League Lunacy II
( League Lunacy is the first problem in this set)
5 friends--Fred, Graham, Harry, Isaac, Jim--are playing badminton. They will each play everyone else once, recording their scores in a league table. Each game is first to 11 points. If the score in a game reaches 10-10, it is left like that and called a draw. However, since all of the players are close in quality, no player won a game by more than 5 points. The "total score" column of the league table is 3 for a win and 1 for a draw. After all 4 games each, the table was complete. Given this starting arrangement as shown below for the table, can you complete the table?
| Wins | Draws | Losses | Points Won | Points Lost | Total Score | |
| Fred | 35 | 6 | ||||
| Graham | 2 | |||||
| Harry | 44 | |||||
| Isaac | 0 | 32 | ||||
| Jim | 1 | 38 |
Input your answer as the total sum of all of the numbers in the grid. If you think that there are multiple permutations or that the table is impossible, input -1 as your answer.
Note:
The table is not ranked in place order, simply by alphabetical order.
The answer is -1.
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1 solution
Harry won 44 points, the max points possible in the league. That only happens when he won all the matches with 44 played points divided by 11 winning points to get 4 matches won and therefore, he didn't draw with anyone else.
We're told that Graham draw twice, and those must be in the matches against Fred and Jim (we're also told about a no-draw Isaac and found out about Harry's lucky strike ourselves). So Fred's 6 points must include at least 1 points from his draw with Graham, but the only two ways to get a 6 from 4 matches with +3 for a win and +1 for a draw are +3+3+0+0 = 6 which is impossible since we know Fred had at least a draw, or +3+1+1+1 = 6 in which Fred draw thrice, which is also impossible because the max players who he can draw with is just two others, namely Graham and Jim.