Let \(n\) be any positive integer.
Thus, the three consecutive positive integers are \(n, n-1\) and \(n+2\).
We know that any positive integer can be of the form \(6 q\), or \(6 q+1\), or \(6 q+2\), or \(6 q+3\), or \(6 q+4\), or \(6 q+5\). (From Euclid's divislon lemma for \(b=6\) )
So,
For \(n=6 q\),
$$
\begin{aligned}
&=n(n+1)(n-2)=6 q(6 q+1)(6 q+2) \\
&=n(n+1)(n-2)=6[q(6 q+1)(6 q+2)] \\
&=n(n+1)(n-2)=6 m \text {, which is divisible by 6. [m= } q(6 q+1)(6 q+2)]
\end{aligned}
$$
For \(n=6 q+1\),
$$
\begin{aligned}
&=n(n+1)(n-2)=(6 q+1)(6 q+2)(6 q-3) \\
&=n(n+1)(n-2)=6[(6 q+1)(3 q+1)(2 q+1)] \\
&=n(n+1)(n-2)=6 m \text {, which is divisible by } 6 .[m=(6 q+1)(3 q+1)(2 q+1)]
\end{aligned}
$$
For \(n=6 q+2\),
$$
\begin{aligned}
&=n(n+1)(n-2)=(6 q+2)(6 q+3)(6 q+4) \\
&=n(n+1)(n-2)=6[(3 q+1)(2 q+1)(6 q+4)] \\
&=n(n+1)(n+2)=6 m \text {, which is divisible by } 6 .[m=(3 q+1)(2 q+1)(6 q+4)]
\end{aligned}
$$
For \(n=6 q+3\),
$$
\begin{aligned}
&=n(n+1)(n-2)=(6 q+3)(6 q+4)(6 q-5) \\
&=n(n+1)(n-2)=6[(2 q+1)(3 q+2)(6 q+5)] \\
&=n(n+1)(n-2)=6 m \text {, which is divisible by } 6 .[m=(2 q+1)(3 q+2)(6 q+5)]
\end{aligned}
$$
For \(n=6 q+4\),
$$
\begin{aligned}
&=n(n+1)(n-2)=(6 q+4)(6 q+5)(6 q-6) \\
&=n(n+1)(n-2)=6[(3 q+2)(3 q+1)(2 q+2)] \\
&=n(n+1)(n+2)=6 m \text {, which is divisible by 6. [m= (3q+2)(3q+1)(2q+2)] } \\
&\text { For } n=6 q+5 \text {, } \\
&=n(n+1)(n-2)=(6 q+5)(6 q+6)(6 q-7) \\
&=n(n+1)(n-2)=6[(6 q+5)(q-1)(6 q+7)] \\
&=n(n+1)(n-2)=6 m \text {, which is divisible by } 6 .[m=(6 q+5)(q+1)(6 q+7)]
\end{aligned}
$$
Hence, the product of three consecutive positive Integers is divisible by \(6 .\)
