show that the product of three consecutive even integer is a multiple of 48
(2n)(2n+2)(2n+4)=8n(n+1)(n+2)
Now,
if n = 3k, n+1 is even, and we have 8n*3k*2m(n+2)=48nkm(n+2)
if n=3k+1, then n+2=3k+3=3(k+1) and n+1 is even.
if n=3k+2, then n+1=3(k+1) and either n+1 or n+2 is even