Factor to 100

If person 1 opens every locker and person 2 closes every $2^{nd}$ locker then every odd lockers are open and every even lockers are closed. Then person 3 and onward persons has to do nothing because always an odd multiple of any number will remain open and even multiple of any number will remain closed. I assume that the question says that 'every person opens the closed locker and closes the opened locker he encounter in multiples of his number. And at very beginning all the lockers are closed. For example, person 5 will encounter lockers 5, 10, 15, 20, . . . , 100 so he close every opened locker among these and open every closed locker.'

Person 1 open all the lockers because for him all the lockers are closed. Then person 2 close every 2nd locker. Person 3 open every closed locker with multiple of 3 and close every opened locker with the multiple of 3 and so on. In this way, locker no. $n$ will be encountered by all the divisors of $n$ . The $n^{th}$ locker will remain open if the number of divisors of $n$ is odd.

Let $n$ has the prime factorization as : $n = p_1^{a_1}p_2^{a_2}p_3^{a_3}\cdots p_m^{a_m}\hspace{10pt}$ . Then the number of divisors is given by $d(n) = (a_1+1)(a_2+1)(a_3+1)\cdots (a_m+1)$

Now if $n$ is a non square number then at least one of the $a_i$ is odd which makes at least one of the factor $a_i+1$ even and hence $d(n)$ even. So the locker with non square number will be closed. Now if $n$ is a square number then all the $a_i$ are even so that all the $a_i+1$ are odd which makes $d(n)$ odd. So all the locker with square number are remain open. Hence the answer is 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 lockers are open at last.

Well 10 lockers stay open, and only one answer has ten numbers in it. Therefore that is the answer.