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Mathematics 4 Online
imqwerty (imqwerty):

smart guys i need your help

imqwerty (imqwerty):

FBI, one of the most feared Law Enforcement Agency on this earth, has captured 100 terrorists. Out of these 100 terrorists, some ‘k’ are high valued targets and the agency plans to torture them. It has marked all of the high valued targets with a sign on their head. Now the problem is, the person who has been tagged is not aware if he was actually tagged or not. The inmates can see each other and the marks on the head of the tagged terrorists but they can’t communicate with each other. So, they have no way to determine if they themselves are tagged or not. All the “high valued” terrorists want an easy death instead of torture. They somehow manage a way to commit suicide, but then only the ones who are tagged will suicide. Also, they can commit suicide only during night. Days pass by and nothing happens, but suddenly, on the night of 37th day, all the “high-valued” terrorists commit suicide. Determine, that out of the 100 terrorists, how many were the “high-valued” terrorists? Additional Info: • k ≥ 1 i.e. atleast one “high-valued” terrorist exists! • Terrorists are aware that k ≥ 1, but they don’t know the value ‘k’. 4 5 6 8

imqwerty (imqwerty):

please tell the method of finding the solution

OpenStudy (anonymous):

this one's tough lemme see... @jim_thompson @Mashy @camacho01erick I don't know the method so...

OpenStudy (anonymous):

This question is really hard!

OpenStudy (anonymous):

start thinking how would each terrorist determine if he is tagged or not... also consider that all of them are applying the same method. for example, suposse only 1 terrorist is tagged, in such case he would see that no one else has a tag, and knowing that k ≥ 1, he would then know he is the one tagged!! so he would kill himself the first day now try to think the general case !

imqwerty (imqwerty):

the options are 4 , 5 ,6, 8 and all are >1 now what case can we make in this.

OpenStudy (anonymous):

i would like you to think about the problem... let's make another example: suppose k=2 (of course terrorists dont know this) lets call the tagged terrorist A and B, in this case tagged terrorist B would think: "i can see 1 other terrorist with a tag, so the possible cases are: a) he is the only one tagged (i dont have a tag) [k=1] b) we are both tagged [k=2] in case a, the tagged terrorist A would see no other tags (as in the example i made before), and would know he is the only one, and would kill himself the first night in case b, tagged terrorist A would not kill himself the first night, and that would let the others know he is not the only one, so tagged terrorist B will now he is the second one, since he can see only another one with a tag if they both think the same way (apply the same method) they will both commit suicide the 2nd night ! OK... that was long.. and english is not my native lang, so... i hope this help you to think the problem!

imqwerty (imqwerty):

thanks @Greg_D

OpenStudy (anonymous):

yw! keep thinking about it, we can make an example for bigger numbers later :)

OpenStudy (baru):

XD sorry...couldn't help myself.. whats the answer? following gregs logic..if terrorist C sees two others tagged and still alive after two nights then hes the last one...if D sees three tagged and alive for three nights then he's the last one...and so on and so on..then k=37 which not an option

imqwerty (imqwerty):

lol i forgot about this question ok lemme think :)

ILovePuppiesLol (ilovepuppieslol):

im a smart guy!!!!

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