Difference between revisions of "2010 AMC 10B Problems/Problem 24"
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n=3, a=[1,2] | n=3, a=[1,2] | ||
n=4, a=1 | 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 | + | Checking each of these cases individually back into the equation <math>a+an+an^2+an^3=4a+6m+1</math>, 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 <math>(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=34</math> |
Revision as of 15:56, 28 January 2011
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 , 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 
