Prove that product of three consecutive positive integer is divisible

1.	Prove that product of three consecutive positive integer is divisible by 6
Answer» 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 6if 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 6if 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}\xa0the product of any three natural numbers is divisible by 6.

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