Prove that the product of three consecutive even numbers is divisible by 48

Answer 1

Without loss of generality, we represent the smallest even number by 2k, where k is an integer.

The product of the three numbers is therefore:
P=2k(2k+2)(2k+4)
=8k(k+1)(k+2)
so 8|k (i.e. k is divisible by 8) for all values of k.

We will now concentrate on the part
k(k+1)(k+2) and identify two cases:
if k is odd, then k+1 is even, therefore 2|k(k+1)(k+2).
if k is even then clearly 2|k(k+1)(k+2).
Therefore we conclude that
16|p.

Examine k(k+1)(k+2) again for divisibility by 3, using 3 cases:
if k mod 3=0, clearly 3|k(k+1)(k+2).
if k mod 3=1, then for some q, k=3q+1
and k(k+1)(k+2)=(3q+1)(3q+2)(3q+3)=3(3q+1)(3q+2)(q+1)
which is again clearly divisible by 3.
if k mod 3=2, then for some q, k=3q+2
and k(k+1)(k+2)=(3q+2)(3q+3)(3q+4)=3(3q+2)(q+1)(3q+4)
which again is clearly divisible by 3.

Therefore we conclude that
k(k+1)(k+2) is divisible by 2 and 3, or
8k(k+1)(k+2) is divisible by 8*2*3=48.

Note: it is also possible to take 6 cases k=6q, 6q+1, 6q+2, .... and proceed similar to above.

Answer 2

Let 2k,2k+2 and 2k+4 be three consecutive even integers.
Then,their product
2k(2k+2)(2k+4)=8k(k+1)(k+2)
Which is divisible by 8.
But k,k+1,k+2 are three consecutive integers and the product of these three integers is divisible by 6.
So,2k(2k+2)(2k+4) is divisible by 8×6,i.e.,48.

Answer 3

X=4+10x