Web Puzzles

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.

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.

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

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

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:

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.

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.

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.

Napoleon's Star Puzzle Solution

To be able to place nine coins, it is necessary to make the 3rd point of each step equal to the start point of the previous step. For example:

a-g; i-a; c-i; f-c; e-f; h-e; b-h; j-b; d-j.


 With such a simple solution, it's hard to believe that Napoleon stayed in charge for so long.

Southern Cross Puzzle Solution

The missing number is 18. The numbers in the bottom row are the square of the numbers in the top row, but with their digits reversed.

4	5	6	7	8	9
61	52	63	94	46	18

Deneb Puzzle Solution

Visit At The White House Puzzle Solution

Person	Buttons
President	X	X	X	X	X	X	X
Vice President	X	X	X	X	X	X	-
President of Senate	X	X	X	-	-	-	X
Secretary of State	X	-	-	X	X	-	X
Chief of Armed Forces	-	X	-	X	-	X	X
Dean of Harvard	-	-	X	-	X	X	X

Frogs Puzzle Solution

The only frog is Amanda. The table below shows, on the left, all possible combinations of frogs (X stands for frog, - stands for non-frog), while on the right pane there is the control of the truthfullness (T for true, F for false) of each given statement, for each combination of frog population. The only combination where every girl tells one true statement and one false one, is the emphasised row. Hence, there's only one frog, and that's Amanda.

Possible Frogs				Statement
A	B	C	D	A		B		C		D
-	-	-	-	F	F	T	F	F	F	F	T
-	-	-	X	F	F	T	F	T	F	F	F
-	-	X	-	F	F	T	F	T	T	F	T
-	-	X	X	F	F	T	T	F	T	T	F
-	X	-	-	F	F	F	F	T	F	F	T
-	X	-	X	F	F	F	F	F	F	T	F
-	X	X	-	F	F	F	T	F	T	T	T
-	X	X	X	F	T	F	T	F	T	F	F
X	-	-	-	T	F	T	F	T	F	F	T
X	-	-	X	T	F	T	F	F	F	T	F
X	-	X	-	T	F	T	T	F	T	T	T
X	-	X	X	T	T	T	T	F	T	F	F
X	X	-	-	T	F	F	F	F	F	T	T
X	X	-	X	T	T	F	F	F	F	F	F
X	X	X	-	T	T	F	T	F	T	F	T
X	X	X	X	T	T	F	T	F	T	F	F

Swapping Art Puzzle Solution

The swap is unfair. If we abbreviate with an E the drawings by Ernst, an M for Magritte's sketches, and a B for Bacon's paintings, we can rewrite the equations as stated by the Italian dealer as:

1. EEEE = MMMMM (both dealers agreed on this)
2. MME = BB (again, both dealers agreed on this too)
3. MMMBB = EEEEM (the French dealer wasn't sure of this)

Now, if for the 3rd equation we substitute BB with MME (from equation 2), we'll have a 4th equation: MMMMME = EEEEM. But the 1st equation states that MMMMM = EEEE, so it is possible to cancel out, in equation 4, MMMMM on the left and EEEE on the right. This would leave us with the equation E = M, which does not fit with equation 1, where E is obviously greater than M.


The problem gives us 3 equations:
    1. 4x = 5y
    2. 2y + x = 2z
    3. 3y + 2z = 4x + y

Equation 2. can be rewritten as:
    4. x = -2y + 2z

Equation 3. can be rewritten as:
    5. 4x = 2y + 2z

Adding equations 4. and 5. gives us:
    6. 5x = 4z

Combining 1. and 6., we get:
    7. 4x = 5y = (16/5)z

If we convert the offer (3.) in terms of y, we get:
    8. 3y + 2(25/16)y = 5y + y
    ie 3y + (25/8)y = 6y
    ie 6.125y = 6y

The last equation is not absolutely true, which proves the swap is unfair.

100 Puzzle Solution

• 111 - 11 = 100
• (3 * 33) + (3 / 3) = 100
• (5 * 5 * 5) - (5 * 5) = 100
• (5 + 5 + 5 + 5) * 5 = 100
• (11 - 1) ^ ( 1 + 1) = 100 [Thanks to Steven Renich for that one!]
• ((2 * 2 * 2) + 2) ^ 2 = 100 [Thanks to David Cohen for this other one!]
• ((99 * 9) + 9) / 9 = 100 [Thanks to Taylor Lowry for this other one!]
• ((22 - 2) / 2) ^ 2 = 100 [Thanks to Karen D. Miller for this other one!]
• 6! / 6 - 6! / (6 * 6) = 100 [Thanks to Karen D. Miller for this other one!]
• 5! - 5 - 5 - 5 - 5 = 100 [Thanks to Karen D. Miller for this other one!]
• 5! - (5 + 5 + 5 + 5) = 100 [Thanks to Jim St. Clair for this other one!]
• 5 * 5 * (5 - 5 / 5) = 100 [Thanks to Rishi Mohan Sanwal for this one!]
• 4! + 4! + 4! + 4! + 4 = 100 [Thanks Karen D. Miller and Saurabh Gupta!]
• 99 + 9 ^ (9 - 9) = 100 [Thanks Gopalakrishnan Thirumurthy for this one!]
• 5! - 5 * (5 - 5 / 5) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
• (4! + 4 ^ (4 - 4)) * 4 = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
• 4! * 4 + 4 - 4 + 4 = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
• (4! * 4) + (4 * 4 / 4) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
• 5 * 5 * (5 - (5 - 5)!) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
• ((9 - 9) */ 9)! + 99 = 0! + 99 = 1 + 99 = 100 [Thanks to Bala Neerumalla!]
• (3 - 3)! + 33 * 3 = 100 [Thanks Bala Neerumalla for this other one!]
• (5 + 5) ^ ((5 + 5) / 5) = 100 [Thanks Bala Neerumalla for this one!]
• ((2 ^ 2) * 2 + 2) ^ 2 = 100 [Thanks Bala Neerumalla for this one!]
• (4! + ((4 - 4) */ 4)!) * 4 = 100 [Thanks Bala Neerumalla for this one!]

Forum member Bealzbob reminds us that, whenever we have a subtraction like A - B, we can rewrite it as an addition with a negative number, like A + (-B). In the case of the first solution, 111 - 11 = 111 + (-11).

Glen Parnell points out that, in any solution in which the digit 2 is used, it could be replaced by 2!, whenever the 2 is not only a digit but also a number, as 2 = 2! (but of course, 22 is not equal to 22!).
Bala Neerumalla has suggested some alternative solutions using trigonometric functions Sin() and Cos():

• Sin(99 - 9) + 99 = Sin(90) + 99 = 1 + 99 = 100
• Cos((9 - 9) */ 9) + 99 = Cos(0) + 99 = 1 + 99 = 100
• Cos(3 - 3) + 33 * 3 = 100
• (4! + (Sin(4-4)*4)!)*4 = 100

Gopalakrishnan Thirumurthy has intelligently played around with cubic root functions. If we were to represent this function with a name like, say, CR(), the following alternative solutions could be achieved:

• 88 + 8 + CR(8) + CR(8) = 100
• (8 + CR(8)) ^ (CR(8) * (8 / 8)) = 100
• ((CR(8) * CR(8) * CR(8)) + CR(8)) ^ CR(8) = 100
• ((8 * (8 / 8)) + CR(8)) ^ CR(8) = 100

However! If the cubic root function is represented as it should be, ie ³Ö, then there is the extra "3" which invalidates the solution. These alternative solutions using CR(), have therefore been given as possible examples that will not be accepted in the future. Unless, of course, the "3" in ³Ö is actually one of the five digits used in a solution. Same goes with any other root or power. This also applies to any usage of the symbol "%", as it really involves the number 100.
Bala Neerumalla smartly noticed that he can produce number 100 by using a different number base than the normal base 10 (b10). The following example is in base 5 (b5):

• 444 - 44 = (4 * 5^2 + 4 * 5^1 + 4 * 5^0) - (4 * 5^1 + 4 * 5^0) = 100

However! The above equation mixes up two different number bases: the left hand side is b5, while the right hand side is b10 (100 b5 is equal to 25 b10). This is the reason why we will not be posting anymore solutions that mix different number bases. We will, however, accept solutions that use number bases different from 10, as long as they are consistent on both sides of the equation. Therefore, we will accept:

• 111 - 11 = 100 [Wolfgang Solfrank pointed out that this works in any number base, thanks!]

Here are some more examples of equations using different number bases:

• b16 4 ^ 4 +- 4 * (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
• b16 (2 * 2 * 2 * 2) ^ 2 = 100 [Thanks to Glen Parnell for this one!]
• b8 4 * 4 * 4 +- (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
• b8 (4 ^ 4) / 4 +- (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
• b8 4 * 4 * 4 * (4 / 4) = 100 [Thanks to Glen Parnell for this one!]
• b8 (2 ^ (2 * 2)) * 2 * 2 = 100 [Thanks to Glen Parnell for this one!]
• b2 11 - 1 + 1 + 1 = 100 [Thanks to Glen Parnell for this one!]

Furthermore, Glen Parnell noticed that there exists a general rule that applies to equations using any number base:

• bX+1 ((X - X) */ X)! + XX = 100

Here are some examples of it:

• b16 ((F - F) */ F)! + FF = 100
• b12 ((B - B) */ B)! + BB = 100
• b10 ((9 - 9) */ 9)! + 99 = 100 [As already shown by Bala Neerumalla]
• b8 ((7 - 7) */ 7)! + 77 = 100
• b4 ((3 - 3) */ 3)! + 33 = 100
• b2 ((1 - 1) */ 1)! + 11 = 100

Jeff Shall cleverly discovered that he can produce 100 using the Roman numeral 'L', which corresponds to number 50:

• ((L / L) + (L / L)) * L = ((50 / 50) + (50 / 50)) * 50 = 100

Unashamedly based on the above solution by Jeff Shall, we found that we can also use the Roman numeral 'C', which corresponds to number 100:

• (C / C) * (C / C) * C = (100 / 100) * (100 / 100) * 100 = 100
• (C / C) - (C / C) + C = (100 / 100) - (100 / 100) + 100 = 100
• CCC - CC = 300 - 200 = 100
• CC - CC + C = 200 - 200 + 100 = 100

Gopalakrishnan Thirumurthy has discovered a lot of new combinations with Roman numerals, including numeral 'X', which is number 10:

• (L + L) * L ^ (L - L) = (50 + 50) * 50 ^ (50 - 50) = 100
• (L + L) / L ^ (L - L) = (50 + 50) / 50 ^ (50 - 50) = 100
• XX * X - X * X = 20 * 10 - 10 * 10 = 100
• (X + (X / X)) * X - X = (10 + (10 / 10)) * 10 - 10= 100
• X * X * (X ^ (X - X)) = 10 * 10 * (10 ^ (10 - 10) = 100
• X * X / (X ^ (X - X)) = 10 * 10 / (10 ^ (10 - 10) = 100
• C * C * C / C * C = 100 * 100 * 100 / 100 * 100 = 100
• (C - C) + (C - C) + C = (100 - 100) + (100 - 100) + 100 = 100
• L * (L - L) + L + L = 50 * (50 - 50) + 50 + 50 = 100
• C * (CC / C) - C = 100 * (200 / 100) - 100 = 100
• C * (C / C) + (C - C) = 100 * (100 / 100) + (100 - 100) = 100

Gopalakrishnan Thirumurthy has also noticed that all the solutions involving five times the number 5, could be rewritten by substituting it with Roman numeral 'V'. This is also true if, instead of number 5 and numeral 'V', we were talking about number 1 and Roman numeral 'I'.
Now, how about some solution using Roman numerals 'D' (500), or 'M' (1000)? If you find any alternative solutions, get in touch with us by email!

Easy!

A Law-Abiding Citizen Puzzle Solution

The electrician packed the tube diagonally into a flat-ish squared package, with sides of less than one metre. More precisely, the sides were about 0.85 metres long, because [squareroot(1.2² / 2) = 0.84852...]