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The night porter

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A hotel floor has $120$ doors numbered $1$ to $120$ , all locked. On night $k$ , for $k$ from $1$ to $120$ , the porter turns the key in every door whose number is a multiple of $k$ , locking an unlocked door and unlocking a locked one. After night $120$ , what is the sum of the numbers of the unlocked doors?

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Hint

First decide which doors end unlocked, then add their numbers.

Solution

Door $n$ is turned once for each divisor of $n$ , so it ends unlocked exactly when $n$ has an odd number of divisors.
Divisors come in pairs $d$ and $\frac{n}{d}$ except when $d^{2} = n$ , so the unlocked doors are those whose numbers are perfect squares.
The squares up to $120$ are $1^{2}$ to $1 0^{2}$ , because $1 1^{2} = 121$ .
Their sum is $10 \times 11 \times \frac{21}{6} = 385$ , so the answer is $385$ .

Answer: $385$

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