Henry Ernest Dudeney (1857 – 1930) was an English author and mathematician who specialized in logic puzzles and mathematical games. He is known as one of the country's foremost creators of mathematical puzzles.
Replace every letter with a different digit between 0 and 9, such that you get a correct calculation.
The answer is 9567 + 1085 = 10652.
You can first see that M = 1. Then S = 8 or 9 and O = 0. MORE is at most 1098, and if S = 8, then E = 9, N is either 9 or 0, but this is impossible. Therefore S = 9, N = E + 1. You can check easily the 6 possibilities for (N, E) and then find the answer.
Tango and Cash are playing the following game: Each of them chooses a number between 1 and 9 without replacement. The first one to get 3 numbers that sum up to 15 wins. Does any of them have a winning strategy?
Place the numbers from 1 to 9 in a 3×3 grid so that they form a magic square. Now the game comes down to a standard TIC-TAC-TOE, and it is well-known that it always leads to a draw when both players use optimal strategies.
David Copperfield and his assistant perform the following magic trick. The assistant offers a person from the audience to pick 5 arbitrary cards from a regular deck and then hands them back to him. After the assistant sees the cards, he returns one of them to the audience member and gives the rest one by one to David Copperfield. After the magician receives the fourth card, he correctly guesses what card the audience member holds in his hand. How did they perform the trick?
Out of the five cards, there will be (at least) two of the same suit; assume they are clubs. Now imagine all clubs are arranged in a circle in a cyclic manner – A, 2, 3, … J, Q, K (clock-wise), and locate the two chosen ones on it. There are two arks on the circle which are connecting them and exactly one of them will contain X cards, with X between 0 and 5. Now the assistant will pass to David Copperfield first the clubs card which is located on the left end of this ark, will return to the audience member the clubs card which is located on the right end of it and, with the remaining three cards, will encode the number X. In order to do this, he will arrange the three extra cards in increasing order – first clubs A-K, then diamonds A-K, then hearts A-K and finally spades A-K. Let us call the smallest card in this order “1”, the middle one “2” and the largest one “3”. Now, depending on the value of X, the assistant will pass the cards “1”, “2” and “3” in the following order:
X=0 ⇾ 1, 2, 3
X=1 ⇾ 1, 3, 2
X=2 ⇾ 2, 1, 3
X=3 ⇾ 2, 3, 1
X=4 ⇾ 3, 1, 2
X=5 ⇾ 3, 2, 1
In this way David Copperfield will know the suit of the audience member’s card and also with what number he should increase the card he received first in order to get value as well. Therefore, he will be able to guess correctly.
In certain society all parents stop having children right after they get their first boy. After 1000 years, approximately what will be the percentage of the women in the society?
The answer is 50%. The reason is that no matter what birth plan the society comes up with, the chance for having a boy or a girl during every birth is 50/50.
A man stands in the center of a circular field which is encompassed by a narrow ring of water. In the water there is a shark which is swimming four times as fast as the man is running. Can the man escape the field and get past the water to safety?
Yes, he can. Let the radius of the field is R and its center I. First the man should start running along a circle with center I and radius slightly less than R/4. His angular speed will be larger than the angular speed of the shark, so he can keep running until gets opposite to it with respect to I. Then, the man should dash away (in a straight line) towards the water. Since he will need to cover approximately 3R/4 distance and the shark will have to cover approximately 3.14R distance, the man will have enough time to escape.
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?
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.
A boy draws 2015 unit squares on a piece of paper, all oriented the same way, possibly overlapping each other. Then the colors the resulting picture in black and white chess-wise, such that any area belonging to an even number of squares is painted white and any area belonging to an odd number of squares is painted black.
Prove that the total black area is at least one.
Draw a grid in the plane which is parallel to the sides of the squares. Then, take the content of each cell of the grid and translate it (move it) to some chosen unit square. The points in that unit square which are covered by odd number of black pieces color in black, the rest color in white. It is easy to see that after doing this, the entire unit square will be colored in black (each of the 2015 squares cover it once completely). This implies that the total black area is no less than 1.
In the Padurea forest there are 100 rest stops. There are 1000 trails, each connecting a pair of rest stops. Each trail has some particular level of difficulty with no two trails having the same difficulty. An intrepid hiker, Sendeirismo has decided to spend a vacation by taking a hike consisting of 20 trails of ever increasing difficulty.
Can he be sure that it can be done?
He is free to choose the starting rest stop and the 20 trails from a sequence where the start of one trail is the end of a previous one.
Place one hiker in each of the rest stops. Now, go through the trails in the forest one by one, in increasing difficulty, and every time you pick a trail, let the two hikers in its ends change places. This way the 100 hikers would traverse 2000 trails in total, and therefore one of them would traverse at least 20 trails.
A battleship starts moving at 12 PM from an integer point on the real line with constant speed, landing on every hour again on an integer point. Every day at midnight you can shoot at an arbitrary point on the real plane, trying to destroy the battleship. Can you find a strategy with which you will eventually succeed to do this?
If we know the starting point of the battleship and its speed, then we can determine its position at any time after 12 PM.
There are countably many combinations (X, Y) of starting point and speed. We can order them in the following way:
(0, 0) – starting point 0, speed 0;
(0, 1) – starting point 0, speed +1;
(1, 0) – starting point 1, speed 0;
(0, -1) – starting point 0, speed -1;
(1, 1) – starting point 1, speed +1;
(-1, 0) – starting point -1, speed 0;
(0, 2) – starting point 0, speed +2;
(1, -1) – starting point 1, speed -1;
(-1, 1) – starting point -1, speed 1;
(2, 0)- starting point 2, speed 0,
and so on. Of course, we can choose the ordering in many different ways.
Now we can start exhausting all possibilities one after another. First we assume the combination is (0, 0), calculate where the battleship would be at midnight during the first day and shoot there. Then we assume the combination is (0, 1), calculate where the battleship would be at midnight during the second day and shoot there. If we continue like this, eventually we will hit the battleship.
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?
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.
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