A school has 100 students and 100 lockers numbered from 1 to 100. Initially, all lockers are closed. The first student passes and opens all the lockers; the second passes and closes all the even-numbered ones; the third passes and changes the position of all multiples of 3, that is, he closes the ones that are open and opens the ones that are closed; the fourth student changes the position of all lockers with numbers that are multiples of 4; and so on until the hundredth student, who changes the position of the lockers with numbers that are multiples of 100. After all the students have passed, how many lockers remain closed?
Solution

The lockers that remain open are those with numbers that have an odd number of divisors, and this only happens with numbers that are perfect squares. From 1 to 100, we have 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100, which are perfect squares, that is, there are 10 perfect squares. Therefore, the number of lockers that remain closed is $100-10=90$.