Question 182270: At hicksville high school, the students celebrate a very unusual tradition. You see there are exactly 100 students at HHS and exactly 100 lockers. Every november 23, the students celebrate what has come to be known as "Locker Day". Each of the 100 students line up and number off from 1 to 100. Student number 1 goes through the hallways and opens each of the 100 lockers. student number 2 then goes through and closes every other locker. Student number 3 follows and "reverses" every third locker. The fourth then goes through and "reverses" every fourth locker. This continues untill each of the 100 students has passed through the hallways of HHS, ending with the 100th student reversing the 100th locker. My question for you: After this is done, which of the lockers will remain open?
I need an equation and the answer to the question above.
so far i have
1---4---9---16---25---36---49---64---81---100
-+3--+5--+7---+9---+11--+13--+15--+17---+19
thanks for taking the time for reading this question and maybe even helping me answer it
P.S. this is only a math problem not a real life situation at my school
Found 2 solutions by solver91311, jim_thompson5910:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
All integers except for perfect squares have an even number of even divisors, including 1 and the integer itself. Perfect squares have an odd number of divisors because the integer square root is a divisor.
Examples:
6 has 4 factors: 1, 2, 3, and 6. Not a perfect square, even number of factors.
16 has 5 factors: 1, 2, 4, 8, 16. Perfect square, odd number of factors.
You could get all fancy and prove this fact by saying something along the lines of: If p is an even divisor of integer n, then there is an integer quotient q that is also a divisor, hence divisors come in pairs. Except for the case of a perfect square where p is the square root of n in which case the quotient is also p, making the exception to the 'divisors in pairs' rule.
If Student number s touches locker n, then s must be an even divisor of n, according to the rules of the game.
Therefore, non-perfect square number lockers are touched an even number of times. That is, if they started out closed, they end up closed. Perfect square number lockers are touched an odd number of times -- if they started closed, they end up open.
John
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website! 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.
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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.
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