Adapted from a Maths Battle in London:
Several children sit around a circular table, on which lies a collection of 100 chocolates. They proceed to take chocolates, after which it turns out that each child has taken either 6 fewer or 3 times as many chocolates as the child to their left. Prove that some chocolate was left on the table.
It’s possible to spend a long time pondering just how to solve this problem in its full generality. The solution below is brief, and feels natural, so you could be forgiven for believing this to be an easy puzzle; but it takes a clever insight to come up with the presented approach.
It is impossible for every child to have eaten 6 fewer chocolates than the child to their left (since, among other issues, you cannot eat a negative number of chocolates). So some child has eaten 3 times as many chocolates as the child to their left, and hence the number of chocolates eaten by this child is divisible by 3. But now, as we move anti-clockwise around the circle, at every point we must either subtract 6 or multiply by 3, both of which preserve divisibility by 3.
It follows that the number of chocolates eaten by each child is divisible by 3; thus the total number of chocolates consumed is divisible by 3, and so cannot equal 100. With this statement, the result is proved.
I am a fan of this puzzle for several reasons, not least of which because it is typically very tough getting a handle on objects arranged in a circle with dependencies between them. The solution demonstrates that it is possible to dispel the apparent complexity very swiftly.