If you want to test which side is which, I’d make use of the fact that gold is heavier than silver. Carefully balance each coin on its edge.
But any test you have to run on every individual coin is going to be painful when you’ve got hundreds of coins.
In the countdown, the only thing I came up with was the programmer’s solution: one pile with zero coins, and another pile with the remaining coins.
I doubt that anyone will make it down to here unspoiled, but just in case, I give a hint:
What would be your first instinct if there were billions of coins?
This problem gets hard if you start thinking you’ve got to think outside the box. It’s easy as long as you realize you are confined to an incredibly small box. Any action other than dividing the coins into exactly two piles and flipping over one pile are at best useless. Once you realize that, you are only left with the question: how many coins in the pile you flip? That makes is very easy indeed.
This puzzle is based on the older water-and-wine puzzle. Imagine have two glasses, one partially filled with water and the other filled with an equal amount of wine. Some wine is poured from the second glass into the first glass, and then some water-wine mixture is poured from the first glass into the second glass until second glass is as full as it was originally. The riddle is, “Which is the greater quantity, the water in the wine glass or the wine in the water glass?”
The water glass received pure wine while the wine glass recieved a water-wine mixture, so common first impression is that the water glass has more wine. But the water glass lost some wine when some of its mixture was poured into the wine glass. Rather assigning variables to the amounts and performing algebra to track the quantities, we have a simple argument. The levels in the glasses are the save as at the beginning. Hence, an equal quantity of water must be swapped with an equal quantity of wine.
This argument works even if the two glasses started with different quantities. And we can use gold coins and silver coins in piles rather than water and wine in glasses. Suppose we have a pile of 100 gold coins and a pile of 20 silver coins. We move all the siver coins to the gold pile, mix the pile throughly, and blindly take 20 random coins to recreate the 20-coin pile. Are there more silver coins in the 100-coin pile or more gold coins in the 20-coin pile? Equal amounts, by the above argument. Next, suppose we had a way of converting each gold coin into silver while simultaneously converting each silver coin into gold, and we apply this to the 20-coin pile. The Ted-Ed puzzle does this by using two-sided coins that can be flipped over. Then our question simplifies to, “Which pile has more silver coins?” Ted-Ed further complicates the puzzle by making the solver decide on the size of the pile.
I guess I didn’t get the spoiler tags right, either.
EDIT: Thank you, Medievalist, for the advice about spoiler tags.
Edit your post, putting a carriage return after the [spoiler] tag and before the [/spoiler] tag, and it’ll fix it.
I know exactly what you mean! I’ve done the same thing.
I believe the idea here is that a fake gold coin would be primarily lead, which is softer than gold. Thus if it’s too easy to bite, it’s not gold. Even though gold is soft, I was not under the impression it was soft enough to leave a bite mark in. That being said, I’ve never bitten gold, and you apparently have, so I’ll take your word for it.
Admittedly the flaw in this plan is that if you bite a fake coin, you’re biting lead.
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