In a 14 team baseball league, each team played each of the other teams 10 times. At the end of the season, the number of games won by each team differed from those won by the team that immediately followed it by the same amount. Determine the greatest number of games the last place team could have won, assuming that no ties were allowed.
Problem
Source:
Tags: arithmetic sequence
LordoftheMorons
27.01.2006 23:40
There are a even number of teams, so the middle two teamss (7th and 8th place) could have won 66 and 64 games, respectively. Then the score is reduced by 2 each time, leaving the last (14th) place team with 52 wins.
catcurio
28.01.2006 19:22
There are $10(13+1)\left(\frac{13}{2}\right)=910$ games played. Letting $n$ be the number of games the last place team won, $n+n+2+n+4\cdots+n+26=910$. So $n=52$.
JesusFreak197
28.01.2006 20:08
Each team played $130$ games, so there were $10\binom{14}{2}=910$ games played. We have an arithmetic sequence, so $a+(a+d)+(a+2d)+...+(a+13d)=14a+91d=7(2a+13d)=910$, so $2a+13d=130$, with both $a$ and $d$ integers. Obviously, $d$ must be even, and we want to minimize $d$. $d=2$, so $a=\boxed{52}$.