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Riddle:

You want to send a valuable object to a friend. You have a box which is more than large enough to contain the object. You have several locks with keys. The box has a locking ring which is more than large enough to have a lock attached. But your friend does not have the key to any lock that you have. How do you do it? Note that you cannot send a key in an unlocked box, since it might be copied.

Answer:

Attach a lock to the ring. Send it to her. She attaches her own lock and sends it back. You remove your lock and send it back to her. She removes her lock.

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Riddle:

If it is 1,800 kilometers to America, 1,200 kilometers to Japan, 2,400 kilometers to New Zealand, and 1,400 kilometers to Brazil- 

How far is Morocco?

Answer:

The answer is 1,700 kilometers, as vowels in the countries' names are worth 300 kilometers and the consonats are worth 200 kilometers. 

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Riddle:

There are 100 light bulbs lined up in a row in a long room. Each bulb has its own switch and is currently switched off. The room has an entry door and an exit door. There are 100 people lined up outside the entry door. Each bulb is numbered consecutively from 1 to 100. So is each person. Person No. 1 enters the room, switches on every bulb, and exits. Person No. 2 enters and flips the switch on every second bulb (turning off bulbs 2, 4, 6, ...). Person No. 3 enters and flips the switch on every third bulb (changing the state on bulbs 3, 6, 9, ...). This continues until all 100 people have passed through the room. What is the final state of bulb No. 64? And how many of the light bulbs are illuminated after the 100th person has passed through the room?

Answer:

First think who will operate each bulb, obviously person #2 will do all the even numbers, and say person #10 will operate all the bulbs that end in a zero. So who would operate for example bulb 48: Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 ........ That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a bulb on there will be someone to switch it off. This willl result in the bulb being back at it's original state. So why aren't all the bulbs off? Think of bulb 36:- The factors are: 1 & 36, 2 & 13, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it's self. Clearly the sixth person will only flick the bulb once and so the pairs don't cancel. This is true of all the square numbers. There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 bulbs remain on.

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Riddle:

While playing with a metal washer shaped like a ring, Dave accidently pushed it on his finger too far and couldn't get it off. Trying to remove it using soap and water didn't work. The hospital sent him to a service station thinking they could cut the metal. Since the ring was made with a specially hardened steel, it couldn't be cut. Just then Bob arrived on the scene and suggested an easy way to remove the washer in just a few minutes. What was his solution?

Answer:

Bob suggested that Dave hold his finger in the air while someone wound a piece of string tightly around his finger just above the metal ring. The string forced the swelling down. As they unwounded the string from the end nearest the ring, someone else slid the ring up. They continued winding and unwinding the string until the ring could be easily removed.

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Riddle:

I have ten or more daughters. I have less than ten daughters. I have at least one daughter. If only one of these statements is true, how many daughters do I have?

Answer:

If I have any daughters, there will always be two statements which are true. Therefore, I have no daughters.

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Riddle:

A hunter met two shepherds, one of whom had three loaves and the other, five loaves. All the loaves were the same size. The three men agreed to share the eight loaves equally between them. After they had eaten, the hunter gave the shepherds eight bronze coins as payment for his meal. How should the two shepherds fairly divide this money?

Answer:

The shepherd who had three loaves should get one coin and the shepherd who had five loaves should get seven coins. If there were eight loaves and three men, each man ate two and two-thirds loaves. So the first shepherd gave the hunter one-third of a loaf and the second shepherd gave the hunter two and one-third loaves. The shepherd who gave one-third of a loaf should get one coin and the one who gave seven-thirds of a loaf should get seven coins.

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Riddle:

You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weight the same or that the dish that falls lower has heavier contents than the other. The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more. Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light

Answer:

Most people seem to think that the thing to do is weight six coins against six coins, but if you think about it, this would yield you no information concerning the whereabouts of the only different coin. As we already know that one side will be heavier than the other. So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8. There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second weighing we would put 1,2 in the left pan and 9,10 on the right. If these balance then the different coin is either 11 or 12. Weigh coin 1 against 11. If they balance, the different coin is number 12. If they do not balance, then 11 is the different coin. If 1,2 vs 9,10 do not balance, then the different coin is either 9 or 10. Again, weigh 1 against 9. If they balance, the different coin is number 10, otherwise it is number 9. That was the easy part. What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings. Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin. For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

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Riddle:

You have two container, a 5 gallon and a 3 gallon container. How do you measure out 4 gallons?

Answer:

Fill up the 3 gallon container and pour the 3 gallons into the 5 gallon container.
Then, fill the 3 gallon container back up, and pour it into the 5 gallon container.
The 3 gallon container will have 1 gallon left.
Empty the 5 gallon container.
Pour the remining 1 gallon into the 5 gallon container.
Then fill the 3 gallon container back up and pour it into the 5 gallon container.
Thus, you have 4 gallons.

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Riddle:

How can this be true? Have a look at the picture. All the lines are straight, the shapes that make up the top picture are the same as the ones in the bottom picture so where does the gap come from?

Answer:

The green triangle has dimensions 2 x 5 and gradient 2 / 5 = 0.4
The red triangle has dimensions 3 x 8 and gradient 3 / 8 = 0.375
Hence the gradient of the green triangle is greater than that of the red triangle.

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Riddle:

Their are three gaurds and three prisoners who need to cross a river. Their boat only holds two people at a time, and the number of prisoners must NEVER be allowed to outnumber the number of gaurds on either side of the river; otherwise, the prisoners will overpower the gaurds and, well, the story will come to an abrupt end.

Determine how many trips it will take to safely transport all of the gaurds and prisoners across the river, list each of the trips that need to be made and who is in the boat and who is on each of the riverbanks during each trip.

Answer:

Inless some one can tell me a way that 2 prisoners, at some point, don't out number the guards whether they are just dropping off and still in the boat or actually on land (because even if they are just dropping off and remain in the boat they are still on the other side of the river) I conclude this to be impossible. Please let me know an alternative if you figure one out because i'm stumped.

thanks

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