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 1 of 14.
Anomie said:
2 months ago
Thanks for the explanation.
(2)
Shyam Ingalagi said:
2 months ago
What is the exponentiation rule? Please explain to me.
Skar said:
11 months ago
@All
Here's my explanation.
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:
(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!
Here's my explanation.
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:
(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!
(20)
Aviii said:
11 months 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!
(4)
Aviii said:
11 months ago
Let, (2^32) = (2^n)
If ( 2^n + 1 ) is divisible by a whole number, then higher powers of that same base will also be divisible by that same number.
(2^nm + 1 ) will also be divisible by the same number, where "m" is an integer.
Since, (2^96 + 1 ) = ( 2^32*3 + 1)
Therefore, (2^96 + 1) is divisible by the same number.
"Option D " is correct!
If ( 2^n + 1 ) is divisible by a whole number, then higher powers of that same base will also be divisible by that same number.
(2^nm + 1 ) will also be divisible by the same number, where "m" is an integer.
Since, (2^96 + 1 ) = ( 2^32*3 + 1)
Therefore, (2^96 + 1) is divisible by the same number.
"Option D " is correct!
(6)
Donbabu said:
1 year ago
I can't understand, anyone can explain this?
(9)
Selvakumar O S said:
1 year 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)
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)
VENKY said:
3 years ago
I can't understand, anyone can explain this?
(60)
Rrai said:
3 years ago
1&2 option will be eliminated as they are less than the denominator, and in 3rd option even * odd = even.
And the denominator is even + odd = odd, and even no. Can't divide the odd so it will also be eliminated.
And the denominator is even + odd = odd, and even no. Can't divide the odd so it will also be eliminated.
(74)
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