Bah, you would ask!
So here's a short version. You'll see on the schematic that these bulbs are actually on the 25V side of the house, rather than the 6V:
LastBall.jpg
A #455 bulb is a 6.5V, 0.5A bulb, so putting a 6.5V bulb in a 25V circuit will burn it out rather quickly. Some of the 25V needs to be consumed to leave 6.5V for the bulb. The amount needed is the difference: 25V-6.5V = 18.5V.
So a resistor is needed in series with the bulb to take that 18.5V. Since the bulb draws 0.5A, so will the resistor. Using Ohm's Law (V=IR) and solving for R yields V/I = R, so 18.5V/0.5A = 37 Ohms.
Now, 37 Ohms is pretty darn close to the 35 Ohms, but not exact. I'll wave my arms a bit here and say that there's another 2 Ohms of resistance in the circuit already (which can be measured), such that 35 Ohms is sufficient.
But that gives the resistance value, and a resistor has both a resistance and a power rating. With the 0.5A and the 18.5V, power in Watts is just 'volt-amps', or 18.5V*0.5A, or 9.25 W, call it 10 Watts or better.
So, a #455 bulb in series with a 35 Ohm, 10 Watt resistor.
(And no, I don't have this memorized; I have to re-derive each time, although this should do for awhile now...)