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

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Jamya Reid · Accepted Answer

Step I: To prove the divisibility by 3.When a number is divided by 3, the possible remainders are 0or1or2.∴n=3por3p+1or3p+2, where p is some integer.Case 1: Consider n=3pThen n is divisible by 3.Case 2: Consider n=3p+1Then n–1=3p+1–1⇒n-1=3p is divisible by 3.Case 3: Consider n=3p+2Then n+1=3p+2+1⇒n+1=3p+3⇒n+1=3(p+1) is divisible by 3.So, we can say that one of the numbers among n,n–1andn+1 is always divisible by 3.⇒n(n–1)(n+1) is divisible by 3.Step II: To prove the divisibility by 2.Similarly, when a number is divided by 2, the possible remainders are 0or1.∴n=2qor2q+1, where q is some integer.Case 1: Consider n=2qThen n is divisible by 2.Case 2: Consider n=2q+1Then n–1=2q+1–1n–1=2q is divisible by 2 andn+1=2q+1+1n+1=2q+2n+1=2(q+1) is divisible by 2.So, we can say that one of the numbers among n,n–1andn+1 is always divisible by 2.∴n(n–1)(n+1) is divisible by 2.Step III: To prove the divisibility by 6Since, n(n–1)(n+1) is divisible by 2and3.Therefore, as per the divisibility rule of 6, the given number is divisible by 6.Therefore,n3–n=n(n–1)(n+1) is divisible by 6.

Prove that for any positive integer n, n^3-n is divisible by 6.

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