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Difference between revisions of "2003 AIME I Problems/Problem 3"
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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>.
 
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>.
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Note: Note that <math>7+6+5+4+3+2+1=\binom{8}{2}</math>, so we have counted all the possible cases.
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~Yiyj1
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== Solution ==
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Thinking of this problem algorithmically, one can "sort" the array to give:
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<cmath>{1, 2, 3, 5, 8, 13, 21, 34}</cmath>
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 +
Now, notice that when we consider different pairs, we are only going to fixate one element and look at the all of the next elements in the array, basically the whole <math>j = i + 1</math> shebang. Then, we see that if we set the sum of the whole array to <math>x,</math> we get out answer to be
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<cmath>(x-1) + (x-3) + (x-6) + (x-11) + (x-19) + (x-32) + (x-53) = 7x - 125</cmath>
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Finding <math>7x</math> isn't hard, and we see that it is equal to <math>609</math>:
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<cmath>609 - 125 = \boxed{484}</cmath>
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 23:09, 8 January 2024

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

Order the numbers in the set from greatest to least to reduce error: $\{34, 21, 13, 8, 5, 3, 2, 1\}.$ 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 $34\cdot7+21\cdot6+13\cdot5+8\cdot4+5\cdot3+3 \cdot2+2\cdot1+1\cdot0=\boxed{484}$.

Note: Note that $7+6+5+4+3+2+1=\binom{8}{2}$, so we have counted all the possible cases.

~Yiyj1

Solution

Thinking of this problem algorithmically, one can "sort" the array to give: \[{1, 2, 3, 5, 8, 13, 21, 34}\]

Now, notice that when we consider different pairs, we are only going to fixate one element and look at the all of the next elements in the array, basically the whole $j = i + 1$ shebang. Then, we see that if we set the sum of the whole array to $x,$ we get out answer to be

\[(x-1) + (x-3) + (x-6) + (x-11) + (x-19) + (x-32) + (x-53) = 7x - 125\]

Finding $7x$ isn't hard, and we see that it is equal to $609$:

\[609 - 125 = \boxed{484}\]

See also

2003 AIME I (Problems • Answer Key • Resources)
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

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