Monday, December 29, 2014

100

Find at least four ways of writing the number 100, each time using only one digit repeated five times.
For example: (999 / 9) - 9 = 102 (but you must get 100, not 102!!!)

Good luck!

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!

Wednesday, December 24, 2014

A Law-Abiding Citizen

"Where do you think you're going with that thing?" asked the bus driver.

"Where do you think I'm going? On this bus, of course. Why, can't I?" replied the electrician.

"No, of course you can't," said the bus driver in a very patronising way. "It is forbidden to bring any object of length, width, or height greater than one metre on any bus. That thing you're carrying is longer than one metre."

"It's got nothing to do with a ticket," screeched the driver. "You could buy a dozen tickets, and I still would not let you ride on this bus!"

Irritation grew rapidly within the electrician. "Listen! I need to take this neon light tube to a ceremony. I don't have a car. The cabbies are on strike. And it's raining. What do you expect me to do!"

"I don't know, and I don't care anyway. You ain't gonna come on this bus with that tube. End of story."

Quickly, the electrician dashed into a shop next to the bus stop and came out with a package containing the neon tube. Smugly, with all thirty-two teeth on display, he showed the package to the bus driver. "Now can I come on the bus?"

With a snort, the bus driver pulled out a folding rule and performed a precise measurement. Scowling, he waved in the smug commuter.

How did the electrician manage to pack a 1.2 metre neon tube into a package less than one metre?

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...]

Friday, December 19, 2014

Tamerlano's Trap II

Alvise Moschin, a Venetian merchant, was dragged into the Hazel Room of Samarcanda Palacethe by a pair of soldiers. Although fairly worried, Moschin felt some confidence due to his knowledge of the East. He knew, through tales heard in wine bars, what was waiting for him and how he should react. For a start, he would find himself in front of two doors guarded by two soldiers, a liar and and truth-teller. That would not be a problem.

The street-smart Venetian was thrown onto the rug before a throne. Despite his predicament, he could not contain a grin, which only widened when he saw Tamerlano enter the room and take a seat upon the throne before him.

"Get up, merchant!" barked the conqueror. "There are two doors behind you--"

"Behind one of them there's a horse, and behind the other there are crocodiles, am I right?" interrupted the merchant.

Tamerlano leaned back. "You are smart and well informed, christian," he said. "However, this time we'll have a slight variation. You will not find two guards, but one. He will be the one to whom you may ask the single question. From that, you must decide which door will lead you to certain death and which to freedom. Also, you will not know whether he always lies or always tells the truth."

With his face pale, as if he had seen a ghost, Moschin turned around and saw that between the two doors, there was indeed only one guard. The guard bore a satanic grin, his piercing eyes staring. Moschin approached the guard slowly, his mind working frantically...

What question must Alvise Moschin ask to determine which is the door to freedom?

Tamerlano's Trap II Puzzle Solution

Moschin asked the guard: "If I had asked you which door leads to freedom, which door would you have pointed me to?"

If the guard was truthful, he would have shown the right door. If he lies, he would have again shown the right door, because he would have given the merchant the opposite answer to what he would have given if he was asked a direct question.

Sunday, December 14, 2014

Tamerlano's Trap

"Now be careful!" Tamerlano warned his prisoner. "You can see that in this room, there are two doors guarded by two soldiers. You can tell by their clothes that they come from two different clans. One of the doors leads to a pool of crocodiles; the other one leads to a healthy horse and a sack of gold. To determine which door leads to certain death and which leads to freedom and wealth, you may ask a single question to one of these soldiers. From that answer, you must make your decision. One more warning; one of these soldiers always tells the truth, and the other one always lies."

The prisoner, an intelligent Greek merchant, meditated for a while, bowed to the great conqueror, and with a grin on his face, approached one of the soldiers.

What question would open the door of freedom?

Tamerlano's Trap Puzzle Solution

The merchant asked one of the soldiers, it didn't matter which one: "If I had asked your colleague which door leads to freedom, which of the two doors would he have pointed me to?"

If the interrogated soldier was the one that tells the truth, he would have pointed him to the door that leads to death, because that's the door that the liar would have showed. But even if the same question was asked the liar, the same door (the one that leads to death) would have been the one pointed at, ie the door opposite the one that would have been shown by the truthful soldier.

Once obtained the answer, the merchant went to the door he was NOT pointed at, and enjoyed his freedom.

Tuesday, December 9, 2014

Drama Galore!

It was a beautiful day, perfect for a stroll. After leaving the cars at the edge of the woods, four couples moved towards the river, reaching its bank after a two-mile walk. The restaurant where they intended to dine was on the other side, partially hidden by trees.

But even on a perfectly planned day, evil could come and spoil it. During the stroll, Albert quietly told Amanda that she shouldn't have dressed quite so promiscuously, whilst she replied that he could have done a better job in refraining from making his oh-so-kind compliments to the other three girls. Bernard whispered menacing words at Barbara The Easy Flirt (as he called her then), and Barbara told him that his relationship with the other girls was of dubious morality. Simillar sorts of exchanges happened between Charles and Corinna, and between Douglas and, err, Diana. Reaching the river and seeing the flowing water did little to tame the souls. On the contrary, when the eight friends noticed that instead of the large boat that would carry them all over to the opposite bank, there was only a little boat that would carry no more than two persons at once, the irritation grew to the point that everybody started arguing with everybody else.

The river was about one hundred yards wide, with a small island in the middle. None of the four men were keen on leaving his girlfriend alone with one or more of his other male friends. On the other hand, the women found out that they could only agree on one point: none of their boyfriends should be alone on the boat when one of the girls, excluding his girlfriend, was all alone on any of the two banks or on the island.

Once the tempers calmed, Bernard and Douglas forumlated a plan involving many trips.

How many trips would it take to ferry everyone across whilst still adhering to the wishes of all the people?

 

Notes:
  • In case you haven't guessed, the couples are, Albert and Amanda, Bernard and Barbara, Charles and Corinna, and Douglas and Diana.
  • No woman should stay on one of the banks, on the island, or on the boat, in the company of one or more other men and without her boyfriend
  • No man should be alone on the boat when one of the girls, except his girlfriend, was all alone on one of the two banks or on the island

Drama Galore Puzzle Solution

They needed seventeen trips, frenquently using the island in the middle of the river as temporary destination. If we denote the names of the men with their initial in upper case, and the names of the women with their initial in lower case (by complete chance, the initials of the men are the same as the initials of their girlfriends), we would get the following table:
Trip # Departure bank Direction Island Direction Arrival bank
1 ABCDcd -
=>
ab
2 ABCDbcd
<=
- a
3 ABCDd
=>
bc a
4 ABCDcd
<=
b a
5 CDcd b
=>
ABa
6 BCDcd
<=
b Aa
7 BCD
=>
bcd Aa
8 BCDd
<=
bc Aa
9 Dd bc
=>
ABCa
10 Dd abc
<=
ABC
11 Dd b
=>
ABCac
12 BDd
<=
b ACac
13 d b
=>
ABCDac
14 d bc
<=
ABCDa
15 d -
=>
ABCDabc
16 cd
<=
- ABCDab
17 - -
=>
ABCDabcd

Thursday, December 4, 2014

Hydra


hydra jigsaw puzzle

A team of four girls and six boys put together a 2200-piece jigsaw puzzle in 4 hours. The same jigsaw puzzle was put together in 8 hours by a team of two boys and five girls.

Who are better at putting jigsaw puzzles together, boys or girls?

Hydra Puzzle Solution

If 6 boys and 4 girls spend 4 hours to complete the jigsaw puzzle, then 3 boys and 2 girls will need 8 hours, which is also the amount of time needed by the team of 2 boys and 5 girls. The reader can see that the input of 1 extra boy on the first team equals the input of 3 extra girls on the other team. The conclusion deduced is that the input from 1 boy is worth as much as the input from 3 girls.

To be more precise, it is possible to demonstate that each boy can put together, in one hour, 75 pieces of the puzzle, compared to the 25 for each girl. This is because, if we say that 6 boys and 4 girls (ie team 1) are equal to 22 girls ((6 * 3) + 4), and since that team puts together 550 pieces per hour (2200 / 4), then the workforce of each team member equals 25 pieces/hour (550 / 22).

Sorry ladies about the "non politically correct" nature of this puzzle, but I'm just translating. I was actually thinking of using Martians and Venusians instead, but then I thought "Who cares!" - Mickey.

Saturday, November 29, 2014

Galactic Expedition

When the scientific community predicted that the Sun would explode into a supernova in ten years time, destroying the entire Solar System, the Earthlings duly prepared for the Doomsday event. The first intergalactic spaceship was already being built, even before that prediction of doom. A sample of the human population was going to travel to the Andromeda Galaxy, where a planet in the BSC14823 solar system was believed to be able to sustain human life. It was hoped that on this planet, the human race would survive and re-establish itself as the superior being in the universe.

The person that would navigate the ship would have to rely on his/her own mathematical abilities, since obviously there could be no assistance from Earth after the explosion. The journey was predicted to last for at least 30 years, so the selection of this person was extremely tough. Only the very finest mathematical and logical mind would be able to successfully navigate the spaceship through intergalactic travel. A test to find this person was issued, and the best mathematicians of the planet competed.

spaceship puzzle

Competition was feirce, for this person would become the first hero of the new world. Of all canditates, only a young woman and a young man reached the very final stage of the selection process. The chairman of the Special Selection Board explained the final question, "There are three different integers, A, B, C, where A × B × C = 900, and A > B > C. One of you will receive either A + B or A + C, but you will not know which one of the two sums it will be. The other candidate will be given the number B. You will take turns, and the winning candidate will be the one that can tell us what the 3 numbers are."

The young man, named Wadzru, was asked to start first. "I don't know," was his answer. Then it was the young woman's turn. Her name was Zaxre, and her answer was also "I don't know." Wadzru's second turn was another "Don't know," which is the same answer given by Zaxre on her second turn. They went on answering "I don't know" for a certain number of times, until Wadzru suddenly grabbed the pen and started writing down the answer. As soon as he started writing, Zaxre cried a "YES!" and also started writing down the answer.

The selection board was confused. Wadzru did indeed answer first, but only because he was one turn ahead. Zaxre also answered correctly when it was her turn. At that moment a young mathematician entered the room. The chairman of the board called him over, and explained the mathematical problem to him. After a short pause, the man said, "If I understand the problem correctly, sir, then the two candidates must have given the correct answer on their fourth turn, that is when each of them were answering for their fourth time. At this point, I can also tell you what the three numbers are. However, there are two different answers, depending on who was given the first chance to answer: whether it was the candidate who was given number B, or the candidate who was given the sum A + B or A + C."

The board were so impressed by this latecomer that he was nominated as the commander of the expedition.

What are the two different combinations of the three numbers?

Galactic Expedition Puzzle Solution

The table below shows all possible combinations (32 of them) of three different integers, the product of which is 900. Next to each combination there is the result of the sum A+B, then the result of the sum A+C, and finally the the number B for that particular combination:
Combination # Factors A+B A+C B
1 450 * 2 * 1 452 451 2
2 300 * 3 * 1 303 301 3
3 225 * 4 * 1 229 226 4
4 180 * 5 * 1 185 181 5
5 150 * 6 * 1 156 151 6
6 100 * 9 * 1 109 101 9
7 90 * 10 * 1 100 91 10
8 75 * 12 * 1 87 76 12
9 60 * 15 * 1 75 61 15
10 50 * 18 * 1 68 51 18
11 45 * 20 * 1 65 46 20
12 36 * 25 * 1 61 37 25
13 150 * 3 * 2 153 152 3
14 90 * 5 * 2 95 92 5
15 75 * 6 * 2 81 77 6
16 50 * 9 * 2 59 52 9
17 45 * 10 * 2 55 47 10
18 30 * 15 * 2 45 32 15
19 25 * 18 * 2 43 27 18
20 75 * 4 * 3 79 78 4
21 60 * 5 * 3 65 63 5
22 50 * 6 * 3 56 53 6
23 30 * 10 * 3 40 33 10
24 25 * 12 * 3 37 28 12
25 20 * 15 * 3 35 23 15
26 45 * 5 * 4 50 49 5
27 25 * 9 * 4 34 29 9
28 30 * 6 * 5 36 35 6
29 20 * 9 * 5 29 25 9
30 18 * 10 * 5 28 23 10
31 15 * 12 * 5 27 20 12
32 15 * 10 * 6 25 21 10

If the first question was asked to the person (Wadzru) that was given A+B or A+C and he answers "don't know", that means that the number he received appears in the A+B and A+C columns more than once. On the other hand, if that number appears in those columns only once, then the solution would be found immediately, as that number would relate to only one of the 32 combinations.

Therefore the first candidate was given one of the following numbers: 65, 61, 37, 65, 29, 28, 27, 25, 23. These numbers appear more than once in columns A+B and A+C, and because of this, they do not yet allow to find the correct combination. But we can now discard, from the 32 combinations, the ones where the 9 numbers listed above are not included in columns A+B and A+C. The combinations left were then:
Combination # Factors A+B A+C B
9 60 * 15 * 1 75 61 15
11 45 * 20 * 1 65 46 20
12 36 * 25 * 1 61 37 25
19 25 * 18 * 2 43 27 18
21 60 * 5 * 3 65 63 5
24 25 * 12 * 3 37 28 12
25 20 * 15 * 3 35 23 15
27 25 * 9 * 4 34 29 9
28 30 * 6 * 5 36 35 6
29 20 * 9 * 5 29 25 9
30 18 * 10 * 5 28 23 10
31 15 * 12 * 5 27 20 12
32 15 * 10 * 6 25 21 10

The second candidate (Zaxre) followed the same reasoning, so she then knew that these were the only possible combinations left, after Wadzru gave his answer. The situation for Zaxre was the same though: if the number she was given appeared only once in column B, then the right combination would be the one containing that number; if the number appeared more than once in column B, then the solution would not yet be within reach. And this is what happened, since she answered "don't know". But the elimination of certain combinations could go on nevertheless, because after she said "don't know", combinations # 11, 12, 19, 21, 28 were automatically discarded. Then it was Wadzru's second turn, and the current situation was:
Combination # Factors A+B A+C B
9 60 * 15 * 1 75 61 15
24 25 * 12 * 3 37 28 12
25 20 * 15 * 3 35 23 15
27 25 * 9 * 4 34 29 9
29 20 * 9 * 5 29 25 9
30 18 * 10 * 5 28 23 10
31 15 * 12 * 5 27 20 12
32 15 * 10 * 6 25 21 10

At this point, the first contestant answered again "don't know", and automatically discarded combinations 9 and 31. Then the second candidate, after following, again, the logical reasoning of Wadzru, answered "don't know", therefore discarding combinations 24 and 25.The third turn started with the first contestant facing the following combinations:
Combination # Factors A+B A+C B
27 25 * 9 * 4 34 29 9
29 20 * 9 * 5 29 25 9
30 18 * 10 * 5 28 23 10
32 15 * 10 * 6 25 21 10

Wadzru, by answering "don't know" on his third turn, discarded the only combination with no alternatives, ie 30, so Zaxre was presented only with combinations 27, 29, 32; her answer "don't know" left only two combinations: 27 and 29. Then it was the first contestant again, and he was faced with the following situation:
Combination # Factors A+B A+C B
27 25 * 9 * 4 34 29 9
29 20 * 9 * 5 29 25 9

If the first contestant now answered "don't know", then the number he was given must have had to be 29; but since he was given number 25, he was able to know the correct combination requested: 20*9*5. When Zaxre saw that Wadzru had the answer, she was also able to find the solution, because she had also been able to follow the selection process. For her, it was only possible to give the answer on her 4th turn, because if Wadzru could have been able to come up with the answer on his 3rd turn (when the numbers appering just once in columns A+B and A+C were more than one), then Zaxre would not have been able to know which of the combinations contained the number given to her opponent.

If instead, the question was first asked to Zaxre, who was given number B, then after her first "don't know" it would have been possible to discard from the table (the one with all 32 combinations) combinations 1, 11, 12. After her opponent also said "don't know", the combinations to be discarded would have been 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23 and 26, leaving the table looking like this:
Combination # Factors A+B A+C B
19 25 * 18 * 2 43 27 18
24 25 * 12 * 3 37 28 12
25 20 * 15 * 3 35 23 15
27 25 * 9 * 4 34 29 9
28 30 * 6 * 5 36 35 6
29 20 * 9 * 5 29 25 9
30 18 * 10 * 5 28 23 10
31 15 * 12 * 5 27 20 12
32 15 * 10 * 6 25 21 10

For the second turn, the "don't know" answer of the contestant answering first (Zaxre in this scenario) would have automatically discarded combinations 19, 25 and 28, and the "don't know" from her opponent would have further discarded another one: 31. For the third turn, the remaining possible combinations are: 24, 27, 29, 30 and 32. If Zaxre could not answer yet, then combination 24 would be discarded, while Wadzru's "don't know" would discard combination 30. The situation would now be:
Combination # Factors A+B A+C B
27 25 * 9 * 4 34 29 9
29 20 * 9 * 5 29 25 9
32 15 * 10 * 6 25 21 10

If Zaxre would now be able to know the solution, she would have been given, as B, number 10 (combination # 32), resulting with the product 15*10*6. But her opponent would also be able to answer then, thanks to Zaxre being able to find the solution. This scenario, like the previous one with Wadzru starting, where both opponents are able to find the solution at the same time, can only happen on the fourth turn. It's worth knowing that a fourth "don't know" from the candidates would have led to the problem being left unsolved.

Monday, November 24, 2014

Top Secret

With his heart rate increasing steadily, James Bents (alias Lt-Colonel Ivanovic Zdanov, as far as the KGB were concerned) lined up behind the scientists who were walking towards the internal gate. Thanks to his forged documentation, he was able to pass through the two previous gates. He was aware that to get right inside the missile launch-pad, he would need to supply a password. He had been informed that the password changed daily. Only his extreme cool and many years of training enabled him to contain the fear.

The two scientists in front of him reached the gate, which was patrolled by machine-gun wielding soldiers. He strained to hear the voices of the people ahead of him in the queue.

"Twelve?" asked the guard.

"Six," replied the first scientist.

The first scientist strode through the gate as the second one walked to the guard.

"Six?" asked the guard.

"Three," replied the second scientist and walked through.

Relief and confidience spread through Bents; the method that drove questions and answers was trivial. He stepped forward.

"Nine?" asked the guard.

Brents hesitated for a split second. This was an unpredicted complication, but his arduous conditioning allowed the secret agent to remain calm and as sharp as a razorblade. "Four and a half," he answered without blinking.

Quite suddenly, the entire area was filled with floodlights. Alarm sirens broke the silence of the otherwise peaceful night. In a fraction of a second the Lt-Colonel realised his mistake. He tried to turn on his heels and run, but instantly felt the cold barrel of a machine-gun pressed against his neck.

What was the secret agent's fatal mistake?

Top Secret Puzzle Solution

The answers given were the the number of letters in the question. When asked "Nine?" the secret agent should have answered "Four".

Sunday, November 23, 2014

Luminarium Labyrinth

Luminarium is a project of British architectural bureau Architects of Air, which is traveling the world for over twenty years. During this time, it had more than five hundred exhibitions in thirty-eight countries. 

Luminarium - a giant inflatable maze with winding passages and pompous domes that resemble masterpieces of Islamic architecture, Archimedean solids and Gothic cathedrals. All this incredible diversity is transformed into an inspiring monument to light and color. The height of the dome may reach ten meters, and the entire structure covers an area of one thousand square meters.







Wednesday, November 19, 2014

Killing Some Time…

  1. The only animals in this house are cats.
  2. Any animal that loves watching the moon is tamable.
  3. When I hate an animal, I stay away from it.
  4. No animal is carnivorous, unless it moves at night.
  5. No cat avoids killing mice.
  6. I do not own any animal, except the ones that live in this house.
  7. Kangaroos are not tamable.
  8. No animal, except the carnivorous ones, kills mice.
  9. I hate animals that I do not own.
  10. All animals that move at night love watching the moon.
From this sequence of sentences, which are based on the foundations of logic, it is possible to reach a conclusion deduced from all 10 sentences.

What is the conclusion?

Killing Time Puzzle Solution

The logical conclusion must necessarily be: I avoid kangaroos. Based on the 10 statements, it is possible to deduce the following conclusions:
Deduction # From statements # Deduction
11 1 and 5
All animals in this house kill mice
12 8 and 11
All animals in this house are carnivorous
13 4 and 12
All animals in this house move at night
14 6 and 13
All animals I own move at night
15 10 and 14
All animals I own love watching the moon
16 2 and 15
All animals I own are tamable
17 7 and 16
I do not own any kangaroo
18 9 and 17
I hate kangaroos
final 3 and 18
I avoid kangaroos

Friday, November 14, 2014

A Struggle For Survival

"There are only two planets in this solar system that would offer a chance of survival to a Gxz," the geoanthropologist told the commander of the starship after having examined the data from the probe. "They are the sixth one from the sun and the eight one towards the sun."

The commander turned towards the astronavigator. "Current position?"

The navigator moved his tentacles quickly on the keyboard, "We are approaching one of the three orbits that are between the two mentioned by the anthropologist; the most internal one of the three, to be precise."

"What is this planet like," the commander asked.

"Uninhabitable," replied the geoantropologist. "The atmosphere is full of lethal gases such as oxygen. Gravity is moderate, and there's bucketloads of a mixture of hydrogen and oxygen without any silicon whatsoever; just thinking about it makes my verrucas crawl!"

"How many planets in this solar system?" asked the commander.

"Less than 12," the astrophysicist replied, peering at the instruments. "The exact number is..."

That's the end of the commander's log, found within the wreckage of the starship, and painstakingly translated. How many planets did that solar system contain?

a solar system

A Struggle For Survival Puzzle Solution

The answer is nine planets. The geoanthropologist states that one of the habitable planets is eighth toward the sun, so there must be at least 8 planets. The navigator mentions that there are 3 planets between the 6th planet and the 8th towards the sun. The only number that can satisfy both statements is the number nine. You will then realise that the navigator was talking about Earth.

Sunday, November 9, 2014

The Greatest Show On Earth

"This is terrible," the customs officer shouted. "It's impossible to count all these people and animals that keep moving around constantly. I can't count the same number twice! There are zebras, lions, giraffes, horses, elephants, rhinos, tigers, cheetas, flamingos, storks, doves and hummingbirds! I can't keep them under control, I just can't!"

"Count the heads, officer," the circus owner advised. "Every animal has got a head and, as the documentation shows, I can count 112 of them."

"Count the legs, officer, that's the way," the clown said. "There are 310 legs; if you subtract the number of heads from the number of legs, you'll be able to tell how many bipeds and how many four-legged animals there are."

"Wait a second, I've just counted the legs just now, and there's 297 of them," the cook said.

"This is driving me mad," the officer muttered. "One of them tells me there are 112 heads, another one talks about 310 legs, and that mentally-disturbed cook tells me that the number of legs is only 297. How many animals, including humans, are there?

To be more precise, how many bipeds and how many four-legged animals are there? And is the cook completely mad?

The Greatest Show On Earth Puzzle Solution

There are 69 bipeds and 43 four-legged animals. If all animals were four-legged, the officer would have counted 448 legs, not 310. Obviously 138 legs are missing, hence 69 subjects are bipeds.

It's worth pointing out that, within the bipeds, there are 13 stilt birds, which include flamingos and storks. The cook must have counted them as they were standing on one leg only, so whilst he is not crazy, he almost drove the customs officer mad.

Tuesday, November 4, 2014

A Thinking Man

Professor Percent was a maths lecturer with an interest for new ways to express mathematical expressions. The traditional symbols (+, -, *, /, etc) were not enough anymore, to convey his superior numeric operations, so he had to invent new symbols, and only a superior brain would be able to understand the need for his new symbols.

The first symbol he invented was §; between two numbers, it meant that, if the first number was greater than the second, then the second should be subtracted from the first one; otherwise the two numbers should be added. Therefore 5 § 2 = 3, while 2 § 5 = 7.

The poor people that had to put up with this were, of course, his students. In the last test they were faced with:

5 ¿ 2 = 27
6 ¿ 3 = 27
8 ¿ 4 = 36

and also with:
5 ¤ 2 = 15
6 ¤ 4 = 12
3 ¤ 8 = 40

What are the meanings of the symbols ¿ and ¤?

 

Notes:
There are at least 3 different solutions for ¿.

A Thinking Man Puzzle Solution

The symbol ¿ means the difference between the number made up of the all digits of the operation, and the mirror of this last number. i.e,
5 ¿ 2 = 52 - 25 = 27
6 ¿ 3 = 63 - 36 = 27
8 ¿ 4 = 84 - 48 = 36

An alternative solution for this symbol (submitted by Alfa Chan... many thanks!) is simply the difference between the two numbers multiplied by 9. i.e,
5 ¿ 2 = (5 - 2) × 9 = 27
6 ¿ 3 = (6 - 3) × 9 = 27
8 ¿ 4 = (8 - 4) × 9 = 36

Another alternative solution submitted by Mickey Kawick... thanks!! We have x ¿ y; If x is odd, then the result is 5x + y, otherwise it's 5x - y. i.e,
5 ¿ 2 = 5 × 5 + 2 = 27 (5 is odd, so we add the 2)
6 ¿ 3 = 6 × 5 - 3 = 27 (6 is even, so we subtract the 3)
8 ¿ 4 = 8 × 5 - 4 = 36 (8 is even, so we subtract the 4)

The symbol ¤ means the difference between the two numbers multiplied by the larger of the two numbers. i.e,
5 ¤ 2 = (5 - 2) * 5 = 3 * 5 = 15
6 ¤ 4 = (6 - 4) * 6 = 2 * 6 = 12
3 ¤ 8 = (8 - 3) * 8 = 5 * 8 = 40

An alternative solution for this symbol, as submitted by Jeff Schall (many thanks!), is the difference between the square of the bigger of the two numbers and their product. i.e,
5 ¤ 2 = (5 ^ 2) - (5 * 2) = 25 - 10 = 15
6 ¤ 4 = (6 ^ 2) - (6 * 4) = 36 - 24 = 12
3 ¤ 8 = (8 ^ 2) - (3 * 8) = 64 - 24 = 40

Friday, October 31, 2014

Scariest Halloween Pumpkins

Just take a look at these terrible and funny at the same time the creations of artist John Neil. The technique of carving the pumpkin is amazing, as well as the artistic vision that reflects the whole spirit of this beautiful holiday - Halloween.










Thursday, October 30, 2014

Three Divine Comedians



divine comedians puzzle

As Dante was reaching River Styx on his way back from the Underworld, he was expecting to hitch a ride back on Charon's boat, and then go to pay a visit in Purgatory. As things went, Charon turned out to be a rather nasty fellow. Instead of giving Dante a nice, hassle-free trip back across the damned river, he called his three best buddies. Dante looked at them. The monsters were pretty ugly overall.

"You see, Dante, nothing is free in life - or death - except suffering," said Charon. "My three friends here, although they all look repulsive, are rather peculiar: one of them always tells the truth, another one always lies, and the last one is a bit of a lunatic: sometimes it tells the truth and sometimes it lies. You have a total of three questions you can ask them to find out which one is which. I'll take you on the other side of the Styx if and only if you can tell me, without a shadow of a doubt, which one tells the truth, which one is the liar, and which one is the lunatic one. Oh, I almost forgot to tell you - silly me - they can only answer yes or no... isn't life great?"

What three questions will enable Dante to cross the River Styx?

 

Notes:
  • Each question is directed to, and answered by, only one creature.
  • The creatures themselves know who is the truth-teller, who is the liar, and who is the lunatic.
  • The solution does not rely on asking them questions that they are not able to answer due to uncertainty. eg, asking the liar or the truth-teller to predict whether the lunatic will say yes or no to a given question.

Three Divine Comedians Puzzle Solution

Dante figured that first of all, he had to find out which of the three monsters was the lunatic one. Let's call the three monsters A, B, C.

In order to do so, he asked one of them (let's say monster A for this example) a question like: "If I asked a question to monster B, would I stand a greater chance of obtaining a truth than if I asked the same question to monster C?"

The possible combinations (where + means the monster that tells the truth, - means the monster that lies, x means the lunatic monster) are:
Answer given Monster A Monster B Monster C
No + - x
Yes + x -
No - + x
Yes - x +
Yes/No x + -
Yes/No x - +

If the answer was a "Yes", then it was safe to say that x couldn't have been monster C; on the other hand, if the answer was a "No", then x couldn't possibly have been Monster B.

Once that Dante obtained this information, he used his second question to find out whether the monster that is definately NOT the lunatic one was + or -, and the easiest way of achieved this was by asking a question with an obvious answer such as, "Is 2 an even number?" After finding out whether the monster is the liar or the sincere one, the third question was used to resolve the other two monsters; a question like "Is monster A the lunatic one" did the trick.

Saturday, October 25, 2014

Barrels 'O' Fun

In the basement of the Italian "cantina", there are 3 small, irregularly-shaped, wine barrels: a 12-litre one, full, and two empty ones, which can contain up to 7 and 5 litres.

Without using any additional tool, how can you get exactly 6 litres of wine in the 7-litre barrel, and have 6 litres left in the 12-litre barrel?

Barrels 'O' Fun Puzzle Solution

There are multiple ways of solving this. One way is given below, and it's probably the fastest one. Each set of 3 numbers separated by hyphens is the amount of wine (in litres) in the 3 barrels after each "pouring operation". The 3 barrels are always in the same order: 12, 7, and 5 litres.
  • 12-0-0
  • 5-7-0
  • 5-2-5
  • 10-2-0
  • 10-0-2
  • 3-7-2
  • 3-4-5
  • 8-4-0
  • 8-0-4
  • 1-7-4
  • 1-6-5
  • 6-6-0.

Monday, October 20, 2014

Three Camels

Three young men travelled across the desert toward the tent of The Great Sage, seeking precious advice.

The eldest of the three moved in front of The Great Sage, who was meditating, and said, "God bless You, Great Sage! Our Father, before dying, left us these camels, and it is his will that I should have a half of the herd, my brother Ali one third, and my brother Ismail one ninth. We've tried, Glorious Sage, we have divided the camels and divided them again until the void opened before us. Help us, Magnificent Sage, we are not gifted with your superior intellect!"
camels riddle


The Great Sage asked the pleading man "How many camels are there?"

"Seventeen, may God bless You!", was the answer.

The Great Sage smiled.

How were the camels divided, strictly observing the fatherly will and without butchering any of them?

Three Camels Puzzle Solution

The Great Sage added his own camel to the other seventeen. He then gave 9 camels (one half of 18) to the eldest of the three, 6 camels (one third) to Ali, and 2 camels (one 9th) to Ismail. Then took his own camel back and sat in front of the tent, thanking God for His generosity.

Another way to explain the solution, perhaps in a more mathematical manner, is the following, courtesy of Gopalakrishnan Thirumurthy.
The three sons are assigned their shares: 1/2, 1/3, and 1/9. The sum of their shares is 1/2 + 1/3 + 1/9 = 17/18. Out of 18 camels, 17 of them are left by their father. So,
  • #1 gets 1/2 of 18 = 9
  • #2 gets 1/3 of 18 = 6
  • #3 gets 1/9 of 18 = 2
9 + 6 + 2 = 17.

Wednesday, October 15, 2014

Cheers To Statistics

Two and a half artists spend two and a half hours painting two and a half models on two and a half canvases.
How many artists are necessary to paint twenty-four models on twenty-four canvases in twenty hours?

Cheers To Statistics Puzzle Solution


Three artist would do the trick. This is because twenty-four artists would paint twenty-four models in two and a half hours. Since the available time increases eight-fold (2.5 * 8 = 20), it is possible to reduce the number of painters by the same number of times (24 / 8 = 3).

Saturday, October 11, 2014

Flying Car Prototype

In Slovakia local engineers and  designers presented the new prototype Aeromobile  (flying car). An impressive projects representing a - two-seater roadster that can reach 160 km/h on the road more than and about 200 km/h in the air. Under the hood we may find an aircraft engine Rotax 912 with a capacity of 100 hp.  A single tank (aircar runs on regular fuel) is enough for 700 kilometers of flight or fort a trip of 875 kilometers. This original automobile unit is extremely lightweight - it weighs only 450 pounds.





Friday, October 10, 2014

Faulty Batch

A little nation in Antarctica has its gold coins manufactured by eight different European companies. The Treasury Minister and his secretary were examining samples just delivered from the eight companies.

"How much should these coins weigh?" the Minister asked.

"Ten grams each, Sir."

"At least one of these coins - this one - is lighter than the others," said the Minister. "Let's check."

He put the coin on the scale, which showed that the coin weighed only nine grams. A bunch of coins, untidily placed on a tray, were frantically searched by the Minister and his secretary. Within the bunch, they found a handful of coins that also weighed one gram less than they should. The two men looked at each other; obviously, one of the manufacturing companies was producing coins with the wrong weight.

"Most of the coins are still packed in the plastic wrappers. It should be easy to tell which company is producing the faulty batch," said the secretary.

The two men placed eight packs of coins on the table, one pack from each company.

"How tedious," sighed the Minister. "Do we really have to use this scale eight more times, just to find the faulty batch of coins?"

"That won't be necessary, Sir," grinned the secretary. "We can find the lighter coins by using the scale only once."

How would they do it?



Notes:
By using the scale once, it means that only one reading can be taken after all the coins to be weighed are placed onto the scale. ie, you cannot read the values as you place the coins on -- that would make the puzzle too easy!

Faulty Batch Puzzle Solution

The secretary placed on the scale 1 coin from the first batch, 2 from the second, and so on until he put 8 from the eighth batch.

If all coins weighed 10 grams each, then the weight displayed on the scale should have been 360 grams ((1 + 2 + ... + 8) × 10). But, since one batch of coins weighs less, the difference between 360 grams and the weight displayed on the scale should point us to the faulty batch. For example, if the faulty batch was the fifth one, then the total weight displayed on the scale would be 355 grams. Or if it was the seventh batch, the weight would have been 353 grams, ie 7 grams less than the theoretical total weight of 360 grams.

An 'optimisation' on this solution is to omit the 8 coins from the eighth batch. In this case, the maximum weight of the coins would be 280 grams, and if it equals 280, then the eighth batch is the faulty one. Thanks to Denis Borris for this observation.

By using the same logic, one could omit the coins from any one of the other batches, instead of the eighth one. For example, if we omit the fourth batch, we'll be left with a theoretical 320 grams and, if it is indeed the total weight, then we will know that the fourth batch was the faulty one. Thanks to Glen Parnell for noticing this.

Sunday, October 5, 2014

Faulty Batches

"This time," said the Treasury Minister, "I ditched those dodgy Europeans, and I have assigned the manufacture of our gold coins to five American companies. Look, they are all shining and beautiful, and they are all exactly the same!"

The secretary looked at the coins, weighed some of them, and cleared his throat. "Ahem, Sir. I would like to point out that here we have at least three different kinds of coin; they all look the same, but their weight is different. Would you please come close to the scale? This coin weights 10 grams, as it should, but this other one is 11 grams, while this one is only 9 grams. Obviously two of our manufacturing companies haven't done a good job."

Sad as he could have been, for having been tricked agin by other dodgy companies, the Minister managed to raise his head. "Well.. it's just a matter of finding the fauly ones using the trick that you've showed me, by using the scale only once..."

"Sir. Actually, this is a different problem altogether, we need to find two sources of errors, rather than just one. One batch is heavier, another is lighter. The method I used before will not be sufficient this time. But we can nevertheless find the two offending batches by using the scale once."

How did they manage to use the scale only once?



Notes:
  • You may assume that each batch is made of a large amount of coins (thousands, millions, up to you! :)
  • All coins of the same batch weight the same amount.
  • The storyline in this puzzle follows from the story in Faulty Batch. It is however NOT necessary to have previously read/solved that puzzle in order to solve this one, even though it may be preferable.

Faulty Batches Puzzle Solution

They had to weigh 1 coin from the 1st batch, 2 from the 2nd, 4 from the 3rd, 8 from the 4th, and 16 from the 5th one.

If all coins weighed 10 grams as they should, the scale would display 310 grams ((1 + 2 + 4 + 8 + 16) * 10). However, since one batch has 9 grams coins, and another 11 grams coins, then the total weight of this combination of coins will be:
Total Weight Number of
9g coins
Number of
11g coins
311 1 2
313 1 4
317 1 8
325 1 16
312 2 4
316 2 8
324 2 16
314 4 8
322 4 16
318 8 16
309 2 1
307 4 1
303 8 1
295 16 1
308 4 2
304 8 2
296 16 2
306 8 4
298 16 4
302 16 8

After seeing the solution to this puzzle, it is clear that it would be a lot easier to simply use the scales up to 5 times rather than go through all this, but where is the fun in that?

Tuesday, September 30, 2014

Orbiting Logic

Colonel Tom Carpenter, during his fifth space mission, was being kept awake by the blabbering of the Cape Canaveral Control Centre operator, who offered him the following puzzle.

"Here's a deck of 52 cards, Tom. I'm taking the Aces and the Royals out of the deck. Do you copy that, Tom?"
cards riddle

"Roger," the yawning voice of the astronaut answered.

"Of the 36 remaining cards, I've drawn 5 of them. These 5 cards have the following properties:
(a) all four suits are represented here;
(b) there is no more than 2 consecutive ranks for each sequence (ie a 2 followed by a 3, or a 7 by an 8, or both, but not 2, 3, 4);
(c) the sum of the even ranks and the sum of the odd ranks produce two numbers: the difference between these two numbers is 9, but I won't specify whether it's the sum of odds being greater than the sum of evens, or viceversa.
(d) the sum of ranks of the red cards is exactly twice the sum of ranks of the black cards.
You awake, Tom?"

After a pause, Tom managed a faint "Roger."
"Ok, you should also know that:
(e) a hearts is a multiple of a clubs;
(f) the rank of a diamond is greater than that of a hearts;
(g) there are no 2 cards with the same rank.

Which cards did I draw? Tom, are you listening? Which cards have I got?"

Deduce which five cards he necessarily holds.

Orbiting Logic Puzzle Solution

During the next orbit, the sleepy but correct answer came forth from the astronaut: 2 of clubs, 5 of hearts, 7 of diamonds, 9 of spades, and 10 of hearts.

Following from (g) - no two cards are the same rank - and (b), the strongest combination (ie highest ranks) possible is 10, 9, 7, 6, 4, which adds up to 36. If we write down all combinations of 2 numbers that have a difference of 9, and a maximum sum of 36, we'll have:
  • 2, 11
  • 3, 12
  • 4, 13
  • 5, 14
  • 6, 15
  • 7, 16
  • 8, 17
  • 9, 18
  • 10,19
  • 11, 20
  • 12, 21
  • 13, 22
The number of odd ranks in the 5-card combination could be 1 or 3 (not 2 and not 4 because their sum would be an even number, and a difference of 9 between two even numbers is inexistent). If there was only 1 odd rank, then there would be 4 even ranks, and the weakest combination of evens would be 2, 4, 6, 8 which add up to 20: the difference, 9, would result in a single number, 11, which does not exist in the deck of cards (the Royals were excluded). Therefore we have 3 odd ranks. The sum of the weakest 3 odd ranks (excluding the Aces) is 15, ie 3 + 5 + 7. So, from the combinations above, we can exclude all combinations that contain an odd number less than 15. We are left with
  • 6, 15
  • 8, 17
  • 10, 19
  • 12, 21
The first one (6, 15) must be excluded because the 3 odd numbers (3, 5, 7) would be joined by the only 2 even numbers which would add up to 6 (2,4), and therefore we would have a sequence of 5 consecutive numbers, which doesn't match the constraint set by (b). We are left with 3 combinations:
  • 8, 17 (and the 5 ranks would be 2, 3, 5, 6, 9)
  • 10, 19 (and the 5 ranks would be 3, 4, 6, 7, 9)
  • 12, 21 (and the 5 ranks would be 2, 5, 7, 9, 10)
But (d) states that the sum of red cards is twice of the sum of black cards, so the first 2 combinations must be excluded, because it's not possible to find 2 sets of numbers, one of which is twice the other; therefore the 5 ranks can only be 2, 5, 7, 9, 10. The red cards must be 5, 7, 10.

Following (e), rank 2 must be clubs and 10 must be hearts, so rank 9 must be spades; following (f), rank 7 has got to be diamonds and rank 5 must be hearts.

Thursday, September 25, 2014

Three Skullcaps

After being captured by the tribe of the forest, the three explorers were taken to the tribe's chief, who declared that the tribe needed a successor of great wisdom. He showed the captives five skullcaps, three of which were red and the remaining two were green. He commanded the three to line up with their faces toward the wall. On each he placed a red skullcap.

"Now", the chief said, "turn around and look at each other. Whoever can tell me, with a logical explanation, the colour of the skullcap you are wearing will be granted freedom."

The three turned around, looked at each other, and after a long pause one of them said, "I don't know." Two natives impaled him on their spears.

After another pause, the second prisoner stuttered, "I don't know." The two natives threw him into a fire pit.

Immediately, the third man turned a cartwheel, announced, "My skullcap is red," and proceeded to explain.

Amidst the applause of the gathered tribesmen, the chief awarded the explorer vice-chiefdom of the tribe for his wisdom

How did the explorer know that his skullcap was red?

Three Skullcaps Puzzle Solution

At first, any explorer could have guessed the colour of his own skullcap only if the other two wore green skullcaps. Unfortunately, the first explorer admits to not being able to work it out, and is killed.

Then, with two people left it is possible for either explorer to know if he wears a red skullcap only if the other wears green. In that case, he could reason, "The other person wears green, so if I also have a green skullcap, then the first man would have deduced that he was wearing a red skullcap, since there are only two green caps. Therefore my skullcap is, without doubt, red." Of course, this is not the case. Stupidly, the second explorer admits he does not know and is killed as punishment.

After seeing that the other two could not deduce their colours, and believing in their deductive capabilities, the third prisoner was then sure he was wearing a red skullcap.

Saturday, September 20, 2014

Red Square, Moscow, 30th April

"And what about these two posters?"

"Those are the posters that will be hung on the south side of Red Square: as you can see they represent comrades Lenin and Marx."

"I can see that by myself. What I meant was the other two posters over there, the one with the Red Star and the one with the Hammer and Sickle."

The four posters were lined up and showed, from left to right, Lenin, Marx, the Red Star, and the Hammer And Sickle.

"Oh, they are nothing but the back of the other two. I wanted you to also see the back-faces of the posters, as these back-faces will be invisible from the inside of the Square."

communist posters riddle

"Hmmm... so enlighten me, which is the front of the poster representing the Hammer And Sickle?"
Nikita Proskoijev grinned, "I would like to test your deduction capabilities, dear comrade; a capability, I might add, which some people have had the guts to doubt. I say that all posters representing Lenin show the Hammer And Sickle on their opposite face. How would you verify this statement, in such a way that leaves no shadow of a doubt?"

"Do you mean, dear tovarisc, that I should turn these gigantic posters around to see which comrade matches the Star and which the Hammer And Sickle?"

"I have said what I have said, dear Ivanovic; it is up to you to decide what is the minimum number of posters to turn around to verify whether my statement was true or false."

Ivanovic felt very cold, as if he was in Siberia. What is the minimum number of posters, out of the four displayed, that he has to turn around to verify the statement of that cunning snake?

Ivanovic had to turn around two posters: the first one (Lenin) and the third one (the Red Star).

All that Nikita Proskoijev said was that all the posters representing Lenin show Hammer And Sickle on their opposite face. Therefore it is needed to check the back of the first poster and the front of the third one, to make sure that the Red Star wasn't linked with Lenin's face, cause if it was, then Proskoijev's statement would have been false.

Checking the front of the fourth poster (which is what Ivanovic did, that's why he's now a lumberjack near Jakutsk) is useless; if the 4th poster shows Lenin face as its front, it would just confirm what Proskoijev stated; but then, if the 4th poster showed Marx, this would not have falsified Proskoijev's statement, because he said that Lenin is linked to Hammer And Sickle, while he didn't state that Hammer And Sickle are linked to Lenin.

Monday, September 15, 2014

Mizar

mizar
The bells of the clock tower signal six o'clock (with 6 loud rings) in 5 seconds.

How long will it take for the bells to signal twelve o'clock (midday)?



Notes: It is not 10 seconds!

Mizar Puzzle Solution:

11 seconds. The 5 seconds needed to signal six o'clock are the 5 silent intermissions between rings. At twelve, the 12 rings are interleaved by 11 silent intermissions, which need 11 seconds to be executed.

Friday, September 5, 2014

Amazing Life-Sized Games

A lot of creative experiments take place every day around the world. IN this gallery we want to present you several board games, computer games or brain teasers made life-sized. These games were created for entertainment, as well as to assist people achieve e better orientation in space. Enjoy the gallery!

1. A group of anonymous artists known as ‘Bored‘ has created a life-size version of Monopoly using the streets of Chicago as their giant game board.


2. In honor of the board game's 60th anniversary in 2009, the crookedest street in the U.S., Lombard Street, San Francisco, was transformed into the sweetest board game around - Candyland.

3. On the 3rd annual Michigan Fallapalooza in 2007 - a day of games, music and sidewalk sales, the life-sized game of Hasbro's popular Chutes and Ladders was perhaps the most attractive event.

4. The world's biggest game of Scrabble was played out on the pitch at Wembley stadium.

5. The world's biggest Chess set by Guinness has a the King that is coming in at 48 inches high.

6. Human-sized Angry Birds was made in the Mount Faber in Singapore and it is super fun!







Thursday, August 7, 2014

Unique Motorcycle Design

Moto art lovers will certainly like this post. These motorcycles are really lovely, they have a unique design and are evaluated at about $500 000. Enjoy the gallery!








Sunday, July 27, 2014

Turning Into a Giant Tarantula

Artist named Emma Faye managed to turn a sexy body model into a real nightmare for all Arachnophobiacs. So, meet the present Spider-Man - or rather, a giant tarantula sized girl!




Tuesday, July 15, 2014

Scotch Tape Artwork

These are unique works of art made by the Dutch artist Max Zorn using scotch tape. He glues the scotch in the right places, while the rest is just a game of light. Enjoy the Gallery.









staffcounter

Friday, July 4, 2014

Japanese Movies Monsters

A collection of Monsters and Unusual Creatures from Vintage and Classic Japanese Horror and Science Fiction Movies. Strange and original monsters that will lead your imagination to unusual and terrific new worlds. IF you enjoyed this gallery solve Monsters Jigsaw Puzzles with Unique Design and Animation.