A horse travels a certain distance each day. Strangely enough, two of its legs travel 30 miles each day and the other two legs travel nearly 31 miles. It would seem that two of the horse's legs must be one mile ahead of the other two legs, but of course this can't be true. Since the horse is normal, how is this situation possible?
The horse operates the mill and travels in a circular clockwise direction. The two outside legs will travel a greater distance than the inside ones.
A man comes to a small hotel where he wishes to stay for 7 nights. He reaches into his pockets and realizes that he has no money, and the only item he has to offer is a gold chain, which consists of 7 rings connected in a row (not in a loop).
The hotel proprietor tells the man that it will cost 1 ring per night, which will add up to all 7 rings for the 7 nights.
"Ok," the man says. "I'll give you all 7 rings right now to pre-pay for my stay."
"No," the proprietor says. "I don't like to be in other people's debt, so I cannot accept all the rings up front."
"Alright," the man responds. "I'll wait until after the seventh night, and then give you all of the rings."
"No," the proprietor says again. "I don't like to ever be owed anything. You'll need to make sure you've paid me the exact correct amount after each night."
The man thinks for a minute, and then says "I'll just cut each of my rings off of the chain, and then give you one each night."
"I do not want cut rings," the proprietor says. "However, I'm willing to let you cut one of the rings if you must."
The man thinks for a few minutes and then figures out a way to abide by the proprietor's rules and stay the 7 nights in the hotel. What is his plan?
The man cuts the ring that is third away from the end of the chain. This leaves him with 3 smaller chains of length 1, 2, and 4. Then, he gives rings to the proprietor as follows:
After night 1, give the proprietor the single ring
After night 2, take the single ring back and give the proprietor the 2-ring chain
After night 3, give the proprietor the single ring, totalling 3 rings with the proprietor
After night 4, take back the single ring and the 2-ring chain, and give the proprietor the 4-ring chain
After night 5, give the proprietor the single ring, totalling 5 rings with the proprietor
After night 6, take back the single ring and give the proprietor the 2-ring chain, totalling 6 rings with the proprietor
After night 7, give the proprietor the single ring, totalling 7 rings with the proprietor
On the game show et´s Make a Deal, Monty Hall shows you three doors. Behind one of the doors is a new car, the other two hide goats. You choose one door, perhaps #1. Now Monty shows you what´s behind door #2 and it´s a goat.He gives you the chance to stay with original pick or select door #3. What do you do?
You should always abandon your original choice in favor of the remaining door (#3). When you make your first choice the chance of winning is 1 in 3 or 33%. When you switch doors, you turn a 2 in 3 chance of losing in the first round into a 2 in 3 chance of winning in the second round.
Search: Monty Hall problem
It was a very large truck. The truck need to cross a 20 mile long bridge. Unfortunately, the bridge can only hold the weight of 12000 lbs. Even a single pound extra, the bridge would collapse. However the weight of the truck is exactly 12000 lbs. The driver carefully drove and crossed almost 85 percent distance of the bridge. He stopped to get a small break. Suddenly, a bird landed on the truck. Did the bridge collapse? Justify your answers with explanation!
No. The bridge doesn't collapse. The truck almost crossed 85 percent of total distance. Equivalent diesel would have been lost. So the extra weight of the bridge doesn't add any extra load to the bridge.
Every day, Jack arrives at the train station from work at 5 pm. His wife leaves home in her car to meet him there at exactly 5 pm, and drives him home. One day, Jack gets to the station an hour early, and starts walking home, until his wife meets him on the road. They get home 30 minutes earlier than usual. How long was he walking? Distances are unspecified. Speeds are unspecified, but constant. Give a number which represents the answer in minutes.
The best way to think about this problem is to consider it from the perspective of the wife. Her round trip was decreased by 30 minutes, which means each leg of her trip was decreased by 15 minutes. Jack must have been walking for 45 minutes.
James ordered a fishing rod, priced at $3.56. Unfortunately, James is an Eskimo who lives in a very remote part of Greenland and the import rules there forbid any package longer than 4 feet to be imported. The fishing rod was 4 feet and 1 inch, just a little too long, so how can the fishing rod be mailed to James without breaking the rules? Ideally James would like the fishing rod to arrive in one piece!
Insert the fishing rod into a box which measures 4 feet on all sides, the fishing rod will fit within the diagonal of the box with room to spare.
There are n coins in a line. (Assume n is even). Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins.
Would you rather go first or second? Does it matter?
Assume that you go first, describe an algorithm to compute the maximum amount of money you can win.
Note that the strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners.
Example 18 20 15 30 10 14
First Player picks 18, now row of coins is
20 15 30 10 14
Second player picks 20, now row of coins is
15 30 10 14
First Player picks 15, now row of coins is
30 10 14
Second player picks 30, now row of coins is
10 14
First Player picks 14, now row of coins is
10
Second player picks 10, game over.
The total value collected by second player is more (20 + 30 + 10) compared to first player (18 + 15 + 14). So the second player wins.
Going first will guarantee that you will not lose. By following the strategy below, you will always win the game (or get a possible tie).
(1) Count the sum of all coins that are odd-numbered. (Call this X)
(2) Count the sum of all coins that are even-numbered. (Call this Y)
(3) If X > Y, take the left-most coin first. Choose all odd-numbered coins in subsequent moves.
(4) If X < Y, take the right-most coin first. Choose all even-numbered coins in subsequent moves.
(5) If X == Y, you will guarantee to get a tie if you stick with taking only even-numbered/odd-numbered coins.
You might be wondering how you can always choose odd-numbered/even-numbered coins. Let me illustrate this using an example where you have 6 coins:
Example
18 20 15 30 10 14
Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64.
Since the sum of even coins is more, the first player decides to collect all even coins. He first picks 14, now the other player can only pick a coin (10 or 18). Whichever is picked the other player, the first player again gets an opportunity to pick an even coin and block all even coins.
