21
Floydian
10d

# 1. 1000 people are standing in line 2. All of them are numbered from 1 to 1000 3. All odds are removed 4. Line moves forward 5. Repeat from step 2 until last person standing Question: what was the original number/position of the last person standing? Consider array starting at 1.

• 8
1000
The last number will always be an even number, but all others will change. Until you reach the last standing person, it is by then his turn to be an odd number
• 9
512 because it is the biggest multiple of 2 in this range so it will divide by 2 the longest
• 14
@gitpush
1000/2 =500
500/2=250
250/2=125 <- odd number after 3 steps
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@unDone aaah that missed me, thanks for explaining it :D
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Congratulations to @unDone and @gitlog for the correct answer :D

@gitpush hahaha that was a near miss maybe?
• 4
Same reduction as 1000 locker problem.
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512 was correct?
Who is gitlog?
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@gitlog Fuck off from here and go study..

You got exams coming up you shit fuck.
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@gitpush probably was thinking about 1024
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@gitpush If anyone would be way of I wouldn't think it'd be you 😅
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@Floydian it was miscalculation 🤦🏼‍♂️
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But what about step 4 (line moving forward)? I'll assume it proceeds by one person, then any power of two will be removed in second loop. Last standing number needs to have binary representation 11…10 so that in every loop after step 3 it will turn to 11...1 and in step 4 to 11…10.

The biggest such number is 111111110 = 510
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@Floydian 4th step says "Line moves forward", doesn't it mean that the first person leaves the line?
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@SgnfcntOverflow I was confused by that too, but I guess he means that people go forward just to fill the gaps
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