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I'm not sure if these questions have been asked before, apologies if they have:

---1---

You have 4 horses on farm A, and you'd like to move them to farm B. The 4 horses take 1 hr, 2hr, 4hr and 6hr to get from A to B respectfully. Now you can ride a horse and take one horse with you during your journey, and you would travel at the speed of the slower horse. What is the shortest possible time that it will take you to get all 4 horses across?

---2---

I have 8 chains, each chain consists of 7 links. Now I would like to connect all 8 chains into a single loop. It costs $1 to open up a link, and $2 to close an opened link again. What is the least amount of money I can spend to create my chain?

---3---

A little girl walks out the house with her little puppy in her arms. She walks 10 meters out. At this instant, her dad walks out the front door to chase after her, she puts her dog down and continues forward. The dad walks at 2m/s and the little girl walks at 1m/s. The dog continuously runs back and forth between the little girl and her dad until the dad catches up to the little girl. The dog runs at 5m/s. How far did the dog run?

---4---

In a room with no windows, only a closed door, there are 3 lightbulbs. You have no way of seeing into the room without opening the door and going inside. Outside there are 3 switches, each corresponds to a lightbulb inside the room but you don't know which. I want to know which switch is for which lightbulb, and I'm only allowed to enter the room once. How would I do it?

---5---

Many years ago, there was a kingdom that had an odd rule. It states that any person on death row has the chance to decide his own fate by pick one of two cards that are faced down. The cards will have one with "Death" written on it, and the other will have "Live" written. If you pick the "Live" card, you get to live. Now one day a person was on death row, the king really really hated him, so he put two "Death" cards faced down on the table for the man to pick. The man still lived afterwards, what did he do?

---6---

(I saw a similar question but this is a little different) There are 13 coins, one is counterfeit. The real coins all weigh the same but the fake is either lighter or heavier than the real ones. You have a balancing scale, you are only allowed to use the scale 3 times, how do you find the counterfeit?

---7---

There's a boat in the harbor, it has a ladder hanging over the edge. The ladder is 10 m long, and the end of the ladder is just touching the surface of the water. The first rung of the ladder is 30 cm above the water, and there's a wrung every 30 cm. The water in the harbor is rising at 50cm/minute. The time now is 10:00 am, when will the water reach the 8th rung?

---8---

There are 10 birds in a tree, I shoot one of them, how many is left? (:D EASY PEASY!!!)

---9---

In ancient China, A general called CaoCao knew an attack was coming in the night, so he wanted to keep his soldiers awake throughout the night. He told them that if they could solve this question, then he would grant them all the riches they will ever need and his beautiful daughter's hand in marriage. The question was that he had a farm with 30 chickens, he wanted to kill them all in a week, each day he only kills an odd number of chickens, how should he go about killing the chickens?

---10---

One day, a father went to his three sons and told them that he would die soon and he needed to decide which one of them to give his property to. He decided to give them all a test. He said, "Go to the market my sons, and purchase something that is large enough to fill my bedroom, but small enough to fit in your pocket. From this I will decide which of you is the wisest and worthy enough to inherit my land." So they all went to the market and bought something that they thought would fill the room, yet was still small enough that they could fit into their pockets. Each son came back with a different item. The father told his sons to come into his bedroom one at a time and try to fill up his bedroom with whatever they had purchased. The first son came in and put some pieces of cloth that he had bought and laid them end to end across the room, but it barely covered any of the floor. Then the second son came in and laid some hay, that he had purchased, on the floor but there was only enough to cover half of the floor. The third son came in and showed his father what he had purchased and how it could fill the entire room yet still fit into his pocket. The father replied, "You are truly the wisest of all and you shall receive my property." What was it that the son had showed to his father?

---11---

Pirate Marcelo had been captured by a Spanish general and sentenced to death by his 50-man firing squad. Marcelo cringed, as he knew the Spanish general's firing squad's reputation for being the worst firing squad in the Spanish military. They were such bad shots that they would often all miss their targets and simply maim their victims, leaving them to bleed to death, as the general’s tradition was to only allow one shot per man to save on ammunition. The thought of a slow painful death made Marcelo beg for mercy.

“Very well, I have some compassion. You may choose where the men stand when they shoot you and I will add 50 extra men to the squad to ensure someone will at least hit you. Perhaps if they stand closer they will kill you quicker, if you’re lucky,” snickered the general. “Oh, and just so you don’t get any funny ideas, they can’t stand more than 20 ft away, they must be facing you, and you must remain tied to the post in the middle of the yard. And to show I’m not totally heartless, if you aren’t dead by sundown I’ll release you so you can die peacefully outside the compound. I must go now but will return tomorrow and see to it that you are buried in a nice spot, though with 100 men, I doubt there will be much left of you to bury.”

After giving his instructions the general left. Upon his return the next day, he found that Marcelo had been set free alive and well. “How could this be?” demanded the general. “It was where Marcelo had us stand,” explained the captain of the squad.

Where did Marcelo tell them to stand?

---12---

Once upon a time, in the small magical village, a servant lived with his master who was a wizard. After service of 30 years, his master became ill and was going to die. One day, the master called his servant and offered him for a wish. It could be any wish but just one, and no wishing for more wishes. The master gave him one day to think about it. The servant became very happy and went to his mother to discuss the wish. His mother was blind and she asked her son to wish for her eye-sight to come back. Next, the servant went to his wife. She became very excited and asked for a son as they were childless for many years. After that, the servant went to his father who wanted to be rich and so he asked his son to wish for a lot of money. The next day he went to his master and made one wish through which all the three (mother, father, wife) got what they wanted. What was his wish?

---13---

One snowy night, Sherlock Holmes was in his house sitting by a fire. All of a sudden a snowball came crashing through his window, breaking it. Holmes got up and looked out the window just in time to see three neighborhood kids who were brothers run around a corner. Their names were John Crimson, Mark Crimson and Paul Crimson. The next day Holmes got a note on his door that read “? Crimson. He broke your window.” Which of the three Crimson brothers should Sherlock Holmes question about the incident?

---14---

An 18-wheeler is crossing a 4 kilometer bridge that can only support 10,000 kilograms and that's exactly how much the rig weighs. Halfway across the bridge a 30 gram sparrow lands on the cab, but the bridge doesn't collapse. Why not?

---15---

Two men working at a construction site were up for a challenge, and they were pretty mad at each other. Finally, at lunch break, they confronted one another. One man, obviously stronger, said "See that wheelbarrow? I'm willin' to bet $100 (that's all I have in my wallet here) that you can't wheel something to that cone and back that I can't do twice as far. Do you have a bet?"

The other man, too dignified to decline, shook his hand, but he had a plan formulating. He looked at the objects lying around: a pile of 400 bricks, a steel beam, the 10 men that had gathered around to watch, his pickup truck, a stack of ten bags of concrete mix, and then he finalized his plan.

"All right," he said, and revealed his object.

That night, the strong man went home thoroughly teased and $100 poorer. What did the other man choose?

---16---

General Custer is surrounded by Indians and he's the only cowboy left. He finds an old lamp in front of him and rubs it. Out pops a genie. The genie grants Custer one wish, with a catch. He says, "Whatever you wish for, each Indian will get two of the same thing." Custer ponders a while and thinks:"If I get a bow and arrow they get two. If I get a rifle they get two!" He then rubs the bottle again and out pops the genie. "Well," the genie asks "have you made up your mind?"

What did Custer ask for to help him get away?

---17---

What is the next number in the sequence?

1

11

21

1211

111221

312211

13112221

---18---

If a bottle and a cork cost a dollar and a nickel, and the bottle costs a dollar more than the cork, how much does the cork cost?

---19---

You have a lighter and two fuses that take exactly one hour to burn, but they don't burn at a steady rate. For example, one fuse could take 59 minutes to burn the first inch and then burn the rest of the fuse in the last minute. How would you use these two fuses to measure 45 minutes?

---20---

You're riding a horse. To the right of you is a cliff and in front of you is an elephant going the same pace as you and you can't overtake it. To the left of you is a hippo running at the same speed and behind you is a lion chasing you. How do you get to safety?

---21---

You have 50 quarters on the table in front of you. You are blindfolded and cannot discern whether a coin is heads up or tails up by feeling it. You are told that x coins are heads up, where 0 < x < 50. You are asked to separate the coins into two piles in such a way that the number of heads up coins in both piles is the same at the end. You may flip any coin over as many times as you want. How will you do it?

---22---

You are a prisoner sentenced to death. The Emperor offers you a chance to live by playing a simple game. He gives you 50 black marbles, 50 white marbles and 2 empty bowls. He then says, "Divide these 100 marbles into these 2 bowls. You can divide them any way you like as long as you use all the marbles. Then I will blindfold you and mix the bowls and marbles up. You can then choose one bowl and remove one marble. If the marble is white, you live, but if the marble is black...you die.

How do you divide the marbles up so that you have the greatest probability of choosing a white marble?

---23---

Suppose you’re on a game show, and you’re given a choice of three doors. Behind one door is a car; behind the others, goats. You pick a door — say, No. 1 — and the host, who knows what’s behind the doors, opens another door — say, No. 3 — which has a goat. He then says to you,"Do you want to pick door No. 2?" Is it to your advantage to switch your choice? Why?

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Put all the marbles but one white one into one bowl and the one white one in the other. Hopefully, even blindfolded you are allowed to feel for the bowls or sense them by sound, weight or by the way they are placed before you, or by some other non-visual cue so you can know which bowl to make your selection from and be sure to get the white marble and live.

Edited by Magi-Ken
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The time it takes the dad to catch his daughter times the avg speed of the dog is correct, but this is not just for the initial 10 meters. By the time dad has gone 10 m his daughter has gone another 5 m. It takes dad another 2.5 sec to cover that distance and in the mean time the girl moves forward half that distance as well. I used a spread sheet to calculate 62 levels of micro seconds and totaled the column. It totaled 10 seconds. In the 10 seconds it took dad to catch his daughter, at 5m/sec the pup would have covered 50 meters total going back and forth between them (with no lag time for turn arounds, this being covered in it average speed).

You didn't need to use excel to do this. The formula for how long it takes for something to catch up to something else is the distance divided by the difference in speeds. If you think about it, assume that both the dad's speed and the girl's speed is reduced by 1m/s, then the girl moves at 0m/s, and the dad moves at 1m/s, and there's 10 meters between them, so its easy to see that it takes 10 seconds for the dad to catch up.

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Put all the marbles but one white one into one bowl and the one white one in the other. Hopefully, even blindfolded you are allowed to feel for the bowls or sense them by sound, weight or by the way they are placed before you, or by some other non-visual cue so you can know which bowl to make your selection from and be sure to get the white marble and live.

Your answer is correct, however you will need to choose a bowl, then pick a marble. So 50% of the time you live for sure, other 50% you have just less than 50% chance of living. Those are your best chances

:)
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I have seen some questions on this but no steps for doing it yet so I think that is in order as part of the answer. However, I am not sure I have THE answer because I need a clarification in what USE THE SCALES 3 times means. Based on that I have 2 answers. The first is 100% and ALWAYS tells if the fake coin is lighter or heavier. The second does not work infallibly.

Solution #1: Start placing one coin on each side of the scale and keep adding one coin to each side, keeping track of the two coins added last. As soon as the balance is unequal one of the two last coins is fake. Remove those two and set them aside. If you hold the odd 13th coin and the scales still balance at 6 coins on each side then you hold the fake. Balance used once; clear the scale. The second time place one of the two suspect coins on one side and any of the non-suspect coins on the other side. If they balance then the other suspect coin is the fake. If they do not balance the suspect coin on the balance is the fake one and is obviously heavier or lighter than the non-deviant/non-suspect one. Use number three of the scale is only if they balance in step 2 and is used to verify (not determine) whether or not the fake is lighter or heavier, but even this would not be needed if one kept track of which side was heavier in step #1.

Solution #2 takes "use the scales" as meaning -- each time a coin or coins are added or removed this counts as one use. In that case, remove one coin and set it aside. Place 6 of the 12 remaining coins on each side. If they balance - Lucky YOU! - You held out the fake and have a good eye or 6th sense. Weigh it against any of the others and you will confirm it is different and know if it is lighter or heavier. However, if the 12 coins do not balance. the coin you held out is a good one. Since you do not know which set of 6 has the light or heavy coin the 2nd use is a 50/50 guess remove the 12 keeping them in distinct piles of 6. Put one group of 6 back on the scale; three on each side. If they balance, you can't do it in three steps by this method. If they do not balance, you have narrowed it down to 1 of 2 groups of 3. If the first test showed this group to be heavier then you are looking for a heavy coin. If test #1 showed it lighter then a lighter fake. The third use is to hold out one of the 3 coins from the group of 3 that has the fake (you know which group of 3 that is because you know it to be lighter or heavier). Set your new choice of one of the three aside and weigh the remaining 2. If these 2 balance you picked out the fake. If they don't balance, then the fake is the one that is consistant with your known lighter or heaver weight. The only time this does not work is if you pick the wrong pile of 6 to test. In that case, your groups of 3 will balance and it will take an extra step to go back and divide the other set of six into 2 sets of 3 etcetera.

If there is a better, more clever solution, I am anxious to hear it. It will be exciting.

Okay so this is quite a famous question.

You cannot add coins to both sides of the scale and not count them as a 'weighing'. So I'll give you a a hint:

Divide the coins up into 3 groups. Each consisting of 4 coins, 4 coins and 5 coins.

Step 1: Weigh the two groups of 4 together, if they are balanced: go to step 2. Else go to step 3.

Step 2: ...........

Step 3: ...........

Go from there :D (Try step 2 first, it's a bit easier)

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Contrary to what some have said, # 6 does have a

13 coins, labeled A B C D E F G H I J K L M. First weighing: A B C D E vs. F G H I J If they balance, then simple,: Second weighing K vs. L Third weighing L vs. M If K vs. L balance, then M is counterfeit and L vs. M determines whether heavier or lighter. If K vs. L not balanced, note which is heavier and use that info to determine relative weight of K if L vs. M balances. If L vs, M does not balance, then L is counterfeit and its relative weight is determined. If the first weighing does not balance, then we know K L & M are real. Second weighing A B C K vs. D E F G If they balance, the bad coin is H I or J and we know whether heavier or lighter Third weighing H vs. I, if it balances, then J is bad and relative weight was determined in First weighing. If they don't balance, then using weight imbalance of First weighing to know if bad coin is heavier or lighter, this identifies counterfeit. Now assuming Second weighing A B K L vs. C D E F G did not balance, if weight reversed, then it is either D or E (the ones we reversed the side.) And knowing the relative weight of the counterfeit, third weighing of C vs. D If balances, coin is E, if not imbalance direction determines counterfeit. If weight is the same, then bad coin has been narrowed to H I J. Third weighing in this scenario: H vs. I if balanced, J is coin and relative weight was determined in First weighing. If not balanced, from knowledge of First weighing, we now can determine counterfeit.

solution.

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Hint by TwinHelix will not give full solution.

Step 2, if first groups of four balance, leaves 5 with no knowledge of previous balance. In this scenario, the best outcome that can be guaranteed is to identify the counterfeit coin, but not its weight differential. My solution above will give weight differential.

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Contrary to what some have said, # 6 does have a

13 coins, labeled A B C D E F G H I J K L M. First weighing: A B C D E vs. F G H I J If they balance, then simple,: Second weighing K vs. L Third weighing L vs. M If K vs. L balance, then M is counterfeit and L vs. M determines whether heavier or lighter. If K vs. L not balanced, note which is heavier and use that info to determine relative weight of K if L vs. M balances. If L vs, M does not balance, then L is counterfeit and its relative weight is determined. If the first weighing does not balance, then we know K L & M are real. Second weighing A B C K vs. D E F G If they balance, the bad coin is H I or J and we know whether heavier or lighter Third weighing H vs. I, if it balances, then J is bad and relative weight was determined in First weighing. If they don't balance, then using weight imbalance of First weighing to know if bad coin is heavier or lighter, this identifies counterfeit. Now assuming Second weighing A B K L vs. C D E F G did not balance, if weight reversed, then it is either D or E (the ones we reversed the side.) And knowing the relative weight of the counterfeit, third weighing of C vs. D If balances, coin is E, if not imbalance direction determines counterfeit. If weight is the same, then bad coin has been narrowed to H I J. Third weighing in this scenario: H vs. I if balanced, J is coin and relative weight was determined in First weighing. If not balanced, from knowledge of First weighing, we now can determine counterfeit.

solution.

Mate please read your solution's last bit, it makes no sense!!

First weighing of A B C D E vs F G H I J is unequal, assume for arguments sake, and without the loss of generality, assume that the side of A B C D E was heavier.

So you're saying second weighing is ABCK vs. DEFG, if they are balanced, then easy, if they not, there's two possible scenarios:

1. The imbalance is reversed (i.e. DEFG is heavier). Then yes, either D or E must be the fake since they are the ones that determines which side is heavier.

2. However if the imbalance still remains, all we know is that D and E are real, but still AB could have the heavier one, OR FG can have a lighter one. Think about it.

I don't know if that is what you're trying to get at, cuz at one stage you said the second weighing is "A B C K vs. D E F G" then later you said it is "Now assuming Second weighing A B K L vs. C D E F G". So I think you may have gotten confused?

My hint definitely works, and provides you with the counterfeit, but doesn't always give you enough info whether it is heavier or lighter. If you find a way to always tell whether it is heavier or lighter, I will be very interested to see the solution. I will post my answer tomorrow if still no one gets it. :)

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Mate please read your solution's last bit, it makes no sense!!

First weighing of A B C D E vs F G H I J is unequal, assume for arguments sake, and without the loss of generality, assume that the side of A B C D E was heavier.

So you're saying second weighing is ABCK vs. DEFG, if they are balanced, then easy, if they not, there's two possible scenarios:

1. The imbalance is reversed (i.e. DEFG is heavier). Then yes, either D or E must be the fake since they are the ones that determines which side is heavier.

2. However if the imbalance still remains, all we know is that D and E are real, but still AB could have the heavier one, OR FG can have a lighter one. Think about it.

I don't know if that is what you're trying to get at, cuz at one stage you said the second weighing is "A B C K vs. D E F G" then later you said it is "Now assuming Second weighing A B K L vs. C D E F G". So I think you may have gotten confused?

My hint definitely works, and provides you with the counterfeit, but doesn't always give you enough info whether it is heavier or lighter. If you find a way to always tell whether it is heavier or lighter, I will be very interested to see the solution. I will post my answer tomorrow if still no one gets it. :)

I humbly apologize to Twinhelix. As I was working it out and typing, was interrupted several times and apparently used a typo to continue from. I must now agree that we can determine which coin is counterfeit in three weighings but can only determine whether heavy or light 12/13 of the time.

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Here is a solution for determining counterfeit in 3 weighings and 92% chance of determining weight differential.

13 coins, labeled A B C D E F G H I J K L M.

First weighing: A B C D vs. E F G H If they don't balance, then,:

Second weighing A I J K L vs. B C D E F

If they don't balance but are reversed then B C or D is counterfeit and third weighing of B vs. C will yield counterfeit and weight differential since the weight differential of all three was determined in First weighing

If second weighing yields same balance as first, then the Counterfeit is either A E or F and, utilizing knowledge of direction of first weighing, Third weighing of E vs F, if Balanced, bad coin is A, If not balanced, using knowledge of First weighing we know which coin is bad and weight.

(It should be noted that a similar process can be used by using a second weighing of A B E J vs C D E F)

If second weighing above is balanced, then Counterfeit is G or H and we already know weight differential and can then determine counterfeit coin by weighing G vs. H

Now for the condition that First weighing is balanced, we know the counterfeit is I J K L or M but no knowledge of relative weight. In this case, Second weighing is I J vs. K A. If not balanced, we can weigh I vs J and utilize knowledge of imbalance to determine whether bad coin is I J or K and weight differential.

if balanced, we weigh A vs. L. If not balanced, then L is counterfeit and weight differential has been determined, however, if balances, then counterfeit is M but we have no knowledge of direction of weight difference.

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Here is a solution for determining counterfeit in 3 weighings and 92% chance of determining weight differential.

13 coins, labeled A B C D E F G H I J K L M.

First weighing: A B C D vs. E F G H If they don't balance, then,:

Second weighing A I J K L vs. B C D E F

If they don't balance but are reversed then B C or D is counterfeit and third weighing of B vs. C will yield counterfeit and weight differential since the weight differential of all three was determined in First weighing

If second weighing yields same balance as first, then the Counterfeit is either A E or F and, utilizing knowledge of direction of first weighing, Third weighing of E vs F, if Balanced, bad coin is A, If not balanced, using knowledge of First weighing we know which coin is bad and weight.

(It should be noted that a similar process can be used by using a second weighing of A B E J vs C D E F)

If second weighing above is balanced, then Counterfeit is G or H and we already know weight differential and can then determine counterfeit coin by weighing G vs. H

Now for the condition that First weighing is balanced, we know the counterfeit is I J K L or M but no knowledge of relative weight. In this case, Second weighing is I J vs. K A. If not balanced, we can weigh I vs J and utilize knowledge of imbalance to determine whether bad coin is I J or K and weight differential.

if balanced, we weigh A vs. L. If not balanced, then L is counterfeit and weight differential has been determined, however, if balances, then counterfeit is M but we have no knowledge of direction of weight difference.

Yes your solution is absolutely correct :)

Step 1: Divide coins into groups of 4, 4 and 5. Weigh the 4 vs 4. If they balance, go to step 2. If they don't go to step 3.

Step 2: So not you know you have a fake in one of the 5 coins you set aside in the beginning, name these coins A B C D E, and we know that all 8 coins from the first weighing must be real, call these R.

Now for the second weighing we weigh AB vs CR. If they balance, D or E is fake, go to step 2.1. If they don't balance, we know A, B or C is fake, go to step 2.2.

2.1: Weigh D against R, if they are imbalanced, we know that D is fake. Also we know if it's heavier or lighter. Else E is the fake. We don't know if it is heavier or lighter.

2.2: Assume that AB was heavier, then we weigh A vs B, if it is imbalanced, we know the heavier one is the fake, else if they are equal, we know C is fake and lighter. The same applies if AB was lighter.

Step 3: If the original 4 vs 4 was imbalanced. Then without the loss of generality, we can call the heavier side ABCD, and the lighter side 1234. We know the 5 coins are real, so call one of them R.

Now for the second weighing, weigh AB1 vs. C2R. If they are balanced, then we know that either D or 34 are fake. Go to step 3.1. If they are imbalanced, go to 3.2.

3.1: We now know D or 34 has a fake in it. From our first weighing, we know that if D is fake, it is heavier, else 3 or 4 is lighter. So weigh 3vs4, and if they are imbalanced, the lighter one is the fake, else if they are balanced, D is fake and is heavier.

3.2 If the imbalance remains the same (i.e. AB1 is still heavier, and C2R is lighter) go to step 3.2.1. Else if the imbalance is reversed (AB1 is now lighter than C2R), go to step 3.2.2.

3.2.1: Now we know either AB is fake and is heavier, or C is fake and lighter. Weight A vs B, same logic as 3.1.

3.2.2: Because we switched around 1 and C, and they affected the outcome of the balancing scale, we know one of them is the fake, so weigh any one against a R (real coin) and we'll know which is fake. If C is fake, it is heavier, if 1 is fake, then it is lighter.

Hope that was clear enough. If you have any questions, please ask! :)

Still got 21 to solve I think :)

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Solution for 21

Count off x coins for first group, remaining group has 50 - x coins. Now flip all the coins in the first group. Explanation: When the two groups of divide, there is a finite number of heads (H) in the first group, making the number in the second group x-H. By flipping the coins in the first group, we have changed the number of heads in the first group to x-h. Ergo, problem solved.

Edited by thoughtfulfellow
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Solution for 21

Count off x coins for first group, remaining group has 50 - x coins. Now flip all the coins in the first group. Explanation: When the two groups of divide, there is a finite number of heads (H) in the first group, making the number in the second group x-H. By flipping the coins in the first group, we have changed the number of heads in the first group to x-h. Ergo, problem solved.

I had thought so at first, but then I thought of a simple counter example that proves this method wrong. Suppose there are 26 heads and somehow 25 of them end up in the first half you split, leaving one head in the second group. Flipping all the coins in the first group will leave no heads, which doesn't fulfill the requirements. It seems there is still more thinking to do.

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I had thought so at first, but then I thought of a simple counter example that proves this method wrong. Suppose there are 26 heads and somehow 25 of them end up in the first half you split, leaving one head in the second group. Flipping all the coins in the first group will leave no heads, which doesn't fulfill the requirements. It seems there is still more thinking to do.

No the solution supplied is perfect. You know there are 26 heads, so you put aside 26 coins, which now can have as you stated 25 heads (and 1 tail), then when you flip it all, you have 25 tails and 1 head. The other group also has 1 head.

You don't divide the groups into halves, but into a group with x coins and a group with 50 - x, where x = number of heads given.

Edited by Twinhelix
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totodeal, it takes longer than that.

After you ride your first two horses to farm B, you have to ride a horse back to farm A in order to get the other horses. So the idea is like this:

1) Take two horses to farm B.

2) Return to farm A with one horse. One horse left at B.

3) Take two horses to farm B. Now there are three horses at B.

4) Return to farm A with one horse. Two horses left at B.

5) Take the remaining two horses to farm B. Now they are all there :)

Now find the shortest time to do these five steps!

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Your answer is correct, however you will need to choose a bowl, then pick a marble. So 50% of the time you live for sure, other 50% you have just less than 50% chance of living. Those are your best chances

:)

I disagree... put the white marbles all in one bowl and the black marbles in another. Turn the bowl of black marbles upside down (on the table so the marbles don't fall out) when you are blindfolded, you will reach into the bowl of white marbles because it is impossible to reach into an upside-down bowl :)

Edited by harpuzzler
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1)

Take 2 hr and 1 hr horse. = 2 hrs

Come back by 1 hr horse = 1 hr

take 4 hr and 6 hr horse = 6hrs

come back by 2 hr horse = 2 hr

Take 1 hr and 2 hr horse = 2 hr

Total = 13 hrs

2)

Cut all links of 1 chain to get 7 links = $7

Using 7 links join the remaining 7 chains = $14

Total = $21

3)

Assuming no halt the dog moved 50m. This is because the dad took 10 secs (net 20 m travelled by girl and dad) to catch the gir;. The speed of dog is 5 m/sec. The answer is simply speed * time.

4)

I'll light 1 bulb for quite some time. Turn it off. Turn on another bulb and go in.

Switch 1 = bulb which is heated

Switch 2 = bulb which is lighted

switch 3 = the remaining bulb

5)

He took 1 card, saw it secretly and said that he got the card for living on. When asked by the king to show the card he simply asked the king to check the other card which was death card of course.

6)

Still trying to work out.

7)

None because boat will rise with the water.

8)

None because all oother would fly away.

9) Can't be done since sum of 7 odd numbers would always be an odd number while the chickens are in even numbers.

10) Even though the conventional answer is a match which fills the room with light when lit and yet stay in a pocket, there may be more answers like a perfume bottle which when sprayed would fill the room with itself.

11)

He asked everyone to stand in a compact circle.

12)

"I want to see my mother's grandson to be playing around in his own garden of gold."

13)

Question Mark (?) Crimson. He broke your window.

14)

Probably because it must have lost a lot of fuel by then.

15)

He asked the strong man to get inside the barrow.

16)

He asked for 1 stone eye.

17)

1113213211 -> row before this contains one 1, one 3, two 1, three 2, one 1 in consecutive sequence.

18)

Half a nickel.

19)

Burn fuse 1 on both sides and fuse 2 on 1 side. As soon as the 1st fuse is burnt, lit the 2nd fuse fron the other side. The total time would be 45 mins.

20)

Get off the merry go round. ;)

21)

Take x coins from the stack and flip them. Make 1 stack of these flipped coins and the remining coins in the other stack.

22)

1 white marble in 1 bowl and rest in the other bowl would give the maximum probability.

23)

Will pick up door 2 as it will have the probability of 2/3 for being correct now.

wink.gifwink.gifwink.gifwink.gifwink.gifwink.gifwink.gif

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1) Open both ends of chain #1 ($2)

2) Hook one end of chain #1 through one end of chains #2 through #8 and close the link ($2)

3) Hook the other end of chain #1 through the other end of chains #2 through #8 and close the link ($2)

You end up with a loop of two chain lengths in circumference, with one half of the loop being 7 chains thick.

Total $6

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1) Open both ends of chain #1 ($2)

2) Hook one end of chain #1 through one end of chains #2 through #8 and close the link ($2)

3) Hook the other end of chain #1 through the other end of chains #2 through #8 and close the link ($2)

You end up with a loop of two chain lengths in circumference, with one half of the loop being 7 chains thick.

Total $6

The trick to the question wasn't in sneakily changing the way the chain will look once it is completed. The chain must look like a normal chain, the trick isn't there. The trick is in how you split the chains. So see the other solutions for what I mean =)

But good thinking!

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the question #23 isn't taken from the Blackjack Movie 21 ? :P

and he says each one(door) make a 33.3% so when he ask u another time he's adding a 33.3 percent so now u have 66.6 % (am i right?)

btw ,can u post the numbers of question unsolved yet,cz i don't want to go around 8 pages :P

:thanks: in advance

Edit:for typo

Edited by Hidden G
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the question #23 isn't taken from the Blackjack Movie 21 ? :P

and he says each one(door) make a 33.3% so when he ask u another time he's adding a 33.3 percent so now u have 66.6 % (am i right?)

btw ,can u post the numbers of question unsolved yet,cz i don't want to go around 8 pages :P

:thanks: in advance

Edit:for typo

Sorry bud, all the questions have been answered already hey... So like there's no more unanswered questions sorry :)But go ahead and try solve them anyways :) Also try the new talent search questions I've posted!

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Sorry bud, all the questions have been answered already hey... So like there's no more unanswered questions sorry :)But go ahead and try solve them anyways :) Also try the new talent search questions I've posted!

ok , thanks ,first :isn't my answer to #23 correct?

second :i found your Talent search today but most of them were solved,the prob is most questions are solved when i'm asleep :(

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