Best riddles


In classic mythology, there is the story of the Sphinx, a monster with the body of a lion and the upper part of a woman. The Sphinx lay crouched on the top of a rock along the highroad to the city of Thebes, and stopped all travellers passing by, proposing to them a riddle. Those who failed to answer the riddle correctly were killed. This is the riddle the Sphinx asked the travellers: "What animal walks on four legs in the morning, two legs during the day, and three legs in the evening?"
This is part of the story of Oedipus, who replied to the Sphinx, "Man, who in childhood creeps on hands and knees, in manhood walks erect, and in old age with the aid of a staff." Morning, day and night are representative of the stages of life. The Sphinx was so mortified at the solving of her riddle that she cast herself down from the rock and perished.
87.11 %
584 votes

Justin Case and Auntie Bellum are fellow con artists who deliver coded messages to each other to communicate. Recently Auntie Bellum was put in jail for stealing a rare and expensive diamond. Only a few days after this, Justin Case sent her a friendly letter asking her how she was. On the inside of the envelope of the letter, he hid a code. Yesterday, Auntie Bellum escaped and left the envelope and the letter inside the jail cell. The police did some research and found the code on the inside of the envelope, but they haven't been able to crack it. Could you help the police find out what the message is? This is the code: llwatchawtfeclocklnisksundialcirbetimersool
The message was "loose bricks in left wall." The message was put backward with words related to time in between. This is how the message looks when separated: ll watch awtfe clock Inisk sundial cirbe timer sool If you take out watch, clock, sundial, and timer, this is what is left: llawtfelniskcirbesool Look at this backwards and this is what you have: loose bricks in left wall Auntie Bellum took out the bricks and escaped in the night. Then, she put the bricks back where they were.
85.67 %
49 votes

There are 5 pirates in a ship. Pirates have hierarchy C1, C2, C3, C4 and C5. C1 designation is the highest and C5 is the lowest. These pirates have three characteristics: a. Every pirate is so greedy that he can even take lives to make more money. b. Every pirate desperately wants to stay alive. c. They are all very intelligent. There are total 100 gold coins on the ship. The person with the highest designation on the deck is expected to make the distribution. If the majority on the deck does not agree to the distribution proposed, the highest designation pirate will be thrown out of the ship (or simply killed). The first priority of the pirates is to stay alive and second to maximize the gold they get. Pirate 5 devises a plan which he knows will be accepted for sure and will maximize his gold. What is his plan?
To understand the answer,we need to reduce this problem to only 2 pirates. So what happens if there are only 2 pirates. Pirate 2 can easily propose that he gets all the 100 gold coins. Since he constitutes 50% of the pirates, the proposal has to be accepted leaving Pirate 1 with nothing. Now let's look at 3 pirates situation, Pirate 3 knows that if his proposal does not get accepted, then pirate 2 will get all the gold and pirate 1 will get nothing. So he decides to bribe pirate 1 with one gold coin. Pirate 1 knows that one gold coin is better than nothing so he has to back pirate 3. Pirate 3 proposes {pirate 1, pirate 2, pirate 3} {1, 0, 99}. Since pirate 1 and 3 will vote for it, it will be accepted. If there are 4 pirates, pirate 4 needs to get one more pirate to vote for his proposal. Pirate 4 realizes that if he dies, pirate 2 will get nothing (according to the proposal with 3 pirates) so he can easily bribe pirate 2 with one gold coin to get his vote. So the distribution will be {0, 1, 0, 99}. Smart right? Now can you figure out the distribution with 5 pirates? Let's see. Pirate 5 needs 2 votes and he knows that if he dies, pirate 1 and 3 will get nothing. He can easily bribe pirates 1 and 3 with one gold coin each to get their vote. In the end, he proposes {1, 0, 1, 0, 98}. This proposal will get accepted and provide the maximum amount of gold to pirate 5.
84.78 %
55 votes

A Panda Bear walked into a restaurant. He sat down at a table and ordered some food. When he was finished eating, he took out a gun and shot his waiter. He then left the restaurant.After the police caught up with him, they asked him why he had killed the waiter.He replied, "Look me up in the dictionary." What did the dictionary say?
When they looked up the word "Panda" in the dictionary, it stated, "Panda: Eats shoots and leaves."
84.75 %
229 votes

You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races. You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in. What is the least number of races you can conduct to figure out which 3 horses are fastest?
You need to conduct 7 races. First, separate the horses into 5 groups of 5 horses each, and race the horses in each of these groups. Let's call these groups A, B, C, D and E, and within each group let's label them in the order they finished. So for example, in group A, A1 finished 1st, A2 finished 2nd, A3 finished 3rd, and so on. We can rule out the bottom two finishers in each race (A4 and A5, B4 and B5, C4 and C5, D4 and D5, and E4 and E5), since we know of at least 3 horses that are faster than them (specifically, the horses that beat them in their respective races). This table shows our remaining horses: A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 B3 C3 D3 E3 For our 6th race, let's race the top finishers in each group: A1, B1, C1, D1 and E1. Let's assume that the order of finishers is: A1, B1, C1, D1, E1 (so A1 finished first, E1 finished last). We now know that horse D1 cannot be in the top 3, because it is slower than C1, B1 and A1 (it lost to them in the 6th race). Thus, D2 and D3 can also not be in the to 3 (since they are slower than D1). Similarly, E1, E2 and E3 cannot be in the top 3 because they are all slower than D1 (which we already know isn't in the top 3). Let's look at our updated table, having removed these horses that can't be in the top 3: A1 B1 C1 A2 B2 C2 A3 B3 C3 We can actually rule out a few more horses. C2 and C3 cannot be in the top 3 because they are both slower than C1 (and thus are also slower than B1 and A1). And B3 also can't be in the top 3 because it is slower than B2 and B1 (and thus is also slower than A1). So let's further update our table: A1 B1 C1 A2 B2 A3 We actually already know that A1 is our fastest horse (since it directly or indirectly beat all the remaining horses). So now we just need to find the other two fastest horses out of A2, A3, B1, B2 and C1. So for our 7th race, we simply race these 5 horses, and the top two finishers, plus A1, are our 3 fastest horses.
84.52 %
54 votes
Page 1 of 116.