IQ Test, Logic test, Fun :)
finzEr .. your solution proves to be pretty logical but it has some flaws in it.First of all.. the counter can come into the room and the swtich A could be turned on because someone turned it on or that it was on randomly.If it's case a then he misses one person and he will count only 22 even with him.:) also the solution is not that good as it might take forever
If switch A is on then he just increments the count by one. It cannot be turned on by default because it was said that both switches are turned off in the beginning.
first of all, finzers solution is correct and the most efficient one for this kind of riddle: http://www.youtube.com/watch?v=wDajqW561KM&feature=youtu.be
it will actually take a very long time to finish but thats logical aswell, because when viewing it as a algorithm or turing machine time-complexity increases by a gaining limitation of space to "save" certain symbols & states. theoretically it could be solved in a linear ammount of time if we had ceiling(log(ammount of prisoners))+1 many switches (5 switches in this case). even if you can only perform one action at the switches you could still count in binary with a little modification everybody has to agree on and remember. 'switch on' would be '1' and 'switch off' stands for '0'... also in this case everybody could tell the director about the correct answer.
anyways, for this scenario it is the best (and probably only, if you exclude periodical iterations of the counting) solution.
hitmanz critism would be justified if the initial state of the 2 switches wouldnt be clear, as remake said. in that case you would have to create a certain "counter-start situation" when the counter goes to the yard for the very first time which gives a little add to finzers algorithm
im quite busy for the next few days. could be sweet if we can keep this thread going
If you used binary (or Gray code, to be more exact) you'd have to have 6 switches (one for those who have already been outside before).
Originally posted by remakeIf you used binary (or Gray code, to be more exact) you'd have to have 6 switches (one for those who have already been outside before).
thats right. i was talking about gray code and i thought i already included the switch for the ones that were outside 1 time but my calculator used the natural logarithm (wrong base) like this:
log(23)=3.135494215929... -> 4 switches + 1 "bin"-switch
ofc you need 5 switches to store a number between 16 and 32 in binary... never rely on cheap calculators.
I remember doing this long ago so I wanted to post it :'>
I got it from one website.
[blockquote]Detective Ixolite of the NYPD was investigating a murder at Chicago.
It was a difficult case, and Ixolite was completely stumped until he noticed a message sent to him by the killer cunningly hidden in a newspaper advertisement selling Car Licence Plates.
Detective Ixolite thought about it for a while, and when he had solved the puzzle, immediately arrested the guilty man.
Q1) How did Ixolite know the advert was a clue for him?
Q2) Solve the code and tell me who Ixolite arrested.
This is the newspaper advert (Car licence plates for sale) that Detective Ixolite saw.
Plates For Sale;
< W 05 NWO >
< H 13 HSR >
< O 05 EBM >
< D 08 UNE >
< U 10 HTY >
< N 04 BRE >
< N 16 TTE>
< I 26 LHC >
< T 10 AEE >
< I 26 CNA >
< X 22 VDA>[/blockquote]
there is some math for you fellas 6-1x0+2:2=?
ppl gettin dum nowdays :( I saw so many wrong answers in fb :/
06/04/2013 11:01Most of the people will answer 5 but it's 7
here is a good one.I bet some of u have seen this one be4, then just be a good spectator :)
There are 2 boys(Dan,Larry) and 2 girls(Alice,Anna) having a sex party, which means each boy is gonna have sex with each girl, 4 times intercourses totally. the problem is that each of these 4 people has a STD,that means 4 different STDs. but there are ONLY TWO condemns(A,B).
Question: How do u arrange the intercourses to avoid any cross infection？
PS: no HJ ,no BJ (I mean handjob and blowjob :lol: )
well, stop laughing and use ur mind to think, this is a serious logic question :D
no one accepts this challenge, lol?
Originally posted by znx< W 05 NWO >
< H 13 HSR >
< O 05 EBM >
< D 08 UNE >
< U 10 HTY >
< N 04 BRE >
< N 16 TTE>
< I 26 LHC >
< T 10 AEE >
< I 26 CNA >
< X 22 VDA>[/blockquote]
left side, from top to bottom:
WHODUNNITIX
right side, bottom to top:
ADVANCEEACHLETTERBYTHENUMBERSHOWN
-> BUTLERDIDIT
Originally posted by Ggs`there is some math for you fellas 6-1x0+2:2=?
ppl gettin dum nowdays :( I saw so many wrong answers in fb :/
7
Originally posted by Mik0here is a good one.I bet some of u have seen this one be4, then just be a good spectator :)
There are 2 boys(Dan,Larry) and 2 girls(Alice,Anna) having a sex party, which means each boy is gonna have sex with each girl, 4 times intercourses totally. the problem is that each of these 4 people has a STD,that means 4 different STDs. but there are ONLY TWO condemns(A,B).
Question: How do u arrange the intercourses to avoid any cross infection？
dan uses both condoms for sex with alice
he pulls off the one on the outside and gives it to larry
larry uses the condom to have sex with alice
dan uses the condom thats still on his dick to sleep with anna, later gives it to larry
larry pulls it over his condom to have sex with anna
Originally posted by k a f f e edan uses both condoms for sex with alice
he pulls off the one on the outside and gives it to larry
larry uses the condom to have sex with alice
dan uses the condom thta still on his dick to sleep with anna, later gives it to larry
larry pulls it over his condom to have sex with anna
damn right kaffee.u make it by thinking wisely or u knew the answer be？
Originally posted by Mik0Originally posted by k a f f e edan uses both condoms for sex with alice
he pulls off the one on the outside and gives it to larry
larry uses the condom to have sex with alice
dan uses the condom thta still on his dick to sleep with anna, later gives it to larry
larry pulls it over his condom to have sex with anna
damn right kaffee.u make it by thinking wisely or u knew the answer be？
the trick with solving logical riddles is to transfer it in a more abstract and theoretical form. once you get a clear representation of the problem the job is almost done. and ofc many of them are quite similiar
someone got a new one?
27/05/2013 13:06(edited 27/05/2013 13:07)Let's start with an easy one :
It was Debbie's first day at school. The teacher suggested that it would be a good idea for each child to meet every other child in the class. The teacher said, "When you meet, please shake hands and introduce yourself by name."
If there were 9 children in the class, how many total handshakes were there?
Solution is written with a small font below :
The class has 9 children. The first child shakes hands with the other 8 children. The second child has already shaken hands with the first child, and so has to shake hands with only the other 7 children. In this manner, the second-last child has to shake hands with only one child, and the last child has already met all the children. Thus, the number of handshakes is
8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36.
If there were 9 children in the class, then there were 36 total handshakes.
That was an easy one, indeed :) Keep them coming Pu9 !
27/05/2013 16:41middle difficulty :
"Dad, where had you been?" asked John.
"I had been to the attic, my son," replied Dad. "And do you know what I saw there? There was a big web with 13 spiders and flies on it."
"How many spiders were there?" asked the little boy with curiosity.
"Well, there were a total of 86 legs on the web," answered Dad with a smile. "Now you can find out how many spiders were there by yourself. Can't you?"
Can you help the little boy find out how many spiders were on the web in the attic?
27/05/2013 17:43Originally posted by PU9makermiddle difficulty :
"Dad, where had you been?" asked John.
"I had been to the attic, my son," replied Dad. "And do you know what I saw there? There was a big web with 13 spiders and flies on it."
"How many spiders were there?" asked the little boy with curiosity.
"Well, there were a total of 86 legs on the web," answered Dad with a smile. "Now you can find out how many spiders were there by yourself. Can't you?"
Can you help the little boy find out how many spiders were on the web in the attic?
Assuming that spiders have 8 legs and flies have 6 legs :
let "s" be the number of spiders and "f" the number of flies.
s + f = 13
8s + 6f = 86
On solving the above two equations, we get
8s + 6(13 - s) = 86; or
2s = 86 - (6 x 13)
s = 8
So there 8 spiders on the web.
Ive been reading some of thoose and I don't understand a single one d:[
29/05/2013 11:56(edited 29/05/2013 11:59)Here is a little puzzle for you :
Shauna was killed one Sunday morning. The police know who they are going to arrest from this bit of information:
April was getting the mail
Alyssa was doing laundry
Reggie was cooking
Mark was planting in the garden
Who killed Shauna and how did the police know who to arrest?
29/05/2013 12:23Originally posted by PU9makerHere is a little puzzle for you :
Shauna was killed one Sunday morning. The police know who they are going to arrest from this bit of information:
April was getting the mail
Alyssa was doing laundry
Reggie was cooking
Mark was planting in the garden
Who killed Shauna and how did the police know who to arrest?
April is lying, there is no mail on Sunday
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