A new school has exactly 1,000 lockers and exactly 1,000 students. On the first day of school, the students meet outside the building and agree on the following plan: the first student will enter the school and open all the lockers. The second student will then enter the school and close every locker with an even number \((2,4,6,8\), etc.). The third student will then reverse every third locker \((3,6,9,12\), etc.). That is if the locker is closed, he or she will open it; if it is open, he or she will close it. The fourth student will then reverse every fourth locker, and so on until all 1000 students in turn have entered the building and reversed the proper lockers. Which lockers will finally remain open?

Step by step solution

01

Understand the pattern and factor relationship

We need to remember that lockers only change their status when a student passes it (based on the student number as a factor of the locker number). So, for a locker to be open at the end of the process, it needs to be passed by students (changed status) for an odd number of times.

02

Examine the factors

Observe the factor pairs of lockers: - Locker 1: (1) - Locker 2: (1,2) - Locker 3: (1,3) - Locker 4: (1,2) and (2,2) - Locker 5: (1,5) - Locker 6: (1,2) and (3,6) - Locker 7: (1,7) - Locker 8: (1,2) and (2,4) Analyzing the pattern here, it is evident that the lockers with an odd number of factors remain open. We also notice that only perfect squares satisfy this condition.

03

Identify perfect squares

List perfect squares up to 1,000: - 1^2 = 1 - 2^2 = 4 - 3^2 = 9 - ... - (the largest perfect square that is less than or equal to 1000) 31^2 = 961

04

Write the solution

Lockers corresponding to the perfect squares will remain open since they have an odd number of factors. In this case, lockers 1, 4, 9, ..., 961 will remain open.

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