Pronunciation Puzzles

The following 2 puzzles rely on misleading phrasing of the questions. Read them aloud to your friends and let them ponder upon them.

  1. What has 4 letters, sometimes 9, and never 5
  2. There are 30 cows and 28 chickens. How many didn’t?
  3. Pronounce the following words: T-W-A, T-W-E, T-W-I, T-W-O
  4. As I was walking across the London Bridge, I met a man.
    He tipped his hat, and drew his cane.
    In this riddle, I said his name. What is it?

The first puzzle is not a question. It is a statement, saying that the word “what” has 4 letters, the word “sometimes” has 9 letters, and the word “never” has 5 letters. There is nothing to solve, so the puzzle is figuring that out!

The second puzzle actually reads as “There are 30 cows and 20 ate chickens. How many didn’t?” Thus, the answer is that 10 cows didn’t eat chickens.

The third question often confuses people and they pronounce TWO as [twou] instead of [tuː].

The fourth riddle actually says: “He tipped his hat, ‘Andrew Hiscane'”. Thus, the name of the man is Andrew Hiscane.

Rapunzel and the Prince

The evil witch has left Rapunzel and the prince in the center of a completely dark, large, square prison room. The room is guarded by four silent monsters in each of its corners.  Rapunzel and the prince need to reach the only escape door located in the center of one of the walls, without getting near the foul beasts. How can they do this, considering they can not see anything and do not know in which direction to go?

The prince must stay in the center of the room and hold Rapunzel’s hair, gradually releasing it. Then, Rapunzel must walk in circles around the prince, until she gets to the walls and finds the escape door.

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.

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.

Discuss this puzzle in the forum.

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?

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.

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.

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.