Web Puzzles

Island X Puzzle Solution

Assume that A is the Truther. If so, then B is the Liar as A's statement asserts. If so, B's second statement is false, so C is the Alternator. This implies that C's first statement is false as also his second statement that A is the Liar, so that C is the Liar which is a contradiction, so that A cannot be the Truther.

Assume that B is the Truther. If so, his first statement is a direct contradiction, implying amongst other things, that B cannot be the Truther.

Assume that C is the Truther. If so, then A is the Liar in conformity with his second statement, so the remaining member B must be the Alternator. This checks out since as a Alternator, B's first statement is true while in his second statement he falsely identified C as the Liar. Both of A's statement are then clearly false, so this establishes the veracity of both the statements of C.

Consequently, (A, B, C) = (Liar, Alternator, Truther)

A Farmer's Good Fortune Puzzle Solution

Here's one combination:

 94 sheep =  $470
  1 pig   =   $30
  5 cows  =  $500
-----------------
100 heads = $1000

Are there anymore?

A Special Old Guy Puzzle Solution

Firstly, we can say that the date of birth (DOB) for the special old guy (SOG) is:
Equation {a}:
DOB(SOG) = x^2 – x

Similarly, for the son of the special old guy (or SOSOG for short):

Equation {b}:
DOB(SOSOG) = y^3 – y

SOG's age when SOSOG was born was:
Equation {c}:
Age(SOG) = DOB(SOSOG) – DOB(SOG) = (y^3 – y) – (x^2 – x)

There are unlimited solutions to equation {c} so we need some assumptions and limits.

We know that SOG must have been between, say 10 years old and 100 years old when SOSOG was born:

Equation {d}:
10 < Age(SOG) < 100

We also know that both SOG and SOSOG were born some time since the year 500 AD:
Equation {e}:
500 < DOB(SOSOG) < 2004.

Based on equation {b} and {e} we can clearly see that there are only 5 solutions for y. They are y = 8, 9, 10, 11 or 12. Any other solutions for y are in breach of equation {e}.
For each of these possible solutions for y there is only a limited number of solutions for x that comply with {d}. They are:

 x     y
--------
21     8
22     8
26     9
27     9
31    10
36    11
41    12

Any other solutions for x and y are in breach of equation {d}.
SOG's age when his son was born can be calculated for each of these possible solutions by using equation {c} as follows:

 x      y     Age(SOG)
----------------------
21      8           84
22      8           42
26      9           70
27      9           18
31     10           60
36     11           60
41     12           76

So there are 7 different solutions, which is why Buzz said he couldn't solve for x and y.
Now comes the lateral part of the puzzle: Alvin informs Buzz that "...he was your age when his son was born". Of course, Buzz knows his own age. He should therefore be able to pick the correct solution from the list of 7 possible solutions shown above. However, he can't. That means that Buzz (and the special old guy) must be 60 because there are multiple solutions for an age of 60, whereas any other age would yield a unique solution. Any other age and Buzz would be able to solve.
A 60 year old father – quite a special old guy.

Switching Logic Puzzle Solution

Turn switch #1 ON. After about five minutes or so, turn switch #1 OFF and turn switch #2 ON.
Then go upstairs and check on the bulbs.

The one ON is obviously #2. The other ones are OFF, but one of them should be very hot by having been ON for five minutes. That's #1, and the remaining bulb is #3.

Boxes, Beads, and a Blindfold Puzzle Solution

The king puts 1 black bead in 3 of the 4 boxes and all the other beads (both black and white) in the fourth box.

In the kings' view, you will just randomly pick a box because you are so stupid. This gives you barely 1 chance out of 8 to pick a white bead (1/4 to pick that one box containing white beads multiplied with almost 1/2 to pick a white bead out that box).

Assuming each of the four boxes are identical, by picking up each box in turn, you will be able to tell by weight or the rattling noises which one of the boxes contains the mixed beads. Picking the box with the mixed beads will mean that you have a slightly better than 50% chance of living.

Thanks to "Ben Leil" and "Kobold" for posting solutions in the forum

Which Chest Is Which Puzzle Solution

Arthur does not need to open any chests.

Since all labels are on the wrong chests, the chest labelled GOLD OR SILVER COINS cannot contain either gold nor silver, so must contain bronze. Thus the chest labelled GOLD COINS must contain silver coins, and SILVER COINS must contain gold.

U2 Gig Puzzle Solution

The trick is to get the two slowest people to cross at the same time. One solution is...

• Bono and Edge cross the bridge for which they take 2 mins (Total time = 2)
• Then Bono comes back with the torch (Total Time = 2 + 1 = 3)
• Then Adam and Larry cross the bridge (Total time = 3 + 10 = 13)
• Then Edge comes back (Total = 13+2 = 15)
• Then both Bono and Edge cross the bridge (Time = 15+2=17)

Duck Hunt Puzzle Solution

Four ducks in a single row would do it.

The Devil's Muse Puzzle Solution

It's a palindrome.

The Diagram Puzzle Solution

Joe counted how many times each of the 7 LEDs lit up for all 10 digits...

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How Many Monks? Puzzle Solution

The way they work this out and the reason it happens on the same day is as follows. It is important to note that not only are the monks intelligent but they all know all their fellow monks are as well.
Start with one monk:

He knows he is the only one who can have the disease so he kills himself on
day 1.

Then two monks:
They know that either one of them has the disease or both do. If monk1
doesn't see the mark of the disease on monk2 then he will realise that he
must have it so will kill himself. If he sees the mark then he knows that
monk2 will following the same reasoning and kill himself. If the next day
the other monk is still alive then he realises that the other monk must have
seen the mark on him and so they both have the disease and he must kill
himself. Both follow the same reasoning and kill themselves on day 2.

Three monks:
If only one of the monks has the disease he will see no mark on the other
two and so diagnose himself. He will kill himself on day one. If two monks
have the disease then they will each see one monk with the mark and one
without. When they see each other again the next day they will deduce that
the monk they see with the mark would only not have killed himself if he
could see someone else with the mark. They know it is not the third monk so
it must be them. Both diseased monks follow the same reasoning and kill
themselves on day two. If all three monks have the disease then they are all
in the same position as the healthy monk in looking on in the previous
example. Each of them can see two diseased men. When they haven't both
killed themselves on day two there can be only one reason - the viewing monk
must have the mark as well. All taking the same reasoning they all kill
themselves on day three.

And so on...
N monks all with the disease will all kill themselves on day N.

The Geneaology Puzzle Solution

Page 1 in a book is the page just inside the front cover. Page 2 is the other side of the same sheet.

Pages 1201 and 1202 are opposite sides of the same sheet of paper, so finding something between these two is highly improbable.

How Old is the Vicar? Puzzle Solution

The vicar is 50.

The way to solve this puzzle, is to first of all write down all the possible permutations of three numbers whose product is 2450.

Starting Numbers	Product	Sum	Choirmaster
1, 1, 2450	2450	2452	1226
1, 2, 1225	2450	1228	614
1, 5, 490	2450	496	248
1, 7, 350	2450	358	179
1, 10, 245	2450	256	128
1, 14, 175	2450	190	95
1, 25, 98	2450	124	62
1, 35, 70	2450	106	53
1, 49, 50	2450	100	50
2, 5, 245	2450	252	126
2, 7, 175	2450	184	92
2, 25, 49	2450	76	38
2, 35, 35	2450	72	36
5, 5, 98	2450	108	54
5, 7, 70	2450	82	41
5, 10, 49	2450	64	32
5, 14, 35	2450	54	27
7, 7, 50	2450	64	32
7, 10, 35	2450	52	26
7, 14, 25	2450	46	23

Since the choirmaster, after being told that the product of the ages is 2450 and that the sum is twice his age, still can't work out the ages, we can deduce that there are two (or more) combinations with the same sum. Those combinations have been highlighted in the table above.

The vicar then claims to be older than all of them. The oldest of the three is 49 in the first remaining combination, and 50 in the other. The choirmaster knows the vicar's age, and after his claim, he deduces everyone's age. The only way he's able to do so is if the vicar is 50, leaving the combination 7, 7, 50 logically impossible (the vicar has to be older, that is at least 1 year older than the others), and therefore learning that the people's ages are 5, 10, and 49.

The Maze Puzzle Solution

If there are only T junctions, then all you have to do to get back is take all of the side passages that you come across, as they are the T junctions that were taken from your original perspective.

Effectively, every side passage you come across was a T junction that was taken earlier.

Of course, defining this the "ultimate" maze was a bit of an exaggeration on our part...

Four Hats Puzzle Solution

C saves the day.

D clearly has the most information at his hands, but seeing one white and one black hat doesn't give him any certainty about his own hat's colour. Would B and C both have been wearing the same colour, D would have been able to provide the answer.

But C is one clever guy and he knows that if D doesn't answer, it means that B is wearing a different colour than him. Because B is wearing white, C knows he's wearing black.

Note that A is redundant: the puzzle could have included only B, C, D. That way, the hats would have been three, with two hats of the same unspecified colour, and one other hat of the opposite colour.

Alan and Bert Puzzle Solution

Starting Numbers	Sum	Product	Product can also be made using
6, 5	11	30	10, 3
7, 4	11	28	14, 2
8, 3	11	24	6, 4 and 12, 2
9, 2	11	18	6, 3 and 9, 2

Bert knows that there are four possibilities for the starting numbers - so it is impossible to work out the starting numbers using the sum alone.

As mentioned in the question, he's quite clever - so he looks at the product that would appear if he used each of the four possible combinations. As shown in the table above, the product that appears can also be made from different numbers.

So, he announces that it is impossible for him or Alan to work out what the original numbers were. He seems to be on fairly safe ground.

Alan is a little bit more devious as well as being clever. Armed with this snippet of information he needs to look for a pair of numbers that give a non-unique sum and also give a non-unique product.
11 is the only non-unique sum which always gives a non-unique product. Alan is very clever and very smug.

Weighing an Elephant Puzzle Solution

Load the elephant onto a boat large enough to carry it. The boat will sink slightly, and you mark the level of the water on the side of the boat. Then you offload the elephant and fill the boat with bags until the boat sinks to the level marked. The bags can be individually weighed using beam scales and the weight of the elephant is the sum of the weight of the bags.

This puzzle is slightly cunning in that it the geographic location of the city is a small clue.

Carpet Layer Puzzle Solution

The only measurement that needs to be taken is the distance between opposite angles of the rectangular dance floor (figure on the left). That distance then will be squared (figure on the right) and doubled to get the sum of the areas of the 4 squares.

The Farmer's Problem Puzzle Solution

There are two solutions:

Solution A:
1) John takes the goat to the other side, and leaves it there.
2) He then takes the wolf to the other side.
3) He brings the goat back.
4) He takes the cabbages across, leaving them with the wolf.
5) John Comes back for the goat.

Solution B:
1) John takes the goat to the other side, and leaves it there.
2) He then takes the cabbages to the other side.
3) He brings the goat back.
4) He takes the wolf across, leaving it with the cabbages.
5) John Comes back for the goat.

Loadsa Coins! Puzzle Solution

You reach out and pick any 20 coins, and turn each of them over. The 20 coins is one group, and the remaining coins the other.

"Bugger," says Greed. "Why didn't I think of that?"

"It's so simple, yet so difficult to think of," you say. "Now can I have my eyesight back? Hello? Hello......!?"

The Snail And The Well Puzzle Solution

After traveling 3 feet up, then sliding 2 feet down, the snail had a net gain of 1 foot per day and night. One would think, 1 foot per day and night, 30 foot deep well, 30 days and nights! But that's wrong.

After 26 days and nights, the snail is 26 feet up the well. So the snail starts the 27th day at 26 feet. At the end of the 27th day, the snail has travelled to 29 feet, but during the 27th night, it slides down to 27 feet. So, on the 28th day, the snail travels 3 feet up and reaches the top of the well.

So, the final answer is 28 days and 27 nights.

Sailor's Delight Puzzle Solution

We asked you readers to send us a solution to this puzzle, and we kept an open mind about it.
The first person with a simple, elegant, and, in our opinion, valid solution, is Saurabh Gupta, a faithful reader and avid puzzler. Here is Saurabh's solution.

How many pirates will be thrown into the sea? None. And the correct distribution is:

Pirate with rank	Number of coins
1	0
2	1
3	0
4	1
5	0
6	1
7	0
8	1
9	0
10	96

Pirate 10 divided the coins. He will get the votes of pirates 2, 4, 6, 8, and himself. This is taking into consideration what each of the pirates will get if this plan is not passed.

Starting with a situation when there is only pirate 1. He keeps all the 100 coins for himself and live happily by passing the division with his only vote.

In case that there are two pirates, pirate 2 divides and he keeps 100 coins for himself while giving none to pirate 1. He still gets the division passed with his vote and live happily ever after.
In case there are three pirates, pirate 3 divides and gives pirate 1 a single coin and keeps the other 99 coins for himself. Pirate 1 would now vote in his favor because if he votes against, then pirate 2 would get a chance to divide and would keep all the loot for himself.

If four pirates are present, pirate 4 divides and now gives pirate 2 a single coin to gain his vote (who otherwise gets nothing if pirate 3 has a chance to divide the coins). In this case, pirates 1 and 3 get nothing.

Therefore, in a similar manner, the distribution when there is an extra pirate is achieved as follows:

With 5 pirates, pirate 5 divides: Pirate with rank Number of coins 1 1 2 0 3 1 4 0 5 98	With 6 pirates, pirate 6 divides: Pirate with rank Number of coins 1 0 2 1 3 0 4 1 5 0 6 98	With 7 pirates, pirate 7 divides: Pirate with rank Number of coins 1 1 2 0 3 1 4 0 5 1 6 0 7 97
With 8 pirates, pirate 8 divides: Pirate with rank Number of coins 1 0 2 1 3 0 4 1 5 0 6 1 7 0 8 97	With 9 pirates, pirate 9 divides: Pirate with rank Number of coins 1 1 2 0 3 1 4 0 5 1 6 0 7 1 8 0 9 96	With 10 pirates, pirate 10 divides: Pirate with rank Number of coins 1 0 2 1 3 0 4 1 5 0 6 1 7 0 8 1 9 0 10 96

This solution holds because all pirates are rational and think through the situation. They understand that they don't get anything if the next ranked pirate gets a chance to make the division.

There is, however, still some doubt about this solution. After all, it was said that pirates are smart. With this solution, based on a method of dividing the booty supposedly approved by all, it is shown that only one (the last one) is very smart, while 4 of them get only a coin, and 5 of them get nothing. This seems to be in conflict with the statement that pirates are smart. Even if the method was approved only by half of them (the ones who get at least a coin), 4 of these voters don't seem very smart if they get only one coin. Certainly, it couldn't have been a dictatorship decision taken by the leader, as he ends up empty-handed.

An alternative, to make the pirates look brighter, is that the dividing pirate actually shares the loot evenly between the pirates likely to give him a positive vote. In that case, the pirates who previously got only one coin, end up getting just as much as the lowest ranked, while the others get nothing again. This would entail the following solution:

Pirate with rank	Number of coins
1	0
2	20
3	0
4	20
5	0
6	20
7	0
8	20
9	0
10	20

This alternative solution, however, defies a bit the mechanism of the division logic and process explained in the main solution. Perhaps, it would be fair to make a re-wording of the puzzle, stating that, although pirates are smart, the method of dividing the booty is forced on them by some external entity, and no pirate, not even the leader, can oppose to it. However, a rewording of the puzzle will NOT take place, as this is how it was presented on Brent's page, and we are going to preserve it.

Also, it might be possible that this is not the only acceptable solution, and another alternative solution might exist, that doesn't raise any doubts with the puzzle premises.

So, we are still asking you: send us your solution!

12 Balls Puzzle Solution

We found a good solution to this puzzle on the Cut The Knot website. They show a solution taken from the book Mathematical Spectrum, by Brian D. Bundy. Bundy shows two solutions: the first one requires different courses of action depending on the outcome of previous weighings, so it is not particularly elegant or easy to remember. The second solution involves a fixed course of action in all circumstances, and is the one that follows.

In this method we weigh four specified balls against four other specified balls in each of the three weighings and note the result. If we observe say the left hand side of the balance, then for an individual weighing there are three possible alternatives: the left hand side is heavy (>), light (<) or equal (=) as compared with the right hand side of the balance.

Since three weighings are allowed, the number of different results that can be obtained is just the number of arrangements (with repetitions allowed) of the three symbols >, <, =, i.e. 3 3 = 27. If we use all twelve balls in the three weighings, and ensure that no particular ball appears on the same side of the balance in all three weighings, the outcomes >>>, <<<, === are not possible. We thus have only 24 possible outcomes and we shall show that it is possible to set up a one to one correspondence between these 24 outcomes and the conclusion that a particular ball among the twelve is heavy or light.

The 24 outcomes can be divided into two groups of twelve in each group. If we call the reverse of an outcome the outcome obtained by replacing > by <, < by > and leaving = unchanged, one group of twelve will be the reverses of the other group and vice versa. We can thus write the 24 outcomes in the form of two arrays, each array having three rows (the three weighings) and twelve columns (the twelve balls), so that each row contains four >'s, four <'s and four ='s. Thus we have, for example,
> > > >   < < < <   = = = =
< = = =   < < > >   > < > =
= > < =   > = < >   > = < <
A B C D   E F G H   I J K L
< < < <   > > > >   = = = =
> = = =   > > < <   < > < =
= < > =   < = > <   < = > >   We consider just the top array, and for each weighing (row) place the balls corresponding to a > in the left, and the balls corresponding to a < in the right of the balance. Thus we would weigh A, B, C, D against E, F, G, H; then G, H, I, K against A, E, F, J; and finally B, E, H, I against C, G, K, L. The results of these three weighings as observed on the left of the balance are noted. If the outcome is >, <, = we conclude that ball A is heavy; if >, =, > ball B is heavy, etc; if =, =, < ball L is heavy. If we obtain an outcome that appears in the lower array, we conclude that the corresponding ball is light. Thus for <, >, = ball A is light, etc; =, =, > means ball L is light.

Hugh's Horses Puzzle Solution

(0.5 + 0.5) + (0.5 + 1.5) + (0.5 + 3.5) = 7.

Or here's a an algebraic solution kindy submitted by Greg Bradshaw. Thanks!

Solve for x where x is the total number of horses:

x = (.5x + .5) + (.5(.5x - .5) + .5) + (.5[.5{.5x - .5} - .5] + .5)
x = .5x + .5 + .25x - .25 + .5 + .5(.25x - .25 -.5) + .5
x = .5x + .5 + .25x + .25 + .5(.25x - .75) + .5
x = .5x + .5 + .25x + .25 + .125x - .375 + .5
x = .5x + .5 + .25x + .25 + .125x + .125
x = .875x + .875
.125x = .875
x = .875/.125
x = 7

The Dog Situation Puzzle Solution

One of the dachshunds is owned by Bob and is named Alec, while the other is owned by Charlie and is called David.

To get to the solution, let's gather all the information we're given:

1 Each boy has named his dogs after two of his brothers
2 Each boy has two doggy namesakes
3 There are 3 cocker spaniels, 3 terriers, and 2 dachshunds
4 None of the four boys owns two dogs of the same breed
5 No two dogs of the same breed have the same name
6 Neither of Alec's dogs is named David
7 Neither of Charlie's dogs is named Alec
8 No cocker spaniel is named Alec
9 No terrier is named David
10 Bob does not own a terrier 

With this info, it follows that:

11 Alec's dogs are named Bob and Charlie (from 6 and 1)
12 Charlie's dogs are named Bob and David (from 7 and 1)
13 David's dogs are named Alec and Charlie (from 11, 12, 1 and 2)
14 Bob's dogs are named Alec and David (from 11, 12, 13, 1 and 2)
15 Bob's dogs are a cocker and a dachshund (from 3, 4 and 10)
16 The 3 cockers are called Bob, Charlie, and David (from 3, 5 and 8)
17 The 3 terriers are called Alec, Bob, and Charlie (from 3, 5 and 9)
18 The 2 dachshunds are called Alec and David, because the names Bob and Charlie are already taken by cockers and terriers (from 2, 3, 16 and 17)
19 Bob's dachshund must be called Alec, because there are no cockers with that name (from 14, 15, 16 and 18)
20 Bob's cocker is called David (from 14, 15, 16, 18 and 19)
21 The dachshund called David must belong to Charlie (from 12, 18 and 20)

Deductions 19 and 21 have given us the solution to the puzzle.

Who Owns The Zebra? Puzzle Solution

The German owns the zebra!

	#1	#2	#3	#4	#5
Nationality	Norwegian	Dane	English	German	Swede
Colour	Yellow	Blue	Red	Green	White
Drink	Water	Tea	Milk	Coffee	Beer
Cigarettes	Dunhill	Blend	Pall Mall	Prince	Blue Master
Pets	Cats	Horse	Birds	Zebra	Dog

Socks And Shops Puzzle Solution

For each of the twenty pairs of socks, they unclipped it and gave one sock each.

Strawberry Ice Cream Puzzle Solution

First, determine all the ways that three ages can multiply together to get 72:

• 72 1 1 (quite a feat for the bartender)
• 36 2 1
• 24 3 1
• 18 4 1
• 18 2 2
• 12 6 1
• 12 3 2
• 9 4 2
• 9 8 1
• 8 3 3
• 6 6 2
• 6 4 3

As the man says, that's not enough information; there are many possibilities.

So the bartender tells him where to find the sum of the ages--the man now knows the sum even though we don't. Yet he still insists that there isn't enough info. This must mean that there are two permutations with the same sum; otherwise the man could have easily deduced the ages.

The only pair of permutations with the same sum are 8 3 3 and 6 6 2, which both add up to 14 (the bar's address). Now the bartender mentions his "youngest"--telling us that there is one child who is younger than the other two. This is impossible with 8 3 3--there are two 3 year olds. Therefore the ages of the children are 6, 6, and 2.

Pedants have objected that the problem is insoluble because there could be a youngest between two three year olds (even twins are not born exactly at the same time). However, the word "age" is frequently used to denote the number of years since birth. For example, I am the same age as my wife, even though technically she is a few months older than I am. And using the word "youngest" to mean "of lesser age" is also in keeping with common parlance. So I think the solution is fine as stated.

Hansel And Gretal Puzzle Solution

One solution is 010011, and is probably the shortest. In a repeating series of this pattern, we may get:

TV Show Puzzle Solution

This puzzle isn't particularly new, but it became very well known in 1990, thanks to the person that allegedly had "the highest IQ ever recorded" -- 228, according to The Guinness Book of WorldRecords. Miss Marilyn vos Savant (yes, it was a female) wrote a solution to this puzzle on her weekly column, published in a popular magazine. This solution led to waves of mathematicians, statisticians, and professors to very heated discussions about the validity of it.

The contestant, according to vos Savant, had a greater chance of winning the car by switching his pick to the 2nd door.

She claimed that by sticking to the first choice, the chances of winning were 1 out of 3, while the chances doubled to 2 out of 3 by switching choices. To convince her readers, she invited her readers to imagine 1 million doors instead of just 3. "You pick door number 1," she wrote, "and the host, who knows what's behind every door, and doesn't want you to win, opens all of the other doors, bar number 777,777. You wouldn't think twice about switching doors, right?"

Most of her readership didn't find it as obvious as she thought it was... She started receiving a lot of mail, much of it from mathematicians, who didn't agree at all. They argued that the chances were absolutely equal, whether or not the contestant switched choice.

The week after, she attempted to convince her readers of her reasoning, by creating a table where all 6 possible outcomes were considered:

Door 1	Door 2	Door 3	Outcome (sticking to door 1)
Car	Nothing	Nothing	Victory
Nothing	Car	Nothing	Loss
Nothing	Nothing	Car	Loss
Door 1	Door 2	Door 3	Outcome (switching to other door)
Car	Nothing	Nothing	Loss
Nothing	Car	Nothing	Victory
Nothing	Nothing	Car	Victory

The table, she explained, shows that "by switching choices you win 2 out of 3 times; on the other hand, by sticking to the first choice, you win only once out of of 3 times".

However, this wasn't enough to silence her critics. Actually, it was getting worse.

"When reality seems to cause such a conflict with good sense," wrote vos Savant, "people are left shaken." This time, she tried a different route. Let's imagine that, after the host shows that there's nothing behind one of the doors, the set becomes the landing pad for a UFO. Out of it comes a little green lady. Without her knowing which door was picked first by the contestant, she is asked to pick one of the remaining closed doors. The chances for her to find the car are 50%. "That's because she doesn't have the advantage enjoyed by the contestant, ie the host's help. If the prize is behind door 2, he will open door 3; if it is behind door 3, he will open door 2. Therefore, if you switch choice, you will win if the prize is behind door 2 or 3. YOU WIN IN EITHER CASES! If you DON'T switch, you'll win only if the car is behind door 1".

Apparently, she was absolutely right, because the mathematicians reluctantly admitted their mistake.

The Package Puzzle Solution

With the object in the box, G locks it with one of his own locks. He then sends the box to H. H attaches his own lock in addition to G's and then sends it back. G removes his lock and then sends it to H again.

Trick Mules Puzzle Solution

This ambiguous "Trick Mules Puzzle" is solved by the realisation that the mule can have two different orientations. Here the same lines and contours have two interpretations, one horizontal and one vertical.

Camel Race Puzzle Solution

The Great Sage told them to swap camels. It is the second camel, not person, to reach the city gates who wins.

Sherwood Forest Puzzle Solution

Since the order of turns was clockwise, Brother Tuck went to position B, and deliberately missed his firts shot (perhaps he should shoot the raven).

Assuming that missing the first shot gives Brother Tuck the best chance that one of the dangerous adversaries gets killed from the other before becoming himself the target (obviously, both Miller and Allan-a-Dale opt to use their first turn to shoot at the most dangerous adversary), it is necessary to consider the chances of shooting from position A and the chances from position B.

By assuming position at A, and missing the first shot (as it is right in the circumstances), Brother Tuck knows that he wouldn't be Miller's target, because Allan-a-Dale is more dangerous. Allan-a-Dale would certainly be killed (Miller's crossbow has 100% chances of hitting the target). At this point, it would again be the monk's turn (with Allan not being in a condition to take his turn), who has a 50% chance of hitting the other survivor, ie Miller, and to therefore end the duel. Obviously, if the religious man misses, then he could only get on his knees and pray, because Miller's next shot would be faultless. So, by choosing position A, Brother Tuck has a 50% chance of survival.

If Brother Tuck chooses position B, and deliberately misses his first shot, the next turn would be Allan-a-Dale's, who would choose the most dangerous adversary, ie Miller. At this point, two different outcomes can happen:

1. Allan misses Miller (25% chances); at this point, Miller will use his turn to shoot Allan. Then it will be Tuck's turn, who will have one chance out of two to pick one of the good arrows (so, it's 50% chances, but it's 50% of 25%, ie 1/8).

2. Allan hits Miller (75% chances); at this point, Brother Tuck will shoot Allan, with 50% chances of success; but this is 50% of 75%, ie 3/8 which, if added to 1/8 of the first outcome, gives Tuck 4/8, ie 50% chances of survival. In case that his shot misses, Tuck would not be a sure victim of Allan's next shot, who might miss with 25% chances. This gives the monk a further 3/32 probabilities (ie 3/4 * 1/2 * 1/4).

The calculation, as shown here, could keep going on and on until the last arrow, and demonstrates that, by choosing position B, Brother Tuck has, besides the 50% chances also offered by position A, a long string of small chances (3/32, 3/256, etc), that are possible if Allan-a-Dale misses at least one (his first, and eventually the next ones) shot at the monk.