2010 AMC 10B Problems/Problem 24
Represent the teams' scores as: and
We have Manipulating this, we can get , or
Since both are increasing sequences, . We can check cases up to because when , we get . When
n=2, a=[1,6] n=3, a=[1,2] n=4, a=1
Checking each of these cases individually back into the equation a+an+an^2+an^3=4a+6m+1, we see that only when a=5 and n=2, we get an integer value for m, which is 9. The original question asks for the first half scores summed, so we must find (a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=34