Question

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

Answer 1

Hi ,

Let x , ( x + 1 ) , ( x + 2 ) are three

consecutive positive Integers .

their product ( p ) = x( x + 1 )( x + 2 )

Any natural number x when divided

by 6 leaves remainders 0 ,1 , 2 , 3 , 4

and 5

Case 1 :if x = 6a

P = 6a( 6a + 1 )( 6a + 2 )

= 6m [ m = a( 6a + 1 )( 6a + 2 ) ]

= Divisible by 6

Case 2 :

If x = 6a + 1

P = ( 6a + 1 ) ( 6a +1 + 1 ) ( 6a + 1 + 2 )

= ( 6a + 1 ) ( 6a + 2 )( 6a + 3 )

= 6( 6a + 1 )( 3a + 1 )( 2a + 1 )

= 6m [ divisible by 6 ]

Case 3:

If x = 6a + 2

P = ( 6a + 2 )( 6a + 2 + 1 )( 6a + 2 + 2 )

= ( 6a + 2 )( 6a + 3 )( 6a + 4 )

= 6( 3a + 1 )( 2a + 1 )( 6a + 4 )

= 6m [ divisible by 6 ]

Similarly , we can prove

If x = 6a + 3 , 6a + 4 , 6a + 5 P is

Divisible by 6.

Hence proved .

I hope this helps you.

: )

Answer 2

Let three consecutive positive 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.

∴ 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, we can say that one of the numbers among 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, the remainder obtained 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, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.

⇒ n (n + 1) (n + 2) is divisible by 2.

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

∴ n (n + 1) (n + 2) is divisible by 6.