A series of lockers, numbered 1 through 100, are all initially closed. Student 1 goes through and opens every locker. Student 3 goes through and "flips" every 3rd locker ("fipping") a locker means changing its state: if the locker is open he closes it, and if the locker is closed he opens it. Student 5 then goes through and "flips" every 5th locker. This process continues with all students with odd numbers $n < 100$ going through and "flipping" every $n$th locker. How many lockers are open after this process?
1. **Understanding the Problem:**
   - We have 100 lockers initially closed.
   - Students with odd numbers (1, 3, 5, ..., 99) will flip the state of lockers at intervals corresponding to their number.
   - We need to determine how many lockers remain open after all students have passed.

2. **Key Insight:**
   - A locker will end up open if it is flipped an odd number of times.
   - A locker is flipped by a student if the student's number is a divisor of the locker's number.
   - Therefore, a locker will be flipped an odd number of times if it has an odd number of divisors.

3. **Odd Number of Divisors:**
   - A number has an odd number of divisors if and only if it is a perfect square. This is because divisors generally come in pairs, except when the number is a perfect square (e.g., 1, 4, 9, 16, etc.).

4. **Finding Perfect Squares:**
   - We need to find all perfect squares between 1 and 100.
   - The perfect squares in this range are: \(1^2, 2^2, 3^2, \ldots, 10^2\).

5. **Listing Perfect Squares:**
   - \(1^2 = 1\)
   - \(2^2 = 4\)
   - \(3^2 = 9\)
   - \(4^2 = 16\)
   - \(5^2 = 25\)
   - \(6^2 = 36\)
   - \(7^2 = 49\)
   - \(8^2 = 64\)
   - \(9^2 = 81\)
   - \(10^2 = 100\)

6. **Counting Perfect Squares:**
   - There are 10 perfect squares between 1 and 100.

Conclusion:
- Therefore, there are 10 lockers that remain open after all students have passed.

The final answer is \(\boxed{10}\)