For any positive integer n , prove that n3−n divisible by 6
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n3−n=(n2−1)=n(n−1)(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. 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 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. ∴ n(n−1)(n+1)=n3–n is divisible by 6. (If a number is divisible by both 2 and 3, then it is divisible by 6) |