The problem
100 closed lockers. Person 1 toggles every locker. Person 2 toggles every 2nd. Person 3 every 3rd, and so on to person 100. Which lockers are open at the end?
The answer
The perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. Ten lockers.
Why
Locker n is toggled once by each person whose number divides n. So locker n ends open exactly when n has an odd number of divisors.
Divisors come in pairs: if d divides n, so does n/d. For 12: (1,12), (2,6), (3,4) - six divisors, even, so locker 12 ends closed.
The pairing fails in exactly one case: when d equals n/d, which means n = d^2. A perfect square has one divisor that pairs with itself, making the total odd.
16 has divisors 1, 2, 4, 8, 16 - five of them, because 4 is its own partner.
Saying it well
The clean one-liner: "Divisors pair up, so the count is even unless the number is a perfect square, where the square root pairs with itself."
That sentence is the whole solution and it demonstrates you understand the structure rather than having spotted a pattern in the first few cases.
Follow-ups
How many for 1000 lockers? floor(sqrt(1000)) = 31.
Which locker is toggled most? The one with the most divisors under 100 - that is 96, with 12 divisors (and 60 and 72 also have 12).
What if people 1 to 50 only? Then divisors above 50 never fire, breaking the pairing argument. Locker n ends open when it has an odd number of divisors at most 50 - a genuinely different and more interesting problem.
That last variant is a good test of whether you understood the mechanism or memorised the answer.
More in brainteasers and combinatorics.