Please gimme answers to this with logical backings and proof to answer.

My brother I am still thinking of what to eat

The 99th person.

jaelz:
Please gimme answers to this with logical backings and proof to answer.
Cc:FARANO
MYND44
LALASTICLALA
SEUN ONILA FARANO MYND44

No 1 is the last person standing, why? he killed 2 and pass the sword to 3, 3 kill 4 and so on. Then it comes back to the beginning and 1 kills 3, till he is the last person standing

No 1

jaelz:
Please gimme answers to this with logical backings and proof to answer.
Cc:FARANO
MYND44
LALASTICLALA
SEUN ONILA FARANO MYND44

77th

jaelz:
Please gimme answers to this with logical backings and proof to answer.

Number one

The survivor is the only person that did not die na. In fact, the persons will be the one to tell the story of how others died...
As FBI concern, I suspect that 73rd person

Survivors for each round n =1 and ( 1+ 2^n : n=1-k)
Where k = total number of last survivors /2
Make the mathematicians do am finish.

the last odd number ll survive ...
1 kills 2 sequence eliminates all even number.
Then d sequence goes again 1remains 1 while others eliminate each other.
therefore 1 ll be d last odd number

1 is the DESIGNATED survival

If you ask @NPF this puzzle dem go arrest you come torture you till you go tell them say na you be the last person wey kee every 1

Back to the kwestshun
Nobody wee survive eggcept number 99 wey go kee number 100
I dey suspect that number 1 sef, e suppose make am

99th and 1

All the odd numbers will be survivors and all the even numbers killed. That is 1,3,5,7,....99 will survive while 2,4,6,8....100 will all die.

Chai, how much make I tell you guys the answer

Nonsensical Nonsense

1 & 99 were the survivors

Let me goan get the answer from my department group chat.. They solved this long ago

OP I dont have time to start going round circle.

73 is the answer!

How?

Consider a case when there 2^numbers in circle. Each time the number reduces by half and finally at the last number 1 will remain.
example – 2^2 = 4
round 1 – 2 and 4 will be killed.
round 2 – 3 will be killed. One Remains

So our aim here should be to reach to a figure when 2^n number will come and person who will be holding the sword at that moment will survive.

when there are 100 people – closet 2^number will be 64.
so here our target it to find the person who is holding the sword when 64 people are remaining.

64 people remaining means 36 people killed.
As every alternate person is being killed so double of 36 i.e 72. So 72 person was killed at that moment and the sword was passed to 73 from 71 after killing 72.

so the moment when 64 people were remaining, 73 was holding the sword,
so 73 will survive the killings!!!
.
.
.
.
.

OR,

Use the Circular Left Shift method

There are 100 people.
100 in binary is 1100100.
Circular left shift 1100100.
We get 1001001.
Convert it back to base 10.
73rd is the last man standing.


OR,

MATHAMATICAL WAY THAT EVERY NIGERIAN WILL UNDERSTAND

First 1,3,5,7,9,11,13,15,17;19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,51,53,55,57,59,61,63,65,67,69,71,73,75,77,79,81,83,85,87,89,91,93,95,97,99(got sword, passed to 1)

Then 1,5,9,13,17,21,25,29,33,37,41,45,49,53,57,61,65,69,73,77,81,85,89,93,97(sword passed to again)

Then 1,9,17,25,33,41,49,57,65,73,81,89,97(sword passed to 9)

Then 9,25,41,57,73,89(sword passed to 9 again)

Then 9,41,73(sword passed to 9 again)

Then 9,73(73 got sword & killed 9)

Thus the answer is “73"


Oh! And for the programmers or coders.

PYTHON CODE

# find the survivor
n = 100
peeps = range(1, n+1)
idx = 0
while len(peeps) > 1:
    peeps.pop((idx + 1) % len(peeps))
    # Trivia: because of changing index and len(peeps),
    # the code below is not exactly equal to "idx = (idx + 1) % len(peeps)"
    idx = 0 if (idx >= len(peeps)-1) else (idx + 1)
    
print peeps[0]

Output: 73

JAVASCRIPT

// populate the array for 100 people
var team = new Array();
for (var i = 1; i <= 100; i++) {
 team[i - 1] = i;
}

var current;
var next;

var j = 0.
while (team.length > 1) { // loop until only 1 exists
 current = team[j];
 team.push(team[j]); // move current to the end
 team.shift(); // remove from existing
 next = team[j];
 team.shift(); // delete next
 document.writeln("Number " + current + " kills number " + next + "<br />";
}
document.writeln("The last one is number " + team[0]);
//alert(team[0]);

Number 1

between 1 and 3

logically its number 1 but technically the mumu for don clear for number 3 eyes he no go wan return sword give number 1 as he go know say na die (imagine diz scenario ...Giving Buhari 70% vote comes 2019,Our eyes don clear we no go return am to office)

jaelz:
Please gimme answers to this with logical backings and proof to answer.

all the odd numbers survive...

Unu Don chop belle full this morning

number one hundred and onety one

Dont think anyperson in his right mind will solve this correctly.....

Without rewards attached

Me

No1