CEMC Fermat, 2013

Suppose that Megan and Shana competed in exactly $n$ races.

Since Shana won exactly 2 races, then Megan won exactly $n-2$ races.

Since Shana won 2 races and lost $n-2$ races, then she received $2x+(n-2)y$ coins.

Thus, $2x+(n-2)y=35$.

Since Megan won $n-2$ races and lost $2$ races, then she received $(n-2)x+2y$ coins.

Thus, $(n-2)x+2y=42$.

If we add these two equations, we obtain $(2x+(n-2)y)+((n-2)x+2y)=35+42$ or $nx+ny=77$ or $n(x+y)=77$.

Since $n$, $x$ and $y$ are positive integers, then $n$ is a positive divisor of 77, so $n=1,7,11$ or $77$.

Subtracting $2x+(n-2)y=35$ from $(n-2)x+2y=42$, we obtain $$((n-2)x+2y)-(2x+(n-2)y)=42-35$$ or $(n-4)x+(4-n)y = 7$ or $(n-4)(x-y)=7$.

Since $n$, $x$ and $y$ are positive integers and $\text{x gt;y}$, then $n-4$ is a positive divisor of 7, so $n-4 = 1$ or $n-4 = 7$, giving $n=5$ or $n=11$.

Comparing the two lists, we determine that $n$ must be 11.

Thus, we have $11(x+y)=77$ or $x+y=7$.

Also, $7(x-y)=7$ so $x-y=1$.

Adding these last two equations, we obtain $(x+y)+(x-y)=7+1$ or $2x=8$, and so $x=4$.

(Checking, if $x=4$, then $y=3$. Since $n=11$, then Megan won 9 races and Shana won 2 races. Megan should receive $9(4)+2(3)=42$ coins and Shana should receive $2(4)+9(3)=35$ coins, which agrees with the given information.)