For any positive integer n, prove that n3 – n is divisible by 6?

For any positive integer n, prove that n3 – n is divisible by 6.

Solution: when (n3 – n) is divided by 6, then (n3 – n) = n (n -1) (n +1) is also divided by 6.

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         (n3 – n) = n (n -1) (n +1)

 If a number is divisible by both 2 and 3, then the number is divisible by 6. When a number (n) is divided by 3, to get q as quotient and r as remainder.

By Euclid’s division lemma, we get

 n = 3q + r, where 0 ≤  r < 3.

   All possible remainder = 0, 1, 2

Case I: When r = 0               

 n = 3q + r = 3q + 0 = 3q is divisible by 3.

Case II: When r = 1

 n = 3q + 1, then n-1 = 3q + 1 – 1 = 3q is divisible by 3.

Case III: When r = 2

 n = 3q + 2, then n+1 = 3q + 2 + 1 = 3q + 3 = 3(q+1) is divisible by 3.

We can say that n, (n-1) and (n+1) is divisible by 3. Then n (n-1) (n+1) is always divisible by 3.

Similarly, when a number is divided by 2 to get q as quotient and r as remainder.

n = 2q + r, where 0 ≤  r < 2.

   All possible remainder = 0, 1

Case I: when r = 0

   n = 2q + 0 = 2q is divisible by 2.

Case II: When r = 1

      n = 2q + 1

      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.

We can say that n, (n-1) and (n+1) is divisible by 2 then n (n-1) (n+1) is also divisible by 2.

Since, n (n-1) (n+1) is divisible by both 2 and 3.

Hence, n (n-1) (n+1) = n3 – n is divisible by 6.

 

SOME OTHER IMPORTANT QUESTIONS

 

  1. Use Euclid’s division lemma to show that the cube of any positive integer is of form 9m, 9m+1, and 9m+8.
  2. State Fundamental theorem of arithmetic.

For any positive integer n

Related Questions 

Prove that for any positive integer n,n3−n is divisible by 6.

How can we prove that n³-n is divisible by 6 if n is any positive integer?

If n is a positive integer, is n³-n divisible by 6?

 

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My Name is Mukesh Kumar. I am a Teacher, Blogger, Educational Content Writer, and Founder of CBSE Digital Education.

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