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 Four Perfect Logicians (Posted on 2002-09-24)
Four perfect logicians, who all knew each other from being members of the Perfect Logician's Club, sat around a table that had a dish with 11 apples in it. The chat was intense, and they ended up eating all of the apples. Everybody had at least one apple, and everyone knew that fact, and each logician knew the number of apples that he ate. They didn't know how many apples each of the other ate, though.

They agreed to ask only questions that they didn't know the answers to.

• Alonso: Did you eat more apples that I did, Bertrand?
• Bertrand: I don't know. Did you, George, eat more apples than I did?
• George: I don't know.
• Kurt: Aha!!

Kurt figured out how many apples each person ate. Can you do the same?

•  See The Solution Submitted by Chris Middlemiss Rating: 3.6154 (13 votes)

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 eliminate the options | Comment 14 of 28 |
First of all, since all four have had at least one apple, you can think of the numbers in terms of extra apples where the total number available in 7. The key is what each logician identifies when they ask a question they dont know the answer to and that each logician will know this because they are perfect.

(A-Alonso, B-Bertrand, G-George, K-Kurt)

Here are the following options of extra apples each logician had, with reasoning:

A - 1 2 3

he'd know the answer to his own question if he had ≥4, thus he must have had <4

B - 2 3

He knows what A might have

he'd know the answer to his own question if he had ≥4 thus he must have <4

G - 3

He knows what A and B might have

He'd know the answer to B's question if he had eaten ≤2, thus he must have eaten >2

He'd know the answer to B's question if he had eaten ≥4, thus he must have eaten <4

K - 1

Since there are only 7 apples unaccounted for, K must have had <2 otherwise there are not enough apples for everyone to have there minimum possible amount

Kurt knows how many each logician has. He would not have known what A and B had if he had 0, thus he must have had 1.

So there are 4 of 7 accounted for with G and K. Therefore the only possibility for A and B is that they have 1 and 2, respectively.

The count, then, is:

A - 1
B - 2
G - 3
K - 1

 Posted by Chaz on 2002-10-18 10:38:56

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