The universe is huge and complex, but it's also amazingly beautiful. Learning about the universe is easy if we break it down into small doses - as easy as drinking a cup of coffee!
Brain Teaser Wednesday #btw
The human brain is an amazing thing. It is capable of coming up with wonderful ideas and solving incredibly difficult problems. But, like everything, the brain needs maintenance. I think one of the best ways to keep your mind healthy is to exercise it regularly, so some months ago I set up Brain Teaser Wednesday (#btw). A new puzzle appeared weekly from April to December 2016, all the puzzles from the competition are shown here with their solution, and occasional hints. Each puzzle has a difficulty level, marked by the number of stars: easy*, medium**, hard***, and challenging****.
**Christmas Special**
To end #btw there was a 'Christmas Special Challenge', with the winner receiving a custom-made mug. The Challenge had several different parts, posted every few days over the holiday period. It ran from Monday 19th to Saturday 31st, and only one person managed to complete the whole challenge. You can read all the parts of the puzzle and the final solution here.
Christmas Special Part 1:
First think of the person who lives in disguise,
Who deals in secrets and tells naught but lies.
Next, tell me what's always the last thing to mend,
The middle of middle and end of the end?
And finally give me the sound often heard,
During the search for a hard-to-find word.
Now string them together and answer me this,
Which creature would you be unwilling to kiss?
Christmas Special Part 3:
49 441 729 529 121 361 16 1089 225 576 4 961 25 1225 676 64 36 484 9 81 289 1156 1 625 841 100 400 256 324 144 1024 169 784 196 900
Part two tells you 'which', part three tells you 'where'. If you seek guidance, find what parts two and three have in common. If you are lost, remember how everything started and how far we've come.
Christmas Special Part 5:
If you correctly combined parts two and three, you should have already found (and solved) part four. For part five, you need to find the next number in each sequence:
1, 1, 2, 3, 5, 8, 13, ??
1, 7, 10, 13, 19, 23, 28, ??
1, 3, 6, 10, 15, 21, 28, 36, ??
Christmas Special Part 6:
You should now have a name (part four) and six digits (part five). Together, they give you access to a hidden place. This is the place you should go, that is where everything ends. If you are lost, remember where everything else is and go home.
Solution to the Christmas Challenge:
The answer to part one was 'spider' (spy-d-er).
Part two is a cipher with codeword 'spider', so you can do a simple letter substitution with:
ABCDEFGHIJKLMNOPQRSTUVWXYZ
SPIDERABCFGHJKLMNOQTUVWXYZ
Once you've used the codeword, you'll see that part two is a list of thirty-five numbers: fiftyseven, eighteen, seventeen, sixteen, two, ...
Part three was a list of numbers, all of them perfect squares. If you take the square root of all the numbers, you'll find the numbers 1-35 in a different order: 7 21 27 23 11 ...
The next clue was to 'find what parts two and three have in common'. They both have 35 numbers. This was followed by 'remember how everything started and how far we've come'. It all started with a Weekly BrainTeaser Challenge (btw), and coincidentally there were exactly 35 brain teasers before the Christmas Special (with Puzzle 35 directly beneath part three on the webpage at the time it was posted).
So you had to combine parts two and three with the previous puzzles, knowing that part two tells you 'which' and part three tells you 'where': part two tells you which word you take from each puzzle, and part three tells you where to put it, so from puzzle one you take word 57 and put it in position 7, from puzzle two you take word 18 and put it in position 21, from puzzle three you take word 17 and put it in position 27, and so on.
By combining one word from each puzzle using the rules given in parts two and three, you find the following riddle (part four): On the move with captured friends, always outnumbered,
They may have life, but remains to be discovered.
The sixth of eight sailors, only one is bigger.
I travel with rings, more so than any other.
The answer to this riddle is Saturn.
Part five contained three well-known sequences of numbers. The first is the Fibonacci sequence: each number is the sum of the previous two, so the sequence reads 1, 1, 2, 3, 5, 8, 13, 21 ... The second sequence is a list of happy numbers. If you take any number and replace it with the sum of the squares of its digits, you can repeat the process until you either reach 1 (happy number) or enter a loop that does not contain one (sad number). So the sequence reads 1, 7, 10, 13, 19, 23, 28, 31 ... The final sequence is a list of triangular numbers: the objects needed to form equilateral triangles of increasing size, giving 1, 3, 6, 10, 15, 21, 28, 36, 45 ...
Finally, in part six you are told to 'find a hidden place' using the solutions to parts four and five, and to 'remember where everything else is'. All other parts of the puzzle, previous puzzles, rules, etc., are here, on the blog. So the best thing to try is the url cupofcosmology.blogspot.com/saturn213145, which takes you here.
On that page you are told to find a four-digit code, marking the end of the puzzle challenge. The code is hidden in the photo of Saturn (look at the 'moons' on the bottom left). And you're done!
Puzzle 35** (14/12/16): After a big family dinner you have 25 empty bottles, so you all decide to play a game to choose who will be crowned 'Puzzle Master'. The 25 bottles are placed in a square at regular distances, forming a 5x5 grid. You also have five circular plastic rings, which you can set to any size you wish. The winner of the game will be the first person who manages to place all five rings on top of the bottles in such a way that every bottle has at least one ring on top of it. Where should you place the rings to receive the title of 'Puzzle Master'?
Solution 35: There are several different solutions to this puzzle. One good way to find the solution is to notice the maximum amount of bottles you can cover with one ring (eight). Start with a ring covering eight bottles, then shift the next ring by the distance between two bottles. Now you will have 14 bottles covered, and you can find several ways to cover the remaining 11 bottles with the three rings you have left. Below are two sample solutions, sent by Patrick S. and Alessandro C.
Puzzle 34** (07/12/16): Anna, Brian, and Candela reach the final of a paint-ball tournament. To decide the winner a Western-style three-way duel is proposed: they will take turns shooting, and when a person is hit they are eliminated and can't shoot any more. Anna will shoot first, followed by Brian, and then Candela, repeating until only one person remains. Candela is a good shot and hits her target every time. Brian isn't so good, and only hits his target two out of three times. Anna is a very bad shot, only hitting her target one third of the time. What strategy should Anna adopt to maximise her chances of winning the tournament?
Solution 34: To maximise her chances of survival Anna should deliberately miss her first shot. Brian and Candela will each shoot at the person who poses the biggest threat, so in the first round Anna deliberately misses, Brian shoots Candela, and if Candela isn't eliminated, she'll shoot at Brian (and hit). The reason Anna should deliberately miss is because whatever she does she will end up in a two person duel with Brian or Candela, but if she misses her first shot, she will have the first shot in the two-way duel. Let's see the maths for a two person duel. Assume $P(X,Y)$ is the probability of X surviving a duel with Y where X has the first shot. Then we have:
The probabilities for Anna in a two way duel are then:
$P(A,B) = \frac{3}{7} = \frac{9}{21}$
$P(A,C) = \frac{1}{3} = \frac{7}{21}$
$P(B,A) = \frac{1}{7} = \frac{3}{21}$
$P(C,A) = 0$
So we can see that when Anna has the first shot, she is much more likely to survive the two-way duel. We can calculate Anna's full chance of surviving with this strategy: Brian and Candela will face off, with the winner set to battle Anna but Anna will have the first shot. Brian will win the duel with Candela 2/3 of the time. We have:
Therefore with this strategy Anna will win the tournament $39.7\%$ of the time. You can find a more detailed solution (and interactive) solution here.
Puzzle 33*** (30/11/16): On a perfectly circular island there are six villages along the coast, evenly spaced so the distance between any two neighbouring villages is the same. There are straight paths through the island connecting every pair of villages. Every day you must journey to the village furthest away from your home, always obeying the following rules. You must travel in a straight line between villages; you can only change direction at the villages. You can stop at any village or villages on the way, but you may not visit the same village twice on one journey. You may not travel to a village further away from your final destination than your current village. At the end of the day, you will take the shortest path back to your home village. Assuming you always follow the rules, how many days will it take you to cover every possible route between your home and the furthest village?
Solution 33: There are 25 different routes you can take between your home (A) and the furthest village (B).
If you don't stop off in any other villages on the way, there is one route: AB
If you stop off in one other village on the way, there are four routes: ACB, ADB, AEB, AFB
If you stop off in two other villages on the way, there are eight routes: ACDB, ACEB, ACFB, ADCB, ADEB, ADFB, AEFB, AFEB
If you stop off in three other villages on the way, there are eight routes: ACDEB, ACDFB, ACEFB, ACFEB, ADCEB, ADCFB, ADEFB, ADFEB
If you stop off in four other villages on the way, there are four routes: ACDEFB, ACDFEB, ADCEFB, ADCFEB
Considering you take one route per day, it will take you 25 days to cover every route (or 24 days if you take into account the fact that you always return via the shortest route, AB, so you don't need an extra day to cover that route).
Puzzle 32** (23/11/16): One day scientists find a strange genetic mutation, referred to as Mutation Z, present in one person out of every 1000. The geneticists design a genetic test that is able to correctly identify everyone who has Mutation Z, but the test also has a false-positive rate of 5 percent, meaning the test will wrongly find Mutation Z in 5 percent of the people who do not have it. If we choose a person at random, without knowing anything else about them, and the test shows they have Mutation Z, what is the probability that this person actually has the mutation?
Solution 32: While intuitively one could think the answer is 95%, the chance a person who tested positive actually has the mutation is only 2%. There are several different ways we could get to this answer.
Method 1 (frequentist): Imagine we take 1000 people. Out of these, we know only one person has Mutation Z (1/1000). We also know that the test will turn out positive for 5% of individuals who don't have the mutation, which means approximately 50 (5% of 999) people will show up as false positives. In total, the test will have found 51 people with the mutation, even though only one actually has it. So the probability that any one person who tested positive actually has the mutation is
$\frac{1}{51} = 0.0196 \approx 2\%$
Method 2 (Bayesian): We can use Bayes' Theorem, which states:
where P(mut|pos test) is the probability of having the mutation having tested positive, P(mut) is the probability of having the mutation, P(pos test) is the probability of testing positive, and P(pos test|mut) is the probability of testing positive when having the mutation. We know:
$P($mut$)=0.001$
$P($no mut$)=1-0.001 = 0.999$
$P($pos test|no mut$) = 0.05$
$P($pos test|mut$)=1$
$P($pos test$)=P($pos test|mut$)*P($mut$)+P($pos test|no mut$)*P($no mut$)$
Putting it all together we have
More information about this puzzle can be found here.
Puzzle 31* (16/11/16): While enjoying a nice wedding reception, the newly weds decide to give an expensive bottle of wine to one of their guests, so they propose a game. First, all 42 guests (over the drinking age) will stand in a circle. The youngest guest will be given a ball, which they will throw at the person standing to their left. This person sits down, and the ball will move clockwise to the next person standing, who will repeat the process (guest 1 eliminates guest 2, guest 3 eliminates guest 4, and so on). The whole process is repeated until only one person remains standing, who will then be given the bottle of wine. As a good logician, you know the game is fully deterministic and the winner can be calculated in advance. How close to the youngest person should you stand to guarantee you are the last person standing?
If we say the youngest guest is number 1 and we proceed clockwise, you should be in position 21 to get the bottle of wine. The picture on the right shows who is eliminated in each round. There's also a really cool way to solve this puzzle (which works for any number of guests). For $n$ initial guests, we write the $n$ in binary (in the case of 42, this is 101010). We now take the first digit and move it to the end (so 101010 becomes 010101). We convert our new binary number back to decimal (010101=21) and we find the number that will win! For a fully mathematical solution, check out this video from Numberphile.
Puzzle 30*** (09/11/16): One day you realise you've had enough of your usual life, so you decide to join a monastery with highly-logical monks. In order to be accepted into the monastery you must prove your logic skills, so the monks sit you down next to a square, rotatable table. After blindfolding you, the monks tell you that they have placed four glasses, one at each corner of the square table, with random orientations (facing up or facing down). Your mission is to get all the glasses with the same orientation, either up or down, while obeying the following rules: you may pick up any two glasses and reverse one, both, or neither of them. Once you put the glasses back down, the table will be rotated a random angle, after which you can repeat the process. As soon as you have all four glasses with the same orientation, the monks will remove your blindfold. Without relying on luck, can you find a strategy that will allow you to succeed with only four rotations of the table (giving you five steps)?
Solution 30: First we can identify the possible configurations: three up and one down (A), three down and one up (B), two up and two down diagonally opposite (C), and two up and two down adjacent to each other (D). We now do the following:
Take two diagonally opposite glasses and place them facing up. You will now be in configuration A or C.
Take any two adjacent glasses and turn them both up. If you were in configuration A, you will either have picked up two glasses already facing up and left them the same, or you will have picked up one facing down and turned it, leaving you with four glasses facing up. If you were in configuration C, you will have picked up one glass facing each way and left them both facing up. You are now in configuration A.
Take two diagonally opposite glasses. If one is facing down, turn it up and you are done. If both already face up, turn one of them facing down. This will leave you in configuration D (two glasses facing up adjacent to each other, and two facing down also adjacent).
Take two adjacent glasses and invert them. If you choose two with the same orientation, you are done. If you choose two with opposite orientation and reverse them both, you will be in configuration C.
Take two diagonally opposite glasses and reverse them both, leaving you with either four glasses facing up or four glasses facing down.
Puzzle 29** (02/11/16): Four bandits ride into town on their horses. They leave the horses tethered outside a bank, which they then go and rob. After robbing the bank, the bandits need to make a quick escape. Each bandit has their own horse, but as they are leaving quickly the first bandit to exit the bank randomly takes one of the four horses. When the second bandit comes out of the bank, if his own horse is not yet taken, he will choose his own horse, but if his horse has already been taken, he will randomly choose another horse. The third bandit will do the same. When the forth bandit leaves the bank, what is the probability that the remaining horse is his own?
Solution 29: The fourth bandit has a 50% chance of getting his own horse. In the case of four bandits we can do a simple brute-force approach and try all the possible combinations. Considering that they will all, except the first bandit, choose their own horse if it is available, there are only eight possibilities, shown in the figure in the right. Out of these 8 only 4 lead to the last bandit getting his own horse, so the probability is 4/8 = 1/2 = 50%.
This puzzle was a shorter version of the "Lostboardingpass" puzzle, in which there are 100 people instead of four. So let's generalise our solution for any number of bandits. Warning: maths ahead!
First, we define the probability $p(n)$ as the probability that the last bandit $(n)$ gets their own horse. This is what we want to calculate, and should work for any $n$ larger than 1. The first bandit will take their own horse with a probability of $\frac{1}{n}$. If this happens, the last bandit will inevitably get their own horse as well. If the first bandit instead takes horse $m$ (with $m$ also larger than one), then bandits $2, 3, 4, ..., m-1$ will also get their own horses. Bandit $m$ will then have the reduced problem of finding $n-(m-1)$ horses to choose from. This is exactly what would happen if we had started with $n-m+1$ bandits, so we can see that the probability will recursively depend on the probability in the case of fewer bandits. So we have to consider the probability of the first bandit choosing their own horse and all other horses this bandit can choose:
This shows us that to solve the case of $p(n)$ we can use the case of $p(n-1)$. In the case of two bandits, it's clear the second bandit has a 50% chance of finding their own horse, as the first bandit only has two horses to choose from. As such, we know that $p(2)=1/2$. So we can use the recursive formula found above and $p(2)=1/2$ to find
Which means that regardless of the number of bandits, the last one ($n$) will get their own horse 50% of the time, which is very cool.
Puzzle 28** (26/10/16): While organising an international cosmology school, Prof. Forgetful wasted coffee on his notes, erasing most of the information. He knows there will be one lecture per day from Monday to Thursday where they will discuss the CMB, Dark Energy, Dark Matter, and gravitational waves. He knows the speakers are called Albert, Emmy, Lisa, and Nikola, and they come from Heidelberg, London, New York, and Rome, but he doesn't know how the information fits together. Now he needs your help to figure out which lecture takes place on which day, who will be giving each lecture, and where each person comes from. He was able to recover the following details from the notes:
The person who will discuss Dark Matter will give a talk one day after Nikola.
The speaker from New York will give their talk sometime after the expert in gravitational waves.
Neither the person from Heidelberg nor the Dark Matter expert is Albert.
The person from Heidelberg will give a talk one day before the CMB expert.
Of the person from London and the person scheduled to talk on Wednesday, one will discuss gravitational waves and the other is Nikola.
Nikola will give his talk two days before the gravitational waves expert.
Lisa won't give a talk on Wednesday.
Solution 28: On Monday Nikola, who comes from London, will discuss Dark Energy. On Tuesday Dark Matter will be discussed by Lisa, who comes from Rome. On Wednesday Emmy, from Heidelberg, will talk about gravitational waves. On Thursday Albert, the New Yorker, will talk about the CMB.
To solve this type of puzzle it is often useful to draw a grid, like the one in the picture to the right, and mark all of the wrong choices with an 'X' and the correct options with a 'O'. If you like these puzzles, you can find lots of them here.
Puzzle 27***(19/10/16): Recently, a group of astronomers detected a strange periodic radio signal containing a combination of short and long pulses, coming from a nearby galaxy. Denoting the long signals as '1' and the short signals as '0', the signal was “11110001111000”, repeating over and over. After a while, the signal changed to “001100000000000101000000000000110000011”. The astronomers believe it's a message from an intelligent civilisation, and that the first part of the signal tells us how to understand the second part of the message. Knowing that you are good at solving puzzles, the astronomers ask you to help answer three questions. What does the message say? What does it mean? How many fingers do you expect the aliens to have? Points: 1 point for each question answered correctly, maximum 3 points.
The first part of the signal tells you that the message will contain four relevant characters followed by three blanks, so for the second part of the message:
001100000000000101000000000000110000011
you can ignore all the red text.
Aliens are unlikely to communicate in any Earth language, they would have to use a more universal system.
Solution 27: The first part of the message tells us how to read the second part of the message: four long signals followed by three short signals tells you that the message will contain four relevant characters followed by three blanks, so the second part of the message becomes
001100000000000101000000000000110000011
This means that we have six groups of four characters, following a clearly binary system. Translating from binary, the message reads as
3 0 5 0 3 3
This number would look very familiar to anyone using a base-6 number system (instead of the usual decimal system we use), as the number 3.05033 is the beginning of the fractional expansion of pi in base six. In our system, this would be 3.14152. So the aliens are telling us this message is not random space noise, they are showing us they understand mathematics, and their default base is base-6. It is generally assumed that humans use base 10 (decimal) because we have 10 fingers, so given that the aliens are using base-6, it is easy to assume they have six fingers!
Comment: I received some really creative answers to this puzzle, some people got a point for creativity!
Puzzle 26* (12/10/16): Marie and you are engaged in a battle of wits over a chess board to decide who eats the last brownie. After defeating you, Marie decides to give you a puzzle. If you can answer the puzzle on the first try, she will share the brownie with you. Her puzzle is: “If we count all the squares of different sizes, there are over 200 squares on a chess board. If I place three pawns in three random squares, what is the maximum amount of squares we could find that do not contain a pawn?”
Solution 26: The maximum number of squares you can find without a pawn is 183. This would happen if the three pawns are placed next to each other in a line, starting at one of the corners. The easiest way to see how many squares wouldn't contain a pawn is to first calculate the total number of squares on an empty chess board:
There are 64 (8*8) squares of size 1x1 There are 16 (4*4) squares of size 5x5
There are 49 (7*7) squares of size 2x2 There are 9 (3*3) squares of size 6x6
There are 36 (6*6) squares of size 3x3 There are 4 (2*2) squares of size 7x7
There are 25 (5*5) squares of size 4x4 There is 1 (1*1) square of size 8x8
So in total there are 204 squares on a chess board. Once you have placed the pawns in a line, starting at the corner, you can count how many squares are occupied:
There are 3 1x1 squares occupied. There are 3 5x5 squares occupied.
There are 3 2x2 squares occupied. There are 3 6x6 squares occupied.
There are 3 3x3 squares occupied. There are 2 7x7 squares occupied.
There are 3 4x4 squares occupied. There is 1 8x8 square occupied.
Which gives a total of 21 occupied squares. Finally, subtract the occupied squares from the total squares to find the amount of squares that don't contain a pawn: 204-21 = 183.
Puzzle 25**** (05/10/16): An evil demon captures 100 people and decides to play a game with them to see if he sets them free or harvests their souls. He assigns each person a different number between 1 and 100, and writes each number on a separate piece of paper. In a separate room he has 100 boxes in a line, and in each box he randomly places one of the numbered pieces of paper. One by one the people enter the room, where they can look in any 50 boxes of their choosing. After opening 50 boxes they have to tell the demon (in private) which box contains the piece of paper with their own number on it. If all of the people manage to find their number, they will all go free, however if any of them fail to find their own number, they will all be killed. The people are allowed to discuss a strategy beforehand, but once the first person has opened a box they can no longer communicate with each other. Which strategy will allow them to survive with more than 30% probability?
Hint 25: The people don't need to decide in advance which 50 boxes they will open, they only need to decide the first box, and how they will choose the other boxes.
Solution 25:
The key to this puzzle is realising that the boxes they have to open depend on the numbers inside the boxes. To maximise their chance of survival, they should use the following strategy:
Number the boxes. As they are in a line, the most logical would be to assign the first box number 1 and the last box number 100
Each person starts by opening the box corresponding to their own number. So the person with number 3 will start at box 3
If the box contains their own number, they are done. If not, they move on to the box corresponding to the number they just found. So if in box 3 the person finds number 77, they then go and open box 77. Then they read the number in box 77, and move on to that box
Continue cycling through the boxes until they either find their own number, or have opened 50 boxes in total
Let's take a smaller example to see why this works. Assume we have ten people, ten boxes, and they can each open 5 boxes. We need to think of this in terms of permutations. Imagine we have the following configuration:
1 2 3 4 5 6 7 8 9 10
3 9 5 8 1 4 10 6 7 2
The number on the top row represents the box number, and the number on the bottom row is the number in the box. So in box 1 we find number 3, in box 3 we find number 5, in box 5 we find number 1, and so on. We can express this in what is called 'cycle notation' as: (1,3,5)(2,9,7,10)(4,8,6). With this configuration and using the strategy described above, each person will find their own number. If you don't believe me, test it out for all of them! However, let's suppose now we have the following configuration:
1 2 3 4 5 6 7 8 9 10
4 5 8 9 2 3 10 1 6 7
which we know can be written as (1,4,9,6,3,8)(2,5)(7,10). Only the people with numbers 2, 5, 7, and 10 will find their on numbers, the others will not succeed. The reason they will not succeed is the cycle containing their number is larger than the amount of boxes they can open. So what is the chance of having a cycle larger than 5? Well, we can see initially that we can only have one cycle larger than 5 at a given time: we can't have a cycle of six and a cycle of seven, for example, so the overall chance of having a long cycle is quite low. This also scales up to 100 boxes: the people will succeed as long as there is not a cycle larger than 50, and by starting on their own number, they guarantee they are looking in the correct cycle. If you don't want the maths, and are convinced that this strategy is going to allow them to survive more than 30% of the time, you can stop here. If you want the maths, read on!
We now want to calculate exactly how likely it is to have a cycle of length 51 or more. A permutation of 100 numbers can contain at most one cycle with $n >50$, and there are exactly ${100 \choose n}$ ways to select the numbers of this cycle. In this long cycle, there are $(n-1)!$ ways you can arrange the numbers, and the remaining numbers (not part of the long cycle) can be arranged in $(100-n)!$ ways. This means that the number of possible cycles larger than 50 is
As we need to consider the possibility of a cycle of length 51 and the possibility of finding a cycle of length 52, 53, 54, and so on, the total probability of finding a deadly cycle is
So, finally, the probability of the people surviving is equal to the probability of them not finding a deadly cycle, so
$P_{survive} = 1-0.69 = 0.31 = 31\%$
The nice thing about this strategy is either they all succeed, or more than half of them fail. They won't be able to blame one person if they all die. For an even more detailed solution, check out this link.
Puzzle 24** (28/09/16): A private detective is following a man suspected of having ties to a criminal network. While buying her usual salad box for lunch, she notices someone slip the suspect a note. With an accidental-coffee-spill trick, she is able to acquire the note, where she reads the following text: “MAATDIETGEAGETLLYHTHEMATMEHOTPEEONEM”.
When and where are the suspects meeting?
Solution 24: The suspects are meeting Monday 8 PM at the Eagle Hotel. The simplest way to find the solution is to notice the text has 36 letters, so you can write it in a 6x6 box as
M A A T D I
E T G E A G
E T L L Y H
T H E M A T
M E H O T P
E E O N E M
and now you just have to read the columns to see the hidden text. This is one of the oldest ciphers know, and is often called "Caesar's Box", which is why she's having a salad box for lunch!
Puzzle 23** (21/09/16): While out camping in a forest, you are captured by a group of witches. They lead you to a sacrificial altar where a pentagram (five pointed star) has been drawn. The witches give you two straight sticks, and tell you “You are to place these two sticks on the pentagram such that there are ten non-overlapping triangles. If you succeed, you will be allowed to go free. If not, you will be sacrificed.” How should you place the sticks to survive?
Solution 23: You can save yourself from the witches by placing the sticks across two of the star's points, as shown in the image below.
Puzzle 22*** (14/09/16): You have two coconuts and access to a 100-floor building. You want to know which is the highest floor from which you can drop a coconut without it breaking. Both coconuts are identical, and if a coconut is dropped and does not break, it is undamaged and can be used again. If the coconut breaks when dropped from floor 'n', it would also break from any floor above that, and if a coconut survives a drop from floor 'n', it would survive the drop from any lower floor as well. What strategy should you adopt to minimise the number of coconut drops it takes to find the solution? With this strategy, what is the worst case for how many drops it will take?
Solution 22: In the best strategy, you can find the floor the coconuts break on in 14 drops or fewer. Imagine we start with the first coconut going up floors ten at a time. We drop it from floor 10, if it survives we go up to floor 20, then floor 30, and so on until the first one breaks. Then we would go back down nine floors, and work our way up one by one. So if it breaks on floor 10, we then try 1, 2, 3, ... worst case is when it breaks on floor nine: we will have used 10 drops. If instead of breaking on floor 10 the first coconut breaks on floor 20, we then try 11, 12, 13... If it breaks on floor 19, we have used 11 drops in total. If the first coconut breaks on floor 30, we then try 21, 22, 23... until the second coconut breaks. If this happens on floor 29, we have used 12 drops. We can see that as the solution goes to higher floors, we are using more drops. So going up the floors in a uniform 'ten at a time' pattern is not working; we need to decrease the step-size for the first coconut by one every time.
The optimum solution comes when we start on floor 14, and then decrease by one floor each time the first coconut survives: start at floor 14, if the coconut survives we move up to 14+13=27. If it survives, we move to 27+12=39, and then to 39+11=50... Once the first coconut breaks, we go back to one above the last floor we are sure the coconut survives from, and we progress through the floors one by one with the second coconut until it breaks. In the worst case scenario, we will need 14 drops to find the last floor from which the coconuts survive.
Mathematically, the first floor we try will be $n$. If the coconut doesn't break, the next floor will be $n+(n-1)$. Again, if the coconut doesn't break, we move up to $n+(n-1)+(n-2)$. We can continue this until we reach the maximum of 100. This would look like:
Solving the second order equation, we find $n=13.65 \approx 14$. So we start with the first coconut at floor 14, then move up 13 floors to 27, then we move up 12 floors to 39, and so on. Once the first coconut has broken, we then go through the floors one by one with the second coconut. You can read a detailed analysis of the puzzle here.
Puzzle 21** (07/09/16): An anthropologist travels to a recently discovered town hidden in a forest, to learn about the town's people and their customs. In the centre of the town she finds three men with the following sign in front of them. “The Three Wise Brothers can answer any question you have, but be warned: the oldest always tells the truth, the youngest always lies, and the middle one will lie or tell the truth depending on the position of the stars and his general mood. You may ask one yes/no question to any of the three brothers, and then you must choose which brother will answer all your remaining questions. The other two brothers will leave. If you pose a paradoxical question, all three brother will leave.” What question should she ask to guarantee she can choose a brother with whom she can communicate?
Solution 21: The objective is to avoid choosing the unpredictable brother, it doesn't matter if she chooses the oldest or youngest brother, she will be able to quickly determine if the person is lying or not, and thus will be able to easily communicate.
To avoid the unpredictable brother, she should ask any of the brothers about the relative ages of the other two, so for example ask the brother in the middle "Is that man (point to brother on the left) older than him (point to brother on the right)?" And then choose whoever they tell you is youngest; choose 'left' if the answer was 'no', choose 'right' if the answer was 'yes'. To see why this works, let's see all the possibly distributions for the brothers. We'll call the brothers Truth (T), Random (R), and Lie (L).
R T L --> Will answer 'yes', choose 'right', get L (youngest)
L T R --> Will answer 'no', choose 'left', get L (youngest)
R L T --> Will answer 'yes', choose 'right', get T (oldest)
T L R --> Will answer 'no', choose 'left', get T (oldest)
T R L --> If he answers 'yes', choose 'right', get L (youngest), if he answers 'no', choose 'left', get T (oldest)
L R T --> If he answers 'yes', choose 'right', get T (oldest), if he answers 'no', choose 'left', get L (youngest)
In each possible scenario, she avoids the unpredictable brother.
Other questions that also work: "Does 'left' tell the truth more often than 'right'?", and "If I were to ask you if the brother on your right is the one who answers randomly, would you answer 'yes'?" (Credits to Nacholo for finding this difficult solution!)
Puzzle 20* (31/08/16): One day you are abducted by hungry aliens who are trying to understand our rules of geometry. They show you a 4 meter high stone cube that has been painted bright blue. They ask you the following: “If you were to cut the cube into smaller cubes, each one meter high, how many small cubes would have no sides painted blue, how many would have one side painted blue, how many would have two sides painted blue, and how many would have three sides painted blue?” Obviously, you only have one chance to get the correct answer, else the aliens will eat you.
Solution 20: In total there will be 64 small cubes (4*4*4). Eight cubes will have no sides painted blue (the middle ones), 24 will have one side painted blue (the centres of each face), 24 will have two sides painted blue (the edges), and eight will have three sides painted blue (the corners).
Puzzle 19*** (24/08/16): You are competing in a series of shenanigans, and for your final task you are in a room wearing a blindfold. You are led to a table with 87 coins, and you are told 24 of them have tails facing up, and the rest have heads facing up. You can flip and move as many coins as you like. Your task is to separate the coins into two piles that contain the same amount of tails facing up. You can't remove your blindfold, nor can you distinguish heads or tails by feeling the coins. How can you do it?
Solution 19: Let's call the initial pile 'A'. Choose 24 coins at random from A and separate them from the others, making pile 'B'. Flip all 24 coins in pile B. Piles A and B will now have the same amount of tails facing up. If you are not convinced, look at the possible situations:
Case 1: You initially chose all 24 tails. After flipping all the coins in B, both A and B will have 0 tails.
Case 2: You chose 1 heads and 23 tails, leaving 1 tails in pile A. After flipping all the coins in pile B both piles will have 1 tails.
Case 3: You chose 2 heads and 22 tails, leaving 2 tails in pile A. After flipping all the coins in pile B both piles will have 2 tails.
...
Case 24: You chose 23 heads and 1 tails, leaving 23 tails in pile A. After flipping all the coins in pile B both piles will have 23 tails.
Case 25: You chose 24 heads and 0 tails, leaving 24 tails in pile A. After flipping all the coins in pile B both piles will have 24 tails.
Puzzle 18*** (17/08/16): After sailing to a strange island, three sailors are captured by cannibals. The cannibals decide to play a game with the sailors to settle their fate. The three sailors are ordered to stand in a line, such that the one at the back can see the other two sailors, the middle sailor can see the one at the front, and the one at the front can't see any of the others. The cannibals inform them that they have three white hats and two black hats, and they will randomly place one hat on each sailor's head. No sailor can see his own hat, and they are not allowed to communicate in any way. At any point, any of the sailors can state the colour of his own hat. If the sailor is correct, all three of them can go free. If the sailor is wrong, all of them will be eaten. If after one hour none of the sailors have said anything, they will all be eaten. How can the sailors guarantee their survival?
Solution 18: Let's label the sailors Back (B), Middle (M) and Front (F).
If B sees two black hats, he immediately knows that his hat is white, as there are only two black hats available. If this is the case, B immediately says 'My hat is white!' and the sailors go free. If B doesn't see two black hats he will stay silent.
If B remains silent for a while, M knows that B sees at least one white hat. If F is wearing a black hat, M knows his own hat must be white, else sailor B would already have said something. Therefore, if B is silent and M sees a black hat, M can safely proclaim 'My hat is white!' and they will go free. If F is wearing a white hat, M can't know if his own hat is white or black, and so he stays silent.
Near the end of the hour, if B and M have not said anything, F knows his own hat must be white, else one of the others would have spoken. F can happily state that his own hat is white, and they will all go free.
Of course, this puzzle assumes the sailors are logical, and there is nothing like the threat of being eaten to make people think logically!
Puzzle 17* (10/08/16): One day you are out having a coffee with two friends. You know Tim always lies on Monday, Tuesday, and Wednesday, and always tells the truth all the other days. Martha always lies on Thursday, Friday, and Saturday, and always tells the truth on all the other days. They make the following statements:
Tim: “Yesterday I was lying.”
Martha: “So was I.”
On which day of the week did this conversation take place?
Solution 17: The only day in which both statements are consistent is Thursday! Tim was telling the truth when he said 'Yesterday I was lying', and Martha was lying when she said 'So was I'.
Puzzle 16** (03/08/16): Three professors and three students are out celebrating the end of exams. They come upon a river they want to cross, but there are no bridges. They have one small row boat that can carry a maximum of two people, but they can use it as many times as they like. All six people know how to row and none of them want to swim. The professors are feeling a bit mean after the exams, and they decide if at any point there is a student outnumbered by professors on either side of the river, all the students will have to take another difficult exam, but the students get to decide how they all cross the river. How can the students get everyone to the other side of the river without having to take another exam?
Solution 16: Two professors cross, one comes back. Again, two professors cross, one comes back, leaving three students and a professor on the near side, and two professors on the far side. Two students cross, and a professor and a student come back. Two students cross, and the remaining professor comes back. You now have three students on the far side and three professors on the near side. Two professors cross, one comes back. Finally, the two remaining professors cross, leaving everyone on the far end, happy that there will be no more exams. You can find an interactive version of the puzzle here.
PSSS --} PP --}
PSSS {-- P {-- P
SSS --} PP --} P
SSS {-- P {-- PP
PS --} SS --} PP
PS {-- PS {-- PS
PP --} SS --} PS
PP {-- P {-- SSS
P --} PP --} SSS
P {-- P {-- SSSP
--} PP --} SSSP
Puzzle 15**** (27/07/16): Three mathematicians are playing a game. Gary thinks of two natural numbers (a and b), both larger than one, and Sarah and Peter both have to guess the numbers. Gary tells Sarah the sum of the two numbers (a+b) and he tells Peter the product of the two numbers (a*b). He tells them both that b is bigger than a (b$\gt$a) and the sum of the two numbers is smaller than 100 (a+b$\lt$100). The following conversation takes place:
Peter: “I don't know the numbers.”
Sarah: “I already knew that.”
Peter: “Now I know a and b!”
Sarah: “In that case, I also know a and b!”
Which numbers a and b was Gary thinking of?
Solution 15: The only possible answer that would allow both Peter and Sarah to know the numbers is a=4 and b=13.
Let's focus on the first two statements:
'I don't know the numbers.' Here Peter is telling us that the product does not have a single factorisation. This means that a and b can't be prime numbers.
'I already knew that'. If Sarah is really sure that Peter couldn't know the numbers it means there is no way you can express her sum as the sum of two primes. This allows us to discard a lot of possibilities; we know Sarah's number is not even (any even number can be written as the sum of two primes), or any other number that can be formed with two primes (5=2+3, 7=2+5, 9=2+7, 13=2+11, 15=2+13 ...). It also allows us to discard other sums that would lead to a unique product, like 51 (51=34+17, 17*34=348 --> can only be written as 17*34).
We know that the sum (S) has to be smaller than 100, and applying the above rules we find 10 possibilities for S: S can only be 11, 17, 23, 27, 29, 35, 37, 41, 47, or 53.
Now we use the third statement: 'Now I know the numbers!'. Peter is telling us that his product, P, only has one factorisation (a and b) that leads to a sum S that is on the list we found. Looking at all the possibilities in the spreadsheet to the right, Peter could have any of the pair of numbers in white/green. For example, P can't be 30, as 30=5*6 and 2*15, and both 5+6=11 and 2+15=17 would satisfy the previous statements. P could be 92, for example, as 92=4*23 and 4+23=27, which is on our list of possible sums.
Finally, we use Sarah's last statement 'If you know, so do I'. This means of all the possibilities we still had (in white) for the sum and product, S can only be expressed in one way (green) and still fulfil the conditions. For example, S can't be 11, as Sarah wouldn't be able to distinguish between 2/9, 3/8, or 4/7.
This means the only possibility is a=4, and b=13, giving the sum S=17 and the product P=52.
Without doing all the calculations, you can also use a 'trial-and-error' approach, were you calculate the first few possibilities for S (11, 17, 23), and then try all possible a and b that give these sums, to see if they satisfy the statements. This is shorter, but it can also be a bit hard to properly see which numbers fulfil all four statements; for example with 2/9, which would allow Peter (P=18) to know the numbers, but wouldn't allow Sarah (S=11) to distinguish between 2/9, 3/8, and 4/7.
Puzzle 14** (20/07/16): You have three pairs of ping pong balls: two blue balls, two white balls, and two red balls. In each coloured pair you know one ball is slightly heavier than the other. All the heavy balls weigh the same, and all the light balls weigh the same. You have a balance scale, but you are only allowed to use it twice. How can you identify the heavier and lighter ball of each colour? Note: the scale only has three positions: balanced, left side heavier, and right side heavier.
Solution 14: The key idea to solve this puzzle is to realise that the second weighing depends on the result of the first weighing, there is not a 'one-size-fits-all' solution. It is also important to consider all possibilities; a lot of wrong answers come from people forgetting some of the possible combinations of balls. For simplicity, I will name the balls B, B', W, W', R, and R'. Step 1: Put one blue ball on either side of the scale, one with a white ball and one with a red ball: BW vs B'R (you could also do WB vs W'R or RB vs R'W). There are eight possible configurations at this point (see diagram on right).
- If the scale is balanced, you know you have one heavy ball and one light ball on either side. Step 2: compare the two blue balls. This allows you to find the heavier one, say B is heavier than B'. By using the result of step 1, you know B is heavy, B' is light, W is light, R is heavy, and therefore R' is light and W' is heavy.
- If the scale tilts in one direction, you know that side has the heavier blue ball. For the remaining balls, you have three possibilities: W and R are both light, W and R are both heavy, or W and R are different (based on which way the scale tilted in Step 1). Step 2: put the two blue balls on one side of the scale (keeping track of which one is heavier), and the other two balls from weighing 1 on the other side. Based on which way the scale tilts (or doesn't) you can identify all the remaining cases. Check the diagram at the right for a complete solution.
Puzzle 13* (13/07/16): One day you are visiting a strange island and you are captured by cannibals. The cannibals love maths and they frequently play games with numbers. They decide to give you a challenge; if you can solve it, you go free, but if you can't solve it, they'll eat you. Your challenge is to find a 10-digit number, where the first digit is how many ones the number has, the second digit is how many twos the number has, .... etc ... , the ninth digit is how many nines the number has, and the tenth digit is how many zeros the number has. You only have one chance to get the correct number, if you get it wrong the first time, the cannibals will eat you.
Solution 13: 2100010006. To find the solution you can start with the easiest number (10 zeros), and gradually adjust each digit to make it fit:
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 9
0 0 0 0 0 0 0 0 1 9
1 0 0 0 0 0 0 1 0 8
2 0 0 0 0 0 1 0 0 7
2 1 0 0 0 0 1 0 0 6
2 1 0 0 0 1 0 0 0 6
Puzzle 12*** (06/07/16): Five pirates find a treasure chest with 100 gold coins, so they decide to split the coins following specific rules: the oldest pirate proposes how to share the coins, and all the pirates (including the proposer) vote for or against the proposal. If the proposal has 50% or more acceptance, the coins are split following the proposal. If the proposal has less than 50% acceptance, the oldest pirate is thrown overboard and killed, and the whole process is repeated with the remaining pirates. The pirates are rational, greedy, and bloodthirsty; they want to maximise their income and if two proposals give them the same income, they will choose the proposal that results in more deaths of their comrades. Assuming all pirates know and follow these rules, what distribution is the oldest pirate going to propose?
Solution 12: The oldest pirate will propose 98 - 0 - 1 - 0 - 1, and the proposal will be accepted with votes in favour from the youngest and the middle pirate. Let's call the pirates A, B, C, D, and E, where A is the oldest and E is the youngest. We can work backwards and analyse the different possible situations:
If only two pirates remain, D will propose a 100 - 0 split and his vote will allow the proposal to be accepted with 50% in favour.
If there are three pirates left, E knows he will get nothing if C is thrown overboard. Knowing this, C would propose 99 - 0 - 1. E will accept because he knows else he'll get nothing, and the proposal is accepted with 66% in favour.
If there are four pirates left, B only needs one other vote. D knows that if they reach situation 2, he gets nothing, so he will accept B's proposal as long as he gets something. B proposes 99 - 0 - 1 - 0, and the proposal is accepted with favourable votes from B and D.
Finally, with five pirates, C and E know that if A is thrown overboard, they will be in situation 3 and receive nothing. So C and E will vote in favour of A's proposal as long as they get something. Knowing this, A proposes 98 - 0 - 1 - 0 - 1, and the proposal is accepted with three votes in favour.
Interestingly, permutations of this solution do not work: if A proposes 98 - 0 - 0 - 1 - 1, D would vote against the proposal because he knows he can also get a coin from B, so he would prefer to kill A.
Puzzle 11* (29/06/16): You find yourself in a house with four light switches in the basement and four light bulbs in the attic. You know each switch controls one bulb, but you don't know which switch controls which bulb. If you can only go upstairs once, how can you find out which switch controls which bulb?
Solution 11: You can use the temperature of the bulbs. Turn on switches 1 and 2, wait 10 minutes, then switch off 2 and turn on 3. Go upstairs. The bulb that is on and hot corresponds to switch 1, the one that is off and hot corresponds to switch 2, the bulb that is on and mostly cold is controlled by switch three, and the cold bulb that is turned off is controlled by switch 4.
Puzzle 10** (22/06/16): One day at 8:00 a monk starts to climb a mountain to reach the temple at the top. He follows a narrow path that winds around the mountain. Sometimes the monk walks fast, sometimes he slows down, and he stops often to eat and enjoy the scenery. Finally, he reaches the temple at 21:00.
After a few days of meditation, the monk leaves the temple. He sets off at 8:00 and takes the same narrow path back down, walking at different speeds and enjoying the scenery. He reaches the bottom of the path at 19:00.
Is there any point on the path that the monk occupied at exactly the same time of the day on both trips? Justify your answer.
Solution 10: Yes, there is a point on the path where the monk will find himself at exactly the same time of the day. This can be proven in several ways, but the simplest way is to consider two monks making the journey on the same day: monk A makes the journey between 8.00 and 21.00, while monk B makes the trip between 8.00 and 19.00. Inevitably, the monks will meet at some point on the path; when they meet they will be in the same spot at exactly the same time of the day, so you can argue that the lone monk will intersect with his past-self at some point on the path.
You can also prove this with a quick plot: however you draw the green and orange lies, they will intersect at some point.
Puzzle 9*** (15/06/16): You have 3000 apples that you want to transport to a city 1000 km away using your camel. The camel can carry a maximum of 1000 apples at a time. Additionally, the camel needs to eat one apple for every kilometre he walks; he walks one kilometre, and then eats an apple. What is the maximum amount of apples you can transport to the city?
Solution 9: You can get 533 apples to the city. Obviously, the camel can't make the whole trip in one go, you need to find stopping points. If the camel can only carry 1000 apples and you need to move 3000 apples, you will need to make the first part of the journey three times (plus two return trips). Once you are left with 2000 apples, you only need to make the journey twice (plus one return trip). Finally, when you have 1000 apples left, you only need to make one trip. It looks something like this:
For the first part of the journey, you want to move 1000 apples (3000-2000) in five trips: 1000/5=200, so the first stop is at the 200 km mark. For the second part of the trip you need to move 1000 apples in three trips: 1000/3=333 (we will ignore the decimals), so the second stop is at 200+333= 533 km. This means our faithful camel friend will have 467 km to go, with 1000 apples left: he eats 467 apples on the last part of the journey, leaving you with 1000-467= 533 apples.
In summary: pick up 1000 apples, walk to 200 km (and eat 200 apples), drop all the apples except the 200 for the return trip. Repeat the process until you have 2000 apples at the 200 km mark. Pick up 1000 apples, walk to 533 km, drop all the apples except those you need for the return trip. Repeat, and you will be at 533 km with 1000 apples left, which you can then move to the city, munching on the way, leaving you with 533 apples.
Puzzle 8** (08/06/16): A farmer wants to start an orchard following very specific rules: he wants five straight rows of trees, each of which must contain exactly four trees; no row can have more than four trees. However, the farmer only has ten trees he can use. What pattern or template should he use to plant the trees? Bonus question: If the rows can have more than four trees, can you find an alternative solution?
Solution 8: This puzzle has several different solutions, here are the ones I've seen so far:
Most people found the pentagram solution, I think it's the most logical one. If you have another solution that isn't listed here, let me know!
Bonus question: If the rows can have more than four trees, you can just plant all trees in one line of ten, and say that this line contains at least five smaller rows of four trees. Example: in a line of five trees (0 0 0 0 0) you have two lines of four; (0 0 0 0) 0 and 0 (0 0 0 0).
Puzzle 7* (01/06/16): You have ten bags, each filled with approximately 100 coins, but you don't know the exact number in each bag. Nine bags contain real coins and the other bag contains counterfeit coins. You know that the real coins weigh 1 gram, while the counterfeit coins weigh 1.1 grams. You have a very precise digital scale, but you are only allowed to use it once. How can you find the bag that contains the counterfeit coins?
Note: you are allowed to remove the coins from the bags, just be sure not to mix them up.
Solution 7: Number each bag from 1 to 10. Take one coin from the first bag, two coins from the second bag, three coins from the third bag, ..., and ten coins from the tenth bag. You should now have 55 coins, which you can weigh all together. If all the coins were real, the scale would read 55g. If the scale reads 55.1g, you know there is only one fake coin, so bag one has the counterfeit coins. If the scale reads 55.2g, there are two fake coins and the second bag has the counterfeit coins, and so on. If the tenth bag has the counterfeit coins, the scale will read 56g. This puzzle can be solved in different ways, as long as you take a different amount of coins from each bag in a way that doesn't lead to an ambiguous result. For an example, you could take 5 from the first bag, 10 from the second bag, 15 from the third, and so on.
Puzzle 6**** (25/05/16): There is an isolated monastery where highly logical monks live. They have all taken a vow of silence and can't communicate with each other in any way. Furthermore, they are forbidden from seeing their own reflection: there are no reflective surfaces in the monastery. Every morning, all the monks sit down together and have breakfast. The monks have either brown eyes or blue eyes. Blue eyes are considered a curse, and any monk who discovers he has blue eyes must leave the monastery after breakfast. One afternoon a tourist visits and announces to all the monks 'At least one of you has blue eyes'. On the sixth morning after the announcement, after breakfast, all the monks with blue eyes leave the monastery. How many monks had blue eyes, and how did they know they had to leave?
Solution 6: There are six monks with blue eyes. To reach this conclusion we can follow an inductive path:
If there is only one monk with blue eyes, at breakfast on Day 1 he would see everyone else has brown eyes, and therefore he would know that he is the blue eyed monk, and would leave after breakfast on Day 1. We know this isn't the case, because no one leaves until the sixth day.
If there are two monks with blue eyes, on Day 2 they would see that the other blue-eyed monk did not leave on Day 1, and so they would both deduce that the other monk didn't leave because they saw someone else with blue eyes. As each can only see one other person with blue eyes they would know they also have blue eyes, and the two monks would leave together after breakfast on Day 2, having both reached the same conclusion. Again, we know this isn't the case.
If there are three blue-eyed monks, on Day 3 they realise the other two monks did not leave on Day 2, implying there must be more than two blue-eyed monks. Again, if they can only see two people with blue eyes, but they know there must be more than two, each blue-eyed monk will deduce that he must be the third blue-eyed monk. The three monks would reach the same conclusion and leave after breakfast on Day 3. We know that no one leaves until day six, so there must be more than three blue-eyed monks.
...
We can follow this reasoning and see that four blue-eyed monks would leave on the fourth day, five blue-eyed monks would leave on the fifth day, and finally six blue-eyed monks would leave on the sixth day. In fact, we could use induction to say that if there are 'n' blue-eyed monks, they will leave on the 'nth' day.
An interesting point in this puzzle is that if there are six blue-eyed monks, they all already know that 'at least one person has blue eyes', and yet no monks leave until the tourist makes his announcement. We can follow the same thought process as before: if there is only one blue-eyed monk, he would be blissfully unaware of that fact until the tourist comes. If there are two blue-eyed monks, each would assume the other one was blissfully unaware of his cursed condition. If there are three blue-eyed monks, A would see B and C and assume that each one assumes the other is blissfully unaware... and so on. That is why they need the tourist to act as a catalyst. This is what is known as common knowledge in logic.
Puzzle 5* (18/05/16): You have two ropes of different lengths and a box of matches. You know it takes exactly one hour for each rope to burn, but the burning rate is completely irregular, meaning you can't say it takes 30 minutes to burn exactly half of the rope. Without using anything else, how can you measure 45 minutes?
Solution 5: If you burn one rope from both ends, it will take 30 minutes to burn completely. To measure 45, minutes burn rope A from both sides and rope B from one side only. Once rope A is completely burnt (after 30 minutes), start burning the other end of rope B. The remaining section of rope B will burn in 15 minutes, which means when rope B has burnt completely, 45 minutes will have passed in total.
Puzzle 4** (11/05/16): Four people are travelling at night and they come upon a narrow bridge that is only big enough for two people at a time to cross. They only have one torch and everyone needs the torch to cross: it's impossible to cross without the torch. Alex can cross the bridge in 1 minute, Beth needs 2 minutes to cross, Charlie takes 5 minutes to cross, and Dany needs 8 minutes. If two people are crossing the bridge, they will only be able to go as fast as the slowest person crossing. What is the minimum time needed for all four travellers to cross the bridge?
Solution 4: The minimum time they need to cross the bridge is 15 minutes. The trick here is realising that Charlie and Dany need to cross together, even though intuitively one assumes that Alex must help everyone else across (which would take 17 minutes).
A+B cross $\rightarrow$ 2 minutes $\rightarrow$ TOTAL: 2 minutes
A comes back $\rightarrow$ 1 minutes $\rightarrow$ TOTAL: 3 minutes
C+D cross $\rightarrow$ 8 minutes $\rightarrow$ TOTAL: 11 minutes
B comes back $\rightarrow$ 2 minutes $\rightarrow$ TOTAL: 13 minutes
A+B cross $\rightarrow$ 2 minutes $\rightarrow$ TOTAL: 15 minutes
Note that steps 2 and 4 are completely interchangeable.
Puzzle 3*** (04/05/16): Two mathematicians, Sarah and Mike, meet up after many years. The following conversation takes place.
Mike: 'I have three children. The product of their ages is 72, and the sum of their ages is equal to the number on this door.'
After looking at the number on the door, Sarah comments: 'I still don't know how old they are, what else can you tell me?'
Mike: 'The oldest one likes cheesecake.'
Sarah: 'Now I know how old they are!'
How old are Mike's children?
Solution 3: The children are 3, 3, and 8. Knowing that the product of the ages is 72, we have the following possibilities:
1 x 1 x 72 sum = 74 1 x 6 x 12 sum = 19 2 x 4 x 9 sum = 15
1 x 2 x 36 sum = 39 1 x 8 x 9 sum = 18 2 x 6 x 6 sum = 14
1 x 3 x 24 sum = 28 2 x 2 x 18 sum = 22 3 x 3 x 8 sum = 14
1 x 4 x 18 sum = 23 2 x 3 x 12 sum = 17 3 x 4 x 6 sum = 13
Now we have to use the fact that Sarah sees the number on the door (so she knows the sum of their ages), but she still doesn't have enough information. The only sum that is repeated is 14, which means the children are either 2, 6, and 6, or 3, 3, and 8. Mike now says 'My oldest likes cheesecake', which means there is definitely an oldest child, thereby discarding the 2-6-6 option. Only one answer remains: the children are 3, 3, and 8.
Puzzle 2** (27/04/16): You have three jugs; one can hold 3 litres, one can hold 5 litres, one can hold 8 litres. The 8-litre jug is full of water, the other two are empty. You have no other measuring devices or jugs. You want to have two jugs with four litres each. How do you do it? (It can't be done by eye or by guesswork.) Points: 1 point if you need 8 steps or more, 2 points if you need 7 steps, 3 points if less than 7 steps
Solution 2: 8-step solution:
8L 5L 3L
8 0 0 Step 1: Pour from the 8L jug to the 3L jug
5 0 3 Step 2: Pour from the 3L jug to the 5L jug
5 3 0 Step 3: Pour from the 8L jug to the 3L jug
2 3 3 Step 4: Pour from the 3L jug to the 5L jug
2 5 1 Step 5: Pour from the 5L jug to the 8L jug
7 0 1 Step 6: Pour from the 3L jug to the 5L jug
7 1 0 Step 7: Pour from the 8L jug to the 3L jug
4 1 3 Step 8: Pour from the 3L jug to the 5L jug
4 4 0 7-step solution:
8L 5L 3L
8 0 0 Step 1: Pour from the 8L jug to the 5L jug
3 5 0 Step 2: Pour from the 5L jug to the 3L jug
3 2 3 Step 3: Pour from the 3L jug to the 8L jug
6 2 0 Step 4: Pour from the 5L jug to the 3L jug
6 0 2 Step 5: Pour from the 8L jug to the 5L jug
1 5 2 Step 6: Pour from the 5L jug to the 3L jug
1 4 3 Step 7: Pour from the 3L jug to the 8L jug
4 4 0
Puzzle 1** (20/04/15): You have 27 coins, one of which is fake. The real coins all weigh the same, but the fake one weighs a bit more than the others. You have a beam balance scale ⚖. What is the minimum number of weighings you need to find the heavier coin? (Don't rely on luck, the answer should work always).
Solution 1: The minimum (assuming no luck) is 3. Divide the coins into three piles of nine, which we'll call A, B, and C. Set one group aside (C), and weigh A versus B. If the scale is balanced, you know the heavier coin must be in C. If the scale tips, you know which pile contains the heavier coin. Collect the nine coins that contain the heavier coin, discard all other coins. Now divide your group of nine into three groups of three. Set one group of three aside, and weigh the other two piles against each other. Like before, if the scale is balanced, the group you didn't weigh has the heavier coin. Finally, take the group of three that contains the heavier coin. Weigh two of them, and leave the other coin aside. If the scale is balanced, the coin you didn't weigh is the heavier. If the scale is tipped, you can directly see which is the heavier coin.
Ahh,the solution to 4 is excellent and obvious now.
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