50(1136). Before the swimming competition, each of the four participants $A, b, B$ and $\Gamma$ was asked what place they expected to finish in. $A$ said: “I will be first,” $B$ said: “I will not be last,” $V$ said: “I will not be first or last,” and $\Gamma$ said: “I will be last.” After the race, it turned out that only one of them had incorrectly predicted the result. Which swimmer was wrong?[^2]
Solved. Let's create a table where we will indicate the assumed results with a "plus" sign.

| Swimmer | Place |  |  |  |
| :---: | :---: | :---: | :---: | :---: |
|  | 1 | 2 | 3 | 4 |
| $A$ | + |  |  |  |
| 5 | + | + | + |  |
| $B$ |  | + | + |  |
|  |  |  |  | + |

Let's assume that $A$ was wrong. Then he could have taken 2nd or 3rd place (4th place was taken by swimmer $\Gamma$, who, if $A$ was wrong, correctly predicted his result: according to the condition, only one swimmer was wrong). In this case, the following distribution of places is possible:

a) $A-2, B-1, B-3, \Gamma-4$;

b) $A-3, 5-1, B-2, \Gamma-4$.

We will prove that the problem has one answer (the swimmer $A$ was wrong). Indeed, if $B$ was wrong (took 4th place), then $\Gamma$ would also be wrong, which contradicts the condition of the problem. If $B$ was wrong, then he should be either first or last. In such a case, another swimmer, $A$ or $\Gamma$, would also be wrong. If $\Gamma$ was wrong, then another swimmer would also be wrong, otherwise, the last place would not be taken by anyone. Since, according to the condition of the problem, only one swimmer could be wrong, $\Gamma$ did not make a mistake.

Answer: the swimmer $A$ was wrong.