1000 people are standing in a circle, each has a name , 1,2,.. to 1000. 2 is on the right of 1, 3 on the right of 2 .. and so on, and finally, on right of 1000 is again 1.
1 has a sword. He kills 2 and hands over the sword to 3, 3 kills 4 and hands it to 5, and so on. Then 999 kills 1000 and hands it to 1. 1 then kills 3 and gives it to 5 and this process keeps on continuing.
So, which number lasts in the end ?

Question

1000 people are standing in a circle, each has a name , 1,2,.. to 1000. 2 is on the right of 1, 3 on the right of 2 .. and so on, and finally, on right of 1000 is again 1.
1 has a sword. He kills 2 and hands over the sword to 3, 3 kills 4 and hands it to 5, and so on. Then 999 kills 1000 and hands it to 1. 1 then kills 3 and gives it to 5 and this process keeps on continuing.
So, which number lasts in the end ?

anonymous · Answer

this is like the open-close door thingy

anonymous · Answer

so i would say the ones in the end are the ones with numbers that are perfect squares

shubhamsrg · Answer

what is open-close door thingy? :|

anonymous · Answer

i am not sure it is the open closed door problem

anonymous · Answer

for example, if instead of 1000 there are 10, i think the last person standing is 5 (if i read the question correctly)

texaschic101 · Answer

all the even numbers are knocked out on the first go around

anonymous · Answer

it does look like it. it looks like 1 is open. 2 is close. 3 is open. and so on

texaschic101 · Answer

then every other odd number starting with 3

anonymous · Answer

here is one version of the door problem, sometimes instead it is lights turning on and off http://www.techinterview.org/post/526370758/100-doors-in-a-row

shubhamsrg · Answer

5 would have been right for 10 people.

anonymous · Answer

i think (although i certainly could be wrong) that the difference here is no one comes back to life

shubhamsrg · Answer

yes, no one comes back to life .-.

anonymous · Answer

whereas in the doors they open and shut repeatedly

anonymous · Answer

well being a bonehead i have tried it with 10, 12 and 14 and got 5, 9 and 13 respectively
so no pattern has jumped out at me yet

anonymous · Answer

and 16 leaves me with 1

ganeshie8 · Answer

looks 1 wil stay alive as long as 'total group count' is even

anonymous · Answer

i don't think so

anonymous · Answer

when i did it with 10, 5 was left

anonymous · Answer

i know for sure 1 is a survivor though

texaschic101 · Answer

after the first round there is nothing left but odd numbers. On the second round, every other odd person is gone, so would you just keep dividing by 2 till you get to the last number ?

ganeshie8 · Answer

i mean, when we do with 10
1 wil stay alive first round cuz 10 is even
1 will die second round cuz 10/2 = 5

anonymous · Answer

ooh

anonymous · Answer

oh wait. that was the answer...my answer is 1...since no one can kill 1...i think

shubhamsrg · Answer

1 is not right.

anonymous · Answer

aw

anonymous · Answer

yeah one always stays around for the first pass, evens go

ganeshie8 · Answer

with 1000, he wil stay alive till the group count becomes 125

anonymous · Answer

moving right along...
10 : 5
12 : 9
14 : 13
18 : 1
20 : 9

anonymous · Answer

22 : 13

shubhamsrg · Answer

missed 16 ?

anonymous · Answer

16 : 1

texaschic101 · Answer

There is probably a formula for this

anonymous · Answer

i think i made a mistake somewhere

anonymous · Answer

10: 5
12 ; 9
14 : 13
16 ; 1
18 ; 5
20 ; 9 
22 : 13

anonymous · Answer

24 : 17
hmm  maybe it is coming ...

ganeshie8 · Answer

10 to 30 :-

ganeshie8 · Answer

no pattern really

shubhamsrg · Answer

I perhaps see a pattern
is it 31 : 31
and 32 : 1 ?
I am just predicting on the basis of a pattern. Please confirm ?

anonymous · Answer

looks pretty patterny
is 32 : 1 ?

ganeshie8 · Answer

Correct !!

anonymous · Answer

oh, what @shubhamsrg  said

shubhamsrg · Answer

hmm
for powers of 2, it comes back to 1 !!
and increment of 2 takes place after that!

shubhamsrg · Answer

so 1 at 512 , and 1024 
guess we should take case of 1024 and try to revert back ?

shubhamsrg · Answer

nah, with 512 is easier

ganeshie8 · Answer

wow !

anonymous · Answer

start over at powers of 2

shubhamsrg · Answer

488 terms after 512
so final ans is 488*2 +1 = 977
voila! that is correct!
too good.

anonymous · Answer

man i am late in the game

anonymous · Answer

of course the real question is not 'what' but "why"

shubhamsrg · Answer

I had asked the same question 1 year ago. a programming solution was proposed then. thats how i know 977 is the right ans. http://openstudy.com/users/shubhamsrg#/updates/4ff72322e4b01c7be8c9ee6a

texaschic101 · Answer

I agree...its is 977.........check it out http://tierneylab.blogs.nytimes.com/2009/07/09/solutions-and-prizes-for-the-circle-of-fire-problem/

anonymous · Answer

but why does it start over at powers of 2?  knowing that, it is not that hard

shubhamsrg · Answer

cool !

shubhamsrg · Answer

maybe because no. people get half after every round. Thus 2 must have some relation to this :|

loser66 · Answer

I construct the series of numbers:
1 round: 1,3,5,7,9,......... with different is 2= 2^1
2nd round: 1, 5,9,13..... with different is 4 =2^2
3rd round : 1,9,17,25,......with different is 8 =2^3
4th round : 1,17,33,49,.........with different is 16 =2^4
so, it may relate to 2 from that??