![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
"You have a balance scale but no weights. You have 12 coins, identical except for one which is lighter or heavier than the others. Identify the defective coin -- find it, and whether it is lighter or heavier, in only 3 (three) weighings."
It took me a day and a half to rediscover the solution, which I'd known before, despite having a half-remembered clue bouncing around my head. So either it's really hard or I was being particularly dense. I'm not going to give any answers or even open clues, but I thought of a series of problems that might make it easier.
Note: sometimes you could get lucky and find the coin in fewer weighings, but we're looking for the number of weighings which guarantees that you'll find the coin.
*** One coin heavier, all others normal
3 coins, one of them heavier than the others. Find it in 1 weighing.
4 coins, one heavier. Find it in 2 weighings.
In fact, you can use 2 weighings for up to 9 coins, all pretty much the same way. Do so.
How high can you go to find a heavy coin in 3 weighings?
*** One coin defective (heavier or lighter), all others normal
2 coins, one defective. Understand why you can't identify it.
3 coins, call them A B and C. A is known to be normal, and either B is light or C is heavy. Identify the defective coin in 1 weighing.
3 coins. A is normal, one of the others is defective. ID it in 2 weighings.
2 normal coins, plus 2 other coins, one defective. ID it in 2 weighings. Find two different methods.
2 normal coins, plus 3, one defective. ID it in 2 weighings. I know only one solution.
4 coins, one defective; ID it in 3 weighings.
5 coins, one defective; ID it in 3 weighings.
Find an alternate method for the 4 and 5 coin cases. This will probably be very hard but more elegant when you find it. If you think your method is already elegant, kudos. This will come in handy near the end.
4 normal coins, plus 4 coins one defective. ID it in 2 weighings.
That's all the prep work; you should then be able to find 3 weighing solutions for from 6 to 12 coins. Unlike the "one coin heavier" family the method isn't just the same method scaled up. I think 6 and 7 are similar but different from 4 and 5 (either method), then 8 needs another trick, 9 needs another, 10 combines 8 and 9, 11 has another trick, and 12 combines everything. But if you did all the prep problems, you've found the tricks, just need to put them together right.
Enjoy!
It took me a day and a half to rediscover the solution, which I'd known before, despite having a half-remembered clue bouncing around my head. So either it's really hard or I was being particularly dense. I'm not going to give any answers or even open clues, but I thought of a series of problems that might make it easier.
Note: sometimes you could get lucky and find the coin in fewer weighings, but we're looking for the number of weighings which guarantees that you'll find the coin.
*** One coin heavier, all others normal
3 coins, one of them heavier than the others. Find it in 1 weighing.
4 coins, one heavier. Find it in 2 weighings.
In fact, you can use 2 weighings for up to 9 coins, all pretty much the same way. Do so.
How high can you go to find a heavy coin in 3 weighings?
*** One coin defective (heavier or lighter), all others normal
2 coins, one defective. Understand why you can't identify it.
3 coins, call them A B and C. A is known to be normal, and either B is light or C is heavy. Identify the defective coin in 1 weighing.
3 coins. A is normal, one of the others is defective. ID it in 2 weighings.
2 normal coins, plus 2 other coins, one defective. ID it in 2 weighings. Find two different methods.
2 normal coins, plus 3, one defective. ID it in 2 weighings. I know only one solution.
4 coins, one defective; ID it in 3 weighings.
5 coins, one defective; ID it in 3 weighings.
Find an alternate method for the 4 and 5 coin cases. This will probably be very hard but more elegant when you find it. If you think your method is already elegant, kudos. This will come in handy near the end.
4 normal coins, plus 4 coins one defective. ID it in 2 weighings.
That's all the prep work; you should then be able to find 3 weighing solutions for from 6 to 12 coins. Unlike the "one coin heavier" family the method isn't just the same method scaled up. I think 6 and 7 are similar but different from 4 and 5 (either method), then 8 needs another trick, 9 needs another, 10 combines 8 and 9, 11 has another trick, and 12 combines everything. But if you did all the prep problems, you've found the tricks, just need to put them together right.
Enjoy!