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, for 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, for 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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+ | Note: The power of two expansion can be found from the binary form of <math>2024</math>, which is <math>11111101000_2</math>. ~cxsmi | ||
==Solution 2== | ==Solution 2== |
Latest revision as of 22:57, 12 May 2024
Contents
[hide]Problem
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, for 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
Note: The power of two expansion can be found from the binary form of , which is . ~cxsmi
Solution 2
Let with .
If the maximum element of is for some , then each element in can be either in or not in . Therefore, the number of such sets is .
Therefore, the total number of sets is
Thus
Now, the problem becomes writing 4048 in base 2, say, . We have .
We have . Therefore, . Therefore, the sum of all elements in is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 |
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