Remove TWO from FIVE
Wordplay: Remove TWO from FIVE and get FOUR.
SOLUTION
Remove TWO letters, “F” and “E”, from the word “FIVE”, and get “IV”, which is FOUR in Roman numerals.
Category: Puzzles
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White and Dirty
I am white when I am dirty, and black when I am clean. What am I?
SOLUTION
The answer is BLACKBOARD.
Death in the Field
A man is lying dead in a field. Next to him, there is an unopened package. There are no other people or animals in the field. How did he die?
SOLUTION
The man was sky diving and his parachute did not open.
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Square Impossible
Take a square (or circle) coaster. Then, cut a hole in a piece of paper with the shape of a square, so that the side of that square is half of the side (or the diameter) of the coaster. Now, your goal is to push the coaster through the hole without tearing the paper (you can fold it).
SOLUTION
Coming soon.
Hungry Lion
A hungry lion runs inside a circus arena which is a circle of radius 10 meters. Running in broken lines (i.e. along a piecewise linear trajectory), the lion covers 30 kilometers. Prove that the sum of all turning angles is at least 2998 radians.
SOLUTION
Imagine the lion is static, facing North, and instead, the center of the arena moves around. Then, each time the lion runs X meters in some direction, this translates into the center moving X meters South. Each time the lion makes a turn of Y radians, this translates into the center moving along an arc of Y radians.
Thus, the problem translates to a point inside the arena alternating between traveling straight South and then moving along arcs around the center of the arena. Since the total distance traveled straight South by the point is 30KM and the distance between the starting and the ending points is at most 20M, the total distance traveled North must be at least 30KM – 20M = 29980M. Therefore, the total length of the arcs traversed by the point is at least 29980M, and since the radius of each arc is at most 10M, the total angle of the arcs must be at least 2998 radians. The sum of all turning angles of the lion is the same, so this concludes the proof.
Awkward Hidden Objects
Warning: this puzzle involves mature themes that are inappropriate for younger audiences. If you are not an adult, please skip this puzzle.
SHOW PUZZLE
Can you find the objects which are hidden in the images below?
SOLUTION
The solution is shown below.
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Houses on a Farm
Is it possible to connect each of the houses with the well, the barn, and the mill, so that no two connections intersect each other?
SOLUTION
No, it is impossible. Here is a convincing, albeit a informal proof.
Imagine the problem is solvable. Then you can connect House A to the Well, then the Well to House B, then House B to the Barn, then the Barn to House C, then House C to the Mill, and finally the Mill to House A. Thus, you will create one loop with 6 points on it, such that houses and non-houses are alternating along the loop. Now, you must connect Point 1 with Point 4, Point 2 with Point 5, Point 3 with Point 6, such that the three curves do not intersect each other. However, you can see that you can draw no more than one such curve neither on the inside, nor the outside of the loop. Therefore, the task is indeed impossible.
More rigorous, mathematical proof can be made using Euler’s formula for planar graphs. We have that F + V – E = 2, where F is the number of faces, V is the number of vertices, and E is the number of edges in the planar graph. We have V = 6 and E = 9, and therefore F = 5. Since no 2 houses or 2 non-houses can be connected with each other, every face in this graph must have at least 4 sides (edges). Therefore, the total number of sides of all faces must be at least 20. However, this is impossible, since every edge is counted twice as a side and 20/2 > 9.
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Game of Coins
Kuku and Pipi decide to play a game. They arrange 50 coins in a line on the table, with various nominations. Then, alternating, each player takes on their turn one of the two coins at the ends of the line and keeps it. Kuku and Pipi continue doing this, until after the 50th move all coins are taken. Prove that whoever starts first can always collect coins with at least as much value as their opponent.
Remark: On the first turn, Kuku can pick either coin #1 or coin #50. If Kuku picks coin #1, then Pipi can pick on her turn either coin #2 or coin #50. If Kuku picks coin #50, then Pipi can pick on her turn either coin #1 or coin #49.
SOLUTION
Let’s assume Kuku starts first. In the beginning, he calculates the total value of the coins placed on odd positions in the line and compares it with the total value of the coins placed on even positions in the line. If the former has a bigger total value, then on every turn he takes the end coin which was placed on odd position initially. If the latter has bigger value, then on every turn he takes the end coin which was placed on even position initially. It is easy to see that he can always do this because after each of Pipi’s turns there will be one “odd” coin and one “even” coin at the ends of the line.
Two Trains and a Bridge
Two trains are crossing the bridge bellow at the same speed on a non-windy day. One of the trains
SOLUTION
The two trains have different lengths, so the longer one needs more time until it crosses completely the bridge.
Chess Peace
Can you add all 16 black pieces on the board (king, queen, 2 knights, 2 bishops, 2 rooks, 8 pawns) and get a chess position, in which no piece is attacked by another?
SOLUTION
Yes, you can, as shown in the diagram below.
