How would you go about solving this math problem?

Question

There are 1000 lockers

Student 1 Opens every locker
Student 2 Closes 2,4,6,8,10,12 etc.
Student 3 Changes the state of lockers 3,6,9,12,15,18 etc.
Student 4 Changes the state of lockers 4,8,12,16,20,24 etc.
Student 5 Changes the state of lockers 5,10,15,20,25,30 etc.
6-100 as above
When all 1000 students are done, what lockers are open? What are the total number of lockers open?

Answer 1

I like this problem!

Notice: the first locker gets hit once, the second locker gets hit twice, the third locker gets hit twice, the fourth locker gets hit three times, the fifth locker gets hit twice, the sixth locker gets hit three times, the seventh once, the eighth four times times, the ninth three times...

Each locker gets hit as many times as it has divisors. Odd number of divisors--the locker will be open; even number of divisors--the locker will be closed...

Actually, the answer is not so difficult to see using the idea of superposition. Someone opens all the lockers. Then someone comes along and, starting from #2, closes every 2nd locker. Then after this happens the number open will be 1000-1000/2. Then someone comes along and starting at the 3rd locker operns every 3rd one. So you have 1000-1000/2+1000/3 (rounded down to the nearest whole number) open lockers. The pattern continues like this.

Answer 2

2 sets of lockers - 1 open & 1 closed. Create a formula to describe the action of student 3*. Then expand the equation to include all 100 studnets.

* Students 1 & 2 don't perform an action on a set of lockers in an undetermined status - they act on lockers with a fixe/known state.

In each case, determine the lockers moved from open to closed status & vice versa.

Or you could just solve the problem, by visually checking all lockers.

Answer 3

I would figure that every door will change by the number of factors it has, assuming they were all closed to start. For example, door #8 will be changed by person 1, 2, 4, and 8, which is an even number of changes, so it will be closed. No one after 8 will change it. So you need to figure out which numbers have an odd number of factors. This only happens for perfect square numbers such as 16 (1, 2, 4, 8, 16) because the square root of that number is times itself. So, all doors will be closed EXCEPT 1, 4, 9, 16, 25, etc.

Does that make sense?

Answer 4

get out a big piece of paper and draw what it would look like using different markers for the lockers and student or i would cheat and look at the other answers

Answer 5

Here's a little Matlab code that will do this job:
-------------------------
A=zeros(1000,1);
for i=1:1000% i = student
for j = i:i:1000% j = door
A(j) = 1-A(j);% change state of j'th door
end
end
disp('Number of open doors')
disp(sum(A))
--------------------------
The answer is 31 open doors

Answer 6

I would flirt with the smart kid in my class to get the answer. If
that person is a girl, I would bribe her.

Answer 7

that's a cool problem but u'll have to spend some time on it!!!