Best riddles

logicshort

Akbar and Birbal

One day, Emperor Akbar posed a question to Birbal. He asked him what Birbal would choose if he offered either justice or a gold coin. "The gold coin," said Birbal without hesitation. On hearing this, Akbar was taken aback. "You would prefer a gold coin to justice?" he asked, not believing his own ears. "Yes," said Birbal. The other courtiers were amazed by Birbal's display of idiocy. They were full of glee that Birbal had finally managed himself to do what these courtiers had not been able to do for a long time - discredit Birbal in the emperor's eyes! "I would have been disappointed if this was the choice made even by my lowliest of servants," continued the emperor. "But coming from you it's not only disappointing, but shocking and sad. I did not know you were so debased!" How did Birbal justify his answer to the enraged and hurt Emperor?
"One asks for what one does not have, Your Majesty." said Birbal, smiling gently and in quiet tones. "Under Your Majesty´s rule, justice is available to everybody. But I am a spendthrift and always short of money and therefore I said I would choose the gold coin." The answer immensely pleased the emperor and respect for Birbal was once again restored in the emperor's eyes.
90.67 %
45 votes

logic

The best archer ever

A duke was hunting in the forest with his men-at-arms and servants when he came across a tree. Upon it, archery targets were painted and smack in the middle of each was an arrow. "Who is this incredibly fine archer?" cried the duke. "I must find him!" After continuing through the forest for a few miles he came across a small boy carrying a bow and arrow. Eventually the boy admitted that it was he who shot the arrows plumb in the center of all the targets. "You didn't just walk up to the targets and hammer the arrows into the middle, did you?" asked the duke worriedly. "No my lord. I shot them from a hundred paces. I swear it by all that I hold holy." "That is truly astonishing," said the duke. "I hereby admit you into my service." The boy thanked him profusely. "But I must ask one favor in return," the duke continued. "You must tell me how you came to be such an outstanding shot." How'd he get to be such a good shot?
The boy shot the arrow, then painted the circle around it.
90.47 %
44 votes

short

Out for a walk

Samuel was out for a walk when it started to rain. He did not have an umbrella and he wasn't wearing a hat. His clothes were soaked, yet not a single hair on his head got wet. How could this happen?
He is bald.
90.47 %
44 votes

logic

Engineers and Managers

You have just purchased a small company called Company X. Company X has N employees, and everyone is either an engineer or a manager. You know for sure that there are more engineers than managers at the company. Everyone at Company X knows everyone else's position, and you are able to ask any employee about the position of any other employee. For example, you could approach employee A and ask "Is employee B an engineer or a manager?" You can only direct your question to one employee at a time, and can only ask about one other employee at a time. You're allowed to ask the same employee multiple questions if you want. Your goal is to find at least one engineer to solve a huge problem that has just hit the company's factory. The problem is so urgent that you only have time to ask N-1 total questions. The major problem with questioning the employees, however, is that while the engineers will always tell you the truth about other employees' roles, the managers may lie to you if they like. You can assume that the managers will do their best to confuse you. How can you find at least one engineer by asking at most N-1 questions?
You can find at least one engineer using the following process: Put all of the employees in a conference room. If there happen to be an even number of employees, pick one at random and send him home for the day so that we start with an odd number of employees. Note that there will still be more engineers than managers after we send this employee home. Then call them out one at a time in any order. You will be forming them into a line as follows: If there is nobody currently in the line, put the employee you just called out in the line. Otherwise, if there is anybody in the line, then we do the following. Let's call the employee currently at the front of the line Employee_Front, and call the employee who we just called out of the conference room Employee_Next. So ask Employee_Front if Employee_Next is a manager or an engineer. If Employee_Front says "manager", then send both Employee_Front and Employee_Next home for the day. However, if Employee_Front says "engineer", then put Employee_Next at the front of the line. Keep doing this until you've called everyone out of the conference room. Notice that at this point, you'll have asked N-1 or less questions (you asked at most one question each time you called an employee out except for the first employee, when you didn't ask a question, so that's at most N-1 questions). When you're done calling everyone out of the conference room, the person at the front of the line is an engineer. So you've found your engineer! But the real question: how does this work? We can prove this works by showing a few things. First, let's show that if there are any engineers in the line, then they must be in front of any managers. We'll show this with a proof by contradiction. Assume that there is a manager in front of an engineer somewhere in the line. Then it must have been the case that at some point, that engineer was Employee_Front and that manager was Employee_Next. But then Employee_Front would have said "manager" (since he is an engineer and always tells the truth), and we would have sent them both home. This contradicts their being in the line at all, and thus we know that there can never be a manager in front of an engineer in the line. So now we know that after the process is done, if there are any engineers in the line, then they will be at the front of the line. That means that all we have to prove now is that there will be at least one engineer in the line at the end of the process, and we'll know that there will be an engineer at the front. So let's show that there will be at least one engineer in the line. To see why, consider what happens when we ask Employee_Front about Employee_Next, and Employee_Front says "manager". We know for sure that in this case, Employee_Front and Employee_Next are not both engineers, because if this were the case, then Employee_Front would have definitely says "engineer". Put another way, at least one of Employee_Front and Employee_Next is a manager. So by sending them both home, we know we are sending home at least one manager, and thus, we are keeping the balance in the remaining employees that there are more engineers than managers. Thus, once the process is over, there will be more engineers than managers in the line (this is also sufficient to show that there will be at least one person in the line once the process is over). And so, there must be at least one engineer in the line. Put altogether, we proved that at the end of the process, there will be at least one engineer in the line and that any engineers in the line must be in front of any managers, and so we know that the person at the front of the line will be an engineer.
90.47 %
44 votes

logic

Paying With Rings

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
90.47 %
44 votes

logicmath

In a bank

A women walks into a bank to cash out her check. By mistake the bank teller gives her rupee amount in change, and her paise amount in rupees. On the way home she spends 5 paise, and then suddenly she notices that she has twice the amount of her check. How much was her check amount ?
The check was for Rupees 31.63. The bank teller gave her Rupees 63.31 She spent .05, and then she had Rupees 63.26, which is twice the check. Let x be the rupees of the check, and y be the paise. The check was for 100x + y paise He was given 100y + x paise Also 100y + x - 5 = 2(100x + y) Expanding this out and rearranging, we find: 98y = 199x + 5 or 199x ≡ -5 (mod 98) or 98*2*x + 3x ≡ -5 (mod 98) 3x ≡ -5 ≡ 93 (mod 98) this quickly leads to x = 31
90.47 %
44 votes

logic

25 Horses

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.
90.47 %
44 votes