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, 2021

Combinatorics Difficulty 5.8 AIME, harder Prove it United States

Problem:

There are $N$ lockers, labeled from $1$ to $N$ , placed in clockwise order around a circular hallway. Initially, all lockers are open. Ansoon starts at the first locker and always moves clockwise. When she is at locker $n$ and there are more than $n$ open lockers, she keeps locker $n$ open and closes the next $n$ open lockers, then repeats the process with the next open locker. If she is at locker $n$ and there are at most $n$ lockers still open, she keeps locker $n$ open and closes all other lockers. She continues this process until only one locker is left open. What is the smallest integer $N > 2021$ such that the last open locker is locker $1$ ?

Solution

Solution:

Note that in the first run-through, we will leave all lockers $2^{n}-1$ open. This is because after having locker $2^{n}-1$ open, we will close the next $2^{n}-1$ lockers and then start at locker $2^{n}-1+2^{n}-1+1=2^{n+1}-1$ . Now we want $1$ to be the last locker that is open. We know that if $N<2046$ , then closing $1023$ lockers after $1023$ will lead us to close locker $1$ . However, if $N=2046$ , then locker $1$ will stay open, $3$ will close, $7$ will stay open, closing the next $10$ and then $1$ stays open and we close locker $7$ , therefore $N=2046$ does work.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.