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.
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.
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.
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
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.
Fernando + Alonso + McLaren = 6
Fernando x Alonso = 2
Alonso x McLaren = 6
McLaren x Fernando = ?
3 or 0.75
Rewriting the last 2 equations in terms of Alonso,
Fernando = 2/Alonso
McLaren = 6/Alonso
Replacing above values in equation "Fernando + Alonso + McLaren = 6"
2/Alonso + Alonso + 6/Alonso =6
(2 + Alonso^2 + 6)/Alonso = 6
8 + Alonso^2 = 6Alonso
Alonso^2 - 6Alonso + 8 = 0
(Alonso - 4) (Alonso - 2) = 0
Alonso = 4 or 2
Let's take value of Alonso as 2
Fernando = 2/2 = 1
McLaren = 6/2 = 3
McLaren x Fernando = 3 x 1 = 3
Let's take value of Alonso as 4
Fernando = 2/4 = 0.5
McLaren = 6/4 = 1.5
McLaren x Fernando = 1.5 x 0.5 = 0.75
Two trains are traveling toward each other on the same track, each at 60 miles per hour. When they are exactly 120 miles apart, a fly takes off from the front of one of the trains, flying toward the other train at a constant rate of 100 miles per hour. When the fly reaches the other train, it instantly changes directions and starts flying toward the other train, still at 100 miles per hour. It keeps doing this back and forth until the trains finally collide.
If you add up all the distances back and forth that the fly has travelled, how much total distance has the fly travelled when the trains finally collide?
The fly has travelled exactly 100 miles. We can figure this out using some simple math. Becuase the trains are 120 miles apart when the fly takes off, and are travelling at 60 mph each, they will collide in exactly 1 hour. This gives the fly exactly 1 hour of flying time, going at a speed of 100 miles per hour. Thus, the fly will travel 100 miles in this hour.
Mick and John were in a 100 meter race. When Mick crossed the finish line, John was only at the 90 meter mark. Mick suggested they run another race. This time, Mick would start ten meters behind the starting line. All other things being equal, will John win, lose, or will it be a tie in the second race?
John will lose again. In the second race, Mick started ten meters back. By the time John reaches the 90 meter mark, Mick will have caught up him. Therefore, the final ten meters will belong to the faster of the two. Since Mick is faster than John, he will win the final 10 meters and of course the race.