In a single round-robin tournament... In a single round-robin tournament (i.e. each participant plays every other participant once), the scoring rules is that 


*

*the winner earns 5 points

*the loser gets 0

*if it's a deuce, each one get 2 points


Now the winner has 29 more points than the runner-up. Question, what's the least total score (the sum of all players' scores)?
I'm totally lost here...
 A: To get the least total score you would need:
1) as few players as possible
2) one player who is winning lots of games
3) all the others have lots of ties playing against each other, for then you'd get only $4$ points between them instead of $5$ ... and that way you keep the distance between the winner and the runner-up at a maximum.
Now, with $n$ players, if all the $n-1$ players keep tying each other, they all get $2(n-2)=2n-4$ points, while the one player that wins all games would get $5n-5$ points. So, since the difference, which is $3n-1$, needs to be $29$, we get that $n=10$
So, there you have it: $10$ players, $1$ player wins all $9$ games for a score of $45$ points, while each of the other $9$ players get $8$ ties and $1$ loss for a total of $16$ points, and they are all runner-ups. So, the total score for all players together is $45+9 \cdot 16=189$.
It is clear that this is the minimum total score, since with fewer than $10$ players you cannot create a distance of $29$ between the winner and the runner-up, and since the winner won all games, while all others tied all games, we have minimized the total number of points between those $10$ players while creating the necessary distance of $29$.
