The White House
If Mr. Green lives in the green house and Mr. Red lives in the red house, who lives in the White House?
SOLUTION
The President.
Category: Puzzles
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$5 Bill
Bern tells Sandy, “This isn’t the $20 bill you left on the table. It is mine – I was keeping it between pages 15 and 16 of my textbook.”
Sandy retorts, “You are lying and I can prove it.” How does she know?
SOLUTION
Pages 15 and 16 are on the same sheet of paper in the textbook (just like pages 1 and 2 are) and therefore the bill couldn’t be between them.
Water with Ice
You have a glass of water and an ice cube floating in it. When the ice cube melts, will the water level increase, decrease or remain the same?
SOLUTION
It will remain the same. The amount of water that the ice cube displaces is equal to its mass. Since the mass does not change and the density of water is equal to 1, the extra water after melting will be the same amount as the displaced water before that.
FEATURED
Heads Up, Heads Down
You are blindfolded and on the table, in front of you, 50 coins are placed. You are told that X of them are heads up and the rest are heads down. Then you are asked to separate the coins into two groups and optionally flip some of them so that the number of heads in both groups becomes the same. How can you do this?
SOLUTION
Separate the coins into one group of X coins and into another group of 50-X coins, then flip every coin in the first group. If in the first group there were Y heads up initially, then after flipping there would be X-Y – exactly the number of heads up in the second group.
The Man in the Photo
A person is looking at somebody’s photo and exclaims:
“Brothers and sisters, I have none,
But this man’s father is my father’s son.”
Who is the person in the photo?
SOLUTION
It is the person’s son.
Mixed Up Pills
One patient has two bottles with 30 pills each and every night has to take one pill from each of the bottles. Unfortunately one night after he takes out a pill from the first bottle and places it on the table, by accident drops two pills from the second bottle right next to it. The pills look identical, so he can not differentiate them. It is very important that he continues his treatment diligently throughout the entire timespan of 30 days. What should the patient do?
SOLUTION
The patient should keep taking one pill from each bottle until there are 4 pills remaining – 1 in the first bottle and 3 on the table. On the 29th day he splits the pills in halves and takes one half from each pill. On the 30th day he takes the remaining halves of the pills.
Thoka’s Rebus 6
Can you figure out which words are depicted by these rebuses?
SOLUTION
The answers are SKYLINE, HEADQUARTERS, RAINBOW, and EYELID.
Ants Maze
Find a non-intersecting closed path, passing through all four ants, which traverses this double-sided maze. You can not switch between the top and the bottom side of the path at any times.
SOLUTION
The solution is shown below.
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Prisoners and Boxes
There are 100 inmates living in solitary cells in a prison. In a room inside the prison there are 100 boxes and in each box there is a paper with some prisoner’s name (all different). One day the warden tells the prisoners that he has aligned next to the wall in a special room 100 closed boxes, each of them containing some prisoner’s name (all different). He will let every prisoner go to the room, open 50 of the boxes, then close them and leave the room the way it was, without communicating with anybody. If all prisoners find their names in the boxes they open, they will be set free, otherwise they will be executed. The prisoners are allowed to come up with a quick plan before the challenge begins. Can you find a strategy which will ensure a success rate of more than 30%?
SOLUTION
The prisoners can assign numbers to their names – 1, 2, 3, … , 100. When prisoner X enters the room, he should open first the X-th box in the line. If he sees inside prisoner’s Y name, he should open next the Y-th box in the line. If he sees in it prisoner’s Z name, he should open next the Z-th box in the line and so on.
The only way which will prevent all prisoners from finding their names is if there is a long cycle of boxes (length 51 or more), such that the first box in the cycle directs to the second box in it, the second box to the third box, the third box to the fourth box and so on.
It is not hard to compute that the probability of having a cycle of length K>50 is exactly 1/K. Then the probability for failure will be equal to the sum 1/51 + 1/52 + … + 1/100, which is very close to ln(100) – ln(50) = ln(2) ~ 69%. Therefore, this strategy ensures a success rate of more than 30%.