8. In a Riemannian ball game, two teams face each other; at any given time, each team has $k>1$ players on the field. At the end of each play, a point is awarded to one of the two teams; moreover, each team can make an arbitrary number of substitutions before the next play begins. Alice and Barbara are part of the Geodetic team. At the end of today's game, Alice observes that while she was on the field, the Geodetic team won

7 more plays than they lost. When Barbara was on the field, however, they lost 2 more plays than they won. Each of the other Geodetic players participated in as many winning plays as losing ones. What is the value of $k$?
(A) 3
(B) 5
(C) 6
(D) 9
(E) The data is insufficient to determine it.
8. The answer is $\mathbf{( B )}$. Let's build a table with a row for each action played and a column for each player of the Geodeticas. We write +1 in a cell if the player corresponding to the row was on the field during the action corresponding to the column and the action was won by the Geodeticas. Similarly, we write -1 for the pairs (player, action) given by the actions lost by the Geodeticas with that player on the field. Finally, we write 0 in the cells corresponding to pairs (player, action) for which the player was not on the field during the corresponding action. Now consider the sum of all the numbers in the table.

By hypothesis, we know that the column for Alice sums to +7, the column for Barbara sums to -2, and all other columns sum to 0. The sum of all the numbers in the table is therefore 5. On the other hand, each row of the table has a sum of $+k$ (if the action was won by the Geodeticas) or $-k$ (if the action was lost): indeed, on each row there are exactly $k$ non-zero numbers (corresponding to the $k$ players on the field), and they are all equal to +1 if the action was won by the Geodeticas, or all equal to -1 otherwise. Let $V$ and $P$ be the number of actions won and the number of actions lost by the Geodeticas, respectively. We have then obtained the equation $5=k(V-P)$. From this it follows that $k$ divides 5, and since $k>1$ we have $k=5$.

Note. It is not difficult to construct examples of the situation described in the text: the scores considered are indeed realizable.