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)

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

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.

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...

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.

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.

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.

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.

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.

Golf Puzzle Solution

The clues given in the puzzle are:

• The Stock Broker is not Levi.
• Jack is not the Lawyer; neither Jack nor the Lawyer owns the Corvette.
• The Musician owns the Porsche; therefore no one else owns the Porsche and the Musician owns no other car.
• Seth is not the owner of the Cadillac, the Corvette, nor is he the Doctor; also, the Doctor does not own either the Cadillac or the Corvette either.

By Reasoning:

• The Doctor, by elimination, must own the Ferrari; the Lawyer must own the Cadillac; the Stock Broker must own the Corvette.
• Tne Stock Broker owns the Corvette. Neither Seth nor Jack own the Corvette, so the Stock Broker must be Robert; since the Stock Broker drives the Corvette, then Robert drives the Corvette.
• The Lawyer owns the Cadillac. Seth does not own the Cadillac; therefore Seth is not the Lawyer; therefore Seth is the Musician; the Musician owns the Porsche; therefore Seth owns the Porsche.
• By elimination, Jack is the Doctor; the Doctor owns the Ferrari; therefore Jack owns the Ferrari.
• By elimination, Levi owns the Cadillac.

The Solution:

• Seth is the Musician and owns the Porsche.
• Levi is the Lawyer and owns the Cadillac.
• Jack is the Doctor and owns the Ferrari.
• Robert is the Stock Broker and owns the Corvette.

Mountaineer Puzzle Solution

Of course there is. To make sure, imagine two mountaineers: one is in the village, and the other one is at the refuge. They'll both leave at eight o'clock, travel along the same path as our mountaineer, and at his same speed. At some point along the path they'll obviously meet.

Triangular Paradox Puzzle Solution

You will have tried to calculate the areas of everything involved. The red triangle has an area of 12 squares, the blue triangle has 5 squares, the orange shape has 7, and the green shape has 8. The sum of all these parts is 12 + 5 + 7 + 8 = 32 squares.

You might have thought that the large, enclosing figure was (5 × 13) / 2 = 32.5, and wondered how there is a 0.5 square discrepency.

However, this is not the case. The enclosing figures are not actually a triangles at all: the blue triangle has a gradient of 2 / 5 = 0.4, and the red triangle has a gradient of 3 / 8 = 0.375. The thick black lines aid in deceiving the eye into seeing the 'hypotenuse' as a straight line.
Figure 3 shows the differing gradients with exaggerated shapes. Figure 4 shows how Figure 2's "bulge" in its hypotenuse made space for the hole.

OK, so it's a bit of a trick. But you may note that I never said those big "triangles" were actually triangles!

Job Interview Puzzle Solution

"You set light to both ends of the rope 1 and just one end of rope 2. It will take half an hour for the two burning ends of rope 1 to meet. Then you set light to the remaining end of rope 2. The time it will take for rope 2 to finish burning will be a further 15 minutes. Hence all together, both ropes burned in this manner will take 45 minutes to burn." Tom leaned back in his chair and folded his arms.

That squinty eye stared at him some more. "Very well," stated Patrick after some time. "We will let you know about the success of your application shortly." He stood and shook Tom's hand.
As Tom left the room, he grinned again. "Elderberries," he muttered to himself.