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.
(In reply to
possible solution by broll)
A neat little puzzle.
I AGREE.
my reasoning was:
Deducting any sum which is 6 mod 11 from 281(the total of first 14 primes) gives a multiple of 11, say 11*q.
IF q is a prime number, and the sum bay be created by
a triplet of distinct primes below 45 - the remaining set of 11 numbers is the answer.
e.g. an average of 20 is obtained by discounting(eliminating) the triplet ( 3,17,41 ). 281-61=220 ==>220/11=20 nice, but not a prime! -the erased triplet must include 2 to get odd result!
BUT: an average 19 is obtained by discounting( the triplet ( 2,29,41 ). 281-72=209 ==>209/11=19 ...nice, but the set includes 19 - forbidden..
An average 23 is obtained by discounting( the triplet ( 2,3,23 ). 281-48=253 ==>253/11=23
so OUR SOLUTION IS : (5,7,11,13,17,19,29,31,37,41, 43).
re: possible solution
