Showing posts with label puzzle games. Show all posts

Saturday, January 3, 2015

Swapping Art

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"Four drawings by Max Ernst are worth as much as five sketches by Magritte, do you agree?" asked Giorgio Parconi, an Italian art dealer. He was tired of arguing over this.

"D'accord!" agreed Cesar Blanchard, who was the director of a Parisian art gallery.

"And we all agree that two sketches by Magritte plus one drawing by Ernst are worth as much as two paintings by Bacon. Right?"

"Bon," nodded the frenchman.

"So, I'm offering you four drawings by Ernst plus one sketch by Magritte, and in return you give me three sketches by Magritte and two paintings by Bacon. It's perfectly fair, isn't it?"

Blanchard remained silent, he had the feeling that something was wrong.

Was the Italian dealer offering a fair swap?

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.

Wednesday, December 24, 2014

A Law-Abiding Citizen

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"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...]

Sunday, December 14, 2014

Tamerlano's Trap

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"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!

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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

Sunday, November 9, 2014

The Greatest Show On Earth

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"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.

Saturday, October 25, 2014

Barrels 'O' Fun

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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.

Friday, October 10, 2014

Faulty Batch

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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.

Saturday, September 20, 2014

Red Square, Moscow, 30th April

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"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.