Difference between revisions of "2003 AIME I Problems/Problem 3"
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== Solution == | == Solution == | ||
− | Each element will appear in <math>7</math> two element | + | 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. | <math>34</math> will be the greater number in <math>7</math> subsets. | ||
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<math>1</math> will be the greater number in <math>0</math> subsets. | <math>1</math> will be the greater number in <math>0</math> subsets. | ||
− | Therefore the sum is: | + | 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> | <math> \displaystyle 34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=484</math> |
Revision as of 18:25, 4 August 2006
Problem
Let the set Susan makes a list as follows: for each two-element subset of
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 two-element subsets, once with each other number.
will be the greater number in
subsets.
will be the greater number in
subsets.
will be the greater number in
subsets.
will be the greater number in
subsets.
will be the greater number in
subsets.
will be the greater number in
subsets.
will be the greater number in
subsets.
will be the greater number in
subsets.
Therefore the desired sum is: