\section*{Task 4 - 051024}

The 1007 participants of a conference are to be distributed among as few buses with 13, 29, and 41 seats for passengers in such a way that no seat remains empty.

How many buses of each type are to be ordered?
}

It holds that

\[
20 \cdot 41 + 6 \cdot 29 + 1 \cdot 13 = 820 + 174 + 13 = 1007
\]

Therefore, there is a solution that uses \(20 + 6 + 1 = 27\) buses.

Assume there is a (non-negative integer) solution with \(a\) buses with 41 seats each, \(b\) buses with 29 seats each, and \(c\) buses with 13 seats each, such that

\[
\begin{array}{r}
a + b + c = n \quad (\text{with integer } n \leq 26) \\
41a + 29b + 13c = 1007
\end{array}
\]

\(41 \cdot (1) - (2)\) results in the equation \(12b + 28c = 41 \cdot n - 1007 \quad\) (3).

For \(n = 26\), (3) simplifies to:

\[
12b + 28c = 41 \cdot 26 - 1007 = 59
\]

Here, the left side is even, but the right side is not. Therefore, such a solution cannot exist.

For \(n \leq 24\), (3) results in the inequality

\[
12b + 28c = 41 \cdot n - 1007 \leq 41 \cdot 24 - 1007 = -23
\]

Clearly, there are no non-negative solutions here.

Thus, only the case \(n = 25\) remains. Here, (3) results in the equation

\[
12b + 28c = 41 \cdot 25 - 1007 = 18
\]

Since \(a\), \(b\), and \(c\) are non-negative integers, \(c\) must be 0, otherwise the left side of (6) would already be too large.

This leaves the equation \(12b = 18\), which has no integer solution.

In all cases, the above assumption led to a contradiction. Therefore, there is no solution that uses fewer than 27 buses.

The solution given above, which uses 27 buses, thus uses the smallest possible number of buses.