# amit_puzzles by RamSingh20

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```									  Lateral Thinking Logic Problems More
Very Easy - Easy - Difficult - Very Difficult

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1. The Camels                                                                Print Solutions

Four tasmanian camels traveling on a very narrow ledge encounter four       Hide All
tasmanian camels coming the other way.                                      Show All

As everyone knows, tasmanian camels never go backwards, especially          Site Map
when on a precarious ledge. The camels will climb over each other, but
only if there is a camel sized space on the other side.                      Send Feedback

The camels didn't see each other until there was only exactly one camel's
width between the two groups.                                                Discussion Board

How can all camels pass, allowing both groups to go on their way,
without any camel reversing?

Show Hint Show Solution

View the Solutions

Hint:
Use match sticks or coins to simulate the puzzle.

etc....

2. The Waiter

Three men in a cafe order a meal the total cost of which is \$15. They
each contribute \$5. The waiter takes the money to the chef who
recognizes the three as friends and asks the waiter to return \$5 to the
men.

The waiter is not only poor at mathematics but dishonest and instead of
going to the trouble of splitting the \$5 between the three he simply gives
them \$1 each and pockets the remaining \$2 for himself.

Now, each of the men effectively paid \$4, the total paid is therefore \$12.
Add the \$2 in the waiters pocket and this comes to \$14.....where has the
other \$1 gone from the original \$15?

Show Solution

View the Solutions

The payments should equal the receipts. It does not make sense to add
what was paid by the men (\$12) to what was received from that payment
by the waiter (\$2)

Although the initial bill was \$15 dollars, one of the five dollar notes gets
changed into five ones. The total the three men ultimately paid is \$12, as
they get three ones back. So from the \$12 the men paid, the owner
receives \$10 and the waiter receives the \$2 difference. \$15 - \$3 = \$10 +
\$2

3. The Boxes
There are three boxes. One is labeled "APPLES" another is labeled
"ORANGES". The last one is labeled "APPLES AND ORANGES". You
know that each is labeled incorrectly. You may ask me to pick one fruit
from one box which you choose.

How can you label the boxes correctly?

Show Solution

View the Solutions

Pick from the one labeled "Apples & Oranges". This box must contain
either only apples or only oranges.

E.g. if you find an Orange, label the box Orange, then change the
Oranges box to Apples, and the Apples box to "Apples & Oranges"

4. The Cannibals

Three cannibals and three anthropologists have to cross a river.

The boat they have is only big enough for two people. The cannibals will
do as requested, even if they are on the other side of the river, with one
exception. If at any point in time there are more cannibals on one side of
the river than anthropologists, the cannibals will eat them.

What plan can the anthropologists use for crossing the river so they don't
get eaten?

Note: One anthropologist can not control two cannibals on land, nor can
one anthropologist on land control two cannibals on the boat if they are
all on the same side of the river. This means an anthropologist will not
survive being rowed across the river by a cannibal if there is one
cannibal on the other side.

Show Solution

View the Solutions

First, two cannibals go across to the other side of the river, then the
rower gets called back. Next, the rowing cannibal takes the second
across and then gets called back, so now there are two cannibals on the
far side.

Two anthropologists go over, then one anthropologist accompanies one
cannibal back, so now there is one anthropologist and one cannibal on
the far side.

The last two anthropologists go over to the far side, so now all the
anthropologists are across the other side, along with the boat and one
cannibal.

In two trips, the cannibal on the far side takes the boat and ferries the
other two cannibals across the river.

5. The Father

A mother is 21 years older than her child. In exactly 6 years from now,
the mother will be exactly 5 times as old as the child.

Where's the father?

Show Solution

View the Solutions

With the mother. If you do the math, you find out the child will be born
in 9 months.

6. The Double Jeopardy Doors

You are trapped in a room with two doors. One leads to certain death
and the other leads to freedom. You don't know which is which.

There are two robots guarding the doors. They will let you choose one
door but upon doing so you must go through it.

You can, however, ask one robot one question. The problem is one robot
always tells the truth ,the other always lies and you don't know which is
which.
What is the question you ask?

Show Hint Show Solution

View the Solutions

Hint: The two robots know each others personality. That they talk when
they're bored, lonely, etc. Try to get the two robots to cancel their evil &
good ways out.

which door was safe. Then go through the other door.

7. The Frog

A frog is at the bottom of a 30 meter well. Each day he summons enough
energy for one 3 meter leap up the well. Exhausted, he then hangs there
for the rest of the day. At night, while he is asleep, he slips 2 meters
backwards. How many days does it take him to escape from the well?

Note: Assume after the first leap that his hind legs are exactly three
meters up the well. His hind legs must clear the well for him to escape.

Show Hint Show Solution

View the Solutions

Hint: Try to think the problem through for a five meter well. Now what
is the solution for the 30 meter well?

Each day he makes it up another meter, and then on the twenty seventh
day he can leap three meters and climb out.

8. The Bobber

You can paddle your canoe seven miles per hour through any placid
lake. The stream flows at three miles per hour. The moment you start to
paddle up stream a fisherman looses one of his bobbers in the water
fourteen miles up stream of you.
How many hours does it take for you and the bobber to meet?

Show Solution

View the Solutions

Ignore the speed of the stream, as the cork will be carried along at three
miles per hour as will you. It takes two hours to travel fourteen miles, at
a rate of seven miles per hour.

10. The Socks

Cathy has six pairs of black socks and six pairs of white socks in her
drawer.

In complete darkness, and without looking, how many socks must she
take from the drawer in order to be sure to get a pair that match?

Show Solution

View the Solutions

Socks do not come in in left and right, so any black will pair with any
other black and any white will pair with any other white. If you have
three socks and they are either colored black or white, then you will have
at least two socks of the same color, giving you one matching pair.

11. There is something about Mary

Mary's mum has four children.
The first child is called April.
The second May.
The third June.
What is the name of the fourth child?

Show Solution
View the Solutions

Mary's mothers fourth child was Mary herself.

12. Petals Around the Rose

The name of the game is Petals Around the Rose, and that name is
significant. Newcomers to the game can be told that much. They can also
be told that every answer is zero or an even number. They can also be
told the answer for every throw of the dice that are used in the game.
And that's all the information they get.

The person who has the dice and knows the game, rolls five dice and
remarks almost instantly on the answer. For example: in Roll #1 the

Roll #1.

"The answer is what?" says the new player.

"Two."

"On that roll?"

"Yes."

"Would it still be two if I moved the dice without turning any of them
over, just rearranging the pattern?"

"I can tell you only three things: the name of the game, the fact that the
answer is always even, and the answer for any particular throw. In this

"So that's how it is. What am I supposed to do?"

"You're supposed to tell me the answer before I tell you. I'll give you all
the time you want, but don't tell me your theory, just the answer. If you
figure it out, you don't want to give the idea away to these other jokers
around you. Make them work for the answers, too. If you get the answer
right on six successive rolls, I'll take that as prima facie evidence that
you understand the game."

"OK, roll again."

Roll #2.

"I give up. What's the answer?"

"Roll again."

Roll #3.

Roll #4.

Roll #5.

Roll #6.

An integral part of the puzzle is that those who have solved it are urged
to keep the solution a secret, so there is no solution posted here. It is not
a hard puzzle to figure out however.

A claim that often accompanies these instructions is that the smarter an
individual, the greater amount of difficulty the individual will have in
solving it. If such a statement is true, it may be attributed to the fact that
"smarter" people tend to be more knowledgeable in a wide range of
information which they may unnecessarily attempt to draw upon to solve
the puzzle.
Very Easy - Easy - Difficult - Very Difficult

More Very Easy Logic Puzzles >>

Lateral Thinking Logic Problems More
Very Easy - Easy - Difficult - Very Difficult

1. The Camels

Four tasmanian camels traveling on a very narrow ledge encounter four tasmanian camels
coming the other way.

As everyone knows, tasmanian camels never go backwards, especially when on a precarious
ledge. The camels will climb over each other, but only if there is a camel sized space on the
other side.

The camels didn't see each other until there was only exactly one camel's width between the two
groups.

How can all camels pass, allowing both groups to go on their way, without any camel reversing?

Show Hint Show Solution

View the Solutions

Hint:
Use match sticks or coins to simulate the puzzle.

etc....

2. The Waiter

Three men in a cafe order a meal the total cost of which is \$15. They each contribute \$5. The
waiter takes the money to the chef who recognizes the three as friends and asks the waiter to
return \$5 to the men.

The waiter is not only poor at mathematics but dishonest and instead of going to the trouble of
splitting the \$5 between the three he simply gives them \$1 each and pockets the remaining \$2
for himself.

Now, each of the men effectively paid \$4, the total paid is therefore \$12. Add the \$2 in the
waiters pocket and this comes to \$14.....where has the other \$1 gone from the original \$15?

Show Solution

View the Solutions

The payments should equal the receipts. It does not make sense to add what was paid by the
men (\$12) to what was received from that payment by the waiter (\$2)
Although the initial bill was \$15 dollars, one of the five dollar notes gets changed into five ones.
The total the three men ultimately paid is \$12, as they get three ones back. So from the \$12 the
men paid, the owner receives \$10 and the waiter receives the \$2 difference. \$15 - \$3 = \$10 + \$2

3. The Boxes

There are three boxes. One is labeled "APPLES" another is labeled "ORANGES". The last one
is labeled "APPLES AND ORANGES". You know that each is labeled incorrectly. You may
ask me to pick one fruit from one box which you choose.

How can you label the boxes correctly?

Show Solution

View the Solutions

Pick from the one labeled "Apples & Oranges". This box must contain either only apples or only
oranges.

E.g. if you find an Orange, label the box Orange, then change the Oranges box to Apples, and
the Apples box to "Apples & Oranges"

4. The Cannibals

Three cannibals and three anthropologists have to cross a river.

The boat they have is only big enough for two people. The cannibals will do as requested, even
if they are on the other side of the river, with one exception. If at any point in time there are
more cannibals on one side of the river than anthropologists, the cannibals will eat them.

What plan can the anthropologists use for crossing the river so they don't get eaten?

Note: One anthropologist can not control two cannibals on land, nor can one anthropologist on
land control two cannibals on the boat if they are all on the same side of the river. This means an
anthropologist will not survive being rowed across the river by a cannibal if there is one
cannibal on the other side.

Show Solution

View the Solutions

First, two cannibals go across to the other side of the river, then the rower gets called back.
Next, the rowing cannibal takes the second across and then gets called back, so now there are
two cannibals on the far side.

Two anthropologists go over, then one anthropologist accompanies one cannibal back, so now
there is one anthropologist and one cannibal on the far side.

The last two anthropologists go over to the far side, so now all the anthropologists are across the
other side, along with the boat and one cannibal.

In two trips, the cannibal on the far side takes the boat and ferries the other two cannibals across
the river.

5. The Father

A mother is 21 years older than her child. In exactly 6 years from now, the mother will be
exactly 5 times as old as the child.

Where's the father?

Show Solution

View the Solutions

With the mother. If you do the math, you find out the child will be born in 9 months.

6. The Double Jeopardy Doors

You are trapped in a room with two doors. One leads to certain death and the other leads to
freedom. You don't know which is which.

There are two robots guarding the doors. They will let you choose one door but upon doing so
you must go through it.

You can, however, ask one robot one question. The problem is one robot always tells the truth
,the other always lies and you don't know which is which.

What is the question you ask?
Show Hint Show Solution

View the Solutions

Hint: The two robots know each others personality. That they talk when they're bored, lonely,
etc. Try to get the two robots to cancel their evil & good ways out.

Answer: Ask one robot what the other robot would say, if it was asked which door was safe.
Then go through the other door.

7. The Frog

A frog is at the bottom of a 30 meter well. Each day he summons enough energy for one 3 meter
leap up the well. Exhausted, he then hangs there for the rest of the day. At night, while he is
asleep, he slips 2 meters backwards. How many days does it take him to escape from the well?

Note: Assume after the first leap that his hind legs are exactly three meters up the well. His hind
legs must clear the well for him to escape.

Show Hint Show Solution

View the Solutions

Hint: Try to think the problem through for a five meter well. Now what is the solution for the 30
meter well?

Each day he makes it up another meter, and then on the twenty seventh day he can leap three
meters and climb out.

8. The Bobber

You can paddle your canoe seven miles per hour through any placid lake. The stream flows at
three miles per hour. The moment you start to paddle up stream a fisherman looses one of his
bobbers in the water fourteen miles up stream of you.

How many hours does it take for you and the bobber to meet?

Show Solution
View the Solutions

Ignore the speed of the stream, as the cork will be carried along at three miles per hour as will
you. It takes two hours to travel fourteen miles, at a rate of seven miles per hour.

10. The Socks

Cathy has six pairs of black socks and six pairs of white socks in her drawer.

In complete darkness, and without looking, how many socks must she take from the drawer in
order to be sure to get a pair that match?

Show Solution

View the Solutions

Socks do not come in in left and right, so any black will pair with any other black and any white
will pair with any other white. If you have three socks and they are either colored black or
white, then you will have at least two socks of the same color, giving you one matching pair.

11. There is something about Mary

Mary's mum has four children.
The first child is called April.
The second May.
The third June.
What is the name of the fourth child?

Show Solution

View the Solutions

Mary's mothers fourth child was Mary herself.
12. Petals Around the Rose

The name of the game is Petals Around the Rose, and that name is significant. Newcomers to
the game can be told that much. They can also be told that every answer is zero or an even
number. They can also be told the answer for every throw of the dice that are used in the game.
And that's all the information they get.

The person who has the dice and knows the game, rolls five dice and remarks almost instantly
on the answer. For example: in Roll #1 the answer is two.

Roll #1.

"The answer is what?" says the new player.

"Two."

"On that roll?"

"Yes."

"Would it still be two if I moved the dice without turning any of them over, just rearranging the
pattern?"

"I can tell you only three things: the name of the game, the fact that the answer is always even,
and the answer for any particular throw. In this case the answer is two."

"So that's how it is. What am I supposed to do?"

"You're supposed to tell me the answer before I tell you. I'll give you all the time you want, but
don't tell me your theory, just the answer. If you figure it out, you don't want to give the idea
away to these other jokers around you. Make them work for the answers, too. If you get the
answer right on six successive rolls, I'll take that as prima facie evidence that you understand the
game."

"OK, roll again."

Roll #2.

"I give up. What's the answer?"

"Roll again."

Roll #3.

Roll #4.

Roll #5.

Roll #6.

An integral part of the puzzle is that those who have solved it are urged to keep the solution a
secret, so there is no solution posted here. It is not a hard puzzle to figure out however.

A claim that often accompanies these instructions is that the smarter an individual, the greater
amount of difficulty the individual will have in solving it. If such a statement is true, it may be
attributed to the fact that "smarter" people tend to be more knowledgeable in a wide range of
information which they may unnecessarily attempt to draw upon to solve the puzzle.

Four people meet in a room. Each person shakes hands once with each other person.

How many hand shakes are there in all?

Message
Posted: Thu Dec 23, 2010 5:44 am by Rag Post subject: Answer

Message
Posted: Wed Mar 16, 2011 5:49 pm by wesmode12 Post subject:

Message
Posted: Fri May 27, 2011 5:57 am by ychick89c Post subject:

ex: there is guy 1,2,3, and 4
guy one shakes hands with:
1-2, 1-3, 1-4
thats 3 hand shakes
continue with guy 2,3, and 4. or do 3 times 4. either way you find 12

Message
Posted: Sun Jun 05, 2011 8:34 am by aamil4u Post subject:
I agree with wesmode. The answer is 6
_________________
"Live as if you will die tomorrow,
Learn as if you will live forever.."
-Mahatama Gandhi

there is the entrance of the place where two guards are there outside,one guard is facing north
and another south, the one on the north smiles and the another one asks why ? how ?
Message
Posted: Thu Jan 06, 2011 1:42 pm by pollpo Post subject:
there facing each other, one north and one south.

I am very useful.
I can be picked up.
I can carry water.
I have holes in me.
I am very common.

Who am I?

(ok I know this is easy but I just want to share this with you)

Message
Posted: Sat Mar 22, 2008 2:28 am by Eontios Post subject:
Not that easy....a watering hole...a bucket....a toilet? lol
_________________
Hi.

Message
Posted: Sun Mar 23, 2008 2:04 pm by Minnie Post subject: hey
You are a SPONGE

Message
Posted: Tue Dec 16, 2008 4:16 pm by leakybeak Post subject:
could be a person

Message
Posted: Mon Dec 22, 2008 7:58 pm by alexonfyre Post subject:
A Watering Can?

Message
Posted: Sat Nov 07, 2009 3:54 pm by shobainfo Post subject: Re: What am I?
sakura wrote:
I am very useful.
I can be picked up.
I can carry water.
I have holes in me.
I am very common.

Who am I?

(ok I know this is easy but I just want to share this with you)

is this bucket...

Message
Posted: Sat Dec 05, 2009 12:08 pm by mdbbarilla Post subject:
This is a watering can.

Message
Posted: Thu Dec 23, 2010 5:46 am by Rag Post subject: Answer
SPONGE

i have a bag, with some balls in it. all but 4 are blue, all but 4 are green, and all but 4 are red.
how many balls do i have in total?

Message
Posted: Mon Nov 23, 2009 5:24 pm by rj82330 Post subject: 2 by 2 by 2
Six
_________________
Forget Critical Thinking, why isn't there a GCSE in LATERAL Thinking?

Message
Posted: Sat Sep 25, 2010 4:39 pm by Legolas_Acer Post subject: answer
5. right?

Message
Posted: Thu Dec 23, 2010 5:45 am by Rag Post subject: answer
12

two fathers gave their two sons some money.one gave his son 150 pesos and other father gave
100 pesos to his son.when the two sons counted their money,they found that all together they had
become richer by only 150 pesos.how come?
Message
Posted: Thu May 20, 2010 6:41 am by rezaru2000 Post subject:
They are

Father -> Son -> Grand Son

A farmer has to sail three, a tiger , a goat and grass across the river and there is only one boat.
The boat can carry only two of the above four at a time. How Farmer manages all this.

Hint.
Neither tiger will consume the goat nor the goat will consume the grass, if farmer is present.
Farmer can have more than one trip.

Message
Posted: Sun Dec 27, 2009 6:53 am by applescript Post subject:
Farmer + goat.
Farmer returns.
Farmer + tiger.
Farmer + goat return.
Farmer + grass.
Farmer returns.
Farmer + goat.
You are on your way to Salalalah and it is midnight and you are in the middle of the dessert,
when your tire goes flat. You remove the four bolts of your tire and keep them on the side of the
road. Suddenly, a truck passes by at a very high speed and your bolts just go flying all over and
you can't find them. What do you do to replace the tyre? You know there is a Tyre Repair Shop
(equipment supplied by TTC ) about 20 KM away. As it is midnight no one will stop to help you.
There are no camels around that you can ride and if you are thinking of using a flash light or wait
till the sun rises hmmm its not the answer .

Message
Posted: Sun Aug 02, 2009 2:54 am by porcelina Post subject: Salahlahlah
Would you remove one bolt from each of the other tires and then use them to put on your spare?

Message
Posted: Sun Dec 27, 2009 6:48 am by applescript Post subject:
erm I thought tires only need nuts to be fixed onto a car? So yeah..its a trick question?

During the lunch hour at school, a group of five boys from Miss Jones home room visited a
nearby lunch wagon. one of the five boys took a candy bar without paying for it. When the boys
were questioned by the school principal, they made the following statements in respective order:
1. Rex: "Neither Earl nor I did it."

2. Jack: "It was Rex or Abe."

3. Abe: "Both Rex and Jack are lying."
4. Dan: "Abe's statement is not true; one of them is lying and the other is speaking the truth."

5. Earl: "What Dan said is wrong."

When Miss Jones was consulted, she said, "Three of these boys are always truthful, but
everything that two of them say will be a lie." Assuming that Miss Jones is correct, can you
determine who took the candy bar?

Message
Posted: Thu Dec 13, 2007 2:14 pm by Absentee Post subject:
Abe?

Message
Posted: Mon Dec 22, 2008 8:27 pm by alexonfyre Post subject:
Since Dan and Earl contradict each other one of them must be lying. If Earl is lying then Abe is
lying and so is one of Rex and Jack, which is not possible (Earl + Abe + one of Jack and Rex = 3
liars). If Earl is telling the truth, then Dan is lying. However, only one part of Dan's statement
has to be false for him to be lying, if the first part is false then that means that both Jack and Rex
are lying, which is impossible again (Dan + Jack & Rex = 3 liars), which means the first part is
definitely true and the second part is surely false. This means that not only is the statement (one
of [jack and rex] is lying and the other telling the truth) false, but so is (both jack and rex are
lying) leaving only one possibility, that they are both telling the truth.

So:
Truth-tellers:
Rex, Jack, Earl
Lairs:
Dan & Abe

So Rex says that neither Earl nor he did it, which we know is true
Jack says it was Rex or Abe, but because what Rex said was true, it cannot be Rex so Abe is the
culprit.
=)

In short, Absentee is correct
you are an oil mogul considering the purchase of drilling rights to an as yet unexplored tract of
land.
the well's expected value to its current owners is uniformly distributed over [\$1..\$100]. (i.e., a
1% chance it's worth each value b/w \$1..\$100, inclusive).
bcause you have greater economies of scale than the current owners, the well will actually be
worth 50% more to you than to them (but they don't know this).
the catch: although you must bid on the well before drilling starts (and hence, before the actual
yield of the well is known), the current owner can wait until *after* the well's actual value is
ascertained before accepting your bid or not.
what should you bid?

Message
Posted: Sat Mar 22, 2008 2:29 am by Eontios Post subject:
You should bid a cent more than the others....i think XD
_________________
Hi.

Message
Posted: Mon Dec 22, 2008 8:03 pm by alexonfyre Post subject:
I would buy a tenth of the land for 15 bucks (enough to put my wells on and roads or a pipeline
to get out), and then get all of the oil on the whole tract through drainage.
There was this clown. He saw the dust and die. Why did he die? Like I mean elaborate.

Hints: He is a migit
His size does not matter.

Message
Posted: Mon Jun 30, 2008 11:45 am by skintc Post subject:
He could see a dust storm coming, but cos he's a clown everyone thought he was joking. So they
all died.

Message
Posted: Thu Jul 03, 2008 1:05 am by smartgirl23 Post subject: Death
Was the dust sawdust? If so, did he die because he killed him self as someone was cutting
sawdust from his cane or such and he died because he thought that he was growing taller? :|
_________________
Intelligence is the key to excellence.

Message
Posted: Mon Dec 22, 2008 7:55 pm by alexonfyre Post subject:
Even if it isn't correct (though I think it should be) I like skintc's answer the best =)

Anyway, here is my ridiculous answer (though this riddle is lateral thinking, not logic)

Midget clowns are perfect for shooting out of cannons, so he was the cannon clown, however
one day the cannon missed its target (the safety net) and he saw the dust on the circus tent floor
(which is generally a dirt floor or covered in sawdust for a variety of reasons) before he landed
and died.
1. You are driving down the road in your car on a wild, stormy night, when you pass by a bus
stop and you see three people waiting for the bus

   An old lady who looks as if she is about to die.
   An old friend who once saved your life.
   The perfect partner you have been dreaming about.

Knowing that there can only be one passenger in your car, whom would you choose?

Show Hint Show Solution

Hint: You can make everyone happy.

Your car can only contain one passenger, so whom should it be?

to your friend, and wait with your perfect partner for the bus

2. Acting on an anonymous phone call, the police raid a house to arrest a suspected murderer.
They don't know what he looks like but they know his name is John and that he is inside the
house. The police bust in on a carpenter, a lorry driver, a mechanic and a fireman all playing
poker. Without hesitation or communication of any kind, they immediately arrest the fireman.
How do they know they've got their man?

Show Hint Show Solution

Hint: The police only know two things, that the criminal's name is John and that he is in a
particular house.

Answer: The fireman is the only man in the room. The rest of the poker players are women.

3. A man lives in the penthouse of an apartment building. Every morning he takes the elevator
down to the lobby and leaves the building. Upon his return, however, he can only travel halfway
up in the lift and has to walk the rest of the way - unless it's raining. What is the explanation for
this?

Show Hint Show Solution

Hint: He is very proud, so refuses to ever ask for help.

Answer: The man is a dwarf. He can't reach the upper elevator buttons, but he can ask people to
push them for him. He can also push them with his umbrella.

4. How could a baby fall out of a twenty-story building onto the ground and live?

Show Hint Show Solution

Hint: It does not matter what the baby lands on, and it has nothing to do with luck.

Answer: The baby fell out of a ground floor window.

5. Bad Boy Bubby was warned by his mother never to open the cellar door or he would see
things that he was not meant to see. One day while his mother was out he did open the cellar
door. What did he see?

Show Hint Show Solution

Hint: His mother was an odd woman.

Answer: When Bad Boy Bubby opened the cellar door he saw the living room and, through its
windows, the garden. He had never seen these before because his mother had kept him all his life
in the cellar.

6. A man and his son are in a car crash. The father is killed and the child is taken to hospital
gravely injured. When he gets there, the surgeon says, 'I can't operate on this boy - for he is my
son!!!' How can this possibly be?

Show Hint Show Solution

Hint: This has nothing to do with adoption or time travel.

Answer: The surgeon can not operate on her own son; she is his mother.
7. There are six eggs in the basket. Six people each take one of the eggs. How can it be that one
egg is left in the basket?

Show Hint Show Solution

Hint: A alternate version of the problem is... Saradhi, Nick & Ted win a raffle contest. The prize
is three hard boiled eggs in a basket. After discussing how to divide the prize, each take one egg.
Nick & Ted get hungry and so eat their eggs. One of the original eggs is still left in original

Answer: The last person took the basket with the last egg still inside.

brown,jones & smith are suspected of income tax evasion.They testify,under oath as follows
Brown: jone is guilty & smith is innocent
Jones: if Brown is guilty,then so is Smith
Smith: I am innocent but at least one of the other's is guilty.
a)assuming every body told the truth who is or are innocent or guilty?
b)assuming the innocent told the truth & guilty lied who is or are innocent or guilty?

Message
Posted: Thu May 19, 2011 4:28 pm by Verglas Post subject:
a) Jones is guilty and the other 2 are innocent
b) Brown and Smith are guilty and Jones is innocent

I think all sane people are cachers
and one third of all cachers are sane
but half of all whackos are cachers
with only one whacko that s sane

If eight whackos are cachers
and ninety are attending my ball
how many cachers are neither
sane nor whacko at all?
Message
Posted: Fri Mar 18, 2011 11:47 pm by ChibiHoshi Post subject: 48
90 total - 8 wackos = 82 cachers
82- 27 sane (third of 82) = 55
1 of the 8 cacher-wackos is one of these sane folk so - 7 cacher-wackos = 48 plain cachers

Message
Posted: Tue May 17, 2011 8:18 am by Dalton Mcmickle Post subject:
hiiiii...............
thanks dear to share knowledge about role of forums...
:lol:

__________________________
ps3 wireless controller, wenzel tents, tents, tent, ps3 games

Achmed and Ali are camel-drivers and on one day they decided to quit their job. They wanted to
become shepherds. So they went to the market and sold all their camels. The amount of
money(dinars) they received for each camel is the same as the total of camels they owned. For
that money they bought as many sheep as possible at 10 dinars a sheep. For the money that was
left they bought a goat.
On their way home they got in a fight and decided to split up. When they divided the sheep there
was one sheep left. So Ali said to Achmed "I take the last sheep and you can get the goat".
"That's not fair" said Achmed, "a goat costs lesser than a sheep". "Ok", Ali said "then I will give
you one of my dogs and then we are even". And Achmed agreed.
What costs a dog?

Message
Posted: Tue Jan 08, 2008 5:53 am by aatifahmed Post subject: Solution to what
costs a dog?!
The dog costs 4 dinar.
The solution I am submitting below is encrypted so as not to take the fun out of your efforts! To
decrypt use: www.rot-n.com
The cipher used is ROT18-ROT5 & ROT13.

Abgr gung Nyv naq Npuzrq unq rdhny ahzore bs pnzryf (yrgf pnyy a). Urapr;
Gbgny ahzore bs pnzryf = 7a.
Gbgny zbarl gurl erprvir sbe pnzryf = 9a^7
Gur furrcf pbhyq abg or qvfgevohgrq rirayl nzbat gurz, urapr vgf na bqq ahzore (yrgf fnl 7z+6).
Naq yrgf fnl gung pbfg bs bar tbng vf t naq bar qbt vf q.
Gurersber, 65*(7z+6)+p = 9a^7
Abgr p unf gb or rira nf gur EUF vf rira. Nyfb abgr gung p vf gur havg qvtvg bs YUF, jurernf
gur EUF vf n fdhner, yrnivat p gb or bayl 9 be ; (fvapr 7 naq = pna arire or havg qvtvg bs n
fdhner ahzore, naq 5 vf abg cbffvoyr).
Abj,
:z + (65+p)/9 = a^7
Guhf p unf gb qvivfvoyr ol 9, juvpu yrnir p gb or bayl ;.
Fvapr p vf ; naq gur qvfgevohgvba vf rdhny fb q unf gb or 9!!

Message
Posted: Mon Dec 22, 2008 9:48 pm by alexonfyre Post subject:
I don't think it ever said they have equal number of camels aatifahmed...though it works if they
do...I am pretty sure it is still true if they aren't equal numbers, however.

Message
Posted: Sat Mar 19, 2011 12:55 am by ChibiHoshi Post subject: 4
interestingly enough, if you look at a table of perfect squares, those with the tens place as an odd
number (the only way there could be odd sheep) are always followed by a 6 (cost of a goat)
16,36,196,256,576,676,1156,1296,1936, etc.
So sheep-goat = dog
10-6 is 4

Two friends, whom we will call Arthur and Robert, were curators at the Museum of American
History. Both were born in the month of May, one in 1932 and the other a year later.

Each was in charge of a beautiful antique clock. Both of the clocks worked pretty well,
considering their ages, but one of them lost ten seconds an hour and the other gained ten seconds
an hour.

On one bright day in January, the two friends set both clocks right at exactly 12 noon.

``You realize,'' said Arthur, ``that the clocks will start drifting apart, and they won't be together
again until---let's see---why, on the very day you will be 47 years old. Am I right?''

Robert then made a short calculation. ``That's right!'' he said.

Who is older, Arthur or Robert?

Message
Posted: Sat Mar 19, 2011 12:29 am by ChibiHoshi Post subject: Robert
The clock would have been set in either 1979 or 1980 for one of them to turn 47. (1932 or 1933
birth year)

The clocks will resync in 90 days. (Simple conversion calculation of 10s/hr would take 90 days
to lose/gain 6 hours which is when both would resync)

The Jan-May gap could only work if it was set 31st January and a non-leap year (1980 was a
leap year and therefore the clock would have synced on April 30th)

So Robert, the one reaching 47 on May 1st of 1979 was the one born in 1932 and older

A person crosses a river on a paddle boat.
He took 4 hours while paddling against the water flow
and 3 hours while paddling in the direction of the water flow.
On still waters, how long would he take?
Message
Posted: Fri Mar 18, 2011 5:15 pm by ChibiHoshi Post subject:
Assuming the distances are all equal and they cancel each other out

Rower+River=3 hours
Rower-River=4 hours
Basic algebra

2xRower=7

Rower alone takes 3.5 hours

The Island of Baal
Of all the islands of knights and knaves, the island of Baal is the weirdest and most remarkable.
This island is inhabited exclusively by humans and monkeys. The monkey, as well as every
human, is either a kngith or a knave.
In the dead center of this islands stands the Temple of Baal, one of the most remarkable temples
in the entire universe. The high priests are metaphysicians, and in the Inner Sanctum of the
temple can be found a priest who is rumored to know the answer to the ultimate mystery of the
universe; why there is something instead of nothing.
Aspirants to the Sacred Knowledge are allowed to visit the Inner Sanctum, provided that they
prove themselves worthy by passing three series of tests. I learned all these secrets, incidentally,
by stealth: I had to enter the temple disguised as a monkey. I did this at great personal risk. Had I
been caught, the penalty would have been unimaginable. Instead of merely annihilating me, the
priests would have changed the very laws of the universe in such a way that I could never have
been born!
Well, a philosopher who was searching for the answer to the question, Wy is there something
rather than nothing?" arrived on the island of Baal and agreed to try the tests. The first series
took place on three consecutive days in a huge room called the Outer Sanctum. In the center of
the rrom a cowled figure was seated on a golden throne. He was either a human or a monkey,
and also a knight or a knave. He uttered a sacred sentence, and from this sentence the
philosopher had to deduce exactly what he was whether a knight or a knave, and whether a
human or a monkey.
The First Test
The speaker said, "I am either a knave or a monkey." Exactly what is he?
The Second Test
The speaker said, "I am a knave and a monkey." Exactly what is he?
The Thrird Test
The speaker said, "I am not both a monkey and a knight." What is he?

Message
Posted: Mon Nov 23, 2009 5:31 pm by rj82330 Post subject:
First Test: Can never be false - or else it becomes contradictory. So it has to be true, so the
speaker cannot be a knave.

So FIRST Subject = MONKEY KNIGHT

Second Test: Can never be true, since the speaker would then be a knave and a liar. So it must be
false, in which can the speaker must be a knave, but not a monkey

So SECOND Subject = HUMAN KNAVE

Third Test: Can never be false, since the speaker would become a lying knight. So it must be
true. In this case, the speaker must be a knight, so cannot be a monkey.

So THIRD Subject = HUMAN KNIGHT
_________________
Forget Critical Thinking, why isn't there a GCSE in LATERAL Thinking?

Message
Posted: Mon Jan 03, 2011 11:23 am by s.b. Post subject:
it would have reeli helped if ud just mentioned that knaves always lie and knights are always
truthful...maybe its obvious to u but to amateurs lyk me it would really help.....thanks! :)

solve this! it only middle school so i don't think it would be TOO hard anyways here it is. reply
---------------------------------------------------------------------------------------

Three geniuses stand in a file (one behind the other). Each can see only to the front, so the rear
person can see the middle and the front, the middle person can see the front, and the genius in the
front cannot see anyone.

You have five hats. Two are white, and three are red. You blindfold the three geniuses, who are
utterly truthful, and put a hat--at random--on the head of each. Then you hide the other two hats
and remove the blindfolds.

You then ask each genius to name the color of his hat (which he cannot see).

The rear one says "I don't know." The middle one says, "I don't know." Then the front one says,
"I know."

WHAT COLOR HAT IS THE FRONT GENIUS WEARING?????? HOW DOES HE KNOW
WITH 100% CERTAINTY WHAT COLOR HAT IT IS?????

Message
Posted: Fri Feb 08, 2008 11:50 pm by toothslooth Post subject: A bit of logic
works wonders!
The front genius is wearing a red hat. Here's how he knows...
The front 2 were NOT both wearing white or the back 1 would know he was red. So the front 2
must either be wearing both red or 1 red 1 white. If the front was wearing white the middle 1
MUST be wearing red. As he does not know, the front 1 MUST THEREFORE be wearing RED.

Enjoy!

Message
Posted: Tue Aug 11, 2009 12:42 pm by mdbbarilla Post subject:
list the possibilities.

then cancel out.

man 1: red / white
man 2: red / white
man 3: red / white

if man 1 : red
then man 2 : red
man 3 will not be able to know his hat because there is more than one possibility
then man 2: white
man 3 will still not know
but if man 1: white
man 2 : white
man 3 will know that his hat is red.

if man 2 was red man3 will not know his hat

a man needs to go through a train tunnel. he starts through the tunnel and when he gets 1/4 the
way through the tunnel, he hears the train whistle behind him. you don't know how far away the
train is, or how fast it is going, (or how fast he is going). all you know is that
if the man turns around and runs back the way he came, he will just barely make it out of the
tunnel alive before the train hits him.
if the man keeps running through the tunnel, he will also just barely make it out of the tunnel
alive before the train hits him.
assume the man runs the same speed whether he goes back to the start or continues on through
the tunnel. also assume that he accelerates to his top speed instantaneously. assume the train
misses him by an infintisimal amount and all those other reasonable assumptions that go along
with puzzles like this so that some wanker doesn't say the problem isn't well defined.
how fast is the train going compared to the man?

Message
Posted: Tue Jan 08, 2008 7:28 am by aatifahmed Post subject: Solution to A man
needs to go through a train tunnel!
The train is moving at 2 times the speed of man.

Use www.rot-13.com to decipher the solution if need be.

Yrg gur genva or ng n qvfgnapr f sebz gur ghaary. Yrg gur ghaary or bs yratgu q naq gur fcrrq bs
zna i, juvyr bs gur genva ki.
Gurersber:
Gvzr gnxra ol gur zna naq gur znpuvar vf fnzr ng gur ragenapr => (f/ki)=(q/4i)=((f+q)/(ki+4i)) --
-- (v) (hfvat onfvp cebcregvrf bs engvb cebcbegvbaf).

Gvzr gnxra ol gur zna naq gur znpuvar vf fnzr ng gur rkvg =>
((f+q)/ki)=(3q/4i) ---- (vv)
hfr (v) & (vv):
(ki+4i)*q/4i=3q*ki/4i
fbyir gb trg k=2.

Message
Posted: Tue Nov 04, 2008 11:03 pm by Axomomma Post subject: Sorry, but the
The train is moving at 1.5 times the speed of the man.

Assume that the tunnel is 400 yards long.
The man hears the train when he is 1/4 of the way inside, or 100 yards in.
If the man travels back 100 yards the train will just miss him.
If he goes forward 300 yards the train will just miss him.
The train was 150 yards away from the entrance to the tunnel when the man first heard it.
If the man goes forward 300 yards the train will travel 450 yards. 300 x 1.5 = 450
If the man goes back 100 yards the train will travel 150 yards. 100 x 1.5 = 150
_________________
I'm not telling you because "I think", I'm telling you because "I know"

Message
Posted: Mon Dec 22, 2008 9:00 pm by alexonfyre Post subject:
Actually the answer is that the train is moving twice the speed of the man, this forum doesn't
have spoiler tags, so I will go ahead and post my work, hope that is okay.
First to Axomomma, for your hypothetical to be true, then the train would have to travel 550
yards while the man traveled 300 (since it has to go the distance to the tunnel, then all the way
through it, 400 + 150 = 550) and then your two ratios wouldn't match, so that cannot be the
answer. It was a decent guess, it took me a while to realise the error, but I think the solution to
this problem is a system of equations).

There are 4 unknowns here: X (the length of the Tunnel), Y(the distance between the train and
the tunnel), M (the speed of the man) and T (the speed of the train).
The ultimate goal here is to eliminate x and y and leave a relationship between m and t.
We have these two conditions given to us:

The time it takes the man to move 1/4 of x is equal to the time it takes the train to go all of y
which gives us:
x/(4m) = y/t

also, the time it takes the man to go 3/4 x is equal to the time it takes the train to go all of y and x
or:

3x/(4m) = (y+x)/t

For simplicity lets call the time it takes for the man to run through all of x, z or:

z = x/m

so

z/4 = y/t

and

3z/4 = (x+y)/t

now we can solve for z in terms of y and t:

z = 4y/t

make another substitution and we will eliminate t (don't worry it will be back):

(4y/t)*3/4 = (x+y)/t

3y/t=(x+y)/t

3y = x+y

x = 2y

(right here you know that the tunnel is twice the length of the distance between the train and
tunnel, which is enough to solve it, but let's finish the elegant, mathematical way)

now with x in terms of y we can eliminate them both from our initial equation:
x/(4m) = y/t

becomes

2y/4m = y/t

y/2m = y/t

(cross multiply)
t = 2m

or the speed of the train is twice the speed of the man.

I couldn't get aatifahmed's decipher link to get his solution, it is broken now, so I don't know how
similar his method was to mine, but I do know that his answer is correct.
A famous restaurant starts a special marriage anniversary promo to the couples on Thursdays
who carry their proof of marriage...50% discount on meals only on Thursdays. A couple enters
the restaurant on Thursday, tells the manager that they got married on a fine Sunday 28years
back hearing this the manager chased away the couple without being served. Why?

Message
Posted: Fri Sep 19, 2008 11:36 am by skintc Post subject:
Cos he worked out that if they were married on a Sunday 28 years ago, that their anniversary
could not be on a Thrusday that year.

Message
Posted: Fri Oct 03, 2008 12:36 pm by rj82330 Post subject: Further
Isn't it the case that, every 28 years, dates will fall on the same day?

It would be every 7 years, but the leap years throw it out. This is regulated every four cycles of 7
years.

So, what happened on a Sunday 28 years ago, should be a Sunday this year.

(The exceptions are if the 28 year period falls over 1700, 1800 or 1900, which are not divisible
by 400 and therefore not leap years)
_________________
Forget Critical Thinking, why isn't there a GCSE in LATERAL Thinking?

You are standing outside a basement door, Down the stairs, there are 3 light bulbs.

Next to you, are 3 switches. You cannot see the lights from upstairs, and you must decide which
switch controls which lightbulb by going downstairs once.

How do you do this?
Message
Posted: Mon Jun 30, 2008 11:39 am by skintc Post subject:
Turn on first switch, leave it on for a few minutes. Then turn it off and flip second switch.

When you go into the basement, the first switch is the bulb that is off, but hot. The second switch
is the bulb that's on. The thrid switch is the bulb that's cold and off.

you have two identical crystal orbs. you need to figure out how high an orb can fall from a 100
story building before it breaks. you know nothing about the toughness of the orbs: they may be
very fragile and break when dropped from the first floor, or they may be so tough that dropping
them from the 100th floor doesn't even harm them.
what is the largest number of orb-drops you would ever have to do in order to find the right
floor? (i.e. what's the most efficient way you could drop the orbs to find your answer?)
you are allowed to break both orbs, provided that in doing so you uniquely identify the correct
floor.

Message
Posted: Tue Dec 18, 2007 3:24 am by Rick Post subject:
It may be less but I think I could do it in at most 14 drops.

Message
Posted: Tue Jan 08, 2008 5:02 am by aatifahmed Post subject: Solution to Falling
Orbs!
The solution I am submitting below is encrypted so as not to take the fun out of your efforts! To
decrypt use: www.rot-n.com

The cipher used is ROT18-ROT5 & ROT13.
Guvax bs vg nf n ovanel ceboyrz, naq pbaireg vg vagb ovanel. Fgneg sebz bayl 65 fgbel uvtu
ohvyqvat. Guhf lbh unir 6655 sbe 65. Rnpu cbfvgvba bs gur qvtvgf va 6655 qrfpevorf gur fgbel
sebz juvpu gur beo vf gb or qebccrq, va gur qbjajneq snfuvba. Fb sbe rknzcyr bs 65, vg fubhyq
or qebccrq sebz fgbel ahzore =, 9, 7 naq 6 erfcrpgviryl.
Abgr - O zrnaf Oernx, AO zrnaf Ab Oernx.
Guhf sbe 65:
Q=-AO-Q65-AO-65 vf gur znk sybbe
Q=-AO-Q65-O-Q>-AO-> vf gur znk
Q=-AO-Q65-O-Q>-O-= vf gur znk sybbe
Q=-O-Q9-AO-Q;-AO-Q<-O-; vf gur znk sybbe
Q=-O-Q9-AO-Q;-AO-Q<-AO-< vf gur znk sybbe
Q=-O-Q9-AO-Q;-O-Q:-AO-: vf gur znk sybbe
Q=-O-Q9-AO-Q;-O-Q:-O-9 vf gur znk sybbe
Q=-O-Q9-O-Q7-AO-Q8-AO-8 vf gur znk sybbe
Q=-O-Q9-O-Q7-AO-Q8-O-7 vf gur znk sybbe
Q=-O-Q9-O-Q7-O-Q6-AO-6 vf gur znk sybbe
Q=-O-Q9-O-Q7-O-Q6-O-5 vf gur znk sybbe

Nf h pna frr sebz nobir, abar bs gurz gnxr zber guna 9 qebcf! Fvzvyneyl vg pna or jbexrq bhg sbe
655, juvpu va ovanel vf 6655655, univat < qvtvgf naq urapr < qebcf!!

Message
Posted: Tue Jan 08, 2008 6:52 am by aatifahmed Post subject: Correction
My solution holds true if infinite number of Orbs are available, for minimum number of drops.
The correct answer to the problem is 14.

Message
Posted: Mon Dec 22, 2008 10:24 pm by alexonfyre Post subject:
yeah, radically different math for the two solutions as well ;)

Message
Posted: Mon Jun 28, 2010 3:26 pm by shrek619 Post subject:
i think the anwser is 7 (minimum no.).
Message
Posted: Thu Jun 02, 2011 5:26 pm by Unni Post subject: Correct answer

Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks
truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely
random matter. Your task is to determine the identities of A, B, and C by asking three yes-no
questions; each question must be put to exactly one god. The gods understand English, but will
answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in
some order. You do not know which word means which. "

It could be that some god gets asked more than one question (and hence that some god is not
What the second question is, and to which god it is put, may depend on the answer to the first
question. (And of course similarly for the third question.)
Whether Random speaks truly or not should be thought of as depending on the flip of a coin
hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.

Message
Posted: Mon Dec 22, 2008 10:30 pm by alexonfyre Post subject:
how would the truth telling god answer if one were to ask him "Would he tell me that he is the
False God?" referring to the Random God?

Message
Posted: Sat Dec 18, 2010 2:30 pm by bugbustter Post subject: Three Gods, A,B
and C called Truth, False and Random
The Truth God would keep silent, because he would not be able to give a truthfull answer.
_________________
bugbustter

Message
Posted: Tue May 17, 2011 4:08 pm by randompattern Post subject:
I imagine that the false good also would remain silent if you asked what the random god would
say bec the false god must be sure that his answer is false.

were blue?" (or any definite truth)

If A will remains quiet:

I know B is Random. I then ask A what C would say if I asked him if the sky were blue.
Whatever the reply, that is the word for "no", because if A is True, then C is False, so A will
truthfully tell me that C will lie and if A is False and C is true, then A will lie about C's truthful
response. Now that I know which word means "no", I simply ask either A or C if the sky is blue.
Let's Say I ask C: If the Answer is the word for "no", C is False and A is True. If it is "yes", C is
True and A is False.

If after my first question, A gives me an answer:
I know that B is not Random. I ask B what C would say if I asked if the sky were blue. If B gives
an answer, then I know A must be Random, and the Answer that B gave means "No". I then ask
B or C if the sky is Blue to determine which one is True and which one is False.

If A answers, but B remains silent:
I know that C is Random, and that what A answered must have meant "No". I then ask A or B if
the sky is blue to determine which one is True and which one is False.

Say you have one string of alphabetic characters, and say you have another, guaranteed smaller
string of alphabetic characters. Algorithmically speaking, what's the fastest way to find out if all
the characters in the smaller string are in the larger string?

For example, if the two strings were:

String 1: ABCDEFGHLMNOPQRS
String 2: DCGSRQPOM

You'd get true as every character in string2 is in string1. If the two strings were:

String 1: ABCDEFGHLMNOPQRS
String 2: DCGSRQPOZ

you'd get false as Z isn't in the first string.

The naive way to do this operation would be to iterate over the 2nd string once for each character
in the 1st string. That'd be O(n*m) in algorithm parlance where n is the length of string1 and m is
the length of string2. Given the strings in our above example, thats 16*8 = 128 operations in the
worst case.

A slightly better way would be to sort each string and then do a stepwise iteration of both sorted
strings simultaneously. Sorting both strings would be (in the general case) O(m log m) + O(n log
n) and the linear scan after that is O(m+n). Again for our strings above, that would be 16*4 +
8*3 = 88 plus a linear scan of both strings at a cost of 16 + 8 = 24. Thats 88 + 24 = 112 total
operations. Slightly better. (As the size of the strings grow, this method would start to look better
and better)

Finally, the best method would simply be O(n+m). That is, iterate through the first string and put
each character in a hashtable (cost of O(n) or 16). Then iterate the 2nd string and query the
hashtable for each character you find. If its not found, you don't have a match. That would cost 8
operations - so both operations together is a total of 24 operations. Not bad and way better than
the other solutions.

Is there another solution?

What if - given that we have a limited range of possible characters - I assigned each character of
the alphabet to a prime number starting with 2 and going up from there. So A would be 2, and B
would be 3, and C would be 5, etc. And then I went through the first string and 'multiplied' each
character's prime number together. You'd end up with some big number right? And then - what if
I iterated through the 2nd string and 'divided' by every character in there. If any division gave a
remainder - you knew you didn't have a match. If there was no remainders through the whole
process, you knew you had a subset. Would that work?
Message
Posted: Mon Jan 10, 2011 1:30 am by harryaskham Post subject:
The primes solution fails if the second string contains, say, 2 Ds when the original string has
only 1 - you would try and divide by the corresponding prime twice when it exists only once as a
prime factor, giving a remainder.

For what it's worth, I think a hashing solution is overkill, given the space of only 26 characters -
collisions would be very unlikely. Instead, a similar solution would be to simply keep a size-26
array of booleans - iterate over the first string, and set booleans[letterCode] to true for each.
Then, you can simply iterate over the second string, returning false if booleans[letterCode] is
false for any character, and true after the iteration has completed without returning.

So basically a hash table, but since the key is the value, it itself can be used as the index (I tried
both in Java and although both were fast, the array-backed solution used about 1/3 the time).

Out of 52 cards 13 are up and 39 are down and are mixed .
How to arrange them in two decks of equal numbers of up.
Condition
You are sitting in a dark room and there is no light.

Message
Posted: Mon Jun 28, 2010 3:17 pm by shrek619 Post subject:
can you give a small hint ?

Message
Posted: Tue Sep 21, 2010 8:55 am by lonewolf Post subject:
Saperate the piles into 2 equal piles of 26 cards each. Flip all the card of on pile. I think this
question is simmilar to the 20 coin question.
Message
Posted: Wed Dec 29, 2010 11:33 am by moshu123 Post subject: Correct solution
The correct solution is like this:

Saperate the piles into 2 piles of 13 & 39 cards each. Flip all the card of smaller pile.

Say you have one string of alphabetic characters, and say you have another, guaranteed smaller
string of alphabetic characters. Algorithmically speaking, what's the fastest way to find out if all
the characters in the smaller string are in the larger string?

For example, if the two strings were:

String 1: ABCDEFGHLMNOPQRS
String 2: DCGSRQPOM

You'd get true as every character in string2 is in string1. If the two strings were:

String 1: ABCDEFGHLMNOPQRS
String 2: DCGSRQPOZ

you'd get false as Z isn't in the first string.

The naive way to do this operation would be to iterate over the 2nd string once for each character
in the 1st string. That'd be O(n*m) in algorithm parlance where n is the length of string1 and m is
the length of string2. Given the strings in our above example, thats 16*8 = 128 operations in the
worst case.

A slightly better way would be to sort each string and then do a stepwise iteration of both sorted
strings simultaneously. Sorting both strings would be (in the general case) O(m log m) + O(n log
n) and the linear scan after that is O(m+n). Again for our strings above, that would be 16*4 +
8*3 = 88 plus a linear scan of both strings at a cost of 16 + 8 = 24. Thats 88 + 24 = 112 total
operations. Slightly better. (As the size of the strings grow, this method would start to look better
and better)

Finally, the best method would simply be O(n+m). That is, iterate through the first string and put
each character in a hashtable (cost of O(n) or 16). Then iterate the 2nd string and query the
hashtable for each character you find. If its not found, you don't have a match. That would cost 8
operations - so both operations together is a total of 24 operations. Not bad and way better than
the other solutions.

Is there another solution?

What if - given that we have a limited range of possible characters - I assigned each character of
the alphabet to a prime number starting with 2 and going up from there. So A would be 2, and B
would be 3, and C would be 5, etc. And then I went through the first string and 'multiplied' each
character's prime number together. You'd end up with some big number right? And then - what if
I iterated through the 2nd string and 'divided' by every character in there. If any division gave a
remainder - you knew you didn't have a match. If there was no remainders through the whole
process, you knew you had a subset. Would that work?

A wily young man wishes to marry the princess Arlena. Unfortunately, her father, the Sultan, is
opposed to the marriage and is willing to buy the young man off. The Sultan, therefore makes the
young man the following offer--"You get to make one statement. If the statement is false, you
will be put to death and receive none of the following three things. If the statement is true, you
may have one of the following three things, but I get to choose! The three things are: (1) Arlena's
hand in marriage; (2) A cup filled with extremely valuable diamonds; (3) A magic lamp with a
genie (who, unfortunately, cannot do marriages)." The wily young man, however, is too wily for
his future father-in-law. He makes a statement such that the only way the Sultan can keep his
promise is by giving the young man Arlena's hand in marriage.

What was the statement?

Message
Posted: Thu Nov 27, 2008 8:38 am by rj82330 Post subject: A variant
Good one, this - only heard it will TWO options in the past.

He says: "You will not give me the diamonds or the genie".
_________________
Forget Critical Thinking, why isn't there a GCSE in LATERAL Thinking?

Message
Posted: Mon Dec 22, 2008 10:08 pm by alexonfyre Post subject:
If that were correct RJ couldn't the Sultan just say "False, I chose the diamonds for you." and
death.

I think the actual answer is:
"If you do not let me marry your daughter, I will be put to death"

Meaning that if the king chooses diamond or genie and lets him live, he will be telling a lie and
the king will have to put him to death, which would in turn create a paradox as he would be
telling the truth at that point. The only resolution to the paradox is to allow the prince to marry
his daughter. If the Sultan chooses to put him to death later, that would be another story, but in
the mean time the Sultan would have to give up his daughter's hand in marriage.

Message
Posted: Tue Dec 23, 2008 11:39 am by rj82330 Post subject:
My thinking was that with the three variable given, there is only one way to prevent creating

I had the young man say "you will not give me certain presents" not "you would not have given
me certain presents". (You have the sultan saying "I chose").

In this case, if the sultan wants to say "False, I will give you the diamonds", he has to actually
fork over a present to prove the young man's statement false: but the scenario above doesn't
allow for the young man to be given a present in the event of his making a false statement and

If the sultan acknowledges that the young man's statement is true - because he is going to be
executed rather than receiving diamonds or genie, then our hero will receive the punishment for
making a false statement - Paradox.

Surely, the only way to make sure that the integrity of the puzzle remains, and to avoid a
paradox, is to make the statement true, for the one remaining true variable - i.e. giving away his
daughter.

___
This is similar to the Pirate King of the Logical Positivists, who decrees that a certain empiricist
is to be executed via the yardarm or the plank, depending on whether a statement he makes is
true (he'll hang) or false (he'll drown).

In that puzzle, the empiricist has to render the entire problem logically impossible (by saying "I
will drown"). In this conundrum, he has to render most of the problem paradoxical, yet still leave
one option within the puzzle open to him.
_________________
Forget Critical Thinking, why isn't there a GCSE in LATERAL Thinking?

Message
Posted: Mon Mar 09, 2009 5:17 pm by Jeffkins Post subject:
Similarly could he not say "You will give me Arlena's hand in marriage or you will put me to
death."
Essentially the reverse of rj82330's answer.

Message
Posted: Tue Nov 30, 2010 9:13 pm by Trenin Post subject: Optimal Solution
"You will give me none of those three things, or you will give me your throne and I will forfeit
my right to ask for a prize."

To analyze this, it is basically two clauses - A or B - where:
A is "You will give me none of those three things"
B is "You will give me your throne and I will forfeit my right to ask for a prize"
By making any one of them true, the entire statement is true.

Lets suppose the sultan decides this statement is false. Then he gives the young man nothing and
kills him. This makes clause A true, thus the entire statement is true, which is a contradiction.

Thus the statement must be true.

The first clause A can't be the only one that is true because then the sultan must give the young
man a prize of the sultan's choosing if the young man wishes it (which he would) which makes
clause A false - another contradiction.

Thus the second clause B must be true - the sultan is forced to give up his throne. Some of you
may say this is still a contradiction because if the statement is true, then the sultan must give the
young man a prize. However, the wording of the puzzle is:

If the statement is true, you may have one of the following three things, but I get to choose!

Thus, it is up to the young man if he wants to get a prize, but it is up to the sultan which prize.
Thus, the young man can refuse the prize and not have a contradiction. He does this explicitly in
his statement, thus making the way clear for what the sultan must do.

When the young man takes the throne, he then takes not only the princess' hand in marriage, but
everything else in the power of the sultan as well.

The leader of a group of dwarfs wants to test his group. So at night he paints on the back of each
dwarf a dot, either red of blue. The next morning he let's the dwarfs gather at a meeting point and
tell's them the rules of the game:

You may never know what the color of your own dot is. (This is accomplished by saying they
may not communicate in any way or use mirrors and such)

He will give them a few minutes to discuss a tactic to solve the puzzle.

The goal is that the number of dwarfs with a red dot, assume this number is X, on their back
gather at the same meetingpoint the number of days, X, after the start of the game. for example if
there are 10 red dots, after 10 days they must gather

Any other entry upon the meetingpoint of any dwarf results in failure.

Now, how must every dwarf think to solve the puzzle?

Message
Posted: Thu May 20, 2010 2:25 pm by sinb Post subject:
well, the group selects a single person as a leader. The leader can see, let's say Y dots. either X =
Y (if his own dot is green) or X = Y+1 (if his own dot is red)
The plan is that the group meets at a place every day at a pre-decided point, but the leader is
supposed to show up only on (Y-1)'th day. So everyone except for the leader know X, that is
they know Y and depending on the leader's dot they can decide on X. When they decide on X
(the total number of red dots), then they can decide whether their own dot is red or blue.
Now on X'th day, every red dot dwarf meets at the same place if the leader's dot was red,
otherwise they all meet at some other pre-decided place to which leader wont come.

I dont know if the guy is allowed to indicate X value to others by going to the meetingplace, (it
would be indirectly telling everyone which dots they have), but this is the solution I came up
with. Let me know if you have some other solution.

Message
Posted: Mon Jun 28, 2010 2:58 pm by shrek619 Post subject: solution
I assume there are at least 2 red dots.

Consider one dwarf with either red or blue dot, let X be total no. of dots out there. If he is red
dotted he sees X-1 dots let it be Y, if he is blue dotted he sees X dots let it be Y. Y is no. of red
dots as he sees on other dwarfs.
All he need to do is go on to the location on Y+1 day. Each one of them should follow this rule.

If he is red dotted he goes to the location on correct day, if he is blue dotted the game is already
over on Y th day so there is no need for going to the meeting place.
am i right ? is there any other solution

Message
Posted: Mon Nov 29, 2010 9:12 pm by sudheerkuttiyat Post subject: Think Red
dot
Every dwarf must think he has a red dot on his back.

Then every one with red dots in the back will meet X days from now, where X-1 is the number
of red dots a dwarf with a red dot behind him can see.

Anyone with blue dots on the back will see X red dots and will come to meet his mates X+1 days
from now.
Supposedly a true story:
A 5'6" man was found hanged (suicide) in an empty cellar room in a huge mansion in England.
He was a prominent man of good stature (lawyer I believe). It was found he was bi/sexual and
couldn't handle the social pressure and mostly his devastated wife, so he decided to go out in
style and baffle everyone. He did for many months. The situation was this: He was found
hanging from a sprinkler system pipe in the middle of the room just 1' from the ceiling from a 4'
piece of one stranded rope. The room (cellar) was 25' square (wall to wall, ceiling to floor).
There were no windows, the room was totally bare, no furnishings whatsoever, no shelves,
nothing. The walls and ceiling were smooth, the floor was smooth with the exception of the
typical drain hole and it had only one entrance which was just a normal sized door and the door
was locked from the inside with two dead bolts and a hinged padlock. It had only one light bulb
in the center of the ceiling, light switch was outside the door. He was found about a month later
decomposing. How did he do it?
_________________
Just a human being having a spiritual experience!

Message
Posted: Fri Jul 17, 2009 12:30 pm by rackner Post subject:
Could he not simply have stood on a block of ice to hang himself? The ice would then have
melted and gone through the drain hole.

Message
Posted: Sun Jul 26, 2009 12:54 am by Callyst Post subject:
maybe he blocked up the drain with either ice or the rope, and then took the rope and tied himself
to the pipe.
_________________
"whoever said the pen was mightier than the sword has obviously never encountered automatic
weapons" -Douglas MacArthur-

Message
Posted: Tue Aug 18, 2009 12:15 am by Callyst Post subject:
i forgot to add and busted the pipe open
_________________
"whoever said the pen was mightier than the sword has obviously never encountered automatic
weapons" -Douglas MacArthur-

This was the Brain Teaser:

A pack of 9 cards is numbered 1 to 9. One card is placed on the forehead of each of four
logicians. Each of them can see the other three numbers, but not their own. In turn, each states
whether or not he knows his own number. If not, he announces the sum of two of the numbers he
can see.

One game went as follows. Alf: “No, 14”. Bert: “Yes”. Chris: “No, 7. Dave: “No”- but before
Dave could continue, Alf had worked out his own number.

What was Dave’s number?

And this was my proof that it can't be done:

Definitions:

There are two (and only two) possible “pairs” that make up the 14 that A can see:
6-8 and 5-9

These will be called the pairs, and 5,6,8,9 will be referred to as pair numbers. All the other
numbers (i.e: 1,2,3,4,7) will be called “anonymous numbers”.

A preliminary general observation:

It is not possible that all four pair cards (5,6,8,9) are on the four foreheads, for there would
clearly be no way that C could nominate a 2-card total of 7.

Logic:

A can see a 14-pair somewhere on B, C & D.
B states that he knows his own number.
So B must have half of a pair – if a whole pair was on C & D then B couldn’t possibly state that
he knew his own number unless he was clairvoyant, it could be anything.
So the other half of the pair will be on C or D, and, importantly, the person who doesn’t have the
other half must have an anonymous number. For if they had one of the other pair then B couldn’t
make his definitive statement. (Unless A conveniently had a third pair card, in which case B
would know that he had to have the missing fourth one, but this is precluded – see general
observation above).

Just to recap by way of example – say B can see C 8 and D 3. He knows his own number, it must
be a 6, because it’s the only way to make the14 that A has called. But if what he saw was C 8
and D 9, he wouldn’t have known his own number – it could be a 6 or a 5.

So we have established: B has one of a pair; C or D has the other half of that pair, and the other
has an anonymous number.

So where is the other half of B’s pair? – is it on C or D?

Lets say it’s on C. So C is looking at half a pair on B and an anonymous number on D, and
would therefore immediately know his own number, which makes a 14-pair with B. So that’s not
it, because C was unable to nominate his own number.

So is it on D? No, for the same reason. D is looking at half a pair on B and an anonymous
number on C, and would therefore know his own number. But he passed, so that’s not it.

So the whole thing is impossible

Message
Posted: Tue Aug 11, 2009 1:10 pm by mdbbarilla Post subject:
they only state the sum of the two numbers they can see. So A can announce the sum of C & D
rather than B&C or B&D

A sees 14 ( 6/8 or 5/9)
C sees 7 ( 6/1 or 5/2 or 4/3)

Let say they saw 6.

A can see a 14-pair somewhere on B, C & D.
B states that he knows his own number.
So B must have half of a pair B has 6. B = 6
So the only pair that could have a sum of 7 is 6 and 1
If C had 1, C can't say that he sees a 7. C =/= 1
If D had the 1, A wouldn't be able to know his number.
Because A doesn't know anything about his number. D =/= 1
If D didn't have the 1, A would now he has the number 1!
Let say they saw 5.

A can see a 14-pair somewhere on B, C & D.
B states that he knows his own number.
So B must have half of a pair. B has 5.
So the only pair that has a sum of 7 is 5 and 2.
If C had 2, C can't say he sees a 7
If D had 2, A won't know his number.
But if D didn't get the 2, A would know his number is 2.
So Dave's number would 3,4 or 7. Doesn't really matter if you use this approach.

Message
Posted: Fri Jun 03, 2011 7:22 pm by Unni Post subject: Correct answer
What about placing the card (6 or 9) upside down?

At a recent trial in Justiceville, the prosecutor, Mr. Jones, called five witnesses to the stand, after
which the defense attorney, Ms. Smith, called five more witnesses. From this information and the
clues, can you determine the first name (one is Diane) and last name (one is Anderson) and
occupation (one is an accountant) of each of the ten witnesses and the order in which they
testified?

* The five called to the stand by Mr. Jones were, in no particular order: Kathy, Mark, Mr.
Ducklow, Ms. Olson, and the computer programmer.
* Ms. Smith's five witnesses were, in no particular order: John, Sandra, Ms. Fuller, Mr. Simpson
and the secretary.
* Neither Mark nor Kathy is the pilot.
* Mary testified just before Miller, who was before the author, but not just before.
* John is not the teacher.
* Thompson, who is not Mary, testified just before the dentist.
* The mechanic, who was not on the stand first, testified before McNeil, but not just before.
* Frank testified just before the teacher who was just before Ms. Fuller.
* The bank teller testified before Zimmer, but not just before.
* Anne, who is a musician, testified just before Williams and just after the pilot.
* The computer programmer was not just before the secretary.
* Ms. Smith did not call Glenn or Betty.
* Mark testified before Betty, but not just before.

Message
Posted: Sat Mar 19, 2011 4:59 pm by ChibiHoshi Post subject:
Mark Thompson Bank Teller
Kathy Anderson Dentist
Betty Zimmer Computer Programmer
Glenn Ducklow Pilot
Anne Olson Musician
Mary William Secretary
John Miller Mechanic
Frank Simpson Accountant
Sandra McNeil Teacher
Diane Fuller Author

in a far away land, it was known that if you drank poison, the only way to save yourself is to
drink a stronger poison, which neutralizes the weaker poison. the king called a logician an a
chemist to make the strongest poison for him. the logician went straight to work, but the chemist
knew that he had no chance, for the logician was wiser than him, so instead he made up a plan to
survive and make sure the logician dies. on the last day the logician suddenly realized that the
chemist would know he had no chance, so he must have a plan. after a little thought the logician
realized what was the chemist's plan must be, and he conducted a counter plan , to make sure he
survives and the chemist die. when the time came, the king summoned both of them. they drank
the poison as planned, the chemist died, the logician survived and the king got what he wanted.
what exactly happen

Message
Posted: Wed Feb 16, 2011 6:15 pm by nomad Post subject:
The chemist made the strongest poison available till time.And he had to drink it to prove it.So he
died.
But the logician made a weaker poison and drank it himselves.and to save his life he again took
the poison prepared by the chemist and thats how he saved himselves

Puzzle:
There are a hundred coins sitting on the table, ten are currently heads and nintey are currently
tails. You are sitting at the table with a blindfold and gloves on. You are able to feel where the
coins are, but are unable to see or feel if they heads or tails. You must create two sets of coins.
Each set must have the same number of heads and tails as the other group. You can only move or
flip the coins, you are unable to determine their current state. How do you create two even
groups of coins with the same number of heads and tails in each group?

Solution:
Create two sets of ten coins. Flip The coins in one of the sets over, and leave the coins in the
other set alone. The first set of ten coins will have the same number of heads and tails as the
other set of ten coins.

Can somebody hep me better understand the solution? It would seem that if by chance I
(blindfolded) create two sets of coins 0h:10t - 0h:10t and flip one set thus making 10h:0t - 0h:10t
that I do not have the same number of heads and tails in each group. There are situations where
this would work, but not 100% of the time. Am I missing a key point in the puzzle maybe?

Many thanks!

Message
Posted: Mon Nov 29, 2010 9:26 pm by sudheerkuttiyat Post subject:
Yes, The question should be ..."there are twenty coins, with 10 showing heads and 10 showing
tails..... "

The solution does not work for 100 coins with only 10 showing heads initially

Splitting the twenty to two groups and flipping all coins in one group will make both groups look
the same.
ou have a chessboard with two opposite squares cut out. How can you cover the remaining 62
squares completely with 31 domino pieces ?(without breaking the dominoes or the board of
course). Each domino covers exactly two squares.

Message
Posted: Tue Apr 15, 2008 11:22 am by tlmarjot Post subject:
the chess board can be folded in half like a lot of chess boards.
so half of the 62 squares is 31 same as the amount of dominos. each domino covers one square
completely but when the board is folded over it covers all 62squares and remains true to the fact
that each domino covers two squares

Message
Posted: Sat May 03, 2008 2:40 pm by dedo Post subject:
tlmarjot, I didn't get it - can you cover it, or not? If yes - how? If not - why?
_________________
In Nature - the only blemish is the mind
None are to be called deform'd but the unkind

Message
Posted: Tue Jun 10, 2008 1:54 pm by steinjim Post subject: checkerboard
If you cut out opposite corners, that will be two of the same color. Since each domino must cover
one black and one white square, you will NOT be able to cover the remaining 62 squares with
dominos (there will be 32 of one color and 30 of the other)

Message
Posted: Mon Dec 22, 2008 9:54 pm by alexonfyre Post subject:
I took it to mean that one square of each color was cut out, at random...though it would be
impossible to show a solution without the actual coordinates and/or a board showing such that
we could draw on.

Message
Posted: Sun Aug 02, 2009 3:05 am by porcelina Post subject:
If I'm not mistaken, the opposite squares could also be squares directly across the middle from
one another. In which case, you'd simply lay the dominoes horizontally, filling up each row
EXCEPT the two middle rows. There would be one square left uncovered on each. You would
then place the last domino vertically, thereby covering both.

A man goes out for a walk. He walks 1mile to the south, then 1 mile to the east and finally 1
mile to the north and ends up in the same place he started. now he dint started at north
pole,for sure. . .so where did he :?:

Message
Posted: Tue Jul 07, 2009 7:59 am by rj82330 Post subject:
Any number of points just over a mile NORTH of the SOUTH POLE.

Provided the circumference of the earth at the point where he starts walking east is an exact
factor of 1 mile (i.e. a circumference of 1/2 a mile, or 1/3 a mile, or 1/4 a mile etc), he can walk
round and round on this latitude until he has walked a mile "east", whereupon he'll be at his point
of origin, and can head 1 mile north to the original start point.

Theoretically, he could start exactly 1 mile north of the South Pole, then "walk east" by standing
on one spot and spinning in a circle until his body had arced 1760 yards, then walk back north:
but this isn't really "walking".
_________________
Forget Critical Thinking, why isn't there a GCSE in LATERAL Thinking?

Message
Posted: Tue Jul 07, 2009 8:31 am by v_jey Post subject:
good job rj82330
a more simpler way to expln is by assuming a perfectly spherical Earth, somewhere one mile
north of the latitude (in the southern hemisphere) that is one mile in circumference. The man
walks south one mile to this latitude and walks one mile east, which takes him all the way around
and back to where he started. The last step (one mile north) retraces the first step he took (one
mile south). 8)

A woman was being robbed and she could only run away with two of her golden balls they each
were 10 pounds an she was 100 pounds in order to get away she had to cross a bridge and the
bridge had a completely accurate sighn that said ony 110 pounds was aloud to cross. if all
together they were 120 pounds how did she get across?

_________________
Sylvia E

Message
Posted: Mon Mar 09, 2009 5:46 pm by Jeffkins Post subject:
The act of throwing a 10 pound ball upwards generates greater than 10 pounds of force
downwards. So the act of throwing one ball up in the air would cause the bridge to collapse.

A possible answer would be to roll or throw the ball across the bridge. But of course that means
this is a lateral thinking puzzle.
Three coworkers would like to know their average salary. how can they do it, without disclosing
their own salaries?

Message
Posted: Sat Mar 01, 2008 10:56 am by aggynd Post subject:
ask each coworker total of other two's salary. total the three figures and divide by two.

Message
Posted: Sun Mar 30, 2008 12:42 pm by Jeebok Post subject:
Person A writes a number that is her salary plus a random amount (AS + AR) and hands it to B,
without showing C. B then adds his salary plus a random amount (BS + BR) and passes to C (at
each step, they write on a new paper and don't show the 3rd person). C adds CS + CR and passes
to A. Now A subtracts her random number (AR), passes to B. B and C each subtract their
random number and pass. After C is done, he shows the result and they divide by 3.

As has been noted already, there's no way to liar-proof the scheme.

It's also worth noting that once they know the average, any of the three knows the sum of the
other 2 salaries.

Message
Posted: Sat May 03, 2008 2:38 pm by dedo Post subject:
Jeebok, you got it!
_________________
In Nature - the only blemish is the mind
None are to be called deform'd but the unkind

```
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