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Hint: In this question the 3 consecutive integers can be taken as n, n+1, n+2 where n belongs to the set of positive real numbers. Now the divisibility by 6 means the number should be divisible by 2 as well as 3. So take two cases first divisibility by 3 and write three consecutive integers which are 3 divisible by 3, and then do the same for divisibility by 2.

Complete step-by-step answer:

Let three consecutive positive integers are (n, n + 1, n + 2) where n is any positive real number.

Proof:So the product of three consecutive positive integers are

$ \Rightarrow n\left( {n + 1} \right)\left( {n + 2} \right)$

According to the divisibility rule of 6 if a number is divisible 2 and 3 then it is also divisible by 6.

As we know if a number is divisible by 3 then the remainder will be (0, 1 and 2).

Therefore n should be in the form of 3k or 3k + 1 or 3k + 2, where k is some integer.

So if n = 3k, then n is divisible by 3.

If n = 3k + 1, then n + 2 = 3k + 1 + 2 = 3k + 3 = 3(k + 1) which is again divisible by 3.

If n = 3k + 2, then n + 1 = 3k + 2 + 1 = 3k + 3 = 3(k + 1) which is again divisible by 3.

So we can say that one of the numbers among (n, n + 1 and n + 2) is always divisible by 3.

Therefore the product of numbers $n\left( {n + 1} \right)\left( {n + 2} \right)$ is always divisible by 3.

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

Therefore n should be in the form of 2m or 2m + 1, where m is some integer.

So, if n = 2m, then n is divisible by 2.

If n = 2m + 1, then n + 1 = 2m + 1 + 1 = 2m + 2 = 2(m + 1) which is again divisible by 2.

So we can say that one of the numbers among (n, n + 1 and n + 2) is always divisible by 2.

Therefore the product of numbers $n\left( {n + 1} \right)\left( {n + 2} \right)$ is always divisible by 2.

Since, $n\left( {n + 1} \right)\left( {n + 2} \right)$ is divisible by 2 and 3.

Hence, $n\left( {n + 1} \right)\left( {n + 2} \right)$ is divisible by 6.

Hence Proved.

Note: Just like the concept that for a number to be divisible by 6 there are some conditions we have these tricks for other numbers as well. For divisible by 2 the last digit must be 2, 4, 6 or 8. For it to be divisible by 3, the sum of all the digits must be divisible by 3. For a number to be divisible by 5, it must have 5 or 0 in the last digit and for 10 it must have a 0 in as its last digit.

Complete step-by-step answer:

Let three consecutive positive integers are (n, n + 1, n + 2) where n is any positive real number.

Proof:So the product of three consecutive positive integers are

$ \Rightarrow n\left( {n + 1} \right)\left( {n + 2} \right)$

According to the divisibility rule of 6 if a number is divisible 2 and 3 then it is also divisible by 6.

As we know if a number is divisible by 3 then the remainder will be (0, 1 and 2).

Therefore n should be in the form of 3k or 3k + 1 or 3k + 2, where k is some integer.

So if n = 3k, then n is divisible by 3.

If n = 3k + 1, then n + 2 = 3k + 1 + 2 = 3k + 3 = 3(k + 1) which is again divisible by 3.

If n = 3k + 2, then n + 1 = 3k + 2 + 1 = 3k + 3 = 3(k + 1) which is again divisible by 3.

So we can say that one of the numbers among (n, n + 1 and n + 2) is always divisible by 3.

Therefore the product of numbers $n\left( {n + 1} \right)\left( {n + 2} \right)$ is always divisible by 3.

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

Therefore n should be in the form of 2m or 2m + 1, where m is some integer.

So, if n = 2m, then n is divisible by 2.

If n = 2m + 1, then n + 1 = 2m + 1 + 1 = 2m + 2 = 2(m + 1) which is again divisible by 2.

So we can say that one of the numbers among (n, n + 1 and n + 2) is always divisible by 2.

Therefore the product of numbers $n\left( {n + 1} \right)\left( {n + 2} \right)$ is always divisible by 2.

Since, $n\left( {n + 1} \right)\left( {n + 2} \right)$ is divisible by 2 and 3.

Hence, $n\left( {n + 1} \right)\left( {n + 2} \right)$ is divisible by 6.

Hence Proved.

Note: Just like the concept that for a number to be divisible by 6 there are some conditions we have these tricks for other numbers as well. For divisible by 2 the last digit must be 2, 4, 6 or 8. For it to be divisible by 3, the sum of all the digits must be divisible by 3. For a number to be divisible by 5, it must have 5 or 0 in the last digit and for 10 it must have a 0 in as its last digit.

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