Here is a new logic puzzle from Presh Talwalkar. A teacher has a puzzle for three students: Alicia, Brittany and Cheryl. The teacher puts the following words on the board: CAT, DOG, HAS, MAX, DIM, TAG.
The teacher then hands each student a slip of paper with a single letter written on it, explaining that the three letters together spell one of the words on the board. The teacher asks Alicia if she knows what the word is, and she answers Yes. Then the teacher asks Brittany if she knows what the word is, and she thinks for a moment and also answers Yes. Finally she asks Cheryl if she knows what the word is, and she also thinks for a moment and says Yes. What is the word?
If you want a hint, click on See Explanation below and read as far as you want to. If you think you have figured out the answer, click See Answer.
See ExplanationNow think about Brittany. She also answered Yes, after looking at her own letter and thinking about Alicia’s answer. So Brittany’s letter, combined with one of Alicia’s letters, must narrow it down to one word. This eliminates two possibilities:
- The word cannot be MAX. If Alicia had X, and Brittany had M, Brittany would not be sure what the word is. From Brittany’s point of view, all she would know is that she has M, and Alicia has a unique letter. But that could be X or I, and the word could be MAX or DIM.
- The word cannot be DIM either. If Brittany had M, the same logic that eliminated MAX would apply: the word could be MAX or DIM and Brittany would not be able to answer Yes. If Brittany had D, from her point of view Alicia could have O or I, the word could be DOG or DIM, and she could not answer Yes.
All this can be figured out by Cheryl without even looking at her own letter. She now knows that the word must be CAT, DOG or HAS. She knows that Alicia’s letter is C, O, H or S. Now what is Cheryl’s letter, and what is the word?
- We can eliminate HAS. If Alicia had H and Brittany had S, or vice versa, then Cheryl would have to have A. But if Cheryl had A, she would not be able to answer Yes, because the word could still be CAT.
- We can eliminate CAT by the same logic. If Alicia had C and Brittany had T, then Cheryl would have to have A, but she would not know whether the word was CAT or HAS.
Therefore…
See Answer
Wrong Answer. Answer would be CAT. As in case HAS was an option, Bernard wouldn’t have had to think when he got S.
Then Cheryl would have easily said DOG.
But watching Bernard think, Cheryl knows that it’s not HAS. So with her letter A she knows that the word is CAT.
Sorry but I find the answer to be has due to it is the only word with two unique letters h and s. If we start with eliminating all words with A you are left with DOG and DIM. But you can’t determine from these due to both have D. So we now eliminate DOG and DIM and are left with CAT, HAS, MAX, TAG. Out of the words left, only HAS has 2 unique letters H and S. Thus the word is HAS!! In my opinion this is the most full proof solution mathamatically!
HAS does in fact have two unique letters, but let’s consider the scenario of person 1, 2, and 3 having the letters C, T, and A, respectively. Person 1 has C, so they know the word is CAT. Person 2 has T, so they know the word can be either CAT or TAG. But If the word were TAG, person 1 would not have claimed to know the word, since person 1 needs to hold a unique letter to know. Therefore, person 2 having a T knows the word is CAT. Then, person 3 sees that the first two people knew the word. Person 3 has an A, but note how this is the exact same situation as the HAS scenario. So person 3 cannot know if the first two people held an H and S, or if they held a C and T. That’s why HAS cannot be the word.
Following this logic, one can deduce that the only possible working scenario is DOG, with person 1 having the O, person 2 having the G, and person 3 having the D.