Difference between revisions of "2010 AMC 10B Problems/Problem 24"
Huangkevin57 (talk | contribs) (Created page with 'Represent the teams' scores as: a, an, an^2, an^3 a, a+m, a+2m, a+3m We have a+an+an^2+an^3=4a+6m+1 Manipulating this, we can…') |
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| − | Represent the teams' scores as: a, an, an^2, an^3 | + | Represent the teams' scores as: (a, an, an^2, an^3) and (a, a+m, a+2m, a+3m) |
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We have a+an+an^2+an^3=4a+6m+1 | We have a+an+an^2+an^3=4a+6m+1 | ||
Revision as of 19:14, 2 September 2010
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
