Difference between revisions of "2024 AIME II Problems/Problem 6"
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==Solution 1== | ==Solution 1== | ||
Let <math>k</math> be one of the elements in Alices set <math>A</math> of positive integers. The number of sets that Bob lists with the property that their maximum element is k is <math>2^{k-1}</math>, since every positive integer less than k can be in the set or out. Thus, fore the number of sets bob have listed to be 2024, we want to find a sum of unique powers of two that can achieve this. 2024 is equal to <math>2^10+2^9+2^8+2^7+2^6+2^5+2^3</math>. We must increase each power by 1 to find the elements in set <math>A</math>, which are <math>(11,10,9,8,7,6,4)</math>. Add these up to get <math>\boxed{055}</math>. -westwoodmonster | Let <math>k</math> be one of the elements in Alices set <math>A</math> of positive integers. The number of sets that Bob lists with the property that their maximum element is k is <math>2^{k-1}</math>, since every positive integer less than k can be in the set or out. Thus, fore the number of sets bob have listed to be 2024, we want to find a sum of unique powers of two that can achieve this. 2024 is equal to <math>2^10+2^9+2^8+2^7+2^6+2^5+2^3</math>. We must increase each power by 1 to find the elements in set <math>A</math>, which are <math>(11,10,9,8,7,6,4)</math>. Add these up to get <math>\boxed{055}</math>. -westwoodmonster | ||
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| + | ==See also== | ||
| + | {{AIME box|year=2024|num-b=5|num-a=7|n=II}} | ||
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| + | [[Category:]] | ||
| + | {{MAA Notice}} | ||
Revision as of 20:52, 8 February 2024
Alice chooses a set 
of positive integers. Then Bob lists all finite nonempty sets 
of positive integers with the property that the maximum element of belongs to . Bob's list has 2024 sets. Find the sum of the elements of A.
Solution 1
Let 
be one of the elements in Alices set of positive integers. The number of sets that Bob lists with the property that their maximum element is k is 
, since every positive integer less than k can be in the set or out. Thus, fore the number of sets bob have listed to be 2024, we want to find a sum of unique powers of two that can achieve this. 2024 is equal to 
. We must increase each power by 1 to find the elements in set , which are 
. Add these up to get 
. -westwoodmonster
See also
| 2024 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
[[Category:]]
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
