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If X and Y are the same, then you must be A, and will answer in one iteration.

Otherwise, you will have 2 possibilities: X+Y or Y-X.
But if you're Y-X, that means you're not A. Y will be the actual A, since they're the largest of the three, and they will be seeing X and (Y-X) in front of them.

It gets recursive here - if that is the case, then that person has 2 possibilities too, and so on.. this will go until we reach point where a person has 2 same numbers in front of them.
Thus: Let $F(X, Y)$ be the number of turns you need to solve your own number, with $Y$ being the larger of the two $(Y >= X)$.

$$F(X,Y) = \begin{cases} \text{1} &\quad\text{if X = Y} \\ \text{$1 + F(X, Y - X)$} &\quad\text{otherwise} \\ \end{cases}$$

The reason the guy with A hat always solves first is because by the nature of this formula, they will always be one step ahead of the others:

Remember that $A = B + C$.
A's case will be $F(B, C)$
B's case will be $F(C, B + C) = 1 + F(B, C)$
C's case will be $F(B, B + C) = 1 + F(B, C)$

As you can see, B and C will reach $F(B, C)$ in one iteration... but that's what A started with! As such, they'll always lag one iteration behind A.

By the way, answering based on the number of iterations is tricky if they're asked one by one. Depending on A's ordering (before or after the others), the answer can change. For every step of the recursion where the A answers after the Y, then the final answer will be one iteration shorter (as he can already know the current iteration Y's answer without waiting for the next iteration).

To solve this, I propose that within one round/iteration of questioning, all of them are asked separately and at the same time, and likewise they can only know of that entire round's result all at once afterwards.

If X and Y are the same, then you must be A, and will answer in one iteration.

Otherwise, you will have 2 possibilities: X+Y or Y-X.
But if you're Y-X, that means you're not A. Y will be the actual A, since they're the largest of the three, and they will be seeing X and (Y-X) in front of them.

It gets recursive here - if that is the case, then that person has 2 possibilities too, and so on.. this will go until we reach point where a person has 2 same numbers in front of them.
Thus: Let $F(X, Y)$ be the number of turns you need to solve your own number, with $Y$ being the larger of the two $(Y >= X)$.

$$F(X,Y) = \begin{cases} \text{1} &\quad\text{if X = Y} \\ \text{$1 + F(X, Y - X)$} &\quad\text{otherwise} \\ \end{cases}$$

The reason the guy with A hat always solves first is because by the nature of this formula, they will always be one step ahead of the others:

Remember that $A = B + C$.
A's case will be $F(B, C)$
B's case will be $F(C, B + C) = 1 + F(B, C)$
C's case will be $F(B, B + C) = 1 + F(B, C)$

As you can see, B and C will reach $F(B, C)$ in one iteration... but that's what A started with! As such, they'll always lag one iteration behind A.

Do note however that the actual answer is actually slightly more complicated, thanks to the effect of asking them in order:

Answering based on the number of iterations is tricky if they're asked one by one, as depending on A's ordering (before or after the others), the answer can change.

For every step of the recursion where the A answers after the Y, then the final answer will be one iteration shorter (as he can already know the current iteration Y's answer without waiting for the next iteration). So it's more like this:
$$F(X,Y) = \begin{cases} \text{1} &\quad\text{if X = Y} \\ \text{$F(X, Y - X)$} &\quad\text{if X $\ne$ Y and you answer after Y} \\ \text{$1 + F(X, Y - X)$} &\quad\text{if X $\ne$ Y and you answer before Y} \\ \end{cases}$$
Still, even though this is the more 'correct' answer, I think the simplified formula is a better answer as it is easier to understand. This issue of less-iterations-due-to-answering-order feels completely unrelated to the main riddle, and only serves to add unnecessary complexity.

To solve this, I propose that within one round/iteration of questioning, all of them are asked separately and at the same time, and likewise they can only know of that entire round's result all at once afterwards.

Let's say you're A (but you don't know that), and you're looking at 2 hats in front of you - call them X and Y.

Case 1: X = 1, Y = 1, You = 2 - Solved in iteration #1
You have 2 possibilities: 2 or 0.
As 0 is impossible, you can immediately answer 2 the first time you were questioned.

Case 2: X = 1, Y = 2, You = 3 - Solved in iteration #2
You have 2 possibilities: 3 or 1.
However, if you were 1, then Y would be seeing Case 1 in front of them, and will solve it within the first iteration. Therefore, if the game is not over yet after the first iteration, you'll know you're 3, and solve it in the second.

Case 3: X = 2, Y = 2, You = 4 - Solved in iteration #1
Again, it's trivial if X and Y is the same.

Case 4: X = 1, Y = 3, You = 4 - Solved in iteration #3
You have 2 possibilities: 4 or 2.
However, if you were 2, then Y would be seeing Case 2 in front of them, and will solve it within 2 iterations. Therefore, if the game is not over yet after 2 iterations, you'll know you're 4, and solve it in the third.

Case 5: X = 2, Y = 3, You = 5 - Solved in iteration #2
You have 2 possibilities: 5 or 1.
However, if you were 1, then Y would be seeing Case 1 in front of them. You know the drill by now.

Case 6: X = 1, Y = 4, You = 5 - Solved in iteration #4
You have 2 possibilities: 5 or 3.
However, if you were 3, then Y would be seeing Case 4 in front of them.

Case 7: X = 2, Y = 4, You = 6 - Solved in iteration #2
You have 2 possibilities: 6 or 2.
However, if you were 2, then Y would be seeing Case 3 in front of them.

And so on and so on.

Let $F(X, Y)$ be the number of turns you need to solve your own number, with $Y$ being the larger of the two $(Y >= X)$.

$$F(X,Y) = \begin{cases} \text{1} &\quad\text{if X = Y} \\ \text{$1 + F(X, Y - X)$} &\quad\text{otherwise} \\ \end{cases}$$

The reason the guy with A hat always solves first is because by the nature of this formula, they will always be one step ahead of the others:

Remember that $A = B + C$.
A's case will be $F(B, C)$
B's case will be $F(C, B + C) = 1 + F(B, C)$
C's case will be $F(B, B + C) = 1 + F(B, C)$

As you can see, B and C will reach $F(B, C)$ in one iteration... but that's what A started with! As such, they'll always lag one iteration behind A.

By the way, answering based on the number of iterations is tricky if they're asked one by one. Depending on A's ordering (before or after the others), the answer can be up to one iteration faster.

To solve this, I propose that within one round/iteration of questioning, all of them are asked separately and at the same time, and likewise they can only know of that entire round's result all at once afterwards.

Let's say you're A (but you don't know that), and you're looking at 2 hats in front of you - call them X and Y.

Case 1: X = 1, Y = 1, You = 2 - Solved in iteration #1
You have 2 possibilities: 2 or 0.
As 0 is impossible, you can immediately answer 2 the first time you were questioned.

Case 2: X = 1, Y = 2, You = 3 - Solved in iteration #2
You have 2 possibilities: 3 or 1.
However, if you were 1, then Y would be seeing Case 1 in front of them, and will solve it within the first iteration. Therefore, if the game is not over yet after the first iteration, you'll know you're 3, and solve it in the second.

Case 3: X = 2, Y = 2, You = 4 - Solved in iteration #1
Again, it's trivial if X and Y is the same.

Case 4: X = 1, Y = 3, You = 4 - Solved in iteration #3
You have 2 possibilities: 4 or 2.
However, if you were 2, then Y would be seeing Case 2 in front of them, and will solve it within 2 iterations. Therefore, if the game is not over yet after 2 iterations, you'll know you're 4, and solve it in the third.

Case 5: X = 2, Y = 3, You = 5 - Solved in iteration #2
You have 2 possibilities: 5 or 1.
However, if you were 1, then Y would be seeing Case 2 in front of them. You know the drill by now.

Case 6: X = 1, Y = 4, You = 5 - Solved in iteration #4
You have 2 possibilities: 5 or 3.
However, if you were 3, then Y would be seeing Case 4 in front of them.

Case 7: X = 2, Y = 4, You = 6 - Solved in iteration #2
You have 2 possibilities: 6 or 2.
However, if you were 2, then Y would be seeing Case 3 in front of them.

And so on and so on.

If X and Y are the same, then you must be A, and will answer in one iteration.

Otherwise, you will have 2 possibilities: X+Y or Y-X.
But if you're Y-X, that means you're not A. Y will be the actual A, since they're the largest of the three, and they will be seeing X and (Y-X) in front of them.

It gets recursive here - if that is the case, then that person has 2 possibilities too, and so on.. this will go until we reach point where a person has 2 same numbers in front of them.
Thus: Let $F(X, Y)$ be the number of turns you need to solve your own number, with $Y$ being the larger of the two $(Y >= X)$.

$$F(X,Y) = \begin{cases} \text{1} &\quad\text{if X = Y} \\ \text{$1 + F(X, Y - X)$} &\quad\text{otherwise} \\ \end{cases}$$

The reason the guy with A hat always solves first is because by the nature of this formula, they will always be one step ahead of the others:

Remember that $A = B + C$.
A's case will be $F(B, C)$
B's case will be $F(C, B + C) = 1 + F(B, C)$
C's case will be $F(B, B + C) = 1 + F(B, C)$

As you can see, B and C will reach $F(B, C)$ in one iteration... but that's what A started with! As such, they'll always lag one iteration behind A.

By the way, answering based on the number of iterations is tricky if they're asked one by one. Depending on A's ordering (before or after the others), the answer can change. For every step of the recursion where the A answers after the Y, then the final answer will be one iteration shorter (as he can already know the current iteration Y's answer without waiting for the next iteration).

To solve this, I propose that within one round/iteration of questioning, all of them are asked separately and at the same time, and likewise they can only know of that entire round's result all at once afterwards.

Let F(X, Y) be the number of turns you need to solve your own number, if you're A.
For simplicity purposes, let Y be the larger of the two, Y >= X.

F(X, Y) = 1 (If X = Y)
F(X, Y) = 1 + F(X, Y - X) (Otherwise)

Let F(X, Y) be the number of turns you need to solve your own number as A, where Y >= X.

F(X, Y) = 1 (If X = Y)
F(X, Y) = 1 + F(X, Y - X) (Otherwise)

The reason the guy with A hat always solves first is because by the nature of this formula, they will always be one step ahead of the others:
Remember that A = B + C.
A's case will be F(B, C)
B's case will be F(C, B + C) = 1 + F(B, C)
C's case will be F(B, B + C) = 1 + F(B, C)
As you can see, B and C will reach F(B, C) in one iteration.. but that's what A started with! As such, they'll always lag one iteration behind A.

By the way, answering based on the number of iterations is tricky if they're asked one by one... Depending on A's ordering (before or after the others), the answer can be up to one iteration faster.

To solve this, i propose that within one round/iteration of questioning, all of them are asked separately and at the same time, and they can only know of that entire round's result afterwards.

Let $F(X, Y)$ be the number of turns you need to solve your own number, if you're A.
Where $X$ and $Y$ are the numbers you see, with $Y$ being the larger of the two $(Y >= X)$.

$$F(X,Y) = \begin{cases} \text{1} &\quad\text{if X = Y} \\ \text{$1 + F(X, Y - X)$} &\quad\text{otherwise} \\ \end{cases}$$
And yes, A will always be able to answer first.

Let $F(X, Y)$ be the number of turns you need to solve your own number, with $Y$ being the larger of the two $(Y >= X)$.

$$F(X,Y) = \begin{cases} \text{1} &\quad\text{if X = Y} \\ \text{$1 + F(X, Y - X)$} &\quad\text{otherwise} \\ \end{cases}$$

The reason the guy with A hat always solves first is because by the nature of this formula, they will always be one step ahead of the others:
 
Remember that $A = B + C$.
A's case will be $F(B, C)$
B's case will be $F(C, B + C) = 1 + F(B, C)$
C's case will be $F(B, B + C) = 1 + F(B, C)$

As you can see, B and C will reach $F(B, C)$ in one iteration... but that's what A started with! As such, they'll always lag one iteration behind A.

By the way, answering based on the number of iterations is tricky if they're asked one by one. Depending on A's ordering (before or after the others), the answer can be up to one iteration faster.

To solve this, I propose that within one round/iteration of questioning, all of them are asked separately and at the same time, and likewise they can only know of that entire round's result all at once afterwards.