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 7 of 14.
Varun said:
10 years ago
I am not understand. Please explain again.
Divya said:
10 years ago
Multiplication of 36*2 gives correct solution. As friends said use hit and trial method.
Vaithi said:
10 years ago
I am not clear about the answer. Please help me.
Harshada said:
10 years ago
Problem is that how we can directly get this option i.e. (2^96+1).
Rajendra said:
10 years ago
Take 2^32 (2 power 32 as x).
Given 2^32 + 1.
So x + 1.
Formula a^3 + b^3 (a cube + b cube) = (a+b) (a^2 - a.b + b^2).
So take x as a, 1 as b. So (x+1) (x^2 - x.1 + 1^2).
1^2 (1 square = 1) and x.1 = x......
So they asked divisible by x+1.
So (x + 1)(x^2 - x + 1)/(x + 1).
So we get (x^2 - x + 1). Because (x + 1) in numerator and denominator gets cancelled.
Given 2^32 + 1.
So x + 1.
Formula a^3 + b^3 (a cube + b cube) = (a+b) (a^2 - a.b + b^2).
So take x as a, 1 as b. So (x+1) (x^2 - x.1 + 1^2).
1^2 (1 square = 1) and x.1 = x......
So they asked divisible by x+1.
So (x + 1)(x^2 - x + 1)/(x + 1).
So we get (x^2 - x + 1). Because (x + 1) in numerator and denominator gets cancelled.
Tarun said:
10 years ago
We know that (x^n + 1) is completely divisible by (x+1), when n is natural number.
So, (2^96 + 1) = ((2^32) ^2 +1).
By using given statement it is divisible by (x+1).
= (2^32 + 1).
So, (2^96 + 1) = ((2^32) ^2 +1).
By using given statement it is divisible by (x+1).
= (2^32 + 1).
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?
Saurabh gupta said:
10 years ago
Dear sir I can't understand your explanation. Please explain clearly from basics.
Siprarani tripathy said:
9 years ago
Dear sir, I can't understand. Kindly explain clearly.
Savitri said:
9 years ago
How to solve 2^64, 3^432?
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