[Daily puzzle] Locker problem \ stacker news ~science

1138 sats \ 12 comments \ @south_korea_ln 2 Nov science

You have 100 lockers, all closed, and 100 students. The first student goes and opens every locker. The second student toggles every second locker (i.e., closes lockers 2, 4, 6, …, 100). The third student toggles every third locker (i.e., opens if closed, closes if open, for lockers 3, 6, 9, …, 99), and so on, until all 100 students have passed.

At the end, which lockers will remain open?

Previous iteration: #749797 (several stackers answered correctly this time). I also updated the full answer to the integral problem from two days ago (see #750702).

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264 sats \ 8 replies \ @Scroogey 2 Nov

Lockers 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 are open at the end, all others are closed.

I.e. the square numbers!

Code

1 sat \ 0 replies \ @cryotosensei 3 Nov

I was about to confidently type 1, but I neglected the description about how people needed to open the lockers when they were closed during their turn. Darn it!

1 sat \ 5 replies \ @Undisciplined 2 Nov

I'd love some intuition for this result.

I get that the primes will all be closed, because only the corresponding prime'th student will toggle that locker, but I don't see a clever argument for just the squares being left.

386 sats \ 2 replies \ @Scroogey 2 Nov

I think the answer is in Why do perfect squares have a odd amount of factors

The squares are exactly those numbers with an odd amount of factors. Because the amount is odd, they are now open.

1 sat \ 1 reply \ @Undisciplined 2 Nov

That makes a ton of sense

12 sats \ 0 replies \ @south_korea_ln OP 2 Nov

Indeed! 👍

10 sats \ 1 reply \ @south_korea_ln OP 2 Nov

There is an intuitive argument. But I'll give it some more time before divulging it.

1 sat \ 0 replies \ @Undisciplined 2 Nov

Fair enough

1 sat \ 0 replies \ @south_korea_ln OP 2 Nov

TIL: a one liner to toggle a variable between values of 0 and 1. Elegant.

a[j] = 1 - a[j]