Question

prove that product of three consecutive integer is divisible by 6



pls answer fast with proper steps

Answer 1

Sol;
Let us three consecutive  integers be, n, n + 1 and n + 2.
Whenever a number is divided by 3 the remainder obtained is either 0 or 1 or 2.
let 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, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So that n, n + 1 and n + 2 is always divisible by 3.
⇒ n (n + 1) (n + 2) is divisible by 3.
 
Similarly, whenever a number is divided 2 we will get the remainder is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.
If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So that n, n + 1 and n + 2 is always divisible by 2.
⇒ n (n + 1) (n + 2) is divisible by 2.
 
But n (n + 1) (n + 2) is divisible by 2 and 3.
 
∴ n (n + 1) (n + 2) is divisible by 6.

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Answer 2

let the three consecutive integers be n, (n+1) & (n+2)
whenever a number is divided by 3, the remainder obtained 0, 1 or 2
let n=3p, where p is some integer
so,
n = 3p, (3p+1), (3p+2)
if , n = 3p, then n is divisible by 3
if, n=(3p+1) , then n+2 =3p + 1 + 2 =3p + 3 = 3 ( p+1 ) , is divisible by 3
if, n=( 3p+ 2), then n+1 = 3p + 2 + 1 =3p +3 = 3 (p + 1) , is divisible by 3
so t, n, n+1 , n + 2 is always divisible by 3
=> n, n+1 , n+2 is divisible by 3

similarly whenever a number os divided by 2 the remainder will be as 0 or 1
.
. . n = 2q , 2q +1, or 2q +2 , where q ia some integer
if, n= 2q, then n and n+2 = 2q +2 = 2(q + 1), os divisible by 2
if n = 2q + 1 then n + 1 =2q + 1 +1 = 2q + 2 = 2 ( q+ 1), os divisible by 2
=> n, n+1, n+2 is always divisible by 2

but n, (n + 1 ), (n + 2) is divisible by 2 and 3
.
. . n, (n+1), (n + 2) is divisible by 6