MATH HELP PLEASE!!! door problem?

Question

ok my algebra teacher gave me this word problem to do, and I AM TOTALLY CONFUSED. everything i try doesnt work. someone help:

Every morning, Mike the security guard at CP high school opens all 100 doors in the building. Let's assume the doors are #'d 1-100. The next security guard closes all even numbered doors. The third security guard touches all doors that are multiples of 3. If a door is open, he'll close it and vice versa. The fourth guard changes the position of every fourth door (if it's open he'll close it etc.,) and the fifth guard changes the position of every fifth door and so on, until the 100th guard changes only door 100.

HOW MANY DOORS ARE LEFT OPEN IN THE END?

you also need to explain which doors would be open if the doors went up to 1000. Assume the last guard changes only door #1000.



whew, so that's the problem
someone help!

Answer 1

The first security guard will touch all lockers.
The second security guard will touch all even lockers (multiples of 2)
The third security guard will touch all lockers that are multiples of 3.
etc.

Now think of a specific locker, say 12.

It will be touched by guard 1, guard 2, guard 3, guard 4, guard 6 and guard 12. Notice how the guard numbers correspond to the factors of the locker number.

And how many factors are there in a number? Generally most numbers have an even number of factors because they pair up. 1 goes with 12, 2 goes with 6 and 3 goes with 4.

So in general most lockers will be opened and closed an even number of times resulting in a locker being closed.

The only exceptions are the perfect squares. Say the number 9. It will have factors of 1, 3 and 9. The 1 and 9 pair up, but the 3 "pairs up" with itself.

So all the lockers 1, 4, 9, 16, 25, etc. will remain open at the end.

Now all you need to do is figure how many perfect squares are less than or equal to 100 and 1000.

Since sqrt(100) = 10, that tells that you will have 10 lockers in the first case.

Since sqrt(1000) = 31.6227766, that tells you that you have 31 lockers in the second case.

1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100

With 100 lockers, 10 will be open

Continuing for the 1000 locker case
11² = 121
12² = 144
13² = 169
...
30² = 900
31² = 961

With 1000 lockers, 31 will be open.

Answer 2

If you look at the factors of the numbers, you see how many times it is "touched." For example, prime numbers such as 17 are only touched once, but other numbers like 21 are touched three times (in addition to itself, it has factors of 3 and 7). So you could write out all the numbers and check how many factors each has. If it has an odd number of factors (like 21 or 17) it will be closed in the end (initially open, then closed for 3, opened for 7, closed for 21). If it has an even number, it will be open in the end.

It is probably not the best way, but it might make the most sense. There is probably some algebraic trick involved. Good luck and have fun!

Answer 3

Draw 100 squares and write O for open and C for close. Just cross out the letter before each time it is changed. I may take a while and may not be solving it algebraically, but it should bring you to an answer.

Good luck with #2!

Answer 4

Good Luck

Answer 5

Stumped try a homework helping web site!

Answer 6

i am so sorry but i would tell ur teach to ******* **** his or her self in the butt