Showing posts with label logic games. Show all posts

Wednesday, August 26, 2015

A Special Old Guy

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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, July 17, 2015

How Many Monks?

how many monks riddle
There's a monastry of fifty monks that have taken a vow of silence. They go through the same routine every day. They wake up, pray. They go to lunch, their only meal of the day, were they all sit round a big circular table, and eat a simple meal. Then they go back to their rooms and pray. Then they fall asleep.

One day, the head monk of the area comes to visit. He's a bit different - he's allowed to talk and has a life outside the monastry. He tells them that there is currently a plague ravaging the land. People everywhere are dying. The disease manifests itself as red blotches on the forehead. The blotches are the only manifestation of the disease for three months, whereupon the next stage starts, a horribly painful death.

The head monk tells them that at least one of their midst will have the disease, probably more. Anybody with the disease should kill themselves, to save all of the pain and suffering. By killing themselves, they will restrict the movement of the disease, and will go to heaven. Anybody with the disease will show the first symptoms within a month. The chief monk then leaves. He returns two months later, to find that all of the infected monks have killed themselves, and they all did it on the same day.

Bear in mind the following:
  • They have no mirrors or any other way of seeing themselves
  • The blotches appear only on the forehead and cannot be seen by the monk
  • Infected monks feel no different - the only manifestation of the first part of the disease are the blotches.
  • The monks cannot talk to each other or in any other way communicate.
  • Any monks with the disease will display the blotches within a month of the head monk leaving.
  • At least one monk definitely has the disease.
  • The monks only see each other once per day, at lunch, when they are all sat round the round table.
  • These monks are brighter than the average monk....
How did they know whether or not they were infected and why did they all kill themselves on the same day?

How Many Monks? Puzzle Solution

The way they work this out and the reason it happens on the same day is as follows. It is important to note that not only are the monks intelligent but they all know all their fellow monks are as well.
Start with one monk:

He knows he is the only one who can have the disease so he kills himself on
day 1.

Then two monks:
They know that either one of them has the disease or both do. If monk1
doesn't see the mark of the disease on monk2 then he will realise that he
must have it so will kill himself. If he sees the mark then he knows that
monk2 will following the same reasoning and kill himself. If the next day
the other monk is still alive then he realises that the other monk must have
seen the mark on him and so they both have the disease and he must kill
himself. Both follow the same reasoning and kill themselves on day 2.

Three monks:
If only one of the monks has the disease he will see no mark on the other
two and so diagnose himself. He will kill himself on day one. If two monks
have the disease then they will each see one monk with the mark and one
without. When they see each other again the next day they will deduce that
the monk they see with the mark would only not have killed himself if he
could see someone else with the mark. They know it is not the third monk so
it must be them. Both diseased monks follow the same reasoning and kill
themselves on day two. If all three monks have the disease then they are all
in the same position as the healthy monk in looking on in the previous
example. Each of them can see two diseased men. When they haven't both
killed themselves on day two there can be only one reason - the viewing monk
must have the mark as well. All taking the same reasoning they all kill
themselves on day three.

And so on...
N monks all with the disease will all kill themselves on day N.

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.

Wednesday, October 15, 2014

Cheers To Statistics

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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).

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.

Thursday, September 25, 2014

Three Skullcaps

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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.

Monday, September 15, 2014


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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.