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 14 of 14.
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)
Donbabu said:
1 year ago
I can't understand, anyone can explain this?
(9)
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)
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)
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)
Shyam Ingalagi said:
2 months ago
What is the exponentiation rule? Please explain to me.
Anomie said:
2 months ago
Thanks for the explanation.
(2)
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