At the end of a soccer season, every player on a team scored a prime number of goals, and the average for the team as a whole was also a prime number. No player’s tally was equal to another’s, and no player’s tally was the same as the average.

Given that nobody scored more than 45 goals, how many goals did each player score?

*** There are 11 players in a soccer team.

 If the players scored 5,7,11,13,17,19,29,31,37,41,and  43 goals, then the average would be 23, with no player scoring 23, as required by the problem.

This solution is unique. {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43} sum to 281. If the 3 smallest numbers are deleted the average is under 25; if the 3 largest numbers are deleted the average is over 14, so the (prime) average must be 17, 19, or 23.

We may write 281 as 25 and 6/11, 24 and 17/11 etc. We are only interested in those numbers which have a prime integral component i.e. 23 and 28/11, 19 and 72/11, 17 and 94/11. Each of these denominators is even, so no player can have scored 2, since the sum of any two other primes is even.

Similarly, no player scored the average, leavng as the sole options: (28-2-23=3), (72-2-19=51), (94-2-17=75), with only the first alternative qualifying as required.

A neat little puzzle.

 

Edited on April 17, 2013, 11:48 pm

Posted by broll on 2013-04-17 10:46:10