Difference between revisions of "2010 AMC 10B Problems/Problem 24"
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| − | Represent the teams' scores as: (a, an, an^2, an^3) and (a, a+m, a+2m, a+3m) | + | Represent the teams' scores as: <math>(a, an, an^2, an^3)</math> and <math>(a, a+m, a+2m, a+3m)</math> |
| − | We have a+an+an^2+an^3=4a+6m+1 | + | We have <math>a+an+an^2+an^3=4a+6m+1</math> |
| − | 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 | + | Manipulating this, we can get <math>a(1+n+n^2+n^3)=4a+6m+1</math>, or <math>a(n^4-1)/(n-1)=4a+6m+1</math> |
| − | Since both are increasing sequences, n>1. We can check cases up to n=4 because when n=5, we get 156a>100. When | + | Since both are increasing sequences, <math>n>1</math>. We can check cases up to <math>n=4</math> because when <math>n=5</math>, we get <math>156a>100</math>. When |
n=2, a=[1,6] | n=2, a=[1,6] | ||
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 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 | ||
Revision as of 22:46, 25 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 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
