Kinesys lands crooked on the rod of a giant abacus and slides down, bead by bead, to the end of an Abakos alley lit only by a few candles. Sitting on a crate, an old Abakian vagabond counts and recounts a little pile of stubs. “From four I melt a new one,” he explains, “and around here, not a single stub goes to waste.”
He wants to light seven candles for the long Abakian night, but the planet’s law is strict: you must collect the bare minimum, not one stub more. Liz, who was only looking for an outlet to recharge the ship, finds herself counting bits of wax.
Dord watches him work by trial and error and suffers in silence. M00N is touched and lights the way with her glowing eye.
The Riddle
A vagabond collects candle stubs: by putting 4 of them together, he makes an (almost) new candle. The (almost) new candles, once burned, also leave a stub. If he manages to light 7 (almost) new candles, what is the minimum number of stubs he must have found, and how many does he have left at the end?
Hint
Remember that the candles made from stubs also leave a stub once they burn down. Try simulating the process starting from a test number, or work backward from the seventh candle.
Solution
The vagabond must have collected 22 stubs, and at the end he has 1 left.
With 22 stubs he makes 5 candles, with 2 stubs left over. Once the first 5 candles burn down, he gets 5 new stubs: together with the 2 left over that makes 7, from which he makes another candle, with 3 left over. When this one (the sixth) burns down, he has 4 stubs, from which he makes the seventh candle. In the end, after the seventh burns down, he has 1 stub left.
With 21 stubs, on the other hand, he would only get to 6 candles: so 22 is the minimum.
The vagabond goes back to hunting for stubs, but only twenty-two: the law of Abakos allows no waste.