I knew the answer from somewhere in the back of my brain but couldn't articulate why fast enough.
Start by clarifying the problem: each person i toggles bulbs that are multiples of i. Recognize that a bulb ends up on if it is toggled an odd number of times, which happens exactly when it has an odd number of divisors. Conclude that only perfect squares have an odd number of divisors, so the bulbs left on are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
Pro tip: Explain the divisor pairing insight: divisors come in pairs (d, n/d), except when n is a perfect square, where the square root is unpaired. This shows why only perfect squares have an odd number of toggles.
Restate the process: person i toggles every i-th bulb. Confirm that toggling means switching between on and off, and that all bulbs start off.
A bulb is toggled once for each divisor of its number. It ends up on if the total number of toggles is odd.
For most numbers, divisors pair up (e.g., 2 and 3 for 6), giving an even count. Only perfect squares have an odd number of divisors because their square root is a divisor that pairs with itself.
The bulbs with an odd number of divisors are the perfect squares up to 100: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
Conclude that these 10 bulbs are on because they are toggled an odd number of times, while all others are toggled an even number of times and remain off.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.