AIME 2024 II · 第 6 题

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)