\section*{Problem 2 - 301032}

As is well known, any sequence of \(n\) numbers of the form

\[
a_{1}=z ; \quad a_{2}=z+d ; \quad a_{3}=z+2 d ; \quad a_{n}=z+(n-1) d
\]

( \(n \geq 1\) natural number; \(z, d\) real numbers) is called a (finite) arithmetic sequence.

Determine the number of all such arithmetic sequences (1), where \(z\) and \(d\) are also natural numbers with \(z \geq 1, d \geq 1\) and for which \(n \geq 3\) and the following holds:

\[
a_{1}+a_{2}+\ldots+a_{n}=1991
\]
}

It is \(11 \cdot 181=1991\), where 11 and 181 are prime numbers. Therefore, 1991 has exactly four divisors: 1, 11, 181, and 1991.

Case 1: \(n\) is odd, i.e., there exists a natural number \(m \geq 2\) such that \(n=2 m-1\). Then, with \(x:=z+(m-1) \cdot d\),

\[
\begin{gathered}
a_{1}=x-(m-1) d, \quad a_{2}=x-(m-2) d, \quad \ldots, \quad a_{m}=x, \quad a_{m+1}=x+d, \\
\ldots \quad a_{n}=a_{m+(m-1)}=x+(m-1) d \quad \text { and } \\
a_{1}+a_{2}+\cdots+a_{n}=(x-(m-1) d)+\cdots+(x+(m-1) d)=(2(m-1)+1) x=n x
\end{gathered}
\]

In particular, \(n\) and \(x\) are divisors and reciprocals of 1991. Since \(x=z+(m-1) d>\frac{n}{2}\), the divisors 181 and 1991 can be ruled out as possibilities for \(n\), as the corresponding reciprocals would be smaller than \(\frac{n}{2}\), and thus also smaller than \(x\). Given \(n \geq 3\), the only remaining possibility is \(n=11\), which means \(181=x=z+5 d\) or \(z=181-5 d\).

Thus, for each integer \(1 \leq d \leq \frac{180}{5}=36\), there is exactly one such arithmetic sequence, as \(z \geq 1\) is uniquely determined. (No further sequences can exist in this case.) Therefore, there are exactly 36 solutions in this case.

Case 2: \(n\) is even, i.e., there exists a natural number \(m \geq 2\) such that \(n=2 m\). Then, for each \(1 \leq i \leq m\),

\[
a_{i}+a_{n+1-i}=(z+(i-1) d)+(z+(n-i) d)=2 z+(n-1) d=: x
\]

thus

\[
a_{1}+a_{2}+\cdots+a_{n}=\left(a_{1}+a_{n}\right)+\left(a_{2}+a_{n-1}\right)+\cdots+\left(a_{m}+a_{m+1}\right)=m \cdot x
\]

so \(m\) and \(x\) must also be divisors and reciprocals of 1991. Since \(x=2 z+(n-1) d \geq 2+(n-1)>n=2 m\), the divisors 181 and 1991 are again ruled out due to the large \(x\), and \(m=1\) is also ruled out because \(m \geq 2\), so \(m=11\) and thus \(181=x=2 z+21 d\) or \(z=\frac{1}{2} \cdot(181-21 d)\) follows. Since \(z\) is an integer, \(d\) must be odd, and because \(z \geq 1\), it follows that \(1 \leq d \leq 7\) since \(9 \cdot 21=189>181\). Therefore, there are exactly 4 solutions in this case.

Together, there are exactly 40 arithmetic sequences that satisfy the conditions of the problem.