Math Trivia
posted by Josh .
1) A school has 1 closed lockers. Larry comes into the school and "toggles" every second locker, then Fred toggles every 3rd, Bob every 4th, etc, up to Zoe who toggles every 100th locker. After all this"toggling" how many lockers are open?
2) If you take any valid time from a 12 hour clock, what is the maximum sum you can obtain by adding the digits? (Eg. For 7:14, the sum is 7+1+4=12)
THANKS!

2)
Isn't it 12:59? 
yah thanks!
i am having a hard time with number one. i know i could sit down and take a long time to do it...but i was wondering if their was a formula to it, or a quicker way to do it... 
make a list of lockers up to whatever you feel like
put c for closed under each one
starting with 2,4,6,8,...switch the c to o, for open
starting with 3,6,9,12,.. switch to c's to o's, and the o's to c's
starting with 4,8,12,16,... switch the c's to o's, and the o's to c's
continue....
you will notice that only the perfect square numbers will be c's
so which are the perfect squares ?
1,4,9,16,25,36,49,64,81,and 100
somebody actually spent time and effort to create an applet that shows this pattern.
only in this problem the first student starts by switching all the lockers, so in their case all the lockers are initially open.
(For some reason on my computer it seems to jump to the final stage almost right away)
(Broken Link Removed) 
I don't know how to do number 1 since it doesn't say how many lockers there are initially. It says "1," but not sure if that should be something else.

Question 1
This has to do with the number of factor of an integer. An even number of factors will cause the door to be toggled (open/close) an even number of times, therefore they will remain closed.
For example, a prime number has two factors:
7=(1,7), 13=(1,13).
Almost all other numbers have an even number of factors:
24=(1,2,3,4,6,8,12,24),
88=(1,2,4,8,11,22,44,88)
You will notice that the product of the first and the last factors give the number itself.
Since the factors are always different, there is an even number of them... except in the case of perfect squares, where the middle factors are identical, so we count them only once, and the locker will be visited an odd number of times, leaving them open.
Examples:
9=(1,3,9)
81=(1,3,9,27,81)
100=(1,2,5,10,20,50,100), etc.
Question 2:
If we were to add all digits separately, I would suggest 9:59 which gives a sum of 23 as the highest sum. 
MathMate is obviously correct. 9:59 yields a greater sum than 12:59.
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