Riddle #751

logic

Fancy restaurant

A fancy restaurant in New York was offering a promotional deal. A married couple could eat at the restaurant for half-price on their anniversary. To prevent scams, the couple would need proof of their wedding date. One Thursday evening, a couple claimed it was their anniversary, but didn't bring any proof. The restaurant manager was called to speak with the couple. When the manager asked to hear about the wedding day, the wife replied with the following: "Oh, it was a wonderful Sunday afternoon, birds were chirping, and flowers were in full bloom." After nearly 10 minutes of ranting, she comes to tell him that today was their 28th wedding anniversary. "How lovely", the manager said, "However, you do not qualify for the discount. Today is not your anniversary, you are a liar". How did the manager know that it wasn't their anniversary?
The calendar repeats itself every 28 years. So, if they were married on a Sunday 28 years ago, the day they were at the restaurant would also have to be a Sunday. Since it was a Thursday, the manager knew they were lying, and abruptly kicked them out of his restaurant.
93.70 %
40 votes

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logicmath

Flipping Lockers

There are 1 million closed school lockers in a row, labeled 1 through 1,000,000. You first go through and flip every locker open. Then you go through and flip every other locker (locker 2, 4, 6, etc...). When you're done, all the even-numbered lockers are closed. You then go through and flip every third locker (3, 6, 9, etc...). "Flipping" mean you open it if it's closed, and close it if it's open. For example, as you go through this time, you close locker 3 (because it was still open after the previous run through), but you open locker 6, since you had closed it in the previous run through. Then you go through and flip every fourth locker (4, 8, 12, etc...), then every fifth locker (5, 10, 15, etc...), then every sixth locker (6, 12, 18, etc...) and so on. At the end, you're going through and flipping every 999,998th locker (which is just locker 999,998), then every 999,999th locker (which is just locker 999,999), and finally, every 1,000,000th locker (which is just locker 1,000,000). At the end of this, is locker 1,000,000 open or closed?
Locker 1,000,000 will be open. If you think about it, the number of times that each locker is flipped is equal to the number of factors it has. For example, locker 12 has factors 1, 2, 3, 4, 6, and 12, and will thus be flipped 6 times (it will end be flipped when you flip every one, every 2nd, every 3rd, every 4th, every 6th, and every 12th locker). It will end up closed, since flipping an even number of times will return it to its starting position. You can see that if a locker number has an even number of factors, it will end up closed. If it has an odd number of factors, it will end up open. As it turns out, the only types of numbers that have an odd number of factors are squares. This is because factors come in pairs, and for squares, one of those pairs is the square root, which is duplicated and thus doesn't count twice as a factor. For example, 12's factors are 1 x 12, 2 x 6, and 3 x 4 (6 total factors). On the other hand, 16's factors are 1 x 16, 2 x 8, and 4 x 4 (5 total factors). So lockers 1, 4, 9, 16, 25, etc... will all be open. Since 1,000,000 is a square number (1000 x 1000), it will be open as well.
93.70 %
40 votes

logicmathprobability

The same birthday

What is the least number of people that need to be in a room such that there is greater than a 50% chance that at least two of the people have the same birthday?
Only 23 people need to be in the room. Our first observation in solving this problem is the following: (the probability that at least 2 people have the same birthday + the probability that nobody has the same birthday) = 1.0 What this means is that there is a 100% chance that EITHER everybody in the room has a different birthday, OR at least two people in the room have the same birthday (and these probabilities don't add up to more than 1.0 because they cover mutually exclusive situations). With some simple re-arranging of the formula, we get: the probability that at least 2 people have the same birthday = (1.0 - the probability that nobody has the same birthday) So now if we can find the probability that nobody in the room has the same birthday, we just subtract this value from 1.0 and we'll have our answer. The probability that nobody in the room has the same birthday is fairly straightforward to calculate. We can think of this as a "selection without replacement" problem, where each person "selects" a birthday at random, and we then have to figure out the probability that no two people select the same birthday. The first selection has a 365/365 chance of being different than the other birthdays (since none have been selected yet). The next selection has a 364/365 chance of being different than the 1 birthday that has been selected so far. The next selection has a 363/365 chance of being different than the 2 birthdays that have been selected so far. These probabilities are multiplied together since each is conditional on the previous. So for example, the probability that nobody in a room of 3 people have the same birthday is (365/365 * 364/365 * 363/365) =~ 0.9918 More generally, if there are n people in a room, then the probability that nobody has the same birthday is (365/365 * 364/365 * ... * (365-n+2)/365 * (365-n+1)/365) We can plug in values for n. For n=22, we get that the probability that nobody has the same birthday is 0.524, and thus the probabilty that at least two people have the same birthday is (1.0 - 0.524) = 0.476 = 47.6%. Then for n=23, we get that the probability that nobody has the same birthday is 0.493, and thus the probabilty that at least two people have the same birthday is 1.0 - 0.493) = 0.507 = 50.7%. Thus, once we get to 23 people we have reached the 50% threshold.
91.22 %
48 votes

logic

Grandma and Cake

You are on your way to visit your Grandma, who lives at the end of the valley. It's her anniversary, and you want to give her the cakes you've made. Between your house and her house, you have to cross 5 bridges, and as it goes in the land of make believe, there is a troll under every bridge! Each troll, quite rightly, insists that you pay a troll toll. Before you can cross their bridge, you have to give them half of the cakes you are carrying, but as they are kind trolls, they each give you back a single cake. How many cakes do you have to leave home with to make sure that you arrive at Grandma's with exactly 2 cakes?
2 Cakes How? At each bridge you are required to give half of your cakes, and you receive one back. Which leaves you with 2 cakes after every bridge.
93.70 %
40 votes

logicshort

Egyptian coin

An archeologist claims he found a Roman coin dated 46 B.C. in Egypt. How much should Louvre Museum pay for the coin? Note: Roman coins can really be found in Egypt
Nothing. That coin is as phony as a three dollar bill. In 46 B.C., they wouldn't have known how many years before Christ it was.
93.39 %
38 votes

logicmathprobability

An infinite basket

You have a basket of infinite size (meaning it can hold an infinite number of objects). You also have an infinite number of balls, each with a different number on it, starting at 1 and going up (1, 2, 3, etc...). A genie suddenly appears and proposes a game that will take exactly one minute. The game is as follows: The genie will start timing 1 minute on his stopwatch. Where there is 1/2 a minute remaining in the game, he'll put balls 1, 2, and 3 into the basket. At the exact same moment, you will grab a ball out of the basket (which could be one of the balls he just put in, or any ball that is already in the basket) and throw it away. Then when 3/4 of the minute has passed, he'll put in balls 4, 5, and 6, and again, you'll take a ball out and throw it away. Similarly, at 7/8 of a minute, he'll put in balls 7, 8, and 9, and you'll take out and throw away one ball. Similarly, at 15/16 of a minute, he'll put in balls 10, 11, and 12, and you'll take out and throw away one ball. And so on....After the minute is up, the genie will have put in an infinite number of balls, and you'll have thrown away an infinite number of balls. Assume that you pull out a ball at the exact same time the genie puts in 3 balls, and that the amount of time this takes is infinitesimally small. You are allowed to choose each ball that you pull out as the game progresses (for example, you could choose to always pull out the ball that is divisible by 3, which would be 3, then 6, then 9, and so on...). You play the game, and after the minute is up, you note that there are an infinite number of balls in the basket. The next day you tell your friend about the game you played with the genie. "That's weird," your friend says. "I played the exact same game with the genie yesterday, except that at the end of my game there were 0 balls left in the basket." How is it possible that you could end up with these two different results?
Your strategy for choosing which ball to throw away could have been one of many. One such strategy that would leave an infinite number of balls in the basket at the end of the game is to always choose the ball that is divisible by 3 (so 3, then 6, then 9, and so on...). Thus, at the end of the game, any ball of the format 3n+1 (i.e. 1, 4, 7, etc...), or of the format 3n+2 (i.e. 2, 5, 8, etc...) would still be in the basket. Since there will be an infinite number of such balls that the genie has put in, there will be an infinite number of balls in the basket. Your friend could have had a number of strategies for leaving 0 balls in the basket. Any strategy that guarantees that every ball n will be removed after an infinite number of removals will result in 0 balls in the basket. One such strategy is to always choose the lowest-numbered ball in the basket. So first 1, then 2, then 3, and so on. This will result in an empty basket at the game's end. To see this, assume that there is some ball in the basket at the end of the game. This ball must have some number n. But we know this ball was thrown out after the n-th round of throwing balls away, so it couldn't be in there. This contradiction shows that there couldn't be any balls left in the basket at the end of the game. An interesting aside is that your friend could have also used the strategy of choosing a ball at random to throw away, and this would have resulted in an empty basket at the end of the game. This is because after an infinite number of balls being thrown away, the probability of any given ball being thrown away reaches 100% when they are chosen at random.
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Cowboy

How could the cowboy travel on friday, then sleep two days and then travel back home on friday.
If the horse was named Friday.
94.48 %
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funnylogic

24 foot chain

A horse is on a 24 foot chain and wants an apple that is 26 feet away. How can the horse get to the apple?
The chain is not attached to anything.
93.39 %
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funnylogic

Hippopotamus, rhino and elephant

Presume that you do not know what a rhino looks like. Now the question goes like this: If one day while walking in a forest with two of your close friends, one friend shows you an elephant and tells that this is a rhino, and another friend shows you a hippopotamus and tells you that this is rhino, who would you believe and why?
I told you that you do not know what a rhino looks like, not that you are unaware of what a hippo and elephant look like. So you shouldn't believe either of them.
93.98 %
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cleanlogicmathshort

Hards maths number

Create a number using only the digits 4,4,3,3,2,2,1 and 1. So I can only be eight digits. You have to make sure the ones are separated by one digit, the twos are separated by two digits the threes are separated with three digits and the fours are separated by four digits.
41312432.
93.22 %
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