X_{1.1}  X_{1.2}  X_{1.3}  X_{1.4}  X_{1.5} 
X_{2.1}  X_{2.2}  X_{2.3}  X_{2.4}  X_{2.5} 
X_{3.1}  X_{3.2}  X_{3.3}  X_{3.4}  X_{3.5} 
X_{4.1}  X_{4.2}  X_{4.3}  X_{4.4}  X_{4.5} 
X_{5.1}  X_{5.2}  X_{5.3}  X_{5.4}  X_{5.5} 
table 1: Solution to the horse race problem. Following the first 5 races each row) every horse is ordered quickest (?.1) to slowest(?.5). A 6^{th }race of the individual winners of the first 5 races allows us to to assign each row a relative ranking and each individual horse as X_{secondrace.firstrace}. Note that the 10 horses X_{?.4}X_{?.5} were eliminated in the first five races by virtue that they cannot place 1^{st}3^{rd}. The 6^{th} race likewise eliminates the remainder of the last 2 rows ( X_{4.?}X_{5.?}), plus the horses labeled X_{3.3}, X_{3.2}, X_{2.3} as they will not be able to place either. With X_{1.1} as the clear winner, X_{1.2},X_{1.3},X_{2.1},X_{2.2} and X_{3.1} race one last time for 2^{nd} and 3^{rd} place. 











table 2: Table of Intermediary winners
following the n^{2} + n^{st }race. A
is the overall 1^{st} place winner, with B's
potential 2^{nd} and 3^{rd} place winners and C's
potential 3^{rd} place winner. The 3 B's race with the n^{2} + n + 1^{st} race with the winner taking 2nd overall,
and the runner up with the winners adjacent C and the runner up racing for 4^{th}
place.
