In a mathematics competition, 5 problems were set. Among the contestants, there were no two who solved exactly the same problems. However, if we ignore any one of the problems, then for any contestant, there is another contestant who solved the same problems from the remaining 4. How many people participated in the competition?
Assuming the text does not state this, we assume there was a competitor.

I. Solution. Ignoring any task, each competitor has only one other competitor who achieved the same result in the remaining four tasks. If there were two, then among the three, there would be two who solved the ignored task, or two who did not solve it, but then these two would have solved the same tasks out of the five (or possibly none), which does not meet the condition.

Ignoring the 1st task, we can pair the competitors so that the pairs solved the same tasks out of the remaining four, and one of them solved the 1st task, while the other did not. Thus, the number of competitors is twice the number of those who solved the 1st task.

If we now form pairs by ignoring the 2nd task, the partner of a competitor who solved the 1st task also solved the 1st task, since they achieved the same result except for the 2nd task; one of them solved the 2nd task, the other did not. Thus, the number of competitors who solved the 1st task is twice the number of all competitors, and therefore four times the number of those who solved both the 1st and 2nd tasks. Similarly, if we form pairs according to the 3rd task, a competitor who solved the first two tasks also has a partner who solved the first two tasks, but only one of them solved the 3rd task, so the number of competitors who solved the first two tasks is twice the number of all competitors, and therefore eight times the number of those who solved the first three tasks. Similarly, it can be seen that the number of competitors is 16 times the number of those who solved the first four tasks, and 32 times the number of those who solved all five tasks. However, there was at most one such competitor, and if we assume there was a competitor, then there had to be one. Thus, there were 32 competitors in total.

II. Solution. We denote the solution of a task with 1 and the non-solution with 0. We write down the results of each competitor in this way, listing the results of the tasks in the order they were attempted, for example, 10110 means that the corresponding competitor solved all tasks except the 2nd and 5th. Let's count the number of possibilities. The first place can be 0 or 1, and either can be followed by 0 or 1 in the second place, so the first two places can be filled in 4 ways. The third place can also be filled in 2 ways, and the number of possible results doubles with each additional task. Thus, in solving the five tasks, $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2=32$ different results are possible, so there could not have been more than 32 competitors, because according to the first condition, there were no two competitors who solved exactly the same tasks.

We will show that if we assume there was a competitor, then there had to be 32, so every result had to occur. Let's select a competitor, $A$, and let their result be denoted by the sequence $abcde$ (each letter represents 0 or 1). For example, someone who achieved a different result from $A$ in the 1st task, their result table starts with 1 - $a$, which we will briefly denote as $\bar{a}$. Then every possible result can be obtained by underlining some of the $a, b, c, d, e$ numbers (possibly none or all), since one of a letter and its underlined version is 0, the other is 1. We need to show that every such denoted result also occurs. We provide a procedure to find the competitor who achieved exactly that result for every result. We demonstrate this with the example of $a \bar{b} \bar{c} \bar{d} \bar{e}$: to $A$, there is a competitor $B$ whose result differs from $A$'s only in the 1st task, so $B$'s result is $\bar{a} b c d e$. The competitor $C$ who differs from $B$ only in the 3rd task has the result $\bar{a} b \bar{c} d e$; the one who differs from this only in the 4th task, $D$, has the result $\bar{a} b \bar{c} \bar{d} e$, and finally, the one who differs from $D$ only in the 5th task, $E$, has the result $\bar{a} b \bar{c} \bar{d} \bar{e}$. So $E$ is the sought competitor. $B, C, D, E$ exist according to the second condition of the problem. It can similarly be shown for any result that it occurs among the competitors, so 32 competitors participated in the competition.

[^0]:    ${ }^{1}$ According to the text of the problem, a solution is also that no competitor started, because then there were no two competitors who solved exactly the same tasks, and the other condition is only not met if we can find a competitor and a task such that, ignoring the task, the competitor achieved a different result from everyone else in the other four tasks. However, if there is no competitor, then we cannot find such a competitor.

    ${ }^{2}$ Read: a with an overline.