Question 182270
Lets say we only have 10 people and 10 lockers. 


If person #1 opens all of the lockers, they're all open. 


Now person #2 goes, and all of the even numbered lockers are shut (2, 4, 6, etc..). 


Now it's #3's turn: locker 3 is shut, locker 6 is open again, and 9 is shut. 



Student four takes a shot and locker 4 and 8 are reopened. 



Student #5 goes and 5 and 10 are shut. 


Students #6, #7, #8, and 10 only shut locker 6, 7, 8, and 10 respectively; while person 9 opens #9. 



---------------------------------


So if we look at say locker #6, person 1 opens it, person 2 closes it, person 3 opens it, and finally person 6 closes it. 


So there are 4 people who interact with it (notice how its an even number). While with locker 4, only person 1,2, and 4 touch it, and it stays open. 



So by this reasoning, if an even number of people touch a given locker, it stays closed. If an odd number of people touch a given locker, it stays open. 





So let's list the factors of each locker number (from 1 to 10):


1: 1
2: 1, 2
3: 1, 3
4: 1, 2, 4
5: 1, 5
6: 1, 2, 3, 6
7: 1, 7
8: 1, 2, 4, 8
9: 1, 3, 9
10: 1, 2, 5, 10


Notice how ALL of the perfect squares (1, 4, 9,...) have an odd number of factors, while all of the other numbers have an odd number of factors.



So using the logic given above, this tells us that ALL of the perfect square lockers will be open. 






Now extend this logic to lockers 1-100


Here are the perfect squares from 1-100: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100



So this means that lockers 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 (all of the perfect squares from 1 to 100) will be open while all of the rest will stay shut.



Let me know if this makes sense.