Prove that product of any three …

Let three consecutive numbers be x, (x + 1) and (x + 2)
Let x = 6q + r 0 {tex}\leq r < 6{/tex}
{tex}\therefore x = 6 q , 6 q + 1,6 q + 2,6 q + 3,6 q + 4,6 q + 5{/tex}
{tex}\text { product of } x ( x + 1 ) ( x + 2 ) = 6 q ( 6 q + 1 ) ( 6 q + 2 ){/tex}
if x = 6q then which is divisible by 6
{tex}\text { if } x = 6 q + 1{/tex}
{tex}= ( 6 q + 1 ) ( 6 q + 2 ) ( 6 q + 3 ){/tex}
{tex}= 2 ( 3 q + 1 ) .3 ( 2 q + 1 ) ( 6 q + 1 ){/tex}
{tex}= 6 ( 3 q + 1 ) \cdot ( 2 q + 1 ) ( 6 q + 1 ){/tex}
which is divisible by 6
if x = 6q + 2
{tex}= ( 6 q + 2 ) ( 6 q + 3 ) ( 6 q + 4 ){/tex}
{tex}= 3 ( 2 q + 1 ) .2 ( 3 q + 1 ) ( 6 q + 4 ){/tex}
{tex}= 6 ( 2 q + 1 ) \cdot ( 3 q + 1 ) ( 6 q + 1 ){/tex}
Which is divisible by 6
{tex}\text { if } x = 6 q + 3{/tex}
{tex}= ( 6 q + 3 ) ( 6 q + 4 ) ( 6 q + 5 ){/tex}
{tex}= 6 ( 2 q + 1 ) ( 3 q + 2 ) ( 6 q + 5 ){/tex}
which is divisible by 6
{tex}\text { if } x = 6 q + 4{/tex}
{tex}= ( 6 q + 4 ) ( 6 q + 5 ) ( 6 q + 6 ){/tex}
{tex}= 6 ( 6 q + 4 ) ( 6 q + 5 ) ( q + 1 ){/tex}
which is divisible by 6
if x = 6q + 5
{tex}= ( 6 q + 5 ) ( 6 q + 6 ) ( 6 q + 7 ){/tex}
{tex}= 6 ( 6 q + 5 ) ( q + 1 ) ( 6 q + 7 ){/tex}
which is divisible by 6
{tex}\therefore {/tex} the product of any three natural numbers is divisible by 6.