Person B eats Person C

Person A eats Person B ]]>

For the first puzzle, give out 1 apple, 1 apple, 1 apple and 1 basket with an apple. They are equally divided and one is still in the basket.

For the second puzzle, give out 2 apples, 1 apple and 1 basket with an apple. The first person has 2 apples which is not more than the others (because between the others they have 2 apples). The second has 1 apple which is not more than the others. The last has 1 apple and basket which again is not more than the others. Again there is one apple left in the basket.

For the second one, the intuitive response is Gregory’s answer – each person takes one apple and leaves one in the basket.

In the scenario in which the three people are good friends, even though they’re holding different numbers of apples, it could be said that they each have the same number of apples. Trivially, they hold three and leave the last in the basket.

In the scenario in which the three people are enemies, they will put one apple in the basket. Then they will fight over how to divide the other apples and the apples will all be discarded in the process.

The two above scenarios are clearly exhaustive, and hence the apples will be evenly divided.

Alternatively, you could leave all four apples in the basket. It can be said that one apple remains in the basket – one apple, and one apple, and one apple, and one apple. The problem does not specify that only one apple must be left behind.

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