## Task 1 - 200711

On the occasion of the award ceremony of a mathematics competition, each prize winner congratulated every other with a handshake. In total, 91 handshakes were performed, and exactly one handshake took place with each congratulation.

Determine the number of prize winners of the competition from this information!
Let $n$ be the number of prize winners. If for each prize winner we count the handshakes he performed with the other prize winners, and add the resulting numbers, we get the sum of $n$ summands $n-1$, that is, the number $n \cdot (n-1)$.

In this number, each of the handshakes that actually occurred is counted exactly twice (namely in the counts for each of the two partners of the respective handshake). Therefore, the number of all handshakes is equal to $\frac{1}{2} n(n-1)$.

Thus, $\frac{1}{2} n(n-1) = 91$, so $(n-1) n = 182$.

First continuation possibility: Due to the prime factorization $182 = 2 \cdot 7 \cdot 13$, 182 has, up to the order, only the factorizations $182 = 1 \cdot 182 = 2 \cdot 91 = 7 \cdot 26 = 13 \cdot 14$ in natural numbers. Of these, only $13 \cdot 14$ is a factorization into two factors of the form $n-1$ and $n$. Therefore, the number sought is $n = 14$.

Second continuation possibility: If $n > 14$, then $(n-1) n > 13 \cdot 14 = 182$. Thus, only the possibility $n = 14$ remains.