Question

The product of $3$ consecutive numbers, when the first number is even is divisible by ___

• A. 24
• B. 12
• C. 6
• D. all of these

Answer 1

Answer: (C)

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

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

$\Rightarrow$ $n=3p$ 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 $3$.

$\Rightarrow n(n+1)(n+2)$ is divisible by $3$.

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

$\therefore n=2q$ or $2q+1$, where $q$ is some integer.

If $n=2q$, then $n$ and $n+2=2q+2=2(q+1)$ are 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 $2$.

$\Rightarrow n(n+1)(n+2)$ is divisible by $2$.

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

$\therefore n(n+1)(n+2)$ is divisible by $6$. Any $3$ consecutive numbers, where the first number is even will be a multiple of $2$ and $3$

i.e., $6$.

Hence, Option D is correct.