2010 AMC 10B Problems/Problem 24

Problem

Solution

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 $(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=34$

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2010 AMC 10B Problems/Problem 24

Problem

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