2024 AIME II Problems/Problem 6
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
\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, .
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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