Battleship

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.

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Soccer Ball

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.

Self-Referential Aptitude Test

The solution to this puzzle is unique, but you don’t need this information in order to find it.

  1. The first question whose answer is B is question:
    (A) 1
    (B) 2
    (C) 3
    (D) 4
    (E) 5
  2. The only two consecutive questions with identical answers are questions:(A) 6 and 7
    (B) 7 and 8
    (C) 8 and 9
    (D) 9 and 10
    (E) 10 and 11
  3. The number of questions with the answer E is:
    (A) 0
    (B) 1
    (C) 2
    (D) 3
    (E) 4
  4. The number of questions with the answer A is:
    (A) 4
    (B) 5
    (C) 6
    (D) 7
    (E) 8
  5. The answer to this question is the same as the answer to question:
    (A) 1
    (B) 2
    (C) 3
    (D) 4
    (E) 5
  6. The answer to question 17 is:
    (A) C
    (B) D
    (C) E
    (D) none of the above
    (E) all of the above
  7. Alphabetically, the answer to this question and the answer to the following question are:
    (A) 4 apart
    (B) 3 apart
    (C) 2 apart
    (D) 1 apart
    (E) the same
  8. The number of questions whose answers are vowels is:
    (A) 4
    (B) 5
    (C) 6
    (D) 7
    (E) 8
  9. The next question with the same answer as this one is question:
    (A) 10
    (B) 11
    (C) 12
    (D) 13
    (E) 14
  10. The answer to question 16 is:
    (A) D
    (B) A
    (C) E
    (D) B
    (E) C
  11. The number of questions preceding this one with the answer B is:
    (A) 0
    (B) 1
    (C) 2
    (D) 3
    (E) 4
  12. The number of questions whose answer is a consonant is:
    (A) an even number
    (B) an odd number
    (C) a perfect square
    (D) a prime
    (E) divisible by 5
  13. The only odd-numbered problem with answer A is:
    (A) 9
    (B) 11
    (C) 13
    (D) 15
    (E) 17
  14. The number of questions with answer D is
    (A) 6
    (B) 7
    (C) 8
    (D) 9
    (E) 10
  15. The answer to question 12 is:
    (A) A
    (B) B
    (C) C
    (D) D
    (E) E
  16. The answer to question 10 is:
    (A) D
    (B) C
    (C) B
    (D) A
    (E) E
  17. The answer to question 6 is:
    (A) C
    (B) D
    (C) E
    (D) none of the above
    (E) all of the above
  18. The number of questions with answer A equals the number of questions with answer:
    (A) B
    (B) C
    (C) D
    (D) E
    (E) none of the above
  19. The answer to this question is:
    (A) A
    (B) B
    (C) C
    (D) D
    (E) E
  20. Standardized test is to intelligence as barometer is to:
    (A) temperature (only)
    (B) wind-velocity (only)
    (C) latitude (only)
    (D) longitude (only)
    (E) temperature, wind-velocity, latitude, and longitude

Remark: The answer to question 20. is (E).

The answers are:

  1. D
  2. A
  3. D
  4. B
  5. E
  6. D
  7. D
  8. E
  9. D
  10. A
  11. B
  12. A
  13. D
  14. B
  15. A
  16. D
  17. B
  18. A
  19. B
  20. E

Fish in a Pond

There are 5 fish in a pond. What is the probability that you can split the pond into 2 halves using a diameter, so that all fish end up in one half?

Let us generalize the problem to N fish in a pond. We can assume that all fish are on the boundary of the pond, which is a circle, and we need to find the probability that all of them are contained within a semi-circle.

For every fish Fᵢ, consider the semi-circle Cᵢ whose left end-point is at Fᵢ. The probability that all fish belong to Cᵢ is equal to 1/2ᴺ⁻¹. Since it is impossible to have 2 fish Fᵢ and Fⱼ, such that the semi-sircles Cᵢ and Cⱼ contain all fish, we see that the probability that all fish belong to Cᵢ for some i is equal to N/2ᴺ⁻¹.

When N = 5, we get that the answer is 5/16.

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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?

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.

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.

Normal Person

I caused my mother’s death and didn’t get convicted.
I married 100 women and never got divorced.
I got born before my father, but I am considered perfectly normal?

Who am I?

A priest, whose mother dies from labor, who marries 100 women to 100 men, and whose father attends his birth.

Crashing Light Bulbs

You are living in a 100-floor apartment block. You know that there is one floor in the block, such that if you drop a light bulb from there or anywhere higher, it will crash upon hitting the ground. If you drop a light bulb from any floor underneath it however, the light bulb will remain intact. If you have two light bulbs at your disposal, how many drop attempts do you need such that you can surely find which the floor in question is?

The answer is 14 drops. You can do this by throwing the first bulb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100 (notice that the difference decreases always by 1) until it crashes and then start throwing the second bulb from the floors in between. For example, if the first bulb crashes at floor 69, you start throwing the second bulb from floors 61, 62, 63, etc. This way the total number of throws would be always at most 14.

Proving that 14 is optimal is done using the same logic. In order to use at most 13 throws, the first throw should be made from floor 13 or lower. The second throw should be made from floor 13+12 or lower, the third throw should be made from floor 13+12+11 or lower, etc. Continuing with the same argument, we conclude that the 13th drop should be made from floor 13+12+…+2+1=91 or lower. However, if the first light bulb does not crash after the last throw, you will not be able to find out which number among 92-100 is X.