Every term in the sequence describes how many 1s, 2s, 3s, etc. were there in the previous term. “11” means that in the previous term there was one 1. “21” means that in the previous term there were two 1s. “1112” means that in the previous term there were one 1 and one 2. “3112” means that in the previous term there were three 1s and one 2. “211213” means that in the previous term there were two 1s, one 2 and one 3. “312213” means that in the previous term there were three 1s, two 2s and one 3. Therefore, the next number in the sequence should be 212223.
After a car accident, the father dies and the boy arrives at the emergency room in the hospital. Upon entering the room however, the surgeon exclaims “This is my son, I can’t operate!” How is this possible?
You are in Monty Hall’s TV show where in the final round the host gives you the option to open one of three boxes and to receive the reward inside. Two of the boxes contain just a penny, while the third box contains $1.000.000. In order to make the game more exciting, after you pick your choice, the rules require the host to open one of the two remaining boxes, such that it contains a penny inside. After that he asks you whether you want to keep your chosen box or to switch it with the third remaining one. What should you do?
SOLUTION
This is the so called “Monty Hall” problem. The answer is that in order to maximize your chances of winning $1.000.000, you should switch your box. The reason is that if initially you picked a box with a penny, then after switching you will get a box with $1.000.000. If initially you picked a box with $1.000.000, then after switching you will get a box with a penny. Since in the beginning the chance to get a penny is 2/3, then after switching your chance to get $1.000.000 is also 2/3. If you stay with your current box, then your chance to get $1.000.000 will be just 1/3.
If you count carefully the number of people before the tiles scramble and after that, you will see that one person disappears. Can you explain how this is possible?
Similarly, in this picture it looks like after changing the places of the tiles in the diagram, their total area decreases by one. Can you explain this?
SOLUTION
If you look carefully, you will notice that every person in the picture of 12 people is slightly taller than the corresponding person in the picture of 13 people. Basically, we can cut little pieces from 12 different people without making noticeable changes and arrange them into a new person.
For the second question, none of the shapes before and after the scrambling is really a triangle. One of them is a bit curved in at the hypotenuse and the other one is a bit curved out. This is barely noticeable, because the red and the blue triangle have very similar proportions of their sides – 5/2 ~ 7/3.
Warning: this puzzle involves mature themes that are inappropriate for younger audiences. If you are not an adult, please skip this puzzle.
SHOW PUZZLE
3 men must have sex with 1 woman, but they have only 2 condoms. Each of the 4 people has some unique STD which they don’t want to transfer to the rest. What can they do?
SOLUTION
They can start by putting the two condoms on top of each other and letting the first man use them. After that, the second man can take the inner condom out and use just the outer condom. Finally, the third man can take the removed inner condom, turn it inside out, and place it back inside the outer condom. Then he, he can use the two condoms simultaneously.
There are 100 prisoners in solitary cells. There is a central living room with one light bulb in it, which can be either on or off initially. No prisoner can see the light bulb from his or her own cell. Every day, the warden picks a prisoner at random and that prisoner visits the living room. While there, the prisoner can toggle the light bulb if he wishes to do so. Also, at any time, every prisoner has the option of asserting that all 100 prisoners have already been in the living room. If this assertion is false, all 100 prisoners will be executed. If it is correct, all prisoners will be set free.
The prisoners are allowed to get together one night in the courtyard and come up with a plan. What plan should they agree on, so that eventually someone will make a correct assertion and will set everyone free, assuming the warden will bring each of them an infinite number of times to the central living room?
SOLUTION
First, the prisoners should elect one of them to be a leader and the rest – followers. The first two times a follower visits the living room and sees that the light bulb is turned off, he should turn it on; after that he shouldn’t touch it anymore. Every time the leader visits the living room and sees that the light bulb is turned on, he should turn it off. After the leader turns off the lightbulb 198 times, this will mean that all followers have already visited the room. Then he can make the assertion and set everyone free.
In which direction is this bus moving – towards the left or towards the right?
SOLUTION
Since we can not see the passenger doors, they should be on the other side of the bus, and therefore the bus should be on the other side of the street. Here we assume the bus is from a right-hand traffic country.
Almost everyone knows what a soccer ball looks like – there are several black regular pentagons on it and around each of them – five white regular hexagons. Around each hexagon there are three pentagons and three more hexagons. However, can you figure out how many pentagons and hexagons are there in total on the surface of the ball?
SOLUTION
Let the number of pentagons is equal to P and the number of hexagons is equal to H. Then the number of edges is equal to (5P + 6H)/2 – that’s because every pentagon has five edges, every hexagon has 6 edges and every edge belongs to 2 sides. Also, the number of vertices is equal to (5P + 6H)/3 – that’s because every pentagon has five vertices, every hexagon has 6 vertices and every vertex belongs to 3 sides. Now using Euler’s Theorem we get P + H + (5P + 6H)/3 – (5P + 6H)/2 = 2, or equivalently P/6=2 and therefore P = 12. Since around every pentagon there are exactly 5 hexagons and around every hexagon there are exactly 3 pentagons, we get H = 5P/3 = 20. Therefore, there are 12 pentagons and 20 hexagons on a soccer ball.
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.
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%.
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