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The Hardest Logic Problem In the World

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posted on Dec, 31 2012 @ 02:49 PM
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Originally posted by sonnny1
reply to post by spy66
 


Am I A Bipolar Opposite?



One of them would say yes.




posted on Dec, 31 2012 @ 03:27 PM
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Originally posted by sirric




Can you move one matchstick and still have an equation that is correct? I think there are at least three answers – can you find all three?




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3-9=6
3+3=6
not sure of the last one


That's a good one.

Here is a math logic question

3 men decide to rent a room for the night, The room is $30

They each pay $10 and go up to the room.

The manager made a mistake and realizes that the room only cost $25 for the night, so he gives the bellboy $5 to give back to the 3 men.

As the bellboy was in the elevator, he knew that he couldn't split $5 between 3 people so he pocket $2 and gave each man $1 back.

How much did each man pay for the room?

Where did the missing dollar go?

sirric
edit on 30/12/12 by sirric because: (no reason given)


got 3+3=6
8-3=5



posted on Dec, 31 2012 @ 03:47 PM
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Two old friends, Jack and Bill, meet after a long time.



Jack: Hey, how are you man?

Bill: Not bad, got married and I have three kids now.

Jack: That’s awesome. How old are they?

Bill: The product of their ages is 72 and the sum of their ages is the same as your birth date.

Jack: Cool… But I still don’t know.

Bill: My eldest kid just started taking piano lessons.

Jack: Oh now I get it.

How old are Bill’s kids?

Spoiler: Solution Below


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Solution:
3,3,8

Lets break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 sum(2, 2, 18) = 22
2, 4, 9 sum(2, 4, 9) = 15
2, 6, 6 sum(2, 6, 6) = 14
2, 3, 12 sum(2, 3, 12) = 17
3, 4, 6 sum(3, 4, 6) = 13
3, 3, 8 sum(3, 3, 8 ) = 14
1, 8, 9 sum(1,8,9) = 18
1, 3, 24 sum(1, 3, 24) = 28
1, 4, 18 sum(1, 4, 18) = 23
1, 2, 36 sum(1, 2, 36) = 39
1, 6, 12 sum(1, 6, 12) = 19

The sum of their ages is the same as your birth date. That could be anything from 1 to 31 but the fact that Jack was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now.

2, 6, 6 sum(2, 6, 6) = 14
3, 3, 8 sum(3, 3, 8 ) = 14

Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8.


Popular Age Problem
edit on 31-12-2012 by TheSubversiveOne because: (no reason given)



posted on Dec, 31 2012 @ 03:52 PM
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reply to post by Byrd
 


I would ask the question "are you a god?" to all three of them

I didn't see anywhere in the rules that you cannot ask the same (1) question multiple times....in which case I would ask each god the same question as many times as needed to identify the random god.

After the random god was identified I would up my odds of guessing the other two gods from 33.333...% to 50% .

which I'm comfortable with lol....



edit on 31-12-2012 by Sly1one because: (no reason given)
edit on 31-12-2012 by Sly1one because: (no reason given)



posted on Dec, 31 2012 @ 04:02 PM
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reply to post by TheSubversiveOne
 

I think Jack was just humouring Bill because Bill was being an asshole again.



posted on Dec, 31 2012 @ 04:07 PM
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Originally posted by john_bmth

Originally posted by Logarock
reply to post by john_bmth
 


If you get a yes answer first it has to be from the random God.

edit on 31-12-2012 by Logarock because: n

50% of the time the random god will answer "no".


Yea but he is the only one that would give you a yes on the first.

Let us say that the random god said no to the first. If the truth God was next and said yes a is a lier, and he must, then the lier god being last is forced to say yes of the second, that the truth god is a lier. No, yes, yes.

If the random god said no again but the lier god is next, he must say of the first no, the random god isnt a lier and then last the truth god would say yes the second god is a lier. no, no, yes.



posted on Dec, 31 2012 @ 04:23 PM
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reply to post by Logarock
 


Using your approach there are many sequences (where the random god isn't first) with multiple god orderings valid for each sequence. Ergo, it is not a solution to the challenge.
edit on 31-12-2012 by john_bmth because: (no reason given)



posted on Dec, 31 2012 @ 04:24 PM
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Originally posted by Ryanssuperman

Originally posted by smithjustinb
reply to post by sirric
 


Each man payed 9 dollars for the room. There is no missing dollar.


9x3 = 27 + 2 (In the bellboys pocket) = 29. Therefor, there's a "missing" dollar.
edit on 31-12-2012 by Ryanssuperman because: (no reason given)


The 30 in the equation does not exist anymore after the hotel gave the bellboy 5 back.
Now 25 was spent for the room.

The bell boy kept 2 and the hotel kept 25....
This adds up to 27 charged.

Each of the three men paid 9.
This equals 27.

There is no longer a 30 to be taken into consideration.
Therefore there is NO missing dollar.



posted on Dec, 31 2012 @ 05:35 PM
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"Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language in which the words for 'yes' and 'no' are 'da' and 'ja', in some order. You do not know which word means which."


To start his little game off; I wouldnt care which god was which, as I would be self serving if a true god did not embrace me honestly and securly as its child of creation. So I could care less for their test..

But as this particular test has a trail of truth to it in comparison to how humanity was tricked by the "Gods" in the first place, which has led us to this controlled version of reality we call life.. I will bite


I stand infront of the first God, who I am suppose to have no reference of which one it is of the three. I am left with one "yes/no - da/ja" question to present them to indentify the truth, the liar and the random God..

I ask the first being if my eye colour is blue. Now its up to me to give an answer to my own question becuase I have no reference of "da/ja" because there arent 3 Gods here to talk or type.. so, this God tells me "da" which I take as no.

My eyes are blue so I know this is the false god who is B.

God 1 - False

I stand infront of the second being self-claimed "God".. I ask it; "are my eyes brown, yellow, red AND purple?" The being answers with "ja" .. I take this as a yes, and thus, I know this God is random because the true answer was da/no. It had a 50/50 chance of getting the answer right, and it failed this particular question.

God 2 - Random

So I assume 2/3 Gods are indentified, I ask the last god.. do you love and embrace me as your own creation? The final true God responds with "yes my child, you are me and I am you"



posted on Dec, 31 2012 @ 05:56 PM
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reply to post by BigBrotherDarkness
 

Lol. My solution was simpler. Shoot one in the head and wait for the responses from the others...

You don't have to say anything. The remaining two will sing like birds.

Edit:I have an old one: You live in a house where all four walls face south. A bear walks by, what color is the bear?
edit on 31-12-2012 by intrptr because: (no reason given)



posted on Dec, 31 2012 @ 06:11 PM
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reply to post by DIDtm
 


I understand the solution ... I was just pointing out the "problem"!



posted on Dec, 31 2012 @ 08:01 PM
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Originally posted by intrptr
reply to post by BigBrotherDarkness
 

Lol. My solution was simpler. Shoot one in the head and wait for the responses from the others...

You don't have to say anything. The remaining two will sing like birds.

Edit:I have an old one: You live in a house where all four walls face south. A bear walks by, what color is the bear?
edit on 31-12-2012 by intrptr because: (no reason given)


White.

Now, along the same lines as Byrd's original problem is this one:


Five pirates arrive back on board ship with 100 gold coins and decide to share them out using a democratic system.

Each of the pirates has a different rank, and the Captain will therefore make the first proposal about how the gold is shared out. The pirates will then have a vote. If it's a draw (two vote yes, two vote no), the Captain gets everything. If they agree to take the Captain's offer, the gold is shared in the way he suggested. If they reject it, they throw him overboard and the First Mate gets to propose a way to share out the gold.

Now, unlike the REAL world, these are Logical and Rational pirates who never act randomly and are in a tiny boat so the can't conspire against each other. They are all trying to get the most amount of gold AND stay alive.

So... got the answer?

Check the Nash Equilibrium page: www.h2g2.com...



posted on Dec, 31 2012 @ 08:41 PM
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First question "do you Always speak the truth?"
Second question "Was the previous question answered honestly?"
Third question "Does Da mean Yes?"

From there is just compare the responses
edit on 31-12-2012 by LucidFusion because: (no reason given)



posted on Dec, 31 2012 @ 08:41 PM
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Originally posted by john_bmth
reply to post by Logarock
 


Using your approach there are many sequences (where the random god isn't first) with multiple god orderings valid for each sequence. Ergo, it is not a solution to the challenge.
edit on 31-12-2012 by john_bmth because: (no reason given)



No kidding? Many? There are only 3 sets 2 sequences, 6 total sequences. One set is already figured out for you. Did you fail alegbra?
edit on 31-12-2012 by Logarock because: n



posted on Dec, 31 2012 @ 09:46 PM
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reply to post by smithjustinb
 


doesn't 3-9=-6? not exactly the same result, but i am not very smart.



posted on Dec, 31 2012 @ 10:42 PM
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reply to post by Indellkoffer
 

Polar to be sure...
I'm sorry to say that even the solution didn't make sense to me. I had to assume that "pirates" are greedy(?) and social climbers (?) and then figure it out? Those traits aren't "logical" to me. I am neither so it made no sense. However...

Thanks for the link in your signature. I read that with relish.

I learned with the dinosaurs too.



posted on Jan, 1 2013 @ 04:03 AM
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Originally posted by Logarock

Originally posted by john_bmth
reply to post by Logarock
 


Using your approach there are many sequences (where the random god isn't first) with multiple god orderings valid for each sequence. Ergo, it is not a solution to the challenge.
edit on 31-12-2012 by john_bmth because: (no reason given)



No kidding? Many? There are only 3 sets 2 sequences, 6 total sequences. One set is already figured out for you. Did you fail alegbra?
edit on 31-12-2012 by Logarock because: n


There are 24 answer sequences using your approach. Your solution is valid for four, the rest are indeterminate. It is not a valid solution because there are 20 sequences it doesn't solve. But by all means explain how your solution is valid when it doesn't actually solve the challenge.
edit on 1-1-2013 by john_bmth because: (no reason given)



posted on Jan, 1 2013 @ 05:58 AM
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OK guys

Let me see if I've got this right

If I have 4 oranges and give Bob 12 grapes, how many tiles can I fit on your roof ?

Answer: None because aliens don't wear hats.

Is that right ?



posted on Jan, 1 2013 @ 07:31 AM
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Originally posted by john_bmth

Originally posted by Logarock

Originally posted by john_bmth
reply to post by Logarock
 


Using your approach there are many sequences (where the random god isn't first) with multiple god orderings valid for each sequence. Ergo, it is not a solution to the challenge.
edit on 31-12-2012 by john_bmth because: (no reason given)



No kidding? Many? There are only 3 sets 2 sequences, 6 total sequences. One set is already figured out for you. Did you fail alegbra?
edit on 31-12-2012 by Logarock because: n


There are 24 answer sequences using your approach. Your solution is valid for four, the rest are indeterminate. It is not a valid solution because there are 20 sequences it doesn't solve. But by all means explain how your solution is valid when it doesn't actually solve the challenge.
edit on 1-1-2013 by john_bmth because: (no reason given)


I am saying 6 because you dont need all that information once you establish some value. As far as 24, the random god can still only give one answer per sequence even though 12 seems like the probable number to deal with. In reality he is limited to 6 as are the others and even there the two gods, truth and lies, have only one possible answer to every single question. Contrary to sequence possibility, it is in fact not possible that the random god can use all of his possible answers.

Here is another set. Lier god first. He must always tell the truth, be true to his nature, and tell a lie.

Are you a lier? No. God 2 is 1 a lier? Yes. God three is 2 a lier? Yes.
Are you a lier? No. God 2 is 1 a lier? No. God 3 is 2 a lier? Yes.

Can you see which is the truth god then? Anyway lots of fun there.

edit on 1-1-2013 by Logarock because: n



posted on Jan, 1 2013 @ 07:52 AM
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reply to post by Logarock
 


Each god will only answer one question (and only once). Take the following example answer sequence from your questions:

NYY

Possible god arrangements for this sequence are as follows:

TRL
RLT

There is no way to deduce which sequence is correct from the questions you have asked.






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