Question

Prove that product of three consicutive positive integer is divisible by 6.​

Answer 1

Answer:

Its very simple..

1 x 2 x 3 = 6 divisible by 6

or one more..

2 x 3 x 4 = 24 divisible by 6

There are so many..

Answer 2

Answer:

Let the 3 consecutive integers be x,x+1,x+2 Product of 3 consecutive integers =x*x+1*x+2 x(x^2+3x+2)=x^3+3x^2+2x Let the 3 consecutive positive integers be n,n+1,n+2 When a number is divided by 3 the reminder obtained is either 0 or 1 or 2 Therefore 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 =>n+2=3p+1+2 =3p+3 Take out 3 as common 3(p+1) is divisible by 3 So we can say that one of the numbers among n,n+1,n+2 is always divisible by 3 n(n+1)(n+2)is divisible by 3 Similarly when a number is divided by 2 the remainder obtained is 0 or 1 Therefore n=2q or 2q+1 where q is some integer If n=2q then n+2=2q+2 Take out 2 as common then 2(q+1) are divisible by 2 If n=2q+1 then n+1=2q+1+1=2q+2 Take out 2 as common then 2(q+1) is divisible by 2 So we can say that one of the numbers among n,n+1,n+2 is always divisible by 2 n(n+1)(n+2) is divisible by 2 Hence n(n+1)(n+2) is divisible by 2 and 3 Therefore n(n+1)(n+2) is divisible by 6 I hope so it might help you.......