    Question

# For any positive integer , n3−n is always divisible by ___?

A

6

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B

12

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C

15

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D

18

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Solution

## The correct option is A 6 n3–n=n(n2–1)=(n−1).n.(n+1) Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2. ⇒n=3p or 3p+1 or 3p+2, where p is some integer. If n=3p, then n is divisible by 3. Ifn=3p+1, thenn–1=3p+1–1=3p is divisible by 3. If n=3p+2, then n+1=3p+2+1=3p+3=3(p+1) is divisible by 3. So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3. ⇒n(n–1)(n+1) is divisible by 3. Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1. ⇒n=2q or 2q+1, where q is some integer. If n=2q, then n is divisible by 2. If n=2q+1,then n–1=2q+1–1=2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2. So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2. ⇒n(n–1)(n+1) is divisible by 2. Since,n(n–1)(n+1) is divisible by 2 and 3. We know that if a number is divisible by both 2 and 3 , then it is divisible by 6. ⇒n(n−1)(n+1)=n3−n is divisible by 6.  Suggest Corrections  0      Similar questions
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