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:
139 comments Page 2 of 14.
Anil said:
5 years ago
It's too hard to understand. Anyone, please help me to get it.
(7)
Mile said:
4 years ago
From where 3 came?
There 2^32 * 3 instead of 3 we can use any other numbers or not. Anyone explain, please.
There 2^32 * 3 instead of 3 we can use any other numbers or not. Anyone explain, please.
(7)
Ronny said:
4 years ago
@Tanvi.
Here (a^3+b^3)identity is used. Please check it.
Here (a^3+b^3)identity is used. Please check it.
(7)
Aviii said:
2 years ago
Given: (2^32 + 1) is completely divisible by a whole number.
Let ( 2^32 + 1 ) = (2^n + 1 ), and since ( 2^n + 1 ) is divisible by a whole number,
we can use the exponentiation rule:
Formula:
( a^m )^n = a^(m * n).
Similarly,
(( 2^n )^ m + 1) is also divisible by the same whole number.
2^96 = ( ( 2^32)^3)+1.
Therefore, Option D: (2^96 + 1) is correct!
Let ( 2^32 + 1 ) = (2^n + 1 ), and since ( 2^n + 1 ) is divisible by a whole number,
we can use the exponentiation rule:
Formula:
( a^m )^n = a^(m * n).
Similarly,
(( 2^n )^ m + 1) is also divisible by the same whole number.
2^96 = ( ( 2^32)^3)+1.
Therefore, Option D: (2^96 + 1) is correct!
(6)
Selvakumar O S said:
2 years ago
I used the logic by considering 32 as 2 and it's 3 times Multiple 96 as 6.
So, I used this value in this equation.
2^6+1/2^2+1.
=> 65/5 = 13.
So, I used this value in this equation.
2^6+1/2^2+1.
=> 65/5 = 13.
(5)
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)
Tanvi said:
5 years ago
Can anyone help me. What identity is used to determine whether a number is divisible or not completely?
(4)
Shyam Ingalagi said:
10 months ago
What is the exponentiation rule? Please explain to me.
(4)
Devi krishna said:
2 years 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).
(2)
SAILAJA said:
3 months ago
Anyone explain me in a simple way.
(2)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers