2024 AIME II Problems/Problem 6: Difference between revisions
Created page with "Alice" |
No edit summary |
||
| Line 1: | Line 1: | ||
Alice | Alice chooses a set <math>A</math> of positive integers. Then Bob lists all finite nonempty sets <math>B</math> of positive integers with the property that the maximum element of <math>B</math> belongs to <math>A</math>. Bob's list has 2024 sets. Find the sum of the elements of A. | ||
==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 | |||
Revision as of 20:21, 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
