# Solving the common knowledge puzzle

I’ve written up the way that I solved the puzzle, step by step:

1. A looks at B and C and says “I don’t know what my number is.”

We assess what information this gives us by considering in which scenario A would have known what their number was, and ruling it out.

Well, for starters, if A saw two different numbers, then A would consider there to be two logically possible worlds, one in which they are the sum of the two numbers, and another in which they are one of the two that are added together. But if A saw the SAME two numbers, they A would know that they must be the sum of those two (since zero is not a possible value for themselves). What this means is that the fact that A doesn’t know their number tells us with certainty that B ≠ C! Furthermore, since B and C will go through this same line of reasoning themselves, they will also know that B ≠ C. And since all of them know that B and C will go through the same line of reasoning, it becomes common knowledge that B ≠ C.

Good! So after A’s statement, we have added one piece of common knowledge, namely that B ≠ C.

2. B thinks a moment and then says “Me neither.”

Ok, one thing this tells us is that A ≠ C, for the exact same reason as before.

But it also tells us more than this, because B knows that B ≠ C and still doesn’t know. So we just need to think of the scenario in which knowing that B ≠ C (and knowing the values of A and C, of course) would tell B the value that they have. Try to figure it out for yourselves!

The answer is, the scenario is that A = 2C! Imagine that B sees that A is 10 and C is 5. This tells them that they are either 5 or 15. But they know they can’t be 5, because C is five and B ≠ C. So they must be 15! In other words, since they would know their value if A equaled 2C and they don’t know their value, this tells us that A ≠ 2C!

So now we have two more pieces of common knowledge: A ≠ C, and A ≠ 2C. Putting this together with what we knew before, we have a total of three pieces of information:

B ≠ C
A ≠ C
A ≠ 2C

3. C thinks a moment and then says “Me neither.”

By exact analogy with the previous arguments, this tells us that A ≠ B, as well as that A ≠ 2B and B ≠ 2A. (We saw previously that B could conclude from B ≠ C that A ≠ 2C. By the same arguments, C can conclude from C ≠ A that B ≠ 2A. And from C ≠ B, C can conclude that A ≠ 2B.)

There’s one more piece of information that we haven’t taken into account, which is that A ≠ 2C. In which situation does the fact that A ≠ 2C tell C their value? Well, if A = 10 and B = 15, then C is either 5 or 20. But C can’t be 5, because then A would be twice C. So C could conclude that they are 20. Since C doesn’t conclude this, we know that 3A ≠ 2B.

Putting it all together, we know the following:

B ≠ C
A ≠ C
A ≠ 2C
A ≠ 2B
B ≠ 2A
3A ≠ 2B

4. A thinks, and says: “Now I know! My number is 25.”

The question we need to ask ourselves is what the values of B and C must be in order that the above six conditions to allow A to figure out their own value. Pause to think about this before reading on…

Let’s work through how A processes each piece of information:

A could figure out their own value by seeing B = C. But we already know that this isn’t the case.

Since A knows that A ≠ C, A could figure out their value by seeing B = 2C. So that’s a possibility… Except that if B = 2C, then A = B + 2C = 3C. But 25 is not divisible by 3. So this can’t be what they saw.

Since A knows that A ≠ B, A could figure out their value by seeing C = 2B. But again, this doesn’t work, since it would imply that A was divisible by 3, which it is not.

Since A knows that A ≠ 2C, A could figure out their value by seeing B = 3C (e.g. B = 15, C = 5). They would rule out themselves being one component of the sum, and conclude that they are 4C. But 25 is not divisible by 4. So this is not the case.

Since A knows that A ≠ 2B, A could figure out their value by seeing C = 3B (e.g. B = 5, C = 15). By the same reasoning as before, this cannot be the case.

Since B ≠ 2A, A could figure out their value by seeing 3B = 2C (e.g. B = 10, C = 15). They would know that they cannot be just one component of the sum, so they would conclude that they must be B + C, or 2.5 B. Now, is there an integer B such that 25 = 2.5 B? You betcha! B = 10, and C = 15!

We can stop here, since we’ve found a logically consistent world in which A figures out that their own value is 25. Since there can only be one such world (as the problem statement implies that this information is enough to solve the puzzle), we know that this must be what they saw. So we’ve found the answer! (A,B,C) = (25,10,15). But if you’re curious, I’d suggest you go through the rest of the cases and show that no other values of B and C would be consistent with them knowing that their own value is 25.

One thing that’s interesting here is the big role that the number 25 played in this. The fact that 25 was not divisible by 3 but was divisible by 2.5, was crucial. For the same puzzle but a different value that 25, we would have come to a totally different answer!

My challenge to anybody that’s made it this far: Consider the set of all integers that A could have truthfully declared that they knew themselves to be. For some such integers, it won’t be the case that A’s declaration is sufficient for us to conclude what B and C are. Which integers are these?