Math riddles

logicmathtrickystoryclever

Three people check into a hotel room. The bill is $30 so they each pay $10. After they go to the room, the hotel's cashier realizes that the bill should have only been $25. So he gives $5 to the bellhop and tells him to return the money to the guests. The bellhop notices that $5 can't be split evenly between the three guests, so he keeps $2 for himself and then gives the other $3 to the guests. Now the guests, with their dollars back, have each paid $9 for a total of $27. And the bellhop has pocketed $2. So there is $27 + $2 = $29 accounted for. But the guests originally paid $30. What happened to the other dollar?
This riddle is just an example of misdirection. It is actually nonsensical to add $27 + $2, because the $27 that has been paid includes the $2 the bellhop made. The correct math is to say that the guests paid $27, and the bellhop took $2, which, if given back to the guests, would bring them to their correct payment of $27 - $2 = $25.
70.72 %
73 votes
logicmathsimpleclean

There are several chickens and rabbits in a cage (with no other types of animals). There are 72 heads and 200 feet inside the cage. How many chickens are there, and how many rabbits?
There are 44 chickens and 28 rabbits in the cage. Let c be the number of chickens, and r be the number of rabbits. r + c = 72 4r + 2c = 200 To solve the equations, we multiply the first by two, then subtract the second. 2r + 2c = 144 2r = 56 r = 28 c = 44
70.43 %
88 votes
logicmathstorycleaninterview

You are somewhere on Earth. You walk due south 1 mile, then due east 1 mile, then due north 1 mile. When you finish this 3-mile walk, you are back exactly where you started. It turns out there are an infinite number of different points on earth where you might be. Can you describe them all? It's important to note that this set of points should contain both an infinite number of different latitudes, and an infinite number of different longitudes (though the same latitudes and longitudes can be repeated multiple times); if it doesn't, you haven't thought of all the points.
One of the points is the North Pole. If you go south one mile, and then east one mile, you're still exactly one mile south of the North Pole, so you'll be back where you started when you go north one mile. To think of the next set of points, imagine the latitude slighty north of the South Pole, where the length of the longitudinal line around the Earth is exactly one mile (put another way, imagine the latitude slightly north of the South Pole where if you were to walk due east one mile, you would end up exactly where you started). Any point exactly one mile north of this latitude is another one of the points you could be at, because you would walk south one mile, then walk east a mile around and end up where you started the eastward walk, and then walk back north one mile to your starting point. So this adds an infinite number of other points we could be at. However, we have not yet met the requirement that our set of points has an infinite number of different latitudes. To meet this requirement and see the rest of the points you might be at, we just generalize the previous set of points. Imagine the latitude slightly north of the South Pole that is 1/2 mile in distance. Also imagine the latitudes in this area that are 1/3 miles in distance, 1/4 miles in distance, 1/5 miles, 1/6 miles, and so on. If you are at any of these latitudes and you walk exactly one mile east, you will end up exactly where you started. Thus, any point that is one mile north of ANY of these latitudes is another one of the points you might have started at, since you'll walk one mile south, then one mile east and end up where you started your eastward walk, and finally, one mile north back to where you started.
70.38 %
80 votes
logicmathstory

A swan sits at the center of a perfectly circular lake. At an edge of the lake stands a ravenous monster waiting to devour the swan. The monster can not enter the water, but it will run around the circumference of the lake to try to catch the swan as soon as it reaches the shore. The monster moves at 4 times the speed of the swan, and it will always move in the direction along the shore that brings it closer to the swan the quickest. Both the swan and the the monster can change directions in an instant. The swan knows that if it can reach the lake's shore without the monster right on top of it, it can instantly escape into the surrounding forest. How can the swan succesfully escape?
Assume the radius of the lake is R feet. So the circumference of the lake is (2*pi*R). If the swan swims R/4 feet, (or, put another way, 0.25R feet) straight away from the center of the lake, and then begins swimming in a circle around the center, then it will be able to swim around this circle in the exact same amount of time as the monster will be able to run around the lake's shore (since this inner circle's circumference is 2*pi*(R/4), which is exactly 4 times shorter than the shore's circumference). From this point, the swan can move a millimeter inward toward the lake's center, and begin swimming around the center in a circle from this distance. It is now going around a very slightly smaller circle than it was a moment ago, and thus will be able to swim around this circle FASTER than the monster can run around the shore. The swan can keep swimming around this way, pulling further away each second, until finally it is on the opposite side of its inner circle from where the monster is on the shore. At this point, the swan aims directly toward the closest shore and begins swimming that way. At this point, the swan has to swim [0.75R feet + 1 millimeter] to get to shore. Meanwhile, the monster will have to run R*pi feet (half the circumference of the lake) to get to where the swan is headed. The monster runs four times as fast as the swan, but you can see that it has more than four times as far to run: [0.75R feet + 1 millimeter] * 4 < R*pi [This math could actually be incorrect if R were very very small, but in that case we could just say the swan swam inward even less than a millimeter, and make the math work out correctly.] Because the swan has less than a fourth of the distance to travel as the monster, it will reach the shore before the monster reaches where it is and successfully escape.
70.36 %
76 votes
logicmathclean

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.
69.25 %
190 votes
logicmathclever

Dean Sam and Castiel are three brothers. Interestingly their current age is prime. What's more interesting that difference between their ages is also prime. How old are they?
Sam : 2 Dean : 5 Castiel : 7 Age diff 7 - 2 = '5' is prime 7 - 5 = '2' is prime 5 - 2 = '3' is prime
68.87 %
109 votes
logiccleanclevermath

Your friend shows you two jars, one with 100 red marbles in it, the other with 100 blue marbles in it. He proposes a game. He'll put the two jars behind his back and tell you to pick one of them at random. You'll then close your eyes, he'll hand you the jar you picked, and you'll pick a random marble from that jar. You win if the marble you pick is blue, and you lose otherwise. To give you the best shot at winning, your friend gives you the two jars before the game starts and says you can move the marbles around however you'd like, as long as all 200 marbles are in the 2 jars (that is, you can't throw any marbles away). How should you move the marbles around to give yourself the best chance of picking a blue marble?
Put one blue marble in one jar, and put the rest of the marbles in the other jar. This will give you just about a 75% chance of picking a blue marble.
68.67 %
101 votes
logicstorymath

You are standing in a house in the middle of the countryside. There is a small hole in one of the interior walls of the house, through which 100 identical wires are protruding. From this hole, the wires run underground all the way to a small shed exactly 1 mile away from the house, and are protruding from one of the shed's walls so that they are accessible from inside the shed. The ends of the wires coming out of the house wall each have a small tag on them, labeled with each number from 1 to 100 (so one of the wires is labeled "1", one is labeled "2", and so on, all the way through "100"). Your task is to label the ends of the wires protruding from the shed wall with the same number as the other end of the wire from the house (so, for example, the wire with its end labeled "47" in the house should have its other end in the shed labeled "47" as well). To help you label the ends of the wires in the shed, there are an unlimited supply of batteries in the house, and a single lightbulb in the shed. The way it works is that in the house, you can take any two wires and attach them to a single battery. If you then go to the shed and touch those two wires to the lightbulb, it will light up. The lightbulb will only light up if you touch it to two wires that are attached to the same battery. You can use as many of the batteries as you want, but you cannot attach any given wire to more than one battery at a time. Also, you cannot attach more than two wires to a given battery at one time. (Basically, each battery you use will have exactly two wires attached to it). Note that you don't have to attach all of the wires to batteries if you don't want to. Your goal, starting in the house, is to travel as little distance as possible in order to label all of the wires in the shed. You tell a few friends about the task at hand. "That will require you to travel 15 miles!" of of them exclaims. "Pish posh," yells another. "You'll only have to travel 5 miles!" "That's nonsense," a third replies. "You can do it in 3 miles!" Which of your friends is correct? And what strategy would you use to travel that number of miles to label all of the wires in the shed?
Believe it or not, you can do it travelling only 3 miles! The answer is rather elegant. Starting from the house, don't attach wires 1 and 2 to any batteries, but for the remaining wires, attach them in consecutive pairs to batteries (so attach wires 3 and 4 to the same battery, attach wires 5 and 6 to the same battery, and so on all the way through wires 99 and 100). Now travel 1 mile to the shed, and using the lightbulb, find all pairs of wires that light it up. Put a rubberband around each pair or wires that light up the lightbulb. The two wires that don't light up any lightbulbs are wires 1 and 2 (though you don't know yet which one of them is wire 1 and which is wire 2). Put a rubberband around this pair of wires as well, but mark it so you remember that they are wires 1 and 2. Now go 1 mile back to the house, and attach odd-numbered wires to batteries in the following pairs: (1 and 3), (5 and 7), (9 and 11), and so on, all the way through (97 and 99). Similarly, attach even-numbered wires to batteries in the following pairs: (4 and 6), (8 and 10), (12 and 14), and so on, all the way through (96 and 98). Note that in this round, we didn't attach wire 2 or wire 100 to any batteries. Finally, travel 1 mile back to the shed. You're now in a position to label all of the wires here. First, remember we know the pair of wires that are, collectively, wires 1 and 2. So test wires 1 and 2 with all the other wires to see what pair lights up the lightbulb. The wire from wires 1 and 2 that doesn't light up the bulb is wire 2 (which, remember, we didn't connect to a battery), and the other is wire 1, so we can label these as such. Furthermore, the wire that, with wire 1, lights up a lightbulb, is wire 3 (remember how we connected the wires this round). Now, the other wire in the rubber band with wire 3 is wire 4 (we know this from the first round), and the wire that, with wire 4, lights up the lightbulb, is wire 6 (again, because of how we connected the wires to batteries this round). We can continue labeling batteries this way (next we'll label wire 7, which is rubber-banded to wire 6, and then we'll label wire 9, which lights up the lightbulb with wire 7, and so on). At the end, we'll label wire 97, and then wire 99 (which lights up the lightbulb with wire 97), and finally wire 100 (which isn't connected to a battery this round, but is rubber-banded to wire 99). And we're done, having travelled only 3 miles!
68.54 %
75 votes
mathcleansimplelogicstory

Farmer Brown came to town with some watermelons. He sold half of them plus half a melon and found that he had one whole melon left. How many melons did he take to town?
Easy, three melons.
68.42 %
71 votes
logicmathclever

Sam has got three daughters. The eldest daughter is the most honest girl in the universe and she always speaks truth. The middle daughter is a modest woman. She speaks truth and lies according to the situations. The youngest one never speaks truth. Not a single word she spoke was true and would never be true. Sam brought a marriage proposal for one of his girls. It was John. John wanted to marry either the eldest or the youngest daughter of Sam as he can easily identify whether the girl speaks truth or lie! John told his desire to Sam. However, Sam laid a condition. He told John that he will not say who the eldest, middle or youngest one is. Also, he allowed John to ask only one question to identify the eldest or youngest so he can marry one. John asked one question and found the right girl. What was the question and whom should he pick?
The question he asked is, 'Is she older than her?' He asks this question to one of the daughters. If he asked this question to older daughter pointing at other two, he probably would know the youngest one! NO matter, she always speaks truth. If he asked the question to middle one, probably he can choose either. If he asked the youngest one, she always lies and he can find eldest one. No matter, he has to choose the youngest one based on the answer.
68.38 %
56 votes