Difference between revisions of "2003 AIME I Problems/Problem 3"

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== Problem ==
 
== Problem ==
Let the set <math> \mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}. </math> Susan makes a list as follows: for each two-element subset of <math> \mathcal{S}, </math> she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list.
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Let the [[set]] <math> \mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}. </math> Susan makes a list as follows: for each two-element subset of <math> \mathcal{S}, </math> she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list.
  
 
== Solution ==
 
== Solution ==
 
Each [[element]] of the [[set]] will appear in <math>7</math> two-element [[subset]]s, once with each other number.
 
Each [[element]] of the [[set]] will appear in <math>7</math> two-element [[subset]]s, once with each other number.
  
<math>34</math> will be the greater number in <math>7</math> subsets.  
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*<math>34</math> will be the greater number in <math>7</math> subsets.
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*<math>21</math> will be the greater number in <math>6</math> subsets.
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*<math>13</math> will be the greater number in <math>5</math> subsets.
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*<math>8</math> will be the greater number in <math>4</math> subsets.
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*<math>5</math> will be the greater number in <math>3</math> subsets.
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*<math>3</math> will be the greater number in <math>2</math> subsets.
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*<math>2</math> will be the greater number in <math>1</math> subsets.
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*<math>1</math> will be the greater number in <math>0</math> subsets.  
  
<math>21</math> will be the greater number in <math>6</math> subsets.  
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Therefore the desired sum is <math>34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=\boxed{484}</math>.
  
<math>13</math> will be the greater number in <math>5</math> subsets.
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== See also ==
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{{AIME box|year=2003|n=I|num-b=2|num-a=4}}
  
<math>8</math> will be the greater number in <math>4</math> subsets.
 
 
<math>5</math> will be the greater number in <math>3</math> subsets.
 
 
<math>3</math> will be the greater number in <math>2</math> subsets.
 
 
<math>2</math> will be the greater number in <math>1</math> subsets.
 
 
<math>1</math> will be the greater number in <math>0</math> subsets.
 
 
Therefore the desired sum is:
 
 
<math> \displaystyle 34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=484</math>
 
 
== See also ==
 
* [[2003 AIME I Problems/Problem 2 | Previous problem]]
 
* [[2003 AIME I Problems/Problem 4 | Next problem]]
 
* [[2003 AIME I Problems]]
 
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]

Revision as of 14:59, 10 June 2008

Problem

Let the set $\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}.$ Susan makes a list as follows: for each two-element subset of $\mathcal{S},$ she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list.

Solution

Each element of the set will appear in $7$ two-element subsets, once with each other number.

  • $34$ will be the greater number in $7$ subsets.
  • $21$ will be the greater number in $6$ subsets.
  • $13$ will be the greater number in $5$ subsets.
  • $8$ will be the greater number in $4$ subsets.
  • $5$ will be the greater number in $3$ subsets.
  • $3$ will be the greater number in $2$ subsets.
  • $2$ will be the greater number in $1$ subsets.
  • $1$ will be the greater number in $0$ subsets.

Therefore the desired sum is $34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=\boxed{484}$.

See also

2003 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions