10 lockers are left open: Lockers #1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.

Each of these numbers are perfect squares. This problem is based on the factors of the locker number.

Each locker is toggled by each factor; for example, locker #40 is toggled on pass number 1, 2, 4, 5, 8, 10, 20, and 40. That's eight toggles: open-closed-open-closed-open-closed-open...

The only way a locker could be left open is if it is toggled an odd number of times. The only numbers with an odd number of factors are the perfect squares. Thus, the perfect squares are left open.

I found it easiest to make this problem smaller, for example 10 lockers.We end up having 3 open and 7 closed.This corresponds with the amount of square numbers that are contained in the amount of lockers.Seeing as 10^2 = 100, this means that there will be 10 lockers open and 90 closed. Q.E.D.

If you do a quick run through of the situation, you will realize that all even-factored numbers will remain closed. Why? Because for every person opening the door, there is an equal number of people closing it.

This narrows us down to the numbers which have only odd factors. Such numbers are all perfect squares, since one of their factors will be repeated (hence, a square).

There are 10 squares between 1 and 100.

python code for that:-

locked=1

open=0

a=[]

for i in range (0,101):

a.append(1)

for i in range (1,101):

for j in range (1,101):

    if (j%i):

        a[j]=a[j]^1

count=0

for i in range (1,101):

if(a[i]==0):

    count=count + 1

print "total number of doors left open is %d" %count