Question

2. Prove that product of three consecutive positive integers is divisible by 6.

Answer 1

(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)