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Can you draw uncountable many non-intersecting “8” shapes in the plane (they can be contained in one another)?
No, you can’t. For each “8” shape you can choose a pair of points with rational coordinates – one in its top loop and one in its bottom loop. Since no two “8” shapes can have the same corresponding pair of rational points, their number should be countable.
A prince decides to get married to the prettiest girl in his kingdom. All 100 available ladies go to the palace and show themselves to the prince one by one. He can either decide to marry the girl in front of him or ask her to leave forever and call the next one in line. Can you find a strategy which will give the prince a chance of 25% to get married to the prettiest girl? Can you find the best strategy?
Remark: Assume that the prince can objectively compare every two girls he has seen.
A strategy which ensures a chance of 25% is the following:
The prince banishes the first 50 girls which enter the palace and then gets married to the first one which is prettier than all of them (if such one arrives). If the prettiest girl in the kingdom is in the second 50, and the second prettiest girl is in the first 50, he will succeed. The chance for this is exactly 25%.
The best strategy is to wait until ~1/e of all girls pass, and then choose the first one which is more beautiful than all of them. This yields a chance of ~37% for succeeding. The proof is coming soon.
Take six coins and arrange them in a triangle as shown in the image. Your goal is to rearrange the coins into a hexagon with only four moves. Every move consists of sliding one coin to a new location where it touches at least two other coins.
The solution is shown below.
Our first puzzle giveaway is over. The winner, Ankush S., is not living in the US, so he decided to donate his prize to a children’s home in Illinois! As an appreciation for his gesture, we created one of our signature avatars for him. Congratulations!
Solve the puzzle below and post the answer on our Facebook page, for the chance of winning the puzzle game Cat Crimes, provided by our friends at ThinkFun.
On the image below you can see 11 points in the plane placed in such
CAT = 3, BOWL = 2, FISH = 1. The scratches represent Roman numerals.
4 (CAT with a FISH) + 6 (VI) + 2 (TWO FISH) = 12
1 (FISH) + 2 (BOWL) + 5 (CAT with a BOWL) = 8
5 (CAT with a BOWL) + 3 (CAT) + 3 (III) = 11
Therefore, the correct answer is 4 + 2 + 3 = 9.
A square has dropped on the ground and broken into ten pieces. Accidentally, an additional, eleventh piece has fallen among the others. Can you figure out which one it is?
The total number of the squares in all pieces is 70 = 8 * 8 + 6. Therefore the extra piece is the one consisted of 6 squares.
Two people play a game of NIM. There are 100 matches on a table, and the players take turns picking 1 to 5 sticks at a time. The person who takes the last stick wins the game. Who has a winning strategy?
The first person has a winning strategy. First, he takes 4 sticks. Then every time the second player takes X sticks, the first player takes 6 – X sticks.
Two very short friends, John and Jack, were living together in an apartment. Since they used to lose their apartment key very often, they decided to leave it on top of the door frame when they leave home. In order to reach the key, John was climbing on Jack’s shoulders and thus taking it down from the frame. However, John was the taller and the heavier guy of the two. Why didn’t Jack climb on top of his shoulders instead?
The reason is that as being taller, John also had longer arms. If Jack climbed on John’s shoulders, he wouldn’t have reached the key.
Two monks are standing on the two sides of a 2-dimensional mountain, at altitude 0. The mountain can have any number of ups and downs, but never drops under altitude 0. Prove that the monks can climb or descend the mountain at the same time on both sides, always staying at the same altitude, until they meet at the same point.
Coming soon.
I have keys without key locks.
I have space without rooms.
You can enter but you cannot go outside.
What am I?
The answer is KEYBOARD.
In a parliament
For each split of the people into 2 groups, compute the animosity level of each person by subtracting the number of enemies in the opposite group from the number of enemies in his own group. Then, split the people into 2 groups so that the total animosity level of all of them is as little as possible. If there is a person, who has more enemies in his own group than the opposite one, then by transferring him to the other group, we will reduce the total animosity level of the people and will get a contradiction.
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