25. At the end of a chess tournament that lasted for $k$ rounds, it turned out that the number of points scored by the participants formed a geometric progression with a natural denominator greater than 1. How many participants could there have been:

a) when $k=1989$

b) when $k=1988$?
89.25. Answer. If \( k=1989 \), there were two participants in the tournament, and if \( k=1988 \), there could have been two or three chess players.

Suppose there were \( n \) participants: the first scored \( A \) points, the second \( A p \) points, ..., and the winner \( A p^{n-1} \) points. The winner cannot score more than \( 2 k(n-1) \) points, while all other participants scored at least \( k(n-1)(n-2) \) points. Comparing these numbers, we see that for \( n \geqslant 4 \), \( k(n-1) \leqslant k(n-1) \times (n-2) \). On the other hand, \( A p^{n-1} > A + A p + \ldots + A p^{n-2} \). Therefore, \( n < 4 \). Moreover, for \( p \geqslant 3 \), the points of the winner are more than twice the sum of the points of all others, which is impossible for \( n=3 \). Hence, \( n=3 \) is possible only for \( p=2 \), i.e., the players scored \( A, 2A \), and \( 4A \) points, meaning the total sum of points, which is obviously \( 6k \), must be divisible by seven. Therefore, \( k \) must be divisible by seven. The number 1989 is not divisible by seven, so \( n=2 \) is the only option in this case. It is quite easy to construct an example for \( k=1988 \) and \( n=3 \).