Logic Loop (10/21-11/3)
Three perfectly logical logicians Kevin, Kuzma, Kyrie are stuck in an infinite loop, a trap set up by King HeeHee. King HeeHee gave each of them a hat with a positive integer (not necessarily distinct) on it so that each logicians cannot know what their number is. However, they can see each other’s numbers.
King HeeHee feeling generous, also tells them that one of the three numbers is the sum of the other two. He tells them that to escape, each of them have to guess the number on top of their heads.
The following conversation proceeds.
Kevin: I don’t know what my number is.
Kuzma: I don’t know what my number is.
Kyrie: I don’t know what my number is.
Kevin: I know what my number is. It’s 50!
What does Kuzma and Kyrie have to guess their number as to escape this loop? (If there are multiple answers, put all of them).
Solution:
Let Kevin, Kuzma and Kyrie’s number be A, B, C respectively. From the fact that two of the numbers sum to the other, \(A = B + C\) or \(A = |B – C|\) and vice versa. Also, we know that \(A, B, C > 0\). Therefore,
From the fact that Kevin doesn’t know his number, we deduce that \(B \neq C\).
From the fact that Kuzma doesn’t know his number, we can deduce that \(A \neq C\) and \(A \neq 2C\).
From the fact that Kyrie doesn’t know his number, we can deduce that \(A \neq B\), \(A \neq 2B\), \(B \neq 2A\), \(2B \neq 3A\).
Now, let’s think about this from Kevin’s perspective. There are two cases.
1. \(|B – C| = 50\)
In this case, Kevin has to have a reason to think \(B + C\) cannot be possible for his number. The cases are: \(B + C = B, B + C = 2B, 2(B + C) = B, 2B = 3(B + C)\). Since otherwise, the three people would’ve guessed the answer in the previous turns. However, all of these are impossible since \(B + C = B, 2(B + C) = B, 2B = 3(B + C)\) is not possible with positive integers and if \(B + C = 2B\), Kevin would’ve realized in the first turn.
2. \(B + C = 50\)
In this case, Kevin has to have a reason to think \(|B – C|\) cannot be possible for his number. The cases are: \(|B – C| = B, |B – C| = 2B, 2|B – C| = B, 2B = 3|B – C|\)
If \(|B-C| = B, B = C\) so Kevin would’ve realized in the first turn.
If \(|B – C| = 2B\), \(C = 3B\) and we cannot have \(B + C = 50\).
If \(2|B – C| = B\), \(B = 2C\) or \(3B = 2C\). In the former, we cannot add up to 50 and the latter, we get (B, C) = (20, 30).
If \(2B = 3|B – C|\), \(B = 3C\) or \(3C = 5B\). Neither of these cases can we add up to 50.
Therefore, the only possible solution is
\[(A, B, C) = (50, 20, 30)\]