Difference between revisions of "2024 AIME II Problems/Problem 6"

Latest revision as of 22:57, 12 May 2024

Problem

Alice chooses a set $A$ of positive integers. Then Bob lists all finite nonempty sets $B$ of positive integers with the property that the maximum element of $B$ belongs to $A$ . Bob's list has 2024 sets. Find the sum of the elements of A.

Solution 1

Let $k$ be one of the elements in Alices set $A$ of positive integers. The number of sets that Bob lists with the property that their maximum element is k is $2^{k-1}$ , 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 $2^{10}+2^9+2^8+2^7+2^6+2^5+2^3$ . We must increase each power by 1 to find the elements in set $A$ , which are $(11,10,9,8,7,6,4)$ . Add these up to get $\boxed{055}$ . -westwoodmonster

Note: The power of two expansion can be found from the binary form of $2024$ , which is $11111101000_2$ . ~cxsmi

Solution 2

Let $A = \left\{ a_1, a_2, \cdots, a_n \right\}$ with $a_1 < a_2 < \cdots < a_n$ .

If the maximum element of $B$ is $a_i$ for some $i \in \left\{ 1, 2, \cdots , n \right\}$ , then each element in $\left\{ 1, 2, \cdots, a_i- 1 \right\}$ can be either in $B$ or not in $B$ . Therefore, the number of such sets $B$ is $2^{a_i - 1}$ .

Therefore, the total number of sets $B$ is \begin{align*} \sum_{i=1}^n 2^{a_i - 1} & = 2024 . \end{align*}

Thus \begin{align*} \sum_{i=1}^n 2^{a_i} & = 4048 . \end{align*}

Now, the problem becomes writing 4048 in base 2, say, $4048 = \left( \cdots b_2b_1b_0 \right)_2$ . We have $A = \left\{ j \geq 1: b_j = 1 \right\}$ .

We have $4048 = \left( 111,111,010,000 \right)_2$ . Therefore, $A = \left\{ 4, 6, 7, 8, 9, 10, 11 \right\}$ . Therefore, the sum of all elements in $A$ is $\boxed{\textbf{(55) }}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/vdj1kCgjHXk

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

@@ Line 3: / Line 3: @@
 ==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, 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
+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==
@@ Line 40: / Line 42: @@
 {{AIME box|year=2024|num-b=5|num-a=7|n=II}}
-[[Category:]]
+[[Category:Intermediate Combinatorics Problems]]
 {{MAA Notice}}

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