Bulb Switcher

Bulb i is toggled once for every positive divisor of i. Most divisors occur in pairs d and i divided by d, so they create an even number of toggles and leave the bulb off. A perfect square is the exception because its square root pairs with itself and is counted once, giving an odd number of divisors; those bulbs remain on. Therefore the answer is simply the number of perfect squares less than or equal to n, which is floor(sqrt(n)). For n = 10, bulbs 1, 4, and 9 remain on, so the answer is 3. Explain the divisor pairing rather than relying only on examples. Be mindful of floating-point square-root rounding for large integer inputs; in fixed-width languages, verify around the computed root if necessary. This solution is O(1) time and O(1) space under the usual arithmetic model, compared with an O(n log n) toggle simulation.