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