Help!!! Due on Monday >< THanks so much!!!?

Question

There are 2000 lockers at this school. They are all shut The first student comes along and opens them all The second student comes along and changes the position of the doors which are multiples of 2 i.e. shuts all the even numbers The third student comes along and alters all the ones that are multiples of 3 i.e. 3 is shut, 6 is opened, 9 is shut, 12 is opened etc The fourth alters all the ones that are multiples of 4and so on till all 2000 students have filed past At the end of all this which ones are still open Show working using diagrams, tables etc make sure you can justify your answer by your reasoning, setting out should be precise and careful headings underlined etc

i noe the answer...i tried to find the answer using a grid but it didnt turn out right...help please??? thanks so much!

Answer 1

Think about this problem using factors.

If a door is visited once, it will be open.

If it visited twice, it will be closed.

3 times=open. 4 times=closed....etc

So, odd # of times is open / even # of times is closed.

NOW...think of it in factors.

A door with a perfect square, i.e. 9, will be visited an odd # of times.

Every first door, third door & ninth door. After that, it will be left open.

If it isn't a perfect square...it will have an even # of factors, and thus will be closed.

Let's find the largest perfect square less than 2000.

40 x 40 = 1600

45 x 45 = 2025

So, it's less than 45.

44 x 44 = 1936.

Therefore, 44 doors will be left open! Those are the doors with perfect squares.

Door # 1, 4, 9, 16, 25, 36 etc...all the way up to door # 1936.

Regards,

Mysstere

Answer 2

are you sure you didnt alternate the locker numbers?