Product=HCFxLCM
hcf (a,b)lcm (a,b) = ab
Proof:
Let d=hcf (a,b), so a=dr, b=ds
If m=ab/d, m=as=br, which is therefore a common multiple
Let c be any common multiple, c=au=bv:
c/m=cd/ab=c(ax+by)/ab=(c/b)x + (c/a)y=vx+uy
i.e. m|c, so m<=c, so m=lcm (a,b)