Brainteaser: Forehead Numbers

There are 3 people placed in a room.  They all have perfect logic.  The 3 people are told by a host that a number has been written on each of their foreheads.  Each of the 3 numbers are unique, they are all positive, and they relate to each other such that A + B = C (i.e. one is the sum of the other two).  In the room, each person can only see the other two people’s numbers, as they cannot see their own foreheads.

Suppose you are one of the 3 and you see one person with “20” on their forehead and the other person with “30.”  The host asks you, then the person with “20,” and then the person with “30” what number is on their heads and all 3 say that they don’t know.  The host then asks, again, you, then the person with “20,” and then the person with “30” what number is on their head, and all 3 answer correctly.  How does this happen?

Answer:

The key to this brainteaser is to calculate the logic of each person’s point of view, i.e. as the first person, while you’re figuring out your own logic, put yourself in each the shoes of each of the two other players.  The annoying part of solving this brainteaser, then, is having to keep track of 3 different points of view.

“First” person: If you see “20” and “30,” that means you are either 10 or 50.  So you don’t know what’s on your forehead among these two numbers.

What the “First” person figures about the “20” person: They see either A.) “30” and “10” or V.) “30” and “50.”  In case A.) they are either 20 or 40.  In case B.) they are either 20 or 80.  Either way, they don’t know.

What the “First” person figures about the “30” person: They see either A.) “20” and “10” or B.) “20” and “50.”  In case A.) they must be 30 because they cannot be 10 since in this case the “First” person is already “10” (the three numbers are unique).  So the key here is that if you see one person with number “x” and another person with number “2x,” you know you cannot also have “x” on your forehead.  You must be “3x.”  So again, in this case, the “30” person would know they have 30 on their head.  In case B.) the “30” person has either 30 or 70, and so they wouldn’t know.

Since after the first round, everyone answered that they did not know, that means that we cannot have the “30” person’s case A.), which is that he sees “20” and “10.”  In other words, the “First” person cannot have 10.  The “First” has 50 on his forehead.  So when the host asks the “First” person the second time, they will answer 50.

The most illuminating and clean part of the problem is just up to here, but in an attempt for completeness, I’ll keep going.

From the “20” person’s point of view, we assume that he or she is able to figure out all of the above logic on his or her own.  We now know that what the “20” person sees is “30” and “50,” which means that he or she is either 20 or 80.  Somehow, the “First” “50” person figured out on their own on the second round of questioning that they have 50 on their head.  The logic is that in order to find out what your number is on the second round, you are using someone’s “I don’t know” answer in the first round of questioning.  So the “20” person can deduce that if they indeed have 20 on his or her head, the “50” person can indeed figure out all the above logic and figure out that their number is 50 on the second round.  If the “20” person has 80 instead, the “First” person sees “80” and “30” and is thus wondering if his or her number is 50 or 110 and the “30” person sees “80” and 50 and wondering if they’re 30 or 130.  In none of these cases where the “20” person has 80 should result in the “First” person figuring out that their own number is 50 on the second round.  If the “20” person has 20, then, the “First” person sees that when the “30” person says that they don’t know their own number in the first number, that is basically the “30” person announcing that they aren’t seeing an “x” and “2x” situation. (If you see two numbers “x” and “2x”, you know immediately that you are 3x. If you (the “First” person) are wondering if someone else (the “30” person) is seeing an “x” and “2x” situation (the “First” person with “x” and the “20” person with “2x”), if they say 3x, you know you have x, and if they say that they don’t know their own number, you know that you do not have x.) This means that the “First” person can’t have 10 and must have 50.  This causes the “20” person to know that his or her number is 20.

Similarly, the “30” person sees “50” and “20” initially and doesn’t know if he or she is 30 or 70.  If it’s 70, the “20” person sees “70” and “50” and the “First” person sees “70” and “20,” which, again, will not cause the situation described above where someone (the “30” person) announces (to the “First” person) that he or she is not seeing an “x” and “2x” situation.  If the “30” person has 30, then everything that’s been discussed happens, and so it must be 30.

The key basically is that if someone sees “x” and “2x,” they should know immediately that they are 3x.  Then, if someone sees “2x” and “3x,” they are immediately on high alert to see if the “3x” person immediately knows that he or she is 3x.  If the “3x” person doesn’t know their own number , that is an announcement by that person that they are not seeing an “x” and “2x” situation, which means that the person we started (the “First” person) with must be 2x + 3x = 5x.