Question

prove that the product of three consecutive positive integers is divisible by 6

Answer 1

Here is the solution :

Theoretical Proof :

Let us assume 3 consecutive numbers be,
x, x+1, x+2,

Now,
Product of these numbers is,

(x)(x+1)(x+2) = Result,
Let result = y !

Divisibility rule of 6 : A number can be divisible by 6 if the number can be divided by both 2 and 3,

Now,
If x is odd :
x+1 is even, So the y can be divided by 2,
If x is Even :
The y can be divided by 2 again !

So,
Point (1) -------> Multiplication of 3 consecutive numbers can always be divided by 2,

Now,
If x = a,
Sum of digits = a + a + 1 + a + 2 = 3a + 3, Which can be divided by 3,

If x = a+1,
Sum of digits = a+1 + a+2 + a+3 = 3a + 6, Which can be divided by 3,

if x = a+2,
Sum of digits = a+2 + a+3 + a+4 = 3a +9, Which again can be divided by 3,

This is the last case, Since according to Euclid's division lemma remainder should always be less than divisor,

From all these, We can conclude that, Multiplication of 3 consecutive numbers can always be divided by both 2 and 3,

So, By which we can say that, Multiplication of 3 consecutive numbers can always be divided by 6,

Therefore : Hence Proved that, Multiplication of 3 consecutive numbers is always be divisible by 6,

Hope you understand, Have a Great day !
Thanking you, Bunti 360 !!