Question

Prove that the product of any three consecutive positive intepers is divisible by 6​

Answer 1

Answer:

Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4, 6q + 5. n (n + 1) (n + 2) = 12 (3q + 1) (2q + 1) (3q + 2), Which is divisible by 6.

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Let the three consecutive positive integers be n, n+1 and n+2.

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

Therefore, 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+2=3p+1+2=3p+3=3(p+1) 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+2 is always divisible by 3 that is:

n(n+1)(n+2) is divisible by 3.

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

Therefore, n=2q or 2q+1, where q is some integer.

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

If n=2q+1, then 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+2 is always divisible by 2.

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

Hence, n(n+1)(n+2) is divisible by 6.