Web Puzzles

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.

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.

General Manoeuvres Puzzle Solution

The manoeuvre was conducted this way:

1. The two boys cross to the opposite bank (10 mins)
   
2. One of them stays there while the other comes back (5 mins)
   
3. The boy gets off the boat, a soldier jumps on board and crosses the river (8 mins)
   
4. The soldier gets off, and the boat returns with the other boy (5 mins)
   

This operation required 28 minutes. The sequence had to be repeated as many times as the number of men in the platoon, ie 39 more times. However, it was needed to subtract 5 minutes from from the total: when the last man of the platoon crossed, the time (5 mins) taken by the second boy to cross back must not be counted, as the last soldier had already reached the other bank of the river.

The total time was therefore [(28 * 40) - 5] = 1115 minutes, which amounts to 18 hours and 35 minutes.

Mike Horan points out that it can be done faster if you leave both boys stranded on one side with the boat on the other. The first 39 soldiers cross at 28 minutes each (1092 minutes). You then have the two boys plus the last solider on one side. The final soldier then rows across himself, hence 1092 + 8 = 1100 minutes.

Poker Results Puzzle Solution

The 5 people and their wins/losses are:

Person #1: Barbara (+ £4).
Person #2: Edward (- £5).
Person #3: Alice (+ £14).
Person #4: Daniel (- £7).
Person #5: Claire (- £6).

This result is obtained from the statements of the 5 people.

• Since the sisters-in-law (#1 and #5) lost, wile #5 and Alice have won, then #3 is Alice.
  
• Alice won £14, because £2 lost by the sister-in-law, and £12 lost by the men.
  
• #1 is Barbara, as she talks about Claire.
  
• Alice is the only child, therefore the sisters-in-law must be sisters of the husband (Daniel).
  
• Daniel lost £7, as the bunch of brothers and sisters have lost a total of £9, of which only £2 was lost by the sisters.
  
• Edward lost £5, as the two men lost a total of £12, of which £7 was lost by Daniel.
  
• Barbara - Edward's wife because Claire is single - won £4, as the total lost by the couple Barbara-Edward is £1.
  
• Claire lost £6, as the total loss of the two sisters is £2, but Barbara won £4.

Snake Puzzle Solution

Counters 1 and 8 need to be in the middle since they each have only one number that they cannot be neighbours with. Hence 7 and 2 must go in the sidewings, and the rest is trivial.

	3	5
7	1	8	2
	4	6

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!