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.
Ambika said:
9 years ago
Was it possible to use power cycle of 2? Can anyone please explain how to use this?
Nishu kumari said:
9 years ago
a^n + b^n is divisible by a + b only when n is odd. Suppose 2^32 is a then (2^32) ^3 = 2^96 i.e. a^3 which is odd means a^3 + b^3 here b=1. i.e, a^3 + b^3 is divisible by a + b.
Sravanthi said:
9 years ago
I don't understand this problem. Please explain it in a simple way.
Sravanthi said:
9 years ago
I can't understand. Please explain it.
Rayanna said:
9 years ago
I can't understand please explain me easily.
Antony mary said:
9 years ago
Confused with this problem. Let me know the shortcut method?
Madhu said:
9 years ago
How you take 2^96 +1 exactly?
Savitri said:
9 years ago
How to solve 2^64, 3^432?
Siprarani tripathy said:
9 years ago
Dear sir, I can't understand. Kindly explain clearly.
Saurabh gupta said:
10 years ago
Dear sir I can't understand your explanation. Please explain clearly from basics.
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