A puzzle.........

[leleram]

1000 persons are standing in a circle (circumference), say they are numbered from 1 to 1000.
1st person kill the 2nd and passes revolver to 3rd.
3rd kills the 4th and passes to 5th
5th kills the 6th ........and so on ......

which number person will be saved at last...(make as many rounds as possible)

[kaliamastaan007]

only one will remain

[leleram]

only one will remain
thats too simple:tt2:..already explained in puzzle...U HAVE TO TELL THE NUMBER OF PERSON......1st...2nd ...or100th...or 1000th:award:

[GandFaadBanda]

1st person will be the last man standing.

[leleram]

1st person will be the last man standing.
PLZ explain...y u think so:blink:

[GandFaadBanda]

Yaar! Ye explanation type karne mein time lagega. Achcha! Koshish karta hoon.

First round:
1st person shoots 2nd, 3rd person shoots 4th and so on. That means that in this round, all the people in the even number position are being shot. That means that 999th person will shoot 1000th person and hand over the gun to 1st person.

Second round:
Since 2nd person is already down in first round, 1st person now shoots 3rd person and hands over the gun to 5th person (because 4th is already down in first round). 5th person shoots 7th person and hands over the gun to 9th person. That means that in this round, all the people in the alternate odd number position are being shot. That means that 997th person will shoot 999th person and hand over the gun to 1st person.

Third round:
1st person now shoots 5th person and hands over the gun to 9th person. In the end, 993rd person shoots 997th person and hands over the gun to 1st person.

It goes on an on. Whatever the sequence, 1st person always gets the gun at the end of each round. So he will be the last person to shoot the second last person.

Right or wrong?

[leleram]

Yaar! Ye explanation type karne mein time lagega. Achcha! Koshish karta hoon.

First round:
1st person shoots 2nd, 3rd person shoots 4th and so on. That means that in this round, all the people in the even number position are being shot. That means that 999th person will shoot 1000th person and hand over the gun to 1st person.

Second round:
Since 2nd person is already down in first round, 1st person now shoots 3rd person and hands over the gun to 5th person (because 4th is already down in first round). 5th person shoots 7th person and hands over the gun to 9th person. That means that in this round, all the people in the alternate odd number position are being shot. That means that 997th person will shoot 999th person and hand over the gun to 1st person.

Third round:
1st person now shoots 5th person and hands over the gun to 9th person. In the end, 993rd person shoots 997th person and hands over the gun to 1st person.

It goes on an on. Whatever the sequence, 1st person always gets the gun at the end of each round. So he will be the last person to shoot the second last person.

Right or wrong?
approach is right...but u cldnt reach to the final answer....u may try make a formula for this one :toast:

[Eshtyle Raja]

GFB BHAI :adore: :adore: :adore:

You are on right track I think, but lele is still demanding something as reply... keep it up...

[GandFaadBanda]

Person at the position 977 will be the last one. Ab theek hai?

[GandFaadBanda]

Person at position 1 will be shot at the end of round 3. My apologies. I made a wild guess looking at the trend.

This will be the scenario:

Initial position:
1,2,3,4,5,6,7,8,9,10,.......,997,998,999,1000
1 shoots 2, 3 shoots 4 and so on.

After first round of firing:
1,3,5,6,7,9,....,997,999 are left
1 shoots 3, 5 shoots 6 and so on.
In the end, gun is passed on to person no. 1

After second round of firing:
1,5,9,13,....,993,997 are left
1 shoots 5, 9 shoots 13 and so on.
In the end, gun is passed on to person no. 1

After third round of firing:
1,9,17,25,....,977,985,993 are left
1 shoots 9, 17 shoots 25 and so on.
In the end, 993 shoots person no. 1 and passes the gun to no. 17

After fourth round of firing:
17,33,49,65,....,945,961,977,993 are left
17 shoots 33, 49 shoots 65 and so on.
In the end, 977 shoots person no. 993 and hands over the gun to 17

After fifth round of firing:
17,49,81,113,....,913,945,977 are left
17 shoots 49, 81 shoots 113 and so on.
In the end, 913 shoots 945 and hands over the gun to 977. 977 shoots 17 and hands over the gun to 81.

After sixth round of firing:
81,145, 209, 273,....,849,913,977 are left
81 shoots 145, 209 shoots 273 and so on.
In the end, 849 shoots 913 and hands over the gun to 977. 977 shoots 81 and hands over the gun to 209.

After seventh round of firing:
209,337,465,593,721,849,977 are left
209 shoots 337, 465 shoots 593 and so on.
In the end, 721 shoots 849 and hands over the gun to 977. 977 shoots 209 and hands over the gun to 465.

After eigth round of firing:
465,721,977 are left
465 shoots 721 and hands over the gun to 977. 977 shoots 465 in the next round.

The End.

[Mastikhor]

lele ... apni ganit ke hisab se ... revolver has 6 bullets

So only 6 people will die :)

:p

[jhoomroo]

gfb ka ganit padh ke apna head spin ker raha hai...lage raho gfb...good going...

[JUNGLEE RAJA]

Lele bhai has again started with his puzzles and riddles, Iam afraid Gurudev will be here shortly to solve this issue

:w00t: :w00t: :w00t:

[leleram]

great GFB...ur answer is corret :thumbup:

Masti bhaiya....revolver mein goli to 6 hi hoti hai....magar use n throw nahi hota,,,it can be reloaded :tt2:

[greatalok]

.................
After eigth round of firing:
465,721,977 are left
465 shoots 721 and hands over the gun to 977. 977 shoots 465 in the next round.

The End.

So far so good but at this stage, 465 will shoot both 721 and 977 if he has an even iota of sense. Why would he hand over the gun to other person to get shot again:D