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MrCoolness

answer this question and make a race with me RIGHT NOW! [ROUND 2]

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A mailman runs the same route of 1000 mail boxes every day on a crescent. One day he gets bored and decides to do his job a little differently. He goes to every single mailbox and opens it, then he returns to the beginning and closes every other (mailbox 2,4,6...) mailbox, then he returns to the beginning and does the same thing to every third mailbox (3,6,9...) and if the mailbox was open he closes it and if the mailbox was closed he opens it, then every 4th mailbox... every 5th... and so forth...

 

He repeats this until he goes to every "1000'th" mailbox, meaning he goes to the last mailbox and opens/closes it and then calls it a day.

 

When the mailman looks back at what he's done, which mailboxes are still open?

 

 

STAFF AREN'T ALLOWED TO HELP

 

 

CHANGING THE QUESTION TO 1000 MAILBOXES

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Think in factors. Well, primes would have to be closed (opened once and then closed whenever you count by itself) and squares would be open (odd number of factors).

So assuming it's all the squares from 1-1000.

1, 4, 9 , 16, 25, 36, etc etc on my phone too lazy to count them all out

Edited by TurtleFrenzy

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Being a computer science major I actually wrote a script for this in java but it was too late. Here it is if you're still interested :c https://www.dropbox.com/s/6mjt70ssqf7bfw7/Sean.java

Full number sequence would be

1

4

9

16

25

36

49

64

81

100

121

144

169

196

225

256

289

324

361

400

441

484

529

576

625

676

729

784

841

900

961

so as a formula, its n^2 where n < 31

Edited by Sean

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