Which of the following divides (n3 – n)? A) 7 B) 2 C) 3 D) 5

1.	Which of the following divides (n3 – n)? A) 7 B) 2 C) 3 D) 5
Answer» Correct option is (C) 3 \(\because\) \(n^3-n=n(n^2-1)\) = n (n-1) (n+1) Whenever a number is divided by 3, the obtained remainder is either 0 or 1 or 2. \(\therefore\) For any number n, n = 3p or 3p+1 or 3p+2, where p is same integer. If n = 3p, then n is divisible by 3. If n = 3p+1 then n-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 for any number n, one of the numbers among n, n-1 and n+1 is always divisible by 3. \(\therefore\) n (n-1) (n+1) is always divisible by 3 for any n. \(\Rightarrow\) \(n^3-n\) is always divisible by 3. Correct option is C) 3

Answer»

Correct option is (C) 3

\(\because\) \(n^3-n=n(n^2-1)\) = n (n-1) (n+1)

Whenever a number is divided by 3, the obtained remainder is either 0 or 1 or 2.

\(\therefore\) For any number n,

n = 3p or 3p+1 or 3p+2, where p is same integer.

If n = 3p, then n is divisible by 3.

If n = 3p+1 then n-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 for any number n, one of the numbers among n, n-1 and n+1 is always divisible by 3.

\(\therefore\) n (n-1) (n+1) is always divisible by 3 for any n.

\(\Rightarrow\) \(n^3-n\) is always divisible by 3.

Correct option is C) 3

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