Math Is Fun Forum

Thanks also for sharing your work!

Maybe I need to clarify that all Chinese, Russians and Mexicans are treated equally. Any Chinese can sit in any of their 5 seats etc.
Sorry for not having clarified this earlier; now that I saw your numbers I realised that you have this confusion.

All questions must be answered by yes or no.

I believe yes.

Thank you, but how about in the general case, where we don't know whether A lies on the circle?

This is the solution I propose:
We have 5 available colors, so, in order to transmit the information to the others, we will use this method:
We have 5 different values of mod(5): 0, 1, 2, 3, 4.
They coordinate and agree the following:
They assign these 5 values to the five hat colors Cyan, Magenta, Yellow, Red, Green.
0: Cyan
1: Magenta
2: Yellow
3: Red
4: Green

They also assign themselves with one of the numbers 0 to 4 (i.e. the 1st prisoner is assigned with number 0, the 2nd with 1 etc).
The above is pre-agreed before they wear the hats.
After they wear the hats, they do the following:
Then each prisoner calculates the sum of the numbers of the hat colors, based on the above table and DEDUCTS this number from his own and then converts the difference to mod(5). Then he finds the color that this number corresponds to, based on the above table.
Example: Suppose the 2nd prisoner sees the following hats to the other four: 1st-3rd-4th-5th: Cyan-Cyan-Red-Green. Let's assume his own hat is Yellow.
Then he calculates the sum 0+0+3+4=7. His own number is 1 (because he is the 2nd and number 1 is assigned to him). He deducts 7 from 1 and converts the difference (6) to mod(5) = 1. Then he guesses that he wears a Magenta hat. Obviously he is wrong, but let's see what the others will guess (because, even if all are wrong and ONLY one guesses correctly, they are saved).
The 1st one sees Yellow-Cyan-Red-Green and calculates the sum 9, which he deducts from his own that is 0, so the remainder is 9 which is 4 mod(5) so he guesses that he wears Green, which is also wrong because he wears Cyan.
Similarly the 3rd guesses Yellow and is wrong because he wears Cyan,
the 4th guesses Red which is CORRECT
and the 5th guesses Magenta and is wrong because he wears Green.
We therefore see that they are saved because always, at least one will guess correctly, based on the above strategy (the key here is that they are each assigned one different number of the 5 mod(5) so at least one will be correct).

Obviously no one knows the sum of all colors, since each of them does not see his own hat, but we know for sure that it will be one of the 5 numbers 0-4 mod(5).
We know, though, that it is sufficient that at least one finds the correct color, then at least one will have the number that will be the same as the correct sum of the hat colors, so he will announce the correct color!

As per the terms, we only have the following condition: A must have a different number of TOTAL handshakes from B or C etc. It is possible, however, that he has 10 handshakes with P and 10 with someone else, provided that his handshakes in TOTAL are not the same with anybody else's.

All the possible pairs of 32 people are 32!/2!30! = 496. If each of them shook hands (i.e. "participated" in handshaking) at least twice and up to X times, then the total number T of handshakes must be from 2*496 to X*496. On the other hand, the total number of "single" handshakes must be a series of different numbers.
Their sum must be 2*T, correct?

Maybe you can combine the above to give a solution through calculations (and also verify your above outcome)?

Much appreciated!!
BTW I don't know the correct answer!!