Brain Puzzlers: January 2013

Saturday, January 26, 2013

Crossing a Bridge

Four men want to cross a bridge. They all begin on the same side. It is night, and they have only one flashlight with them. At most two men can cross the bridge at a time, and any party who crosses, either one or two people, must have the flashlight with them.
The flashlight must be walked back and forth: it cannot be thrown, etc. Each man walks at a different speed. A pair must walk together at the speed of the slower man. Man 1 needs 1 minute to cross the bridge, man 2 needs 2 minutes, man 3 needs 5 minutes, and man 4 needs 10 minutes. For example, if man 1 and man 3 walk across together, they need 5 minutes.

How can all four men cross the bridge in 17 minutes?

Answer:

First man 1 and man 2 walk across the bridge. This takes 2 minutes.
After this, man 1 walks back with the flashlight. This takes 1 minute.
Then man 3 and man 4 walk across the bridge. This takes 10 minutes.
After this, man 2 walks back with the flashlight. This takes 2 minutes.
Then man 1 and man 2 walk across the bridge. This takes 2 minutes as before.
In total: 2+1+10+2+2=17 minutes.

Brain Workout: Variety Puzzles to Boost Your Memory and Brainpower

Frogs

Here you see a pool with six happy frogs. You must exchange the positions of the brown and green frogs, using the following rules:
a frog can only jump to an empty water-lily;
a frog can only jump over at most one other frog, that has another color;
a frog can only jump forward (so the green frogs jump right and the brown frogs jump left).

How can this be done?

Answer:

Number the water-lilies from left to right from 1 up to 7. Then jump as follows:
the green frog on place 3,
the brown frog on place 5,
the brown frog on place 6,
the green frog on place 4,
the green frog on place 2,
the green frog on place 1,
the brown frog on place 3,
the brown frog on place 5,
the brown frog on place 7,
the green frog on place 6,
the green frog on place 4,
the green frog on place 2,
the brown frog on place 3,
the brown frog on place 5,
the green frog on place 4.
Note: it is also possible to mirror the above solution, starting with the brown frog on place 5.

Eureka

There is a story that Archimedes, the Greek mathematician, was asked to find out if the new golden crown of the king was made of pure gold, while keeping the crown intact. Sitting in a public bath and thinking about it, Archimedes noticed the displacement of the water caused by sinking his body lower into the water. He suddenly realized that he had found the solution: if the crown was made of pure gold, it should displace the same volume of water as a bar of pure gold with an equal weight. Excited, he jumped out of the bath and ran home shouting "Eureka!" ("I've found it!"), forgetting that he was still naked...

We do not know if the story is true. But we do know that Archimedes discovered the first law of hydrostatics: when a body is immersed in a fluid, it experiences an upward buoyant force which is equal to the weight of the fluid displaced by the immersed part of the body.

Can you solve the following questions, and have your "Eureka!" moments, using this famous law?

In an aquarium filled with water, a block of ice floats. We mark the current water level.

When the ice has molten completely, will the water level be higher, lower, or still the same?

Answer:

The upward force, needed to keep a block of ice floating, is equal to the weight of the ice. According to Archimedes, the weight of the ice therefore equals the weight of the displaced water. But when the ice has melted, the weight of the resulting water still equals the original weight of the ice. So the amount of water from the melted ice equals the amount of displaced water.

Conclusion: the water level stays the same.

A stack of 23 cards

You have a stack of 23 cards. Each card has an image of a cross on one side, and an image of a circle on the other side. You know that 14 cards in the stack are laying with the cross up, and therefore 9 cards are with the circle up. But you don't know in which order the cards are. Moreover, you are in a completely dark room, so you can't see anything.

How can you, without seeing anything, divide the stack of cards in two smaller stacks, in such a way that both stacks have the same number of cards with a cross up?

Answer:

Take 14 cards from the stack. Of these 14 cards, k cards are with the cross up, and therefore 14-k cards are with the circle up. In the remaining stack of 9 cards are 14-k cards with a cross up.

Then turn the stack of 14 cards round. Now there are 14-k cards with a cross up in both stacks.

Note that we do not know exactly how many cards are laying with a cross up in both stacks (or how many cards are laying with a circle up). We only know that both stacks have the same number of cards with a cross up.

Cat and Mice

Four white pieces (the mice) are placed on one side of a chessboard, and one black piece (the cat) is placed at the opposite side. The game is played by the following rules:

Black wins if it reaches the opposite side.
White wins if it blocks black in such a way that black can not make any move anymore.
Only diagonal moves (of length 1) on empty squares are allowed.
White only moves forward.
Black can move backward and forward.
Black may make the first move, then white makes a move, and so on...

Is this game computable (i.e. is it possible to decide beforehand who wins the game, no matter how hard his opponent tries to avoid this)?

Answer:

There are three possible outcomes:

Yes, the game is computable and white can always win.
Yes, the game is computable and black can always win.
No, the game is not computable.

We define W_white(game situation) and W_black(game situation) which denote whether a player has won in a certain game situation:

W_white(game situation) holds if and only if white has won in "game situation" (i.e., black cannot make any move anymore).
W_black(game situation) holds if and only if black has won in "game situation" (i.e., black has reached the opposite side).

Now we define C_white(player whose move it is,game situation) and C_black(player whose move it is,game situation) which denote whether a player can always win in a game situation in which the move is with a certain player:

C_white(white,game situation) holds if and only if W_black(game situation) does not hold and there exists a white move w in "game situation" for which C_white(black,"game situation after move w") holds.
C_white(black,game situation) holds if and only if W_white(game situation) holds or for all possible black moves b in "game situation", C_white(white,"game situation after move b") holds.
C_black(white,game situation) holds if and only if W_black(game situation) holds or for all possible white moves w in "game situation", C_black(black,"game situation after move w") holds.
C_black(black,game situation) holds if and only if W_white(game situation) does not hold and there exists a black move b in "game situation" for which C_black(white,"game situation after move b") holds.

Now holds:

The game is computable and white can always win if C_white(black,begin situation).
The game is computable and black can always win if C_black(black,begin situation).
The game is not computable if both C_white(black,begin situation) and C_black(black,begin situation) do not hold.

Because there are only five pieces which can each be on at most 32 fields, there are at most 32⁵=33554432 possible game situations. Therefore, a computer program can, for each game situation, calculate C_white(player whose move it is,game situation) and C_black(player whose move it is,game situation) within reasonable time. With the help of such a program one finds that C_white(black,game situation) holds, and therefore that white can always win.

Friday, January 25, 2013

3 Pawn Puzzles

Add 3 pawns on the following 7 * 7 board; such that there will be only one pawn in any vertical, horizontal, and diagonal line!

Answer:

Find Numbers for Letters

In the following equation, each number is replaced with its corresponding letter. Can you find the right number for each letter?

Answer:

Re-position to form 5 bird names!

Re-position the following letters to form 5 bird names!

L O W: _ _ _
T O R P A R: _ _ _ _ _ _
O O O E P H: _ _ _ _ _ _
B A R I L L P S H: _ _ _ _ _ _ _ _ _
F E E W R R N N: _ _ _ _ _ _ _ _

Answer:

1. OWL
2. PARROT
3. HOOPOE
4. SHARPBILL
5. FERNWREN

Monday, January 21, 2013

Pulleys and Pinions

In the following figure, the diameter of pulley A is twice that of pulley B, and pinions C, D and E have 15, 5, and 10 teeth, respectively. If pulley A rotates four times, how many times will pinion E rotate?

Pulleys and Pinions Puzzle

Answer:

12 times.
C rotates 8 times where it has 15 teeth (8 * 15). E has 10 teeth: (8 * 15) / 10 = 12.

Friday, January 18, 2013

More triangles ...

Which of the following two figures does contain more triangles?

Figure A

Figure B

Answer:

Both figures contain 24 triangles!

Which string ...

Which of the following strings of letters is odd one out?

1) SUVXY
2) BNOPL
3) CGPQR
4) MPTVZ

Answer:

No. 2. In the other strings, letters are in the alphabetically ascending order .

Word Puzzle: Find an eleven-letter ...

Word Puzzle: Find An Eleven-Letter English Word

In the above figure, an eleven-letter English word can be found! To find it, start from any circle going along the lines without crossing any letter more than once!

Answer: Arrangement

Sunday, January 13, 2013

Who was the killer?

You have 30 seconds to solve the following riddle:

A lady named 'Himanshi' was found shot and they already had a list of suspects- Ankit, Tarun, Harish, Manoj and Manish.

The killer is a fan of Sherlock and challenge him by leaving notes at various places.

•The first was found in a toilet room
•The second was found in an art room
•The third was in a restroom
•The fourth was in an underwater room
•The fifth at the no smoking room

All the notes read the same thing.'The clues are where you find the notes'. 'Yet nothing was found at anyplace the notes were.
Sherlock the genius, immediately solved the case.

Who was the killer?

Answer:

Tarun

Brain Workout: Variety Puzzles to Boost Your Memory and Brainpower

Which symbol will balance ...

Which symbol will balance the third scale?

Answer:

11 players lasting for 90 minutes ...

In a game of 11 players lasting for 90 minutes, 4 reserves alternate equally with each player; it means that all players, including the reserves, play for the same length of time. For how long?

Answer:

66 minutes
66 = (90 * 11)/(11 + 4)

If the first gets 333 from the second ...

Two numbers are such that if the first gets 333 from the second they will be in the ratio 4 to 1, but if the second receives 408 from the first, their ratio will be 1 to 10. What are these two numbers?

Answer

503 and 542

Saturday, January 12, 2013

Missing Box

Answer:

C: each row and column contains six complete lines and six broken lines.

Series

In 30 seconds identify what number should replace the question mark?

16, 23, 19, 19, 22, 15, 25, ?

Answer:

11: there are two alternate sequences: +3 and –4.

LOPER

Using the five letters below only, create a nine-letter word.

LOPER

Answer:

PROPELLER

What Circle?

Answer:

C: each circle is repeated rotated.

Saturday, January 5, 2013

Missing Tile

You have only 20 seconds to identify which is the missing tile?

Answer:

C: looking across and down the triangles turn through 90°.

Brain Workout: Variety Puzzles to Boost Your Memory and Brainpower

How Many Eggs?

On taking delivery of a consignment of eggs the market stall owner was furious to find that several were cracked. In fact, on counting them up in order to assess the damage he found that 72 were cracked, which was 12 per cent of the total consignment. How many eggs in total were in the consignment?

Puzzle Answer

Answer:

600: 72 ÷ 12 × 100