\section*{Problem 5 - 211035}

In the 1st stage of the Mathematics Olympiad, there was the following problem in the 9th grade Olympiad class in 1976:

"Someone claims that it is possible to produce exactly 1976 pieces of paper from 7 pieces of paper in the following way: divide some of the 7 pieces of paper into exactly 7 parts each, then again divide some of the now available pieces of paper into exactly 7 parts each, and so on.

Is it possible that by following this procedure, one can obtain exactly 1976 pieces of paper after continuing the process for a sufficient amount of time?"

The solution required proving that it is not possible to obtain exactly 1976 pieces of paper. Now, we want to start with any number \(n \geq 1\) of pieces of paper and divide these in the described manner, each into exactly \(n\) parts.

Determine all natural numbers \(n\) with \(1 \leq n < 1976\) for which it is possible to obtain exactly 1976 pieces of paper in this way!
}

With each division, the number of paper pieces increases by \(n-1\). After \(t\) divisions, there are thus \(n+t(n-1)\) paper pieces.

We are looking for all \(n\) with \(1 \leq n<1976\), for which there exists a \(t \in \mathbb{N}\) such that \(n+t(n-1)=1976\).

Since \(1976=n+t(n-1)=(n-1)(t+1)+1\), this is the case if and only if \(n-1<1975\) is a divisor of \(1977=3 \cdot 659\) (659 is prime), i.e., if and only if \(n-1 \in\{1,3,659\}\).

Therefore, such a decomposition is possible if and only if \(n \in\{2,4,660\}\).