A bad king has a cellar of 1000 bottles of delightful and very expensive wine. A neighboring queen plots to kill the bad king and sends a servant to poison the wine.
Fortunately (or say unfortunately) the bad king's guards catch the servant after he has only poisoned one bottle.
Alas, the guards don't know which bottle but know that the poison is so strong that even if diluted 100,000 times it would still kill the king. Furthermore, it takes one month to have an effect.
The bad king decides he will get some of the prisoners in his vast dungeons to drink the wine. Being a clever bad king he knows he needs to murder no more than 10 prisoners – believing he can fob off such a low death rate – and will still be able to drink the rest of the wine (999 bottles) at his anniversary party in 5 weeks time.
Explain what is in mind of the king, how will he be able to do so?
Think in terms of binary numbers. (now don’t read the solution, give a try).
Number the bottles 1 to 1000 and write the number in binary format.
bottle 1 = 0000000001 (10 digit binary)
bottle 2 = 0000000010
bottle 500 = 0111110100
bottle 1000 = 1111101000
Now take 10 prisoners and number them 1 to 10, now let prisoner 1 take a sip from every bottle that has a 1 in its least significant bit. Let prisoner 10 take a sip from every bottle with a 1 in its most significant bit. etc.
prisoner = 10 9 8 7 6 5 4 3 2 1
bottle 924 = 1 1 1 0 0 1 1 1 0 0
For instance, bottle no. 924 would be sipped by 10,9,8,5,4 and 3. That way if bottle no. 924 was the poisoned one, only those prisoners would die.
After four weeks, line the prisoners up in their bit order and read each living prisoner as a 0 bit and each dead prisoner as a 1 bit. The number that you get is the bottle of wine that was poisoned.
1000 is less than 1024 (2^10). If there were 1024 or more bottles of wine it would take more than 10 prisoners.
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 standing in a pitch-dark room. A friend walks up and hands you a normal deck of 52 cards. He tells you that 13 of the 52 cards are face-up, the rest are face-down. These face-up cards are distributed randomly throughout the deck.
Your task is to split up the deck into two piles, using all the cards, such that each pile has the same number of face-up cards. The room is pitch-dark, so you can't see the deck as you do this.
How can you accomplish this seemingly impossible task?
Take the first 13 cards off the top of the deck and flip them over. This is the first pile. The second pile is just the remaining 39 cards as they started.
This works because if there are N face-up cards in within the first 13 cards, then there will be (13 - N) face up cards in the remaining 39 cards. When you flip those first 13 cards, N of which are face-up, there will now be N cards face-down, and therefore (13 - N) cards face-up, which, as stated, is the same number of face-up cards in the second pile.
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.
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.
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 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!
A grandfather's clock chimes the appropriate number of times to indicate the hour, as well as chiming once at each quarter hour. If you were in another room and hear the clock chime just once, what would be the longest period of time you would have to wait in order to be certain of the correct time?
You would have to wait 90 minutes between 12:15 and 1:45. Once you had heard seven single chimes, you would know that the next chime would be two chimes for 2 o'clock.