1. In a chess tournament, $n$ chess players participate. Each has played $k$ games with every other. Determine all possible values for $n$ and $k$ if a total of 224 games have been played.
1. The number of pairs of different chess players is $\frac{n(n-1)}{2}$, so since each played $k$ games with each other, it follows that $\frac{n(n-1)}{2} k=224$, i.e., $n(n-1) k=448=2^{6} \cdot 7$. Among the numbers $n$ and $n-1$, one must be odd, and the only odd divisors of 448 are 1 and 7. It is impossible for $n=1$, because then the product on the left side would be 0. For $n-1=1$ we have $n=2$ and then $k=224$, i.e., only two chess players played in the tournament and they played 224 games. The case $n=7$ is also impossible because then we have $n-1=6$, and the right side is not divisible by 3. Finally, for $n-1=7$ we have $n=8$ and then $k=\frac{448}{8 \cdot 7}=8$, i.e., 8 chess players played in the tournament and each played 8 games with each other.

Thus, it is possible that $n=2$ and $k=224$, or $n=8$ and $k=8$.