Question

Prove that the product of three consecutive positive integer is divisible by $6$.

Answer 1

STEP 1 : Assumption

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

STEP 2 : Proving that the product of three consecutive positive integers is divisible by $2$

We know that if $n$ is divided by $2$ then it is of the form $2q$ or $2q+1$. So, we have the following cases:

Case I When $n=2q$: In this case

$n=2q$ and $n+2=2q+2=2\left(q+1\right)$ is divisible by $2$ but

$n+1$ is not divisible by $2$ in this case.

Case II When $n=2q+1$: In this case

$n+1=\left(2q+1\right)+1=2q+2=2\left(q+1\right)$ is divisible by $2$ but

$n$ and $n+2$ are not divisible by $2$ in this case.

Since, $n,n+1$ and $n+2$, one out of these are always divisible by $2$.

Therefore, the product of three consecutive positive integers is always divisible by $2$.

STEP 3 : Proving that the product of three consecutive positive integers is divisible by $3$

We know that if $n$ is divided by $3$ then it is of the form $3q,3q+1$ or $3q+2$. So, we have the following cases:

Case I When $n=3q$: In this case

$n$ is divisible by $3$ but $n+1$ and $n+2$ are not divisible by $3$ in this case.

Case II When $n=3q+1$: In this case

$n+2=\left(3q+1\right)+2=3q+3=3\left(q+1\right)$ is divisible by $3$ but

$n$ and $n+1$ are not divisible by $3$ in this case.

Case III When $n=3q+2$: In this case

$n+1=\left(3q+2\right)+1=3q+3=3\left(q+1\right)$ is divisible by $3$ but

$n$ and $n+2$ are not divisible by $3$ in this case.

Since, $n,n+1$ and $n+2$, one out of these are always divisible by $3$.

Therefore, the product of three consecutive positive integers is always divisible by $3$.

STEP 4 : Proving that the product of three consecutive positive integer is divisible by $6$

We have seen that the product of three consecutive positive integers is always divisible by $2$ and $3$ both.

Therefore by divisibility rule, the product of three consecutive positive integers is divisible by $6$.

Hence Proved.