write,
Formula of (x^3+1) = (x+1)(x^2-x+1) which is divisible by n nis natural no As like 0.1.2.....n
since x+1 divisible by n.
Here,
Let suppose x+1 divisible by n where n is natural number as like 0. 1.2.3....
If put x=1 then,
x+1=1+1=2.
Which is divisible by n.
As like,
2/2=0.
Thus,
2^32=x suppose.
Then we can write,
x+1= 2^23+1.
Now (2^23)^3+1) = 2^96+1.
Now we can write,
Formula of (x^3+1) = (x+1)(x^2-x+1) which is divisible by n nis natural no As like 0.1.2.....n
since x+1 divisible by n.
D.raja said:
1 decade ago
I don't understand this answer please explain briefly. How to replace this answer (2^96+1)?
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