Numbers on heads

The solution to the ‘numbers on heads’ puzzle is that they are 65 (person A), 26 (person B) and 39 (person C).

One way to work this out is as follows:

All of A, B and C know from outset that they have one of two numbers. The sum or difference of the numbers they can see.

Forgetting the actual numbers initially, A’s first statement says that for any positive integer n it is not possible for A to have 2n, B to have n and C to have n. Let’s denote this arrangement as (2n,n,n) where position 1 is A, position 2 B, and position 3 C.

B’s statement also means that (n,2n,n) is not possible. And, here’s the critical bit, because B knows that (2n,n,n) is not possible, if he could see A=2n and C=n, then he would know that B=n is not possible, so given that his only other option is 2n+n=3n, he would be able to say he knew his number and it is 3n. Since he doesn’t know his number then he can’t see A=2n, C=n, and (2n,3n,n) is also not possible. All can deduce this once B makes his statement.

C’s statement says (n,n,2n) not possible and also using the same logic as above, that the other option for C from all those already discounted are also not possible. I.e. (2n,3n,5n), (n,2n,3n), and (2n,n,3n).

Finally, if A could not now deduce his number then the additional arrangements would now not be possible (same logic again): (4n,3n,n), (3n,2n,n), (8n,3n,5n), (5n,2n,3n), (4n,n,3n), (3n,n,2n). But, since A can deduce his number, then one of these new situations is possible. Since A says 65 then the only arrangement where the first number is divisible by 5 must be the answer I.e. (5n,2n,3n) with n=13, and the numbers he sees are B 2*13=26 and C 3*13=39!!