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 5 of 14.
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.
Nasreen.A said:
4 years 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.
(13)
Arjhun said:
7 years ago
@All. Please refer the following.
Formula : (a^3+b^3)=(a+b)(a^2-ab+b^2).
Substitute : (x^3+1^3)=(x+1)(x^2-x*(1)+1^2).
Formula : (a^3+b^3)=(a+b)(a^2-ab+b^2).
Substitute : (x^3+1^3)=(x+1)(x^2-x*(1)+1^2).
(1)
Deependra said:
1 decade ago
It is very easy question guys. Try again to understand the logic of mathematics which applied in this problem.
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)
Aman said:
10 years ago
How we can solve 2^64, 3^432 means?
What is the best way to solve high exponential/powers of a number?
What is the best way to solve high exponential/powers of a number?
Tanvi said:
4 years ago
Can anyone help me. What identity is used to determine whether a number is divisible or not completely?
(4)
RITIK GUPTA said:
4 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.
(10)
Akriti said:
7 years ago
I can't understand last step how it will completely divisible?
Please explain again in simple step.
Please explain again in simple step.
Pari said:
10 years ago
Please give a simple method with which one can get answer in like just 40 secs. Please do help.
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