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:
129 comments Page 1 of 13.
VENKY said:
6 months ago
I can't understand, anyone can explain this?
(4)
Rrai said:
8 months 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.
(5)
Ronny said:
11 months ago
@Tanvi.
Here (a^3+b^3)identity is used. Please check it.
Here (a^3+b^3)identity is used. Please check it.
(1)
Nasreen.A said:
1 year ago
(X^3+1)=(x+1)(x^2-x+1).
((2^32)^3+1)=(2^32+1)((2^32)^2-2^32+1).
= 2^32+1-2^64+2^32-1
= 2^32-2^32 = 0.
((2^32)^3+1) = 0.
((2^96+1)) = 0.
((2^32)^3+1)=(2^32+1)((2^32)^2-2^32+1).
= 2^32+1-2^64+2^32-1
= 2^32-2^32 = 0.
((2^32)^3+1) = 0.
((2^96+1)) = 0.
(2)
Mile said:
1 year 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.
(2)
Tanvi said:
2 years ago
Can anyone help me. What identity is used to determine whether a number is divisible or not completely?
RITIK GUPTA said:
2 years ago
@All.
You have to remember a concept that (x+1) is always a factor of x^n + 1 if n is an odd integer.
You have to remember a concept that (x+1) is always a factor of x^n + 1 if n is an odd integer.
(1)
Anil said:
2 years ago
It's too hard to understand. Anyone, please help me to get it.
Vivek Singh said:
2 years ago
This question is to solve using the formula:- a^3+b^3 = (a+b)(a^2-ab+b^2).
Only option D satisfies the above formula.
We can write (2^96+1) as ((2^32)^3 + 1). So if we solve ((2^32)^3 + 1) using the above formula then we get:
((2^32)^3 + 1) = (2^32+1)(2^64-2^32+1).
Now, divide (2^32+1)(2^64-2^32+1) by (2^32+1). If we divide these two then (2^32+1) gets cancelled and we get the result as (2^64-2^32+1). Hence option D is correct.
Only option D satisfies the above formula.
We can write (2^96+1) as ((2^32)^3 + 1). So if we solve ((2^32)^3 + 1) using the above formula then we get:
((2^32)^3 + 1) = (2^32+1)(2^64-2^32+1).
Now, divide (2^32+1)(2^64-2^32+1) by (2^32+1). If we divide these two then (2^32+1) gets cancelled and we get the result as (2^64-2^32+1). Hence option D is correct.
(2)
Vilaz said:
2 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.
(1)
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