i) The fake coin

A rather intriguing riddle which pops up in different mathematics-related interviews in a more simple form is the following:

You have twelve coins. You know that one is false. The only thing that distinguishes the fake coin from the real one is that its weight is imperceptibly different, but not necessarily lighter or heavier. You have a perfectly balanced scale, but it only tells you which side weighs more than the other side. What is the smallest number of times you must use the scale in order to always find the fake coin?

ii) The 100 coins

This one is easier, but has a creative solution:

There are 10 sets of 10 coins. You know how much the coins should weigh. You know that all the coins in one set of ten are exactly a tenth of a gramme off, making the entire set of ten coins a gramme off. You also know that all the other coins weight the correct amount. You are allowed to use an extremely accurate digital weighing machine only once. How do you determine which set of 10 coins is faulty?

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I’ve got to admit… These riddles are really interesting. Especially because they can be solved by anyone, regardless of age. Here’s my guess at the answers:

For the 1st one: i think the minimum number is 3. I won’t say the method i used in respect to others that want to solve the problem by themselves.

For the 2nd one (which i found to be harder than the first, actually): we take 1 coin from the first set, 2 from the 2nd, 3 from the 3rd… and so on. In the end, we weigh all of the coins we selected. if it’s one tenth of a gramme less than it’s supposed to be, then the first set is the one we’re looking for. If it’s two tenths of a gramme less, then it’s the second set. And so on.

I don’t know if I’m right, but this is how I would solve them.

What do you guys think?

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