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

 
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== Problem ==
 
== Problem ==
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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 will appear in <math>7</math> two element subsets. (Once with each other number.)
 +
 +
<math>34</math> will be the greater number in <math>7</math> subsets.
 +
 +
<math>21</math> will be the greater number in <math>6</math> subsets.
 +
 +
<math>13</math> will be the greater number in <math>5</math> subsets.
 +
 +
<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.
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<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.
 +
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<math>1</math> will be the greater number in <math>0</math> subsets.
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Therefore the sum is:
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<math> \displaystyle 34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=484</math>
  
 
== See also ==
 
== See also ==
 
* [[2003 AIME I Problems]]
 
* [[2003 AIME I Problems]]

Revision as of 20:00, 15 July 2006

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 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 sum is:

$\displaystyle 34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=484$

See also

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