Showing posts with label puzzles online. Show all posts

Saturday, September 5, 2015

Island X

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There are three categories of tribes in Island X; a Truther, who always speaks truthfully; a Liar, who always speaks falsely; and an Altetnator, who makes statements that are alternatively truthful and false, albeit not necessarily in that order.

A visitor approaches three inhabitants and asks who is a Truther. They answer as follows:

A says:
1. I am a Truther
2. B is a Liar

B says:
1. I am a Alternator
2. C is a Liar

C says:
1. I am a Truther
2. A is a Liar

Determine the identity of each of the three inhabitants from the information provided in the above statements.

Island X Puzzle Solution

Assume that A is the Truther. If so, then B is the Liar as A's statement asserts. If so, B's second statement is false, so C is the Alternator. This implies that C's first statement is false as also his second statement that A is the Liar, so that C is the Liar which is a contradiction, so that A cannot be the Truther.

Assume that B is the Truther. If so, his first statement is a direct contradiction, implying amongst other things, that B cannot be the Truther.

Assume that C is the Truther. If so, then A is the Liar in conformity with his second statement, so the remaining member B must be the Alternator. This checks out since as a Alternator, B's first statement is true while in his second statement he falsely identified C as the Liar. Both of A's statement are then clearly false, so this establishes the veracity of both the statements of C.

Consequently, (A, B, C) = (Liar, Alternator, Truther)

Wednesday, August 26, 2015

A Special Old Guy

1 comment :
mathematical nerd riddle
Alvin and Buzz are nerds and like doing nerdy things. So Alvin called Buzz one day...

"Buzz, I've finished tracing my family tree back from the year 500 AD, and I found one quite special guy".

"What's so special about him?" asked Buzz.

"Well, he was x years old in the year x^2 (x squared) and he had a son who was y years old in the year y^3 (y cubed)".

Buzz looked perplexed "Sorry Alvin, but I can't solve for x or y".

"Well, he was your age when his son was born." said Alvin.

"You're right" said Buzz "He was a special old guy! But I still can't solve for x or y".

How old was the old guy when his son was born?


Assume that the nerds have the conversation this year, ie 2004 AD. 

A Special Old Guy Puzzle Solution

Firstly, we can say that the date of birth (DOB) for the special old guy (SOG) is:
Equation {a}:
DOB(SOG) = x^2 – x

Similarly, for the son of the special old guy (or SOSOG for short):

Equation {b}:
DOB(SOSOG) = y^3 – y

SOG's age when SOSOG was born was:
Equation {c}:
Age(SOG) = DOB(SOSOG) – DOB(SOG) = (y^3 – y) – (x^2 – x)

There are unlimited solutions to equation {c} so we need some assumptions and limits.

We know that SOG must have been between, say 10 years old and 100 years old when SOSOG was born:

Equation {d}:
10 < Age(SOG) < 100

We also know that both SOG and SOSOG were born some time since the year 500 AD:
Equation {e}:
500 < DOB(SOSOG) < 2004.

Based on equation {b} and {e} we can clearly see that there are only 5 solutions for y. They are y = 8, 9, 10, 11 or 12. Any other solutions for y are in breach of equation {e}.
For each of these possible solutions for y there is only a limited number of solutions for x that comply with {d}. They are:
 x     y
21     8
22     8
26     9
27     9
31    10
36    11
41    12

Any other solutions for x and y are in breach of equation {d}.
SOG's age when his son was born can be calculated for each of these possible solutions by using equation {c} as follows:
 x      y     Age(SOG)
21      8           84
22      8           42
26      9           70
27      9           18
31     10           60
36     11           60
41     12           76

So there are 7 different solutions, which is why Buzz said he couldn't solve for x and y.
Now comes the lateral part of the puzzle: Alvin informs Buzz that "...he was your age when his son was born". Of course, Buzz knows his own age. He should therefore be able to pick the correct solution from the list of 7 possible solutions shown above. However, he can't. That means that Buzz (and the special old guy) must be 60 because there are multiple solutions for an age of 60, whereas any other age would yield a unique solution. Any other age and Buzz would be able to solve.
A 60 year old father – quite a special old guy.

Friday, August 21, 2015

Switching Logic

1 comment :
bulbs logic game
You are in a basement. In front of you are three light switches, all in their OFF state. These switches are connected to three lightbulbs that are on a wall in the attic, so there's no chance you can see them unless you climb the stairs all the way up. There is a 1-to-1 mapping between the switches and the bulbs.

You are given plenty of time to play around with the switches in the basement, where you can put each individual switch in either its ON or OFF state.

However, you can only go upstairs once to check on the state of the bulbs!

Once you've gone upstairs and checked on the bulbs, you must be able to tell with 100% certainty which bulb is connected to which switch, without having to go down again.

How can you tell which switch is connected to each bulb?


  • You can't put a switch halfway between ON and OFF, hoping that this would make the bulb flicker like a bad neon light...
  • You can't control the switches from a distance, eg with a string or whatever other form of remote control.
  • You can't have anyone cooperating with you on the other floor, and that includes your dog who knows how to bark once for a 'yes' and twice for a 'no'.

Switching Logic Puzzle Solution

Turn switch #1 ON. After about five minutes or so, turn switch #1 OFF and turn switch #2 ON.
Then go upstairs and check on the bulbs.

The one ON is obviously #2. The other ones are OFF, but one of them should be very hot by having been ON for five minutes. That's #1, and the remaining bulb is #3.

Wednesday, July 22, 2015

The Diagram

Bob looked up from his book and noticed that Joe was staring at the VCR clock, holding a pencil and a pad of paper. He knew the clock was set correctly, having set it himself. But he noticed that Joe would occasionally write something down.

Finally Bob's curiosity got the best of him: "What are you doing, Joe?"

"Just another minute" came the reply.

"How long are you going to stare at the clock like that?" As Bob finished he noticed the clock advance a minute.

At this, Joe started scribbling some more until he produced this diagram:
6 8
4 9
"Why, just ten minutes, of course", Joe beamed as he showed the diagram to Bob.

What was Joe doing, and what does his diagram mean?

The Diagram Puzzle Solution

Joe counted how many times each of the 7 LEDs lit up for all 10 digits...

Wednesday, June 17, 2015

Weighing an Elephant

elephant riddle
Hundreds of years ago, a king of an Asian country who lived in a port city, received a visit from a king of adjacent country, his friend.

The visitor brought a present with him. It was an elephant.

The visitor gave it and said, "Can you measure the elephant's rough weight in a day?"

The king of the port city consulted with the retainers. "We just have beam scales weighing bags. Do you have any ideas?"

One vassal said, "I can make the relevant measuring equipment assembling large levers and pulleys, your majesty."

"Can you make it in a day?"

"......I can't."

Another vassal said, "How about weighing in pieces after killing the elephant?"

"I won't."

At last they found the method and measured the elephant's approximate weight without sophisticated devices.

What was the method?

Weighing an Elephant Puzzle Solution

Load the elephant onto a boat large enough to carry it. The boat will sink slightly, and you mark the level of the water on the side of the boat. Then you offload the elephant and fill the boat with bags until the boat sinks to the level marked. The bags can be individually weighed using beam scales and the weight of the elephant is the sum of the weight of the bags.

This puzzle is slightly cunning in that it the geographic location of the city is a small clue.

Friday, June 12, 2015

Carpet Layer

1 comment :
Walter Wall is a carpet layer. He and his two apprentices are asked by a nightclub owner to give a quote on laying carpet.

The owner indicates an oblong dance floor (figure on the right) and tells them that he wants a square of carpet adjacent each of the sides and running its entire length, making four squares in all (figure below left).

Walter asks both apprentices how many measurements must be made to calculate the total area of carpet needed in order to give a quote.
Sam, the slower of the two, replies that eight measurements are needed: two sides of each square.

Walter reprimands him, reminding him that these are squares and therefore have all sides the same length, and that they are in identical pairs, "So we only need to take two measurements - one side of one of the large squares and one side of one of the smaller squares".

Brian, the bright apprentice, points out that they can give the quote after taking only one measurement.

 How can the total area (that is, the sum of the areas of the four red squares) be calculated by taking just one measurement?

Carpet Layer Puzzle Solution

The only measurement that needs to be taken is the distance between opposite angles of the rectangular dance floor (figure on the left). That distance then will be squared (figure on the right) and doubled to get the sum of the areas of the 4 squares.

Tuesday, April 28, 2015

Socks And Shops

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socks colors puzzle
Sarks and Mencer's was a brand new shop built in the basement floor of a shopping centre. To mark their opening day, they had many special offers. The most popular was the buy ten pairs of socks for half price. Each pair of socks were attached together with a plastic clip that doubled as a miniture hanger whilst on display. There was only a single size and only ten colours to choose from, which is why both Andrew and Bob each chose one of each colour. Each armed with ten pairs of differently coloured socks, they were moving towards the checkout when there was a power failure. They bumped into each other and dropped all twenty pairs of socks.

Without being able to see the colour of the socks, how could they divide them so that they both got 10 pairs of uniquely coloured socks each?

Socks And Shops Puzzle Solution

For each of the twenty pairs of socks, they unclipped it and gave one sock each.

Thursday, April 23, 2015

Strawberry Ice Cream

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A man walks into a bar, orders a drink, and starts chatting with the bartender.

After a while, he learns that the bartender has three children. "How old are your children?" he asks.
"Well," replies the bartender, "The product of their ages is 72."

The man thinks for a moment and then says, "That's not enough information."

"All right," continues the bartender. "If you go outside and look at the building number posted over the door to the bar, you'll see the sum of the ages."

The man steps outside, and after a few moments he reenters and declares, "Still not enough!"

The bartender smiles and says, "My youngest just loves strawberry ice cream."

strawberry ice cream riddle

How old are the children?

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.

Wednesday, April 8, 2015

The Package

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G answered his ringing mobile phone. "G here."

"It's H. Do you have the package?"

"Yes. I have the object, in its box."

"Very good," said H. "So send the box to me using the lock I gave you last time we met."

"I'm afraid I can't," said G hestitantly.

"Why not? The locking ring on the box is more than large enough to fit a lock on."

"I seem to have um, misplaced the lock."

"You what!" snarled H. "Do you have any other locks?"

"Yes, but you don't have the keys to them. We can't risk sending an unsecured box, nor an unsecured key. Sending an unlocked lock is out of the question, because they re-lock themselves automatically after a 2 minute timeout."

"Ah, all is not lost... I have an spare lock here... How much time do we have?"

"Plenty," replied G.

"Good. There is a way…
adaugi alt

How can they get the box and the object from G to H without any security risk?


  • They can't meet in person -- that would render the puzzle pointless.
  • No, G can't go and look for the misplaced lock!
  • You cannot send a key by itself or in an unlocked box.

The Package Puzzle Solution

With the object in the box, G locks it with one of his own locks. He then sends the box to H. H attaches his own lock in addition to G's and then sends it back. G removes his lock and then sends it to H again.

Thursday, February 12, 2015

Napoleon's Star

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Napoleon had an obsession: a star. His star. He would talk about it to everyone, and whoever would listen to him out of respect, would point at the star in the sky. Napoleon even talked about the star during the Russian campaign, while his troops were receding.

It seems like Talleyrand sent him the game - Napoleon's Star - on the evening of 17th June, 1815, the day before the Battle of Waterloo. It has been said that the great general spent the entire night and the following day, until sunset, trying to solve the game, without hearing the noise of the battle and without listening to his officers pleading for help. When he came out of his tent to breath some fresh air, looking tired and unshaven, but with the solution in his grasp, Waterloo had already been won by the English, and his troops were fleeing with no order or hope.

napoleon star

Here's the game: start from any of the ten points, marked with a letter, and follow - in a staight line - to the third point from the starting position (eg from a to g); place a coin on this third point. Then pick another point unoccupied by any coin, and again go to a third unoccupied point in a straight line (a coin on the second point doesn't matter), and place a coin on it. Repeat the procedure until you've placed nine coins.

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.

napoleon star soution

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

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.

Sunday, October 5, 2014

Faulty Batches

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"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?

  • 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

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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.

Wednesday, May 21, 2014

Paul Lawrence: The Enigma

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Paul Lawrence is a showman with blue puzzle pieces tattooed all over his body who has made a name for himself performing  tricks such as sword swallowing, pushing a moving power drill up his nose and swallowing various liquids, pumping them out of his stomach and swallowing them again. He is also known as The Enigma and is an accomplished musician as well.

Wednesday, April 16, 2014

Marty Cooper Artwork

1 comment :

Animator artist from San Francisco Marty Cooper, got tired of everything that’s happening around him. He took a transparent celluloid film, a pen and white pencil, and began to change the world in which he lives. The results of his activities, he fixes on the iPhone and puts on his Tumblr. Sometimes funny , sometimes sad , sometimes - quite unexpected pictures on the verge of "Who Framed Roger Rabbit " and "another world ." Very original and creative artwork expressed in an unusual manner. '

Monday, February 10, 2014

Sony World Photography Awards 2014 – Best Works

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Here is a collection of an amazing photo gallery of works that were presented at Sony World Photography Awards 2014. Discover and admire these amazing pictures and find other great photography and imagery by solvent online jigsaw puzzles.