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Difference between revisions of "2010 AMC 10B Problems/Problem 24"
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== Problem ==
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== Solution ==
 
Represent the teams' scores as: <math>(a, an, an^2, an^3)</math> and <math>(a, a+m, a+2m, a+3m)</math>
 
Represent the teams' scores as: <math>(a, an, an^2, an^3)</math> and <math>(a, a+m, a+2m, a+3m)</math>
  
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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  
 
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  
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          n=2, a=[1,6]
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* <math>n=2, a=[1,6]</math>
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          n=3, a=[1,2]
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*<math>        n=3, a=[1,2]</math>
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          n=4, a=1
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*<math>          n=4, a=1</math>
 
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>
 
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>
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== See also ==
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{{AMC10 box|year=2010|ab=B|num-b=23|num-a=25}}

Revision as of 17:08, 28 January 2011

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

See also

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
All AMC 10 Problems and Solutions
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