2010 AMC 10B Problems/Problem 24

Represent the teams' scores as: $(a, an, an^2, an^3)$ and $(a, a+m, a+2m, a+3m)$

We have $a+an+an^2+an^3=4a+6m+1$ Manipulating this, we can get $a(1+n+n^2+n^3)=4a+6m+1$ , or $a(n^4-1)/(n-1)=4a+6m+1$

Since both are increasing sequences, $n>1$ . We can check cases up to $n=4$ because when $n=5$ , we get $156a>100$ . 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 $(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=34$

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2010 AMC 10B Problems/Problem 24