₦airaland Forum

Remimadrid:
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

not number 1 because number 100 will definitely kill him

Enouwem:
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]);

This is actually so amazing. Thumps up guy

in as much as i respect 73 as the mathematical answer following logical principle, in a real life situation where:
9, 41, and 73 are left(sword passed to 9)
the 9th will kill d 41st, but would not "stupidly" pass the sword to d 73rd so that it would be killed..
the 9th person would rather kill d 73rd, and thus become d last man standing...
answer=9

Enouwem:
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]);

Guy u are a genius I will like us to be friends if u don't mind. pls reply so I can send u my email

.

73 is the only survivor.

.

Enouwem:
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]);

Bros I hail thee o... The only method I know is that nigeria method.....I go like make you teach me o

As long as the number is power of 2, the person survived will always be the one who starts. If the number is not power of 2, find the greatest power of 2 which is less than the number i.e. 64 now if 100-64=36 people are killed, the one who will start after that would be the one who will survive.
36 people will be killed as 2,4,6, .... ,72 and the sword will be handed over to 73 who is the first person to start in remaining 64. thus only 73 will survive.

No 1 is the survivor here. He started the killing and by the time the knife goes round, no 100 would have been down and there will be no one remaing to take no 1 down.

1 and 99( where we have only one person to stand out that means when no 99 collect d sword he is to go and kill no 1(that means last man standing is 99)

100 person wen they go round the 100 person kills the 1st

Enouwem:
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]);

m highly impressed bro...excellent perfecto

Remimadrid:
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 way, its not as easy as u think.
1 kills 2 gives the sword to 3 to kill 4 3 after killing 4 gives d sword to 5 and so on.
After the first circle is completed, 1 kills 3 gives the sword to 5 to kill 7 and gives the sword to 9

89 is the last man standing

That was what i thought, like 50 survived and 50 died.

cremedelacreme:
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.

No 99.. Cos after he kills no 100.. He still has the sword to kill no 1

Phi001:
You are correct.


I just finished solving by mathematical induction and behold, I got 73 too...

Enouwem:
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]);

bro you get time oh

Hmmm.... Only no 100 should survive cos he will get the sword and he will kill no 1 since they are standing in a round circle

73...

The odd will be the one to survive, first number one kills 2 gives to three who kills 4 gives to five who kills 6 gives to seven, continuously till 99, which will eventually kill Hundred.

hmmm

Chai. N.a. wa o. This your analysis plus codes n.a. die. This is brilliant. Do you code? Can we be friends, ?

Enouwem:
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]);

Enouwem:
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]);

You are a genius. Keep up the good work. I am dazzled at your level of intelligence.

The answer is actually 69

dejt4u:
77th

Deji, call my number or message me on whatsapp. Very urgent! Are you in Lagos??

naturalwaves:
Deji, call my number or message me on whatsapp. Very urgent! Are you in Lagos??

sir, I lost my phone that I saved your number couple of months ago.. pls message me, 07069***** on WhatsApp.. thank you bros..