Math, asked by llamadrama69, 1 year ago

prove that the product of any 3 consecutive positive integers is divisible by 6​

Answers

Answered by Yashraj1234
2

Answer:Let us three consecutive  integers be, x, x + 1 and x + 2 respectively.


[• Whenever a number is divided by 3 the remainder obtained is either 0 or 1 or 2. ( it's proved )]


Let x = 3p ...................................(i)

where p belongs to an integer and does not equal to zero ( 0 ).


=> x is divisible by 3.


If x = 3p + 1

where p belongs to an integer and does not equal to zero ( 0 ).


then x + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) 

which is divisible by 3.


If x = 3p + 2

where p belongs to an integer and does not equal to zero ( 0 ).


then x + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) 

which is divisible by 3.


So that x, x + 1 and x + 2 is always divisible by 3.

=>  x (x + 1) (x + 2) is divisible by 3.


Similarly, whenever a number is divided 2 we will get the remainder is 0 or 1.


Let x = 2q

where q belongs to an integer and does not equal to zero ( 0 ).


=> x is divisible by 2


If x = 2q + 1

where q belongs to an integer and does not equal to zero ( 0 ).


then x + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1)

which is divisible by 2.

 


So that x, x + 1 and x + 2 is always divisible by 2.

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




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

 

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


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