Solution : - incomplete data
Here is one problem.
On chess tournament there were 3 players from USA and some players from Russia. Three players from USA won all together 7 points and every player from Russia won same number of points. How many players were from Russia?
In this tournament does everyone play everyone or is it only 1 game each against a random partner?
In chess each game rewards a total of 1 point, either someone wins and the winner gets 1 and the loser gets 0, or it is a draw and they both get 0.5 points.
There is far too much left out. Are we to assume this is a "round robin" tournament in which every player plays every other player exactly once? Are we to assume that a player gets 1 point for a win, 1/2 point for a draw, and 0 for a loss? Were there no players from countries other than the USA and Russia? Assuming all of that then, if we let "n" be the number of Russian players, there were a total of n+ 3 players and so (n+3)(n+2)/2 games (each of the n+ 3 players plays each of the remaining n+3- 1= n+ 2 players but each game accounts for 2 players). Further, under this system, each game add "1" to the total score (0+ 1 or 1/2+ 1/2) so there were a total of (n+3)(n+2)/2 points in the tournament. If the USA players gained 7 points, the Russian players gained (n+3)(n+2)/2- 7 points. If all the Russian players got the same score, say, m, then nm= (n+3)(n+2)/2- 7. n and m, of course, must be integer values so we have the Diophantine equation or .
By the quadratic formula . Of course, since n must be a (positive) integer, must be a perfect square.
if every US player plays every russian player exactly once then (x- the number of USA players, y-the number of Russian players, k-the number of points that won each russian player):
x*y=7+k*y
3y=7+ky, k=0 y=7/3, k=1/2 y=14/5, k=1 y= 7/2, k=3/2 y=14/3, k=2 y=7, k=5/2 y=14
So y=7 or 14
I'm not a chess player, but I thought that in round robin chess tournaments, each player played every other player twice, once with the white pieces and once with the black pieces.
There is a considerable advantage in having white rather than black and to rely on the luck of the draw as to who you played as white and who you played as black is unacceptable. Further, if there are an even number of contestants, implying an odd number of opponents, playing each one just once means that some will have more games as white than black while others more games as black than white, again unacceptable.
If this is the case however, the algebra, with the information that we are given, seems to lead to the conclusion that there is only one Russian.