Question

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

Answer 1

For every natural number k greater than or equal to 1, there exists such a natural number a that:

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

This can be proved by induction, for k=1:

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

For k=n+1 the theorem states:

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

To use the induction hypothesis, expand the multiplication using the last bracket:

(n+1)(n+2)(n+3)=
(n+1)(n+2)n+(n+1)(n+2)3=
n(n+1)(n+2)+3(n+1)(n+2)

By the induction hypothesis, n(n+1)(n+2) is equal to 6b for some natural number b. (n+1)(n+2) is a product of an odd number and even number, hence a number of the form 2c for some other natural number c:

n(n+1)(n+2)=6b
(n+1)(n+2)=2c

Now just substitute this into the previous expansion:

(n+1)(n+2)(n+3)=
n(n+1)(n+2)+3(n+1)(n+2)=
6b+3*2c=
6b+6c=
6(b+c)
a=b+c

So, finally:

(n+1)(n+2)(n+3) = 6[(n(n+1)(n+2))/6+((n+1)(n+2))/2]

With (n(n+1)(n+2))/6+((n+1)(n+2))/2 being a natural number, which was to be proven. This proof also illustrates the fact that, having a triple of consecutive numbers and their product, adding to this product the product of the last two numbers and the number 3 results in the product of the next three consecutive numbers, e.g.:

1*2*3=6
2*3*4=24=6+3*(2*3)
3*4*5=60=24+3*(3*4)
4*5*6=120=60+3*(4*5)
5*6*7=210=120+3*(5*6)
...