Question

prove that the product of any 3 consecutive positive integer is divisible by 6​

Answer 1

Let the three consecutive positive integers be x, (x + 1), (x + 2).

For their product to be divisible by 6, it has to be divisible by 2 and 3, since 6 = 2 × 3

Divisibility test for 2 ;

When a number is divided by 2, it will either leave a remainder of 0 or 1.

So, x = 2q or 2q + 1, where q is some integer.

If x = 2q, then

x = 2q is divisible by 2 ;

x + 2 = 2q + 2 = 2(q + 1) is divisible by 2

If x = 2q + 1, then

x + 1 = 2q + 2 = 2(q + 1) is divisible by 2

So, we can say that, among x, (x + 1) and (x + 2) at least one number is always divisible by 2.

=> x(x + 1)(x + 2) is divisible by 2. ---(1)

Divisibility test for 3 ;

When a number is divided by 3, the remainder can either be 0 or 1 or 2.

So, x = 3m or 3m + 1 or 3m + 2, where m is some integer.

If x = 3m, then

x = 3m is divisible by 3

If x = 3m + 1, then

x + 2 = 3m + 3 = 3(m + 1) is divisible by 3

If x = 3m + 2, then

x + 1 = 3m + 3 = 3(m + 1) is divisible by 3

So, we can say that, among x, (x + 1) and (x + 2) at least one number is always divisible by 3.

=> x(x + 1)(x + 2) is divisible by 3. ---(2)

From statements (1) & (2), we can conclude that, x(x + 1)(x + 2) is divisible by 2 and 3. Hence, x(x + 1)(x + 2) is divisible by 6.

Hope, it'll help you.....