The hundred lockers, step by step
A hundred lockers in a row, all shut.
In the end, how many doors are left open?
Open only where the count of divisors is odd
- Start from the only number given: 56 doors in a row, every one of them closed. Nothing is open yet.
- Each door numbered L is changed once for every number that divides L — once per divisor of L. Divisors come in pairs, d with L ÷ d, so the count is even and the door ends closed. That is the trap: it looks as if every door pairs off and NONE stays open — the answer 0.
- But one kind of number breaks the pairing: a perfect square. Its middle divisor, the √, pairs with itself and is counted once, so a square has an ODD number of divisors and stays open. So the open doors are exactly the perfect squares up to 56: 1, 4, 9, 16, …, 49.
- Count them: 1×1, 2×2, and so on up to 7×7 = 49, which is the largest square that fits. That is 7 of them, so 7 doors stay open.
How it works.
- 01
Start from the only number given: 56 doors in a row, every one of them closed. Nothing is open yet.
- 02
Each door numbered L is changed once for every number that divides L — once per divisor of L. Divisors come in pairs, d with L ÷ d, so the count is even and the door ends closed. That is the trap: it looks as if every door pairs off and NONE stays open — the answer 0.
- 03
But one kind of number breaks the pairing: a perfect square. Its middle divisor, the √, pairs with itself and is counted once, so a square has an ODD number of divisors and stays open. So the open doors are exactly the perfect squares up to 56: 1, 4, 9, 16, …, 49.
- 04
Count them: 1×1, 2×2, and so on up to 7×7 = 49, which is the largest square that fits. That is 7 of them, so 7 doors stay open.