Puzzles
30 of my favourite maths and logic puzzles, gathered over the years. Some you’ll know, many you won’t!
Complete the jigsaw
Every puzzle you crack adds a piece to the picture. Solve all 30 to see what it shows. Tap any piece to open its puzzle.
Solve a puzzle to place your first piece.
The Circle of Death
Not arranged by difficulty: that depends on the solver! Many of these first appeared on the New York Times Numberplay blog, which is well worth exploring for more.
The Circle of Death
There are one hundred people standing in a circle. They count off beginning at one and ending at one hundred. Since they are…
Open puzzleThere are one hundred people standing in a circle. They count off beginning at one and ending at one hundred. Since they are in a circle, ONE is next to TWO and ONE HUNDRED.
ONE has a sword and kills TWO. He passes the sword to THREE, who kills FOUR, and so forth. NINETY-NINE kills ONE HUNDRED and passes the sword to ONE. Then ONE kills THREE and passes the sword to FIVE. This goes on until only one person is left standing.
Which number is he?
Try it with smaller circles of 2, 3, 4, 5 … 16 people. Who survives when the number of people is a power of 2?
73. When the number of people is a power of 2, person ONE always survives. 100 = 64 + 36, so once 36 people have gone (the last of them is 72), 64 people remain and the sword is with 73. From there, 73 is the one left standing.
The 4 Coins Problem
You’re creating a new coin system for your country. You must use only four coin values, and you must be able to make every…
Open puzzleYou’re creating a new coin system for your country. You must use only four coin values, and you must be able to make every value from 1 to 10 using at least one coin and at most two coins.
Which 4 coins do you choose? Can you find a second set of 4 coins that also works?
You definitely need a 1 coin. Then think about which values a pair of coins can make, including two of the same coin.
For example 1, 2, 3 and 7: 4 = 1 + 3, 5 = 2 + 3, 6 = 3 + 3, 8 = 1 + 7, 9 = 2 + 7 and 10 = 3 + 7. A second set is 1, 3, 4 and 5.
There are eight sets in all: {1, 2, 3, 7}, {1, 2, 4, 5}, {1, 2, 5, 7}, {1, 2, 5, 8}, {1, 3, 4, 5}, {1, 3, 4, 6}, {1, 3, 4, 9} and {1, 3, 5, 6}.
The Wedding Ring
In certain parts of rural Russia, a would-be bride would gather six long pieces of straw or grass and grasp them in her…
Open puzzleIn certain parts of rural Russia, a would-be bride would gather six long pieces of straw or grass and grasp them in her hand. She would then randomly tie pairs of ends together at the top and at the bottom. Since there are six blades sticking out above and below her hand, she ties three knots at the top and three knots at the bottom.
The story goes that if she formed one big ring, she would get married soon.
What is the probability that she will get married?
Do the top knots first: they simply turn six blades into three long strands. Now tie the bottom ends one knot at a time.
8/15, just over a half. After the top knots there are effectively three long strands. Pick any bottom end: it must be tied to an end of a different strand, and 4 of the other 5 ends work (4/5). That joins two strands into one, leaving two; the next knot must join different strands (2 of the 3 remaining ends, so 2/3). The last knot then closes one big ring.
4/5 × 2/3 = 8/15.
The Averages Problem
Consider a list of all the ways you could take four different digits from 1 to 9 and arrange them to make the sum of two…
Open puzzleConsider a list of all the ways you could take four different digits from 1 to 9 and arrange them to make the sum of two 2-digit numbers. Some answers will appear several times: 134, for example, is 93 + 41, and also 91 + 43.
What are the mean, median and mode of all the numbers on this list?
(The mean is the total of the numbers divided by how many there are, the median is the middle value when they are in order, and the mode is the number that appears most often.)
How often does each digit appear in the tens place compared with the units place? Symmetry is your friend.
The mean is 110 and the median is 110. The mode is a three-way tie between 99, 110 and 121, which each appear 48 times out of the 3,024 arrangements.
By symmetry, every digit appears equally often in every position, and the average digit is 5, so the mean is 2 × (10 × 5 + 5) = 110.
Avoiding the Troll
Every day you make a trip from A to B, 2 miles away. But your trip is unpredictable. On 50% of days, you can walk between A…
Open puzzleEvery day you make a trip from A to B, 2 miles away. But your trip is unpredictable.
On 50% of days, you can walk between A and B without any obstruction.
The other 50% of the time, a troll appears at the halfway point, blocking your trip. The troll also switches on an invisible barrier that stretches 1 mile at right angles to your path in both directions, forcing you to walk around it.
You cannot see the troll or the barrier until you are halfway from A to B, so you can’t plan in advance whether you need to walk around it. You do know the troll appears at random 50% of the time.
If you can walk in straight lines to any points between A and B, what is your best strategy to walk the least distance on average? What is that distance?
Don’t walk straight at the halfway point. Aim slightly to one side, so that if the troll does appear you are already part of the way around the barrier.
About 2.62 miles on average. Head for a point on the barrier line √2/4 ≈ 0.35 miles to one side of the halfway point. If there’s no troll, carry straight on to B; if there is, walk to the end of the barrier and then on to B.
The average distance is (1 + 3√2)/2 ≈ 2.62 miles, beating the 2.71 miles you’d average by walking straight at the halfway point.
Ali Baba’s Cave
Ali Baba found a cave full of gold and diamonds. A bag full of gold weighs 200 kilograms, and a bag full of diamonds weighs…
Open puzzleAli Baba found a cave full of gold and diamonds. A bag full of gold weighs 200 kilograms, and a bag full of diamonds weighs 40 kilograms. Ali Baba can carry only 100 kilograms at a time. A kilogram of gold is worth $20, and a kilogram of diamonds is worth $60.
What is the most money Ali Baba can make from the gold and diamonds he carries out in one trip, in one bag?
The bag has a fixed size. What if part of it is gold and the rest is diamonds? A full bag of gold is worth $4,000 and a full bag of diamonds $2,400.
$3,000. Fill 3/8 of the bag with gold (75 kg, worth $1,500) and the other 5/8 with diamonds (25 kg, worth $1,500). That’s exactly 100 kg.
By comparison, half a bag of gold is worth only $2,000 and a full bag of diamonds $2,400.
The Drinking Game
The game starts with 6 empty glasses in a row, numbered 1 to 6. You roll an ordinary dice. If the glass with that number is…
Open puzzleThe game starts with 6 empty glasses in a row, numbered 1 to 6. You roll an ordinary dice. If the glass with that number is empty, it gets filled up. If it is full, you drink it, so it becomes empty.
There is a special rule when 5 glasses are full. If you roll the number of the one empty glass, it gets filled and you have to drink all 6 glasses. Then the game ends.
From the start of the game, what is the average number of rolls until the game ends?
Only the number of full glasses matters. Write E(k) for the expected number of rolls left when k glasses are full, and set up an equation for each k.
83.2 rolls (exactly 416/5). With k glasses full, you move to k + 1 full with probability (6 − k)/6 and back to k − 1 with probability k/6. With 5 full, the game ends with probability 1/6. Solving the six equations gives E(0) = 416/5 = 83.2.
The Rajah’s Pearls
A rajah left his daughters some pearls in his will, with instructions to divide them as follows. The oldest daughter was to…
Open puzzleA rajah left his daughters some pearls in his will, with instructions to divide them as follows. The oldest daughter was to receive 1 pearl plus 1/7 of what was left. The next oldest was to receive 2 pearls plus 1/7 of what was then left. The next was to receive 3 pearls plus 1/7 of what was left, and so on. The youngest daughter received whatever was left after all the others.
At first, the youngest daughter thought this was unfair, but after some careful calculation she realised that all the daughters would receive the same number of pearls.
How many pearls and how many daughters were there?
The oldest and the second oldest daughters get the same number of pearls. Call the total P and write that as an equation.
36 pearls and 6 daughters, with 6 pearls each. The oldest gets 1 + 35/7 = 6, leaving 30. The next gets 2 + 28/7 = 6, leaving 24, and so on, until the youngest takes the last 6.
The St Petersburg Lottery
Suppose we play the following game. I toss a coin repeatedly until it comes up heads. If heads appears on the first throw, I…
Open puzzleSuppose we play the following game. I toss a coin repeatedly until it comes up heads. If heads appears on the first throw, I pay you $2. If it appears on the second throw, I pay you $4. On the third, I pay $8, and so on, doubling each time.
How much would you be willing to pay me to play this game?
Work out the expected winnings. How much does each possible outcome contribute?
In theory, any amount! Heads first time contributes 1/2 × $2 = $1 to the expected payout, heads on the second throw 1/4 × $4 = $1, and so on, so the expected payout is 1 + 1 + 1 + …, which never ends.
Yet very few people would pay even $20. This is the famous St Petersburg paradox. In real life the bank can’t pay out an infinite amount: if I can pay at most about $1 million, the game is only worth about $21.
Cheryl’s Birthday
Albert and Bernard have just become friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a…
Open puzzleAlbert and Bernard have just become friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates:
May 15, May 16, May 19
June 17, June 18
July 14, July 16
August 14, August 15, August 17
Cheryl then tells Albert the month of her birthday, and tells Bernard the day.
Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard doesn’t know either.
Bernard: At first I didn’t know when Cheryl’s birthday was, but I know now.
Albert: Then I also know when Cheryl’s birthday is.
When is Cheryl’s birthday?
Albert knows the month. For him to be sure Bernard doesn’t know, his month can’t contain a day that appears only once in the list.
July 16. The days 18 and 19 appear only once, so Albert’s month can’t be May or June. That leaves July 14, July 16, August 14, August 15 and August 17.
Bernard now knows, so his day isn’t 14: that leaves July 16, August 15 and August 17. Albert then knows too, so his month must have just one date left: July 16.
Ages of Children
Two mathematicians meet in the street, and one says to the other, “I bet you can’t guess the ages of my 3 children.” (All…
Open puzzleTwo mathematicians meet in the street, and one says to the other, “I bet you can’t guess the ages of my 3 children.” (All the ages are whole numbers.)
The second mathematician says, “I’ll take your bet, but only if you give me enough information to solve it.”
The first agrees, and says, “The product of my 3 children’s ages is 36.”
“That isn’t enough information.”
“Right you are!” He looks around, spots the number on a building and says, “The sum of my children’s ages is the same as the number on that building.”
After thinking a little: “That still isn’t enough information.”
“Right you are again! My eldest son has red hair.”
After some thinking, the second mathematician won the bet. What are the ages?
List every set of three ages with a product of 36, and find each sum. Why wasn’t the sum enough?
2, 2 and 9. Only one sum appears twice: 13, from 1, 6, 6 and 2, 2, 9. That’s why the number on the building wasn’t enough. Having an eldest child rules out 6 year old twins being the oldest, so the ages are 2, 2 and 9.
Band Around the Earth
Imagine there is a steel band around the equator of the Earth, nice and snug (about 25,000 miles long). If you add one foot…
Open puzzleImagine there is a steel band around the equator of the Earth, nice and snug (about 25,000 miles long).
If you add one foot of steel to this band, how far is it raised off the ground all the way round? High enough to slip a hair under it? Your hand? High enough to walk under it?
Circumference = 2πr. How much does r change when the circumference goes up by one foot? Does the size of the Earth matter?
About 1.9 inches, all the way round. The radius goes up by 1/(2π) feet ≈ 0.16 feet, whatever the size of the circle. That’s plenty for a hair, and enough to slide a flat hand under, but you won’t be walking under it.
The School Lockers
A school has 1,000 students and 1,000 lockers, all closed. On the first day of term, the headteacher asks the first student…
Open puzzleA school has 1,000 students and 1,000 lockers, all closed. On the first day of term, the headteacher asks the first student to go along and open every locker. She asks the second student to go to every second locker and close it, the third to go to every third locker and close it if it is open or open it if it is closed, the fourth to do the same with every fourth locker, and so on. The process ends with the thousandth student.
How many lockers are open at the end?
Locker n is changed once for each factor of n. Which numbers have an odd number of factors?
31. Factors usually come in pairs, so most lockers are changed an even number of times and end up closed. Square numbers have an odd number of factors (in 36, 6 pairs with itself), so lockers 1, 4, 9, …, 961 stay open. There are 31 of them.
The Devil’s Wager
You die, and the devil says he’ll let you go to heaven if you beat him in a game. He sits you down at a round table and…
Open puzzleYou die, and the devil says he’ll let you go to heaven if you beat him in a game. He sits you down at a round table and divides a large pile of quarters (coins) into two equal piles, one each. He says, “We’ll take turns putting quarters down on the table. No overlapping is allowed, and the quarters must lie flat on the table. The first one who can’t put a quarter down loses. You can’t shift or squeeze any quarter into a space if it moves another quarter.” The devil says he wants to go first.
You realise that if the devil goes first, he probably has a winning strategy. You ask for some time to think, and he gives you 15 minutes. By the end of that time you know how to beat him, but you also know that you must go first for your strategy to work.
You convince him to let you go first. What is your winning strategy?
A round table has one very special point. Think symmetry.
Put your first quarter exactly in the centre of the table. After that, whenever the devil puts down a quarter, put yours in the mirror-image spot directly opposite, through the centre. If his spot was free, yours always is too, so he is the one who runs out of moves.
Half an Egg
A woman came to market to sell some eggs. The first buyer took half of her eggs plus half an egg. The same happened with the…
Open puzzleA woman came to market to sell some eggs. The first buyer took half of her eggs plus half an egg. The same happened with the eggs that were left: the second buyer took half of them plus half an egg. A third buyer bought what was left over: 1 egg. No eggs were broken.
How many eggs were there at the start?
Work backwards. Before the third buyer she had 1 egg. How many did she have before the second buyer?
7 eggs. Working backwards: if x eggs are left after a sale, she had 2x + 1 before it. So 1 → 3 → 7.
Check: the first buyer takes 3½ + ½ = 4, leaving 3. The second takes 1½ + ½ = 2, leaving 1.
Make 6
Add just two kinds of maths symbol to each line to make it true. You can use symbols more than once, and factorials (!) and…
Open puzzleAdd just two kinds of maths symbol to each line to make it true. You can use symbols more than once, and factorials (!) and square roots count as symbols. Brackets are free.
0 0 0 = 61 1 1 = 62 2 2 = 63 3 3 = 64 4 4 = 65 5 5 = 66 6 6 = 67 7 7 = 68 8 8 = 69 9 9 = 6
6 = 3! and 6 = 2 + 2 + 2. For 0 and 1, factorials will help. For 4, 8 and 9, think about roots.
(0! + 0! + 0!)! = 6(1 + 1 + 1)! = 62 + 2 + 2 = 63 × 3 − 3 = 6√4 + √4 + √4 = 65 ÷ 5 + 5 = 66 + 6 − 6 = 67 − 7 ÷ 7 = 6∛8 + ∛8 + ∛8 = 6√9 × √9 − √9 = 6
Other answers are possible, for example 6 × 6 ÷ 6.
Three Liars Tossing a Coin
Three people each lie 1/3 of the time. They all watch a coin being flipped, and they all say it landed Heads. What’s the…
Open puzzleThree people each lie 1/3 of the time. They all watch a coin being flipped, and they all say it landed Heads.
What’s the probability that it really is Heads?
Compare two worlds: it really was Heads and all three told the truth, or it was Tails and all three lied.
8/9. P(Heads and all three tell the truth) = 1/2 × (2/3)³ = 8/54. P(Tails and all three lie) = 1/2 × (1/3)³ = 1/54. So the chance it is Heads is 8/(8 + 1) = 8/9.
Free the Prisoners
The warden meets 23 new prisoners when they arrive. He tells them, “You may meet today and plan a strategy. But after today…
Open puzzleThe warden meets 23 new prisoners when they arrive. He tells them, “You may meet today and plan a strategy. But after today, you will be in separate cells and will have no communication with one another.
“In the prison is a switch room, which contains two light switches labelled 1 and 2. Each can be either up or down. I am not telling you how they are set now. The switches are not connected to anything.
“After today, whenever I feel like it, I will choose a prisoner at random and take him to the switch room. He must choose one of the two switches and flip it. He must flip exactly one switch. Then he’ll be led back to his cell.
“Given enough time, everyone will visit the switch room as many times as everyone else. At any time, anyone may say to me, ‘We have all visited the switch room.’ If it is true, you will all be set free. If it is false, you will all be punished horribly.”
What strategy should they use to guarantee their freedom?
Choose one prisoner to be the counter. Switch 2 can be the “do nothing” switch.
Appoint one prisoner as the counter. Everyone else follows one rule: if switch 1 is down and you have pushed it up fewer than two times before, push it up; otherwise flip switch 2.
The counter: if switch 1 is up, push it down and add one to your count; otherwise flip switch 2. When the count reaches 44 (2 × 22), declare.
Everyone signals twice because nobody knows how switch 1 started: it might have been up at the beginning. Even so, a count of 44 is only possible once all 22 others have visited.
Burning Ropes to Measure Time
Warm-up: You have a box of matches and a piece of rope. The rope takes exactly an hour to burn, but it may not burn evenly…
Open puzzleWarm-up: You have a box of matches and a piece of rope. The rope takes exactly an hour to burn, but it may not burn evenly. For example, it might be that the first tenth of the rope takes 50 minutes to burn and the other nine tenths take only 10 minutes. How can you measure exactly 30 minutes?
The real problem: You have a box of matches and two of these ropes, which are not necessarily the same. How can you measure exactly 15 minutes?
Lighting a rope at both ends halves the time it takes to burn, however unevenly it burns.
Warm-up: light the rope at both ends. It burns out after 30 minutes.
15 minutes: at the same moment, light rope A at both ends and rope B at one end. When A burns out, 30 minutes have passed and B has 30 minutes of burning left. Now light the other end of B: it burns out exactly 15 minutes later.
Minimum Number of Aircraft
On Bagshot Island, there is an airport. It is home to an unlimited number of identical aeroplanes. Each plane holds enough…
Open puzzleOn Bagshot Island, there is an airport. It is home to an unlimited number of identical aeroplanes. Each plane holds enough fuel to fly exactly halfway around the world, along a great circle. The planes can refuel each other in flight without losing speed or spilling fuel. There is unlimited fuel, but the island is the only place to get it.
What is the smallest number of aircraft needed to get one plane all the way around the world, if every aircraft must get back safely to the airport?
Notes: every plane must take off from and land at the same airport. Ignore the time and fuel used for refuelling.
Planes can top each other up on the way out, and others can fly out the other way to meet the returning plane.
3 aircraft. Measure fuel in distance: a full tank flies half the world. All three, A, B and C, take off together. An eighth of the way round, C gives A and B an eighth each and flies home. A quarter of the way round, B gives A an eighth and flies home. A, now full, flies on to the three-quarter mark.
Meanwhile B and C refuel and fly out the other way. C tops B up at the seven-eighths mark, B meets A at the three-quarter mark and shares its fuel, and C (refuelled again) meets them both at seven-eighths. Every plane gets home safely.
Minimum Number of Weights
A wholesale merchant came to me one day with this problem. Every day he has to weigh amounts from 1 pound to 121 pounds, to…
Open puzzleA wholesale merchant came to me one day with this problem. Every day he has to weigh amounts from 1 pound to 121 pounds, to the nearest pound, using a set of balance scales.
What is the smallest number of weights he needs, and how heavy should each weight be?
You can put weights on both pans of the scales.
5 weights: 1, 3, 9, 27 and 81 pounds (a total of 121). Because weights can go on either pan, each one can add, subtract or sit out, so every whole number from 1 to 121 can be made. For example, 2 = 3 − 1 and 5 = 9 − 3 − 1.
If weights could only go on one pan, he’d need 7: 1, 2, 4, 8, 16, 32 and 64.
The Relatively Prime Bet
A box holds lots of balls with different numbers on them. You pick two balls at random. Someone offers you a bet on whether…
Open puzzleA box holds lots of balls with different numbers on them. You pick two balls at random.
Someone offers you a bet on whether the two numbers are relatively prime (they have no common factor other than 1):
If you pick, say, 6 and 13 (relatively prime), you lose $1.
If you pick, say, 5 and 25 (not relatively prime), you win $2.
Will you accept the bet?
What’s the chance that two random numbers are both even? Both multiples of 3? Neither can happen if they are relatively prime.
Yes, take the bet (as long as the box holds a wide range of numbers). The chance that two numbers picked at random are relatively prime is 6/π² ≈ 61%. So you lose $1 about 61% of the time and win $2 about 39% of the time: on average you gain about 18 cents per go.
Crossing the Bridge
Four people need to cross a rickety bridge at night. They have only one torch, and the bridge is too dangerous to cross…
Open puzzleFour people need to cross a rickety bridge at night. They have only one torch, and the bridge is too dangerous to cross without it. The bridge can only hold two people at a time, and a pair has to walk at the slower person’s speed.
The four people take 1 minute, 2 minutes, 7 minutes and 10 minutes to cross.
What is the shortest time needed for all four of them to cross the bridge?
Send the two slowest people across together, and make sure neither of them has to bring the torch back.
17 minutes. 1 and 2 cross (2 minutes), 1 comes back (1), 7 and 10 cross (10), 2 comes back (2), and 1 and 2 cross again (2). 2 + 1 + 10 + 2 + 2 = 17.
25 Horses, 5 Tracks
Mr John has 25 horses, and he wants to find the fastest 3. He has only 5 tracks, so only 5 horses can race at a time, and he…
Open puzzleMr John has 25 horses, and he wants to find the fastest 3. He has only 5 tracks, so only 5 horses can race at a time, and he doesn’t have a stopwatch.
What is the smallest number of races he needs to find the 3 fastest horses?
Five heats, then a race of the five winners. Which horses could still be in the top three after that?
7 races. Run five heats of 5 (races 1 to 5), then race the five winners (race 6). The winner of race 6 is the fastest overall.
Only five horses can still be 2nd or 3rd overall: the 2nd and 3rd from the overall winner’s heat, the horse that came 2nd in race 6 and the runner-up from its heat, and the horse that came 3rd in race 6. Race those five (race 7): the first two home are 2nd and 3rd overall.
The Flipping Coins Puzzle
There are twenty coins on a table: ten heads and ten tails. You are blindfolded and wearing gloves. You can feel where the…
Open puzzleThere are twenty coins on a table: ten heads and ten tails. You are blindfolded and wearing gloves. You can feel where the coins are, but you can’t see or feel whether they are heads or tails.
You must make two groups of coins, with the same number of heads in each group. You can move coins and flip them over, but you can’t find out which way up any of them is.
How do you do it?
Split the coins into two groups of 10. If the first group has h heads, how many heads does the second group have? What would flipping do?
Take any 10 coins and turn every one of them over. If your 10 contain h heads, the other 10 contain 10 − h heads. After flipping, your 10 contain 10 − h heads too.
Random Aeroplane Seats
People are queueing to board a 100-seat aeroplane. Steve is first in the queue. He gets on the plane but can’t remember his…
Open puzzlePeople are queueing to board a 100-seat aeroplane. Steve is first in the queue. He gets on the plane but can’t remember his seat number, so he picks a seat at random. After that, each person who boards sits in their own seat if it’s free; otherwise they pick a free seat at random.
The flight is full, and you are last in the queue.
What is the probability that you get to sit in your own seat?
Try it with 2, 3 and 4 seats. And ask yourself: by the time you board, which seats could possibly be left?
1/2. The last free seat is always either Steve’s or yours. Every time someone has to pick at random, they are just as likely to pick Steve’s seat (which ends the chaos, so you get yours) as yours (so you don’t). Picking any other seat just passes the problem on to someone else.
Proportion of Girls and Boys
In a country where everyone wants a boy, each family keeps having babies until they have a boy. After some time, what is the…
Open puzzleIn a country where everyone wants a boy, each family keeps having babies until they have a boy. After some time, what is the proportion of boys to girls in the country?
(Assume a boy and a girl are equally likely each time.)
Does a family’s rule about when to stop change the chance of any single birth?
1 : 1, so equal numbers of boys and girls. Every birth is still a 50:50 chance, whatever rule families use to decide when to stop, so half of all babies are boys and half are girls. On average, each family has one boy and one girl.
The Hats and the Monster
An island has 10 inhabitants. One day a monster arrives and says he intends to eat every one of them, but will give them a…
Open puzzleAn island has 10 inhabitants. One day a monster arrives and says he intends to eat every one of them, but will give them a chance to survive.
In the morning, the monster will line everyone up in single file, so the last person sees the other 9, the next person sees the 8 in front, and so on, down to the person at the front, who sees nobody. The monster will then put a black or white hat on each person’s head at random (they could be all white, all black or any mixture).
Starting with the last person (who sees everyone else’s hat), the monster will ask each person to guess the colour of their own hat. They may answer with only one word: “white” or “black”. The monster will eat all ten as soon as a second person answers wrongly. Everyone can hear each guess, but not whether it was right.
Devise a strategy that guarantees everyone survives.
The last person can’t be sure of their own hat, but their one word can carry information to everyone else. Think odd and even.
Use odd and even. The last person says “white” if they can see an even number of white hats, and “black” if they can see an odd number. (They have a 50% chance of being right themselves.)
Each person after that counts the white hats in front of them, plus the white hats already called out behind them (after the first call), and works out their own colour. Everyone from the second call onwards is guaranteed to be right, so at most one person is ever wrong and nobody is eaten.
Eggs from a Building
A building has 100 floors. You are given 2 identical eggs. Every egg has a “threshold” floor, N: the egg will not break if…
Open puzzleA building has 100 floors. You are given 2 identical eggs. Every egg has a “threshold” floor, N: the egg will not break if dropped from any floor below N, but it will definitely break if dropped from floor N or any floor above.
These are very strong eggs, so they can be dropped again and again without breaking, as long as they are dropped from below floor N.
What strategy should you use to find floor N with as few drops as possible? What is the smallest number of drops that guarantees you find it, however unlucky you are?
Dropping the first egg every 10 floors needs up to 19 drops. Can you balance things so that each jump is one floor shorter than the last?
14 drops in the worst case. Drop the first egg from floor 14, then 27, 39, 50, 60, 69, 77, 84, 90, 95, 99 and 100: each jump is one floor shorter than the one before. When it breaks, go back to the floor above the last safe drop and use the second egg one floor at a time.
14 + 13 + 12 + … + 1 = 105, which covers 100 floors, but 13 + 12 + … + 1 = 91 does not, so 14 is the best possible.
5 Pirates, 100 Gold Coins
Five pirates have found 100 gold coins and must divide up the loot. The pirates are all extremely clever, treacherous and…
Open puzzleFive pirates have found 100 gold coins and must divide up the loot. The pirates are all extremely clever, treacherous and selfish (especially the captain).
The captain proposes how to share out the coins, and all the pirates vote. If half the crew or more vote “Aye”, the coins are shared as proposed, as no pirate would take on the captain without a majority behind him.
If the captain doesn’t get the support of at least half his crew (including himself), there is a mutiny: the pirates turn against him and make him walk the plank. They then start again, with the next most senior pirate as captain.
What is the largest number of coins the captain can keep without risking his life?
Start at the end. What happens with just two pirates? Then three?
98 coins. With 2 pirates, the captain keeps all 100, since his own vote is half. With 3, he needs one more vote, so he gives 1 coin to the pirate who would otherwise get nothing (99, 0, 1). With 4, he needs one more vote, and buys the pirate who gets nothing in the 3-pirate case (99, 0, 1, 0).
With 5, he needs two more votes. He buys the two pirates who would get nothing with 4 pirates left, at 1 coin each, and keeps 98.
No puzzles match. Try “coins” or “probability”, or .