Maths question

[devayaani]

Can anyone please help me? Is there any simple mathematical formula to answer this question (or question of this type)?
QUESTION: You are one amongst a thousand people captured by a tribal chief in a jungle. There are exactly one thousand posts arranged in a large circle. Each captive is made to stand at one of these posts. Now the chief orders his assistant to start shooting and killing alternate person. Killing starts at number one post, followed by 3, 5, 7, 9 ,11 and so on in a circular fashion again and again. This continues until at last only one person is standing alive. Which number post is the lucky one to survive when all other 999 captives are killed?
Thanks for the answer and if possible the method to do it.
devayaani

[factor-fiction]

512 is left

[factor-fiction]

After one round: 2,4, 6, 8.... survive (the first survivor is 2 to power 1).
After 2 rounds: 4, 8, 12...survive (first survivor is 2 squared)
Then 3 rounds : 8, 16, 24.... (first survivor is 2 cubed)
Then after 4 rounds: 16, 32, 48, (2 to power 4)
..............

After 8 rounds the fist survivor is 2 to power 8= 256, followed by 512 and 768. So only 3 survivors

The middle one of the 3 survives. That's 512

[devayaani]

factor 30,
Absolutely thrilled. Thank you so much.
I wasn't expecting an answer so soon! I was just trying to check if my question is posted.
I couldn't believe you have answered it and very lucidly.
Thanx again.
devayaani

[Gizmonster]

Sorry but this answer is wrong.

The answer that factor30 provides, does not take into account the last number (1000 in this case).

After round 1:
2,4,6,8 survive ..... right up to 992, 994, 996, 998, 1000.

After round 2:
4, 8, 12, 16, 20 survive ..... right up to 984, 988, 992, 996, 1000.

After round 3:
8, 16, 24, 32, 40 survive ..... right up to 968, 976, 984, 992, 1000.

After round 4:
16, 32, 48, 64, 80 survive ..... right up to 928, 944, 960, 976, 992.

Note that the 1000 has now been eliminated - so on the next round, the first number will now be kept (the number 16) and the number after it will be eliminated instead (the number 32).

After round 5:
16, 48, 80, 112, 144 survive ..... right up to 848, 880, 912, 944, 976.

After round 6:
16, 80, 144, 208, 272, 336, 400, 464, 528, 592, 656, 720, 784, 848, 912, 976 survive .

Keep going - and the last one standing is 976.

More to follow ..........

[Gizmonster]

Starting with lower numbers of people and working out where the survivor is, there is a pattern:

Number of people in group ----------- position of survivor
2 ________________________________2
3________________________________2
4________________________________4
5________________________________2
6________________________________4
7________________________________6
8________________________________8
9________________________________2
10_______________________________4
11_______________________________6
12_______________________________8
13_______________________________10
14_______________________________12
15_______________________________14
16_______________________________16
17_______________________________2

If the number of people in the group can be expressed as 2^n, where n is a whole number (i.e. the numbers, 2, 4, 8, 16, 32, etc), then the survivor is the last one in the group.

For any other number, find the highest number below your number, which can be expressed as 2^n. Work out the difference between this number and the number of people in the group and double this number - this is where the survivor will be.

So, for 1000 people, the highest number below 1000 in the form 2^n is 512 ( 512 = 2^9).
1000 - 512 = 488
488 x 2 = 976 = position of survivor.

This can obviously be used for any number of people, so say you have a group of 10 000 people ..........

highest number in the form 2^n is 8192 (8192 = 2^13)
10 000 - 8192 = 1808
1808 x 2 = 3616 = position of survivor.

[factor-fiction]

On first reading I agree with gizmonster. I had assumed in my quickfire response that the routine would start again from the beginnining after each round but gizmonster's interpretation makes more sense.
Have you been working on it all week gizmonster?

[society]

Gizmonster - I think you're a Wizmonster :-)
Very eloquent and simple explanation. Great!

[Gizmonster]

It took me a while to figure out - I was meaning to post the answer earlier, but just never got round to it =/

[devayaani]

A big thank you Gizmonster, you have saved my paper, ink and time .
Before I posted this question I had wasted a lot of A-4 size papers and writing the thousand numbers and crossing one by one! Yes, I did that but somewhere got distracted and ended getting the number 976 one time and after a few weeks when I tried the same method got 972 as answer. So, one of these two answers was wrong. A few weeks later tried again and got entirely different answer. Frustrated, I tried to figure out a maths formula by using small number of killings such as 7, 9, 20 and thirty. I could not come out with a proper answer.
So, I was thrilled to find the simple answer given by Factor 30.
Believe it or not, after understanding that formula, again I tried to test it...yes, A-4 size papers and 1000 numbers written and crossing off each round with a different coloured ink pen! (otherwise it gets confused if distracted in the middle of this killing of one thousand!)
Many times I crossed of 512 and was disappointed. I had blamed myself.
So, your answer and Factor 30 agreeing with your explanation has made me smile (greatly relieved, becasue at one attempt I had got the answer 976).
Thank you for both.
devayaani Sunday 15 March 2009