A physics question about electrical power

The power supply in Hong Kong is 220V,
The resistance of the 60W lamp is P=V^2/R,60=220^2/R,R=806.7ohm
The resistance of the 100W lamp is P=V^2/R,100=220^2/R,R=484ohm

When they connect in series,by V=IR,220=I(806.7+484),I=0.170A,
Since it is connect in series the current is constant.
The power of the 60W lamp in series circuit,P=I^2R,P=0.17^2*806.7,
P=23.3W
The power of the 100W lamp in series circuit,P=I^2R,P=0.17^2*484
P=14.0W

Since the power of the 60W lamp is greater than the 100W lamp,
60W lamp is brighter than a 100W lamp when they are connected in series　

say home power line voltage is 100V
P=VI=VxV/R ---> R=VxV/P
resistance of 60W lamp = 100x100/60 = 166 ohm
resistance of 100W lamp = 100x100/100 = 100 ohm

We can see here that the 60W lamp has a higher resistance than the 100W lamp.
If the 2 lamps are connected in series, same current I passes thru both lamps, but the voltage across the 60W lamp(166xI) is higher than the voltage across 100W lamp(100xI). Since the power consumed by the 60W lamp is higher (166xII) than the 100W lamp (100xII), The 60W lamp is brighter.

2008-12-06 20:58:24 補充：
Congratulation!!you are 11 minutes faster than me, ha, ha, :)