Difference between revisions of "2003 AIME I Problems/Problem 3"
(→Solution) |
|||
Line 16: | Line 16: | ||
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>. | ||
+ | |||
+ | |||
+ | == Solution == | ||
+ | Thinking of this problem algorithmically, one can "sort" the array to give: | ||
+ | <cmath>{1, 2, 3, 5, 8, 13, 21, 34}</cmath> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | <cmath>(x-1) + (x-3) + (x-6) + (x-11) + (x-19) + (x-32) + (x-53) = 7x - 125</cmath> | ||
+ | |||
+ | Finding <math>x</math> isn't hard, and we see that it is equal to <math>609</math>: | ||
+ | |||
+ | <cmath>609 - 125 = \boxed{484}</cmath> | ||
== See also == | == See also == |
Latest revision as of 23:25, 4 March 2021
Contents
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
Order the numbers in the set from greatest to least to reduce error: 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 .
Solution
Thinking of this problem algorithmically, one can "sort" the array to give:
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 shebang. Then, we see that if we set the sum of the whole array to we get out answer to be
Finding isn't hard, and we see that it is equal to :
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.