GAME: MindTrap

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Originally Posted by Camaro80z

Also, there is more suface area of the bottom of the boat displacing the water than the area of the nut and bolt, so less water is displaced.

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Close...


BTW, the point for the explanation is open to anyone.

I know 96 still doesn't have an official explanation for it, but I want to get to the next question now.


There is a nightclub in Truth town called the Truth Club which is made up entirely of knights and liars.

Sometimes they start singing a song. One person sings "At least one of us is a liar", the next person sings "At least two of us are liars", continuing on like this such that each person says one more person than the last person; each person singing exactly one line. (If there were 10 people in the club, the only person who hadn't sung a line would sing the last line, "at least 10 of us are liars" and then the song would be done.)

One day when you know there was a prime number of people in the club, you hear the start of the song "At least...", but don't hear the middle; all you know is that they sang the song through completely. Even though you only hear those two words at the start of the song, you can tell how many people are in the club. How many people were there?


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Two (one knight and one liar).


The key is to realize that it has to be an even number of people, or else the song is paradoxical.

If an even number 2m are singing, then one can easily verify that the first m are Knights and the second m are liars.

If there is an odd number 2n+1, however, then consider the n+1th singer. He sings "At least n+1 of us are liars". If he is telling the truth, then so was everyone before him (since they all sang "At least x of us are liars" for some x less than n), so there are n+1 knights, which means there could be at most n liars. Thus, he lied, resulting in a contradiction. Similarly, if he were lying, then everyone after him would also be lying, but this would mean that there were n+1 liars, resulting in his having told the truth. Again, contradiction. Once we know that it has to be an even number of singers, the behaviour of the bouncers guarantees that it must be two singers, the only even prime.

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Originally Posted by a d e p t

I know 96 still doesn't have an official explanation for it, but I want to get to the next question now.


There is a nightclub in Truth town called the Truth Club which is made up entirely of knights and liars.

Sometimes they start singing a song. One person sings "At least one of us is a liar", the next person sings "At least two of us are liars", continuing on like this such that each person says one more person than the last person; each person singing exactly one line. (If there were 10 people in the club, the only person who hadn't sung a line would sing the last line, "at least 10 of us are liars" and then the song would be done.)

One day when you know there was a prime number of people in the club, you hear the start of the song "At least...", but don't hear the middle; all you know is that they sang the song through completely. Even though you only hear those two words at the start of the song, you can tell how many people are in the club. How many people were there?

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There are two people in the club. The knight sang first saying "At least one of us is a liar." This being true. The liar sang next saying "At least two of us are liars.", which is false. All other prime numbers would be odd, thereby making the situation impossible.

Quote:

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Originally Posted by a d e p t

I know 96 still doesn't have an official explanation for it, but I want to get to the next question now.


There is a nightclub in Truth town called the Truth Club which is made up entirely of knights and liars.

Sometimes they start singing a song. One person sings "At least one of us is a liar", the next person sings "At least two of us are liars", continuing on like this such that each person says one more person than the last person; each person singing exactly one line. (If there were 10 people in the club, the only person who hadn't sung a line would sing the last line, "at least 10 of us are liars" and then the song would be done.)

One day when you know there was a prime number of people in the club, you hear the start of the song "At least...", but don't hear the middle; all you know is that they sang the song through completely. Even though you only hear those two words at the start of the song, you can tell how many people are in the club. How many people were there?

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Quote:

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Originally Posted by Deity

There are two people in the club. The knight sang first saying "At least one of us is a liar." This being true. The liar sang next saying "At least two of us are liars.", which is false. All other prime numbers would be odd, thereby making the situation impossible.

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In that case wouldn't it be easier just to count the number of people in the club??? What's the deal with having to hear the silly song??? :devil: :p

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Originally Posted by Deity

There are two people in the club. The knight sang first saying "At least one of us is a liar." This being true. The liar sang next saying "At least two of us are liars.", which is false. All other prime numbers would be odd, thereby making the situation impossible.

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That's it. III more IV you.

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Originally Posted by CW_WD_RIOT

In that case wouldn't it be easier just to count the number of people in the club??? What's the deal with having to hear the silly song??? :devil: :p

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Entertainment value? Or maybe they were samurai knights out in a karaoke bar? https://forums.windrivers.com/images.../2005/03/1.gif

You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?

Multiple answers possible with this one. https://forums.windrivers.com/images.../2005/03/1.gif


__________
Find possible answers below.

Quote:

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Originally Posted by a d e p t

You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?

Multiple answers possible with this one. https://forums.windrivers.com/images.../2005/03/1.gif

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10.

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Originally Posted by a d e p t

You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?

Multiple answers possible with this one. https://forums.windrivers.com/images.../2005/03/1.gif

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6

If you were lucky enough to weigh each of the 6 coins of one type, you would then know that the 6 left were all of the other wight. :thumbs2:

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Originally Posted by Camaro80z

6

If you were lucky enough to weigh each of the 6 coins of one type, you would then know that the 6 left were all of the other wight. :thumbs2:

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You could do it in 5 if you were lucky.

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Originally Posted by edball

10.

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No, sorry.

And there seems to be something misssing...

..oh yeah - no explanation given!

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Originally Posted by a d e p t

No, sorry.

And there seems to be something misssing...

..oh yeah - no explanation given!

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If I'm wrong, why should I explain it ? This question is making my brain hurt anyway !

Quote:

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Originally Posted by a d e p t

You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?

Multiple answers possible with this one. https://forums.windrivers.com/images.../2005/03/1.gif

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11 assuming the first 2 coins were different and each coin weighed against one of the first 2 went into a different pile until you got to the last coin which would go into the pile with only 5 coins.

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Originally Posted by a d e p t

Yes. But if you can elaborate as the "why" part, that would give you another point. https://forums.windrivers.com/images.../2005/03/1.gif

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Steel is denser than air, therefore it would take a higher volume of air to displace the weight of the steel in the boat over the surface of the water when the boat was upright.

Quote:

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Originally Posted by a d e p t

You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?

Multiple answers possible with this one. https://forums.windrivers.com/images.../2005/03/1.gif

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I like Ed's initial answer of 10. You pick one coin at random and place it on one side of the scale. You have 11 coins remaining. After measuring 10 of the coins against the initial coin and separating them based on your results, you will know what the weight of the last coin must be so there's no need to measure it. You will also know which pile the initial coin goes in so no need to weigh that one.

You could get lucky and do it in fewer tries but that sounds like the smallest number of measurements to guarantee they are all sorted.

The answer is C

Just had to post in a thread that had been going on for so long

Hrm, I now think the answer is eight. You place six coins on each side and note the level of each side of the scale. Swap one coin from each side to the other side. Worst case, five measurements each result in the two coins being identical so you now have six piles, each consisting of two identical coins.

Two more weighings of two of those piles against each other will result in one of three scenarios

1-two piles of four, each of which is known to contain heavy or light coins
2-two piles of four, neither of which is known to be heavy or light
3-one pile of four identical coins and two piles of two coins known to be heavy or light

Scenario 1, one more weighing of any weighed two coin pile against either unweighed pile will sort them all out

Scenario 2, same method as scenario 1

Scenario 3, the two unweighed piles must be identical. weigh either of them against either of the two separate weighed two coin piles that are known to be heavy or light.

So worst case with this method is five weighings plus two weighings plus one weighing equals eight.

Quote:

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Originally Posted by a d e p t

You have 12 coins. They are completely identical except six of them weigh 24g and the other six weigh 25g. You have only a balance scale to sort them out. What is the minimum number of weighings which guarantees all the coins to be sorted?

Multiple answers possible with this one. https://forums.windrivers.com/images.../2005/03/1.gif

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I say eight also, but I used a different weighing method.

Simple logic can eliminate some of these steps depending on the situation, but this is given worst case scenario.

Step 1: Split coins up into four groups of three, labeled A, B, C and D.
Step 2: Weigh groups A and B (1), this will determine either even or one side is heavy.
Step 3: Weigh 2 coins from group A (2), then weigh one of those and the third from group A (3). This should determine light and heavy for the three coins in group A, using logic on the three weighing results. Three down, 9 to go.
Step 4: Weigh 2 coins from group B (4), the weigh one of those and the third from group B (5). This should determine light and heavy for the three coins in group B, using the same logic as in step 3. 6 down, 6 to go.
Step 5: Weigh group C and group D (6), they will be either even or heavy.
Step 6: Weigh 2 coins from group C (7).
Step 7: At this point you will either weigh the third coin from group C against one of the other two, or you will weigh two coins from group D (8). Anyway it goes, you should be able to determine placement with a little logic.

I have used this method repeatedly and have come up with 5 to 8 weighings, depending on the situation. I'm not sure I've explained this properly, but it's difficult given the various possible situations. Using this method, however, 8 is the maximum number of weighings that guarantees separation.

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Originally Posted by edball

Steel is denser than air, therefore it would take a higher volume of air to displace the weight of the steel in the boat over the surface of the water when the boat was upright.

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Extra point goes to you.

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Originally Posted by cisco2

Hrm, I now think the answer is eight. You place six coins on each side and note the level of each side of the scale. Swap one coin from each side to the other side. Worst case, five measurements each result in the two coins being identical so you now have six piles, each consisting of two identical coins.

Two more weighings of two of those piles against each other will result in one of three scenarios

1-two piles of four, each of which is known to contain heavy or light coins
2-two piles of four, neither of which is known to be heavy or light
3-one pile of four identical coins and two piles of two coins known to be heavy or light

Scenario 1, one more weighing of any weighed two coin pile against either unweighed pile will sort them all out

Scenario 2, same method as scenario 1

Scenario 3, the two unweighed piles must be identical. weigh either of them against either of the two separate weighed two coin piles that are known to be heavy or light.

So worst case with this method is five weighings plus two weighings plus one weighing equals eight.

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3 points.

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Originally Posted by Deity

I say eight also, but I used a different weighing method.

Simple logic can eliminate some of these steps depending on the situation, but this is given worst case scenario.

Step 1: Split coins up into four groups of three, labeled A, B, C and D.
Step 2: Weigh groups A and B (1), this will determine either even or one side is heavy.
Step 3: Weigh 2 coins from group A (2), then weigh one of those and the third from group A (3). This should determine light and heavy for the three coins in group A, using logic on the three weighing results. Three down, 9 to go.
Step 4: Weigh 2 coins from group B (4), the weigh one of those and the third from group B (5). This should determine light and heavy for the three coins in group B, using the same logic as in step 3. 6 down, 6 to go.
Step 5: Weigh group C and group D (6), they will be either even or heavy.
Step 6: Weigh 2 coins from group C (7).
Step 7: At this point you will either weigh the third coin from group C against one of the other two, or you will weigh two coins from group D (8). Anyway it goes, you should be able to determine placement with a little logic.

I have used this method repeatedly and have come up with 5 to 8 weighings, depending on the situation. I'm not sure I've explained this properly, but it's difficult given the various possible situations. Using this method, however, 8 is the maximum number of weighings that guarantees separation.

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3 points here too.

The Boldavian navy has a rigorous training program.

In order to graduate from the naval academy, cadets must enter a pool and fight sharks.

Today, a group of cadets was tested. Each cadet put on his mask, snorkel, and two flippers; and armed himself with a knife. Then the cadets jumped into the pool. The sharks were then released from their pens.

After 10 minutes of gruelling hand-to-fin combat, all of the sharks were killed. Miraculously, none of the cadets were killed. Exiting the pool, the cadets began to haul the shark bodies out of the water. The dead sharks were soon lying in a row.

One of the cadets noted that during the fight, there were 4 times as many cadets in the water as there were sharks.

A second cadet then stated that the number of legs they now had was equal to the square of the number of sharks.

A third said that the number of cadet legs was 14 more than the total number of cadet and shark heads.

How many sharks were killed?


__________
7 sharks were killed.

Since there were 4 times as many cadets, there were 28 cadets. The number of cadet legs is equal to the square of the number of sharks, which would be 49 legs. 49 legs is 14 more than the number of cadet and shark heads: 7 sharks plus 28 cadets equals 35 heads.

Note that although all of the cadets survived, nothing was said about possible injuries. In fact, each shark had bitten off one leg from each of 7 cadets. This explains the shortage of legs, since 28 cadets should have had 56 legs, not 49. Also note the statement of the second cadet, who said 'the number of legs they NOW had ...'.


Check out deity's post for the math.

Quote:

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Originally Posted by a d e p t

The Boldavian navy has a rigorous training program.

In order to graduate from the naval academy, cadets must enter a pool and fight sharks.

Today, a group of cadets was tested. Each cadet put on his mask, snorkel, and two flippers; and armed himself with a knife. Then the cadets jumped into the pool. The sharks were then released from their pens.

After 10 minutes of gruelling hand-to-fin combat, all of the sharks were killed. Miraculously, none of the cadets were killed. Exiting the pool, the cadets began to haul the shark bodies out of the water. The dead sharks were soon lying in a row.

One of the cadets noted that during the fight, there were 4 times as many cadets in the water as there were sharks.

A second cadet then stated that the number of legs they now had was equal to the square of the number of sharks.

A third said that the number of cadet legs was 14 more than the total number of cadet and shark heads.

How many sharks were killed?

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7 sharks were killed.

legs = x
sharks = y
cadets = z

x = y^2 (number of legs = sq of sharks)
z = 4y (number of cadets is 4 times sharks)
x = 14 + z + y (legs was 14 more than the total of heads)

Substitution:

y^2 = 14 + 4y + y
0 = y^2 -5y - 14
0 = (y - 7)(y + 2)
Therefore:
y = 7 or y = -2

Quote:

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Originally Posted by Deity

7 sharks were killed.

legs = x
sharks = y
cadets = z

x = y^2 (number of legs = sq of sharks)
z = 4y (number of cadets is 4 times sharks)
x = 14 + z + y (legs was 14 more than the total of heads)

Substitution:

y^2 = 14 + 4y + y
0 = y^2 -5y - 14
0 = (y - 7)(y + 2)
Therefore:
y = 7 or y = -2

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You baaaaaaad.

3 more points to the non-human. https://forums.windrivers.com/images.../2006/04/1.gif

Quote:

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Originally Posted by Deity

7 sharks were killed.

legs = x
sharks = y
cadets = z

x = y^2 (number of legs = sq of sharks)
z = 4y (number of cadets is 4 times sharks)
x = 14 + z + y (legs was 14 more than the total of heads)

Substitution:

y^2 = 14 + 4y + y
0 = y^2 -5y - 14
0 = (y - 7)(y + 2)
Therefore:
y = 7 or y = -2

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How can 7 squared be 49 and be the number of legs the cadets have ? :confused:

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Originally Posted by edball

How can 7 squared be 49 and be the number of legs the cadets have ? :confused:

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Some of them apparently got chewed off. The question said nobody was killed, it said nothing about not being dismembered.

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Originally Posted by edball

How can 7 squared be 49 and be the number of legs the cadets have ? :confused:

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Answer is now up to explain that.

I had assumed that deity was aware of that... particularity, and his last post proved it.

1 bonus point for edball for the question though, since I had overlooked it, and to fend off the "number of the beast" thing. https://forums.windrivers.com/images.../2005/03/1.gif

Quote:

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Originally Posted by Deity

Some of them apparently got chewed off. The questions said nobody was killed. Said nothing about not being dismembered.

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So that's why it said how many they "now had"...

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Originally Posted by edball

So that's why it said how many they "now had"...

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MWAAAHAAAAHHAAAAHAAAHAAAA! https://forums.windrivers.com/images.../2005/03/1.gif

This is a sick, sick game. I hate feeling stupid in the face of all this genius.

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Originally Posted by Kymera

This is a sick, sick game...

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[Elvis Mode=ON]

Why, thank you - thank you very much. https://forums.windrivers.com/images.../2005/03/1.gif

[Elvis Mode=OFF]


You didn't actually go through the entire 17 pages, did you? https://forums.windrivers.com/images.../2005/03/1.gif

It shows up as 45 pages on in my view, and yes, yes I did.

On the anniversary of the 100th question, this one is worth 5 points. :D



In operating, a cycle.
In duelling, a thrust.
In playing, a throw.
In bidding, a decline.
In a bar, a try.
In racing, a move.
In scaling, a way.

QUESTION: What is it?


__________
A PASS.

A pass is one cycle in the operating room.
It is a thrust with a sword in fencing.
It is a throw of a ball to a teammate in basketball or other games.
It is a decline to bid in auctions.
It is a try to get a girl met in a bar or other place.
It is a move in made by drivers or track athletes in races.
It is a way in mountains.

Quote:

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Originally Posted by a d e p t

In operating, a cycle.
In duelling, a thrust.
In playing, a throw.
In bidding, a decline.
In a bar, a try.
In racing, a move.
In scaling, a way.

QUESTION: What is it?

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A pass.

A pass.

The playing a throw and duelling a thrust gave it away, and the in racing a move confirmed it.

Quote:

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Originally Posted by a d e p t

On the anniversary of the 100th question, this one is worth 5 points. :D



In operating, a cycle. to interate through a set of commands
In duelling, a thrust. a type of thrust in fencing
In playing, a throw. a forward pass in football
In bidding, a decline. to pass on a deal
In a bar, a try. to make a pass at a woman
In racing, a move. to pass another car
In scaling, a way. a mountain pass

QUESTION: What is it?

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So, how is that?

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Originally Posted by edball

A pass.

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That's it. 5 points to you. https://forums.windrivers.com/images.../2006/04/1.gif


...and 3 to Kymera 'cuz she took the time to explain it for the peasants and it's her first time here and she only missed answering by a couple of minutes and I like her.

If

7x7 = 4
11x7 = 6
13x11 = 6

Then

27x28 = ?


__________
12.

Multiply how many syllables are in the word.

ie: se-ven (2) x E-lev-en (3) = 6 (2x3=6)

Quote:

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Originally Posted by a d e p t

If

7x7 = 4
11x7 = 6
13x11 = 6

Then

27x28 = ?

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27x28=4
4 straight lines in the numbers of the equation.

Quote:

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Originally Posted by edball

27x28=4
4 straight lines in the numbers of the equation.

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Not quite.

Quote:

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Originally Posted by a d e p t

Not quite.

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I got it!
twenty-seven x twenty-eight = 12