\section*{Exercise 2 - 300932}

Determine all representations of the number 1991 as a sum of at least three consecutive positive integers.
}

It is \(1991=1001+990=11 \cdot(91+90)=11 \cdot 181\), where 11 and 181 are prime numbers. Thus, 1991 has exactly the four divisors 1, 11, 181, and 1991.

Let \(n \geq 3\) be the number of consecutive natural numbers whose sum is 1991.

Case 1: \(n=2 m+1\) is an odd number. Let \(a-m\) be the smallest summand. Then the others are \(a-(m-1), a-(m-2), \ldots, a-1, a, a+1, \ldots, a+(m-1), a+m\), and we get

\(1991=(a-m)+(a-(m-1))+\cdots+(a-1)+a+(a+1)+\cdots+(a+(m-1))+(a+m)=(2 m+1) \cdot a\)

Thus, \(n=2 m+1\) and \(a\) must be divisors of 1991 with \(0<a-m=\frac{1991}{n}-\frac{n-1}{2}\). This excludes \(n=1991\) and \(n=181\); \(n=1\) is excluded because \(n \geq 3\), leaving only \(n=11\) and thus \(a=181\). We obtain

\[
1991=176+177+178+179+180+181+182+183+184+185+186
\]

Case 2: \(n=2 m\) is an even number with \(m \geq 2\). Let \(a-m\) be the smallest summand. Then the others are \(a-(m-1), \ldots, a+(m-1)\), so we get

\[
1991=(a-m)+(a-(m-1))+\cdots+(a+(m-1))=2 m \cdot a-m=m \cdot(2 a-1)
\]

This time, \(m\) and \(2 a-1\) must be divisors and reciprocals of 1991, so \(2 a=\frac{1991}{m}+1=\) \(\frac{1991+m}{m}\) and thus \(a=\frac{1991+m}{2 m}\).

Again, \(0<a-m=\frac{1991+m-2 m^{2}}{2 m}\), so \(2 m^{2}-m<1991\) must hold. This excludes \(m=1991\) and \(m=181\), while \(m=1\) is excluded because \(m \geq 2\). The only remaining solution is \(m=11\), from which \(2 a-1=181\) and \(a=91\) follow. We obtain the representation

\[
1991=80+81+\cdots+100+101
\]