Question

For any positive integer n,use E.D.L to prove that n^3-n is divisible by 6

Answer 1

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Answer 2

||✪✪ QUESTION ✪✪||

For any positive integer n,use E.D.L to prove that n^3-n is divisible by 6 ?

|| ✰✰ ANSWER ✰✰ ||

n³-n can be written as = n(n²-n) = n(n-1)(n+1)

we know that now, A number is divisible by 6 if it is divisible by both 2 and 3.

Now, when any number is divided by 3 the remainder is 0, 1, 2..

So n can be,

So n can be,(Number=Divisor × Quotient + Remainder)

n = 3p or 3p + 1 or 3p + 2 (where p is any integer.)

n³ - n = n (n - 1) (n + 1)

→ 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.

=> n (n – 1) (n + 1) is divisible by 3.

So we can conclude, any one of the positive integers, n, n-1, n+1 is divisible by 3.

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Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.

So, 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

→ n + 1 = 2q + 1 + 1 = 2q + 2 = 2(q+ 1) is divisible by 2.

=> n (n – 1) (n + 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.

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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.