Question

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

Answer 1

hi mate

Step-by-step explanation:

Let n be any positive integer.Let n (n +1) and (n + 2) are three consecutive positive integers. Then their product is n (n +1)(n+2).

Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4, 6q + 5.

Case : 1

If n = 6q,

n (n + 1) (n + 2)  = 6q (6q + 1) (6q + 2), which is divisible by 6

Case : 2

If n = 6q + 1

n (n + 1) (n + 2)  = (6q + 1)  (6q + 2)  (6q + 3)

n (n + 1) (n + 2)  = (6q + 1) 2(3q + 2) 3(2q + 1)

n (n + 1) (n + 2)  = 6 (6q + 1) (3q + 1) (2q + 1) ,Which is divisible by 6

Case : 3

If n = 6q + 2

n (n + 1) (n + 2)  = (6q + 2) (6q + 3) (6q + 4)

n (n + 1) (n + 2)  = 2(3q + 1) 3(2q + 1) 2(3q + 2)

n (n + 1) (n + 2)  = 12 (3q + 1) (2q + 1) (3q + 2), Which is divisible by 6.

Similarly, n (n + 1) (n + 2) is divisible by 6 if n=  6q + 3 or 6q + 4, 6q + 5.  

Hence it is proved that the product of three consecutive positive integers is divisible by 6.

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Answer 2

Step-by-step explanation:

In every 3 consecutive natural or whole numbers , there will be a multiple of 3.

And in every 3 consecutive natural or whole numbers , there will be a even number.

The product of a multiple of 3 and a even number will definitely divisible by 6.

So every 3 three consecutive positive integers is divisible by 6.