Whispers from the Teacher
The teacher whispers positive integer A to Anna, B to Bob, and C to Chris. The students don't know one another's numbers but they do know that the sum of their numbers is 14. Anna says, "I know that Bob and Chris have different numbers". Then Bob says, "I have already knew that all three of our numbers were different". Finally, Chris announces, "Now I know all three of our numbers". What is the product of A, B, and C?
The answer is 42.
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1 solution
Anna indirectly told her friends (and us) that her number is odd (since she knew that hers —that's A— is odd → 14 – A = odd → B + C = odd → B & C have opposite parity → B & C are distinct positive integers). Bob indirectly told his friends (and us) that his number is odd AND is ≥ 7 (using the same argument as we did with Anna's and the fact that his number must have been big enough for the other odd number that by now we already know is Anna's A can't match up within a total sum of 14 between the three). Chris indirectly told his friends (and us) that his number is just the right size(?) to put their two previously spoken clues to suit around the information that he has about his own even number. Listing all the possibilities, Chris's C of 6 is unique enough for him to be able to say with certainty that he knows what the other two numbers must be (if C was any other even number less than 7, whether it's 2 or 4, then Chris would have two or more possibilities of A–B pair, with 3 maybes for A + B = 12 and 2 maybes for A + B = 10).
Answer
= A × B × C
= 1 × 7 × 6
= 42