Question

Show that the product of three positive integers is divisible by $6$.

Answer 1

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

when a number is divided 2, the remainder obtained is 0 or 1.

$\therefore a=2c$ or $2c+1$, where c is some integer.

If $a=2c\Rightarrow a$ and $a+2=2c+2=2(c+1)$ are divisible by 2.

If $a=2c+1\Rightarrow a+1=2c+1+1=2c+2=2(c+1)$ is divisible by 2.

So, we can say that one of the numbers among a, a + 1 and a + 2 is always

divisible by 2.

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

Similarly,

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

$\therefore a=3b$ or $3b+1$ or $3b+2$, where p is some integer.

If $a=3b$, then a is divisible by 3.

If $a=3b+1,\Rightarrow \Rightarrow a+2=3b+1+2=3b+3=3(b+1)$ is divisible by 3.

If $a=3b+2,\Rightarrow a+1=3b+2+1=3b+3=3(b+1)$ 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.

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

Hence a (a + 1) (a + 2) is divisible by 2 and 3.

$\therefore a(a+1)(a+2)$ is divisible by 6.