product=hcfxlcm

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)