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If 'n' be any natural number, then by which largest number (n3 – n) is always divisible? (SSC CGL 2nd Sit. 2010)
- a)3
- b)6
- c)12
- d)18
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
If n be any natural number, then by which largest number (n3 – n...
The largest number less than (n^3 + 3n^2 + 3n) is (n+1)^3 - 1.
We can expand (n+1)^3 as n^3 + 3n^2 + 3n + 1. Subtracting 1 from this gives us n^3 + 3n^2 + 3n, which is the expression we started with. Therefore, we know that (n+1)^3 - 1 is the largest number less than (n^3 + 3n^2 + 3n).
To prove that this is the largest number, we can assume that there exists a larger number, say x, that is less than (n^3 + 3n^2 + 3n). This means that x is of the form (n^3 + 3n^2 + 3n - k), where k is a positive integer. We want to show that x cannot be greater than (n+1)^3 - 1.
Expanding (n+1)^3 - 1 gives us n^3 + 3n^2 + 3n + 1 - 1 = n^3 + 3n^2 + 3n, which is the same as our expression for x. Therefore, if x is greater than (n+1)^3 - 1, then it must also be greater than (n^3 + 3n^2 + 3n), which is a contradiction. Thus, (n+1)^3 - 1 is the largest number less than (n^3 + 3n^2 + 3n).
We can expand (n+1)^3 as n^3 + 3n^2 + 3n + 1. Subtracting 1 from this gives us n^3 + 3n^2 + 3n, which is the expression we started with. Therefore, we know that (n+1)^3 - 1 is the largest number less than (n^3 + 3n^2 + 3n).
To prove that this is the largest number, we can assume that there exists a larger number, say x, that is less than (n^3 + 3n^2 + 3n). This means that x is of the form (n^3 + 3n^2 + 3n - k), where k is a positive integer. We want to show that x cannot be greater than (n+1)^3 - 1.
Expanding (n+1)^3 - 1 gives us n^3 + 3n^2 + 3n + 1 - 1 = n^3 + 3n^2 + 3n, which is the same as our expression for x. Therefore, if x is greater than (n+1)^3 - 1, then it must also be greater than (n^3 + 3n^2 + 3n), which is a contradiction. Thus, (n+1)^3 - 1 is the largest number less than (n^3 + 3n^2 + 3n).
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Community Answer
If n be any natural number, then by which largest number (n3 – n...
n3 – n = (n2 – 1)
⇒ n (n +1) (n – 1)
For n = 2, n3 – n = 6
23 – 2 = 6
i.e. n3 – n is always divisible by 6.
⇒ n (n +1) (n – 1)
For n = 2, n3 – n = 6
23 – 2 = 6
i.e. n3 – n is always divisible by 6.
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