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.

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.

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.

The Majority Name

In a long list of names, one of the names appears more than half of the time. You will be read the names one at a time, without knowing how many they are, and without being able to write them down. If you have a very weak memory, how can you figure out which is the majority name?

Remember the first name and then keep track of whether it has been repeated more than half of the time. To do that, simply add 1 if you hear the name or subtract one when you hear another name. If the list finishes and your counter is positive, then the first name is the majority. If your counter drops to 0, simply restart the procedure with the next name you hear.

This algorithm, invented by R. Boyer and J. Moore, works, because if the counter ends up at 0, then each of the names up to that moment has been read at most half of the time. Therefore, the majority name appears more than half of the time in the remainder of the list.

Vectors -1, 0, 1

Consider all 1024 vectors in a 10-dimensional space with elements ±1. Show that if you change some of the elements of some of the vectors to 0, you can still choose a few vectors, such that their sum is equal to the 0-vector.

Denote the 1024 vectors with ui and their transformations with f(ui). Create a graph with 1024 nodes, labeled with ui. Then, for every node ui, create a directed edge from ui to ui-2f(ui). This is a valid construction, since the vector ui-2f(ui) has elements -1, 0, and 1 only. In the resulting graph, there is a cycle:

v1 ⇾ v2 ⇾ … ⇾ vk ⇾ v1.

Now, if we pick the (transformed) vectors from this cycle, their sum is the 0-vector:

f(v1) + f(v2) + … + f(vk) = (v2 – v1)/2 + (v3 – v2)/2 + … + (v1 – vk)/2 = 0.

Einstein’s Puzzle

There are 5 houses and each of them has a different color. Their respective owners have different heritages, drink different types of beverages, smoke different brands of cigarettes, and look after different types of pets. It is known that:

  1. The Brit lives in the red house.
  2. The Swede keeps dogs as pets.
  3. The Dane drinks tea.
  4. Looking from in front, the green house is just to the left of the white house.
  5. The green house’s owner drinks coffee.
  6. The person who smokes Pall Malls raises birds.
  7. The owner of the yellow house smokes Dunhill.
  8. The man living in the center house drinks milk.
  9. The Norwegian lives in the leftmost house.
  10. The man who smokes Blends lives next to the one who keeps cats.
  11. The man who keeps a horse lives next to the man who smokes Dunhill.
  12. The owner who smokes Bluemasters also drinks beer.
  13. The German smokes Prince.
  14. The Norwegian lives next to the blue house.
  15. The man who smokes Blends has a neighbor who drinks water.

The question is, who owns the pet fish?

The German owns the pet fish.

Since the Norwegian lives in the leftmost house (9) and the house next to him is blue (14), the second house must be blue. Since the green house is on the left of the white house (4), the person living in the center house drinks milk (8), and the green house’s owner drinks coffee (5), the fourth house must be green and the fifth one must be white. Since the Brit lives in the red house (1) and the Norwegian lives in the leftmost house (9), the leftmost house must be yellow and the center house must be red. Therefore, the colors of the houses are: YELLOW, BLUE, RED, GREEN, WHITE.

Since the Norwegian from the yellow house smokes Dunhill (7), the man from the blue house must keep a horse (11). The person smoking Blends cannot be in the red house, because this would imply that the person in the green house keeps cats and the Swede keeps dogs in the white house (2, 10). However, in this case the Dane must be drinking tea in the blue house (3) and the person smoking Blends does not have a neighbor drinking water (5), which is a contradiction (15). Also, the person smoking Blends cannot be in the green house, because this would imply that the person in the white house drinks water (15), the Dane lives in the blue house (3), and the German and the Swede live in the last two houses. Since the German smokes Prince (13) and the Swede keeps dogs (2), there is nobody who could smoke Bluemaster and drink beer (12). The person smoking Blends cannot be in the white house either, because this would imply that the person in the green house drinks water (15), when in fact he drinks coffee (5).

Therefore, the person smoking Blends must be in the blue house, and then the German and the Swede must live in the last two houses (2, 13). Since the person who smokes Bluemasters drinks beer (12), this must be the Swede with his dogs in the white house (2). The only option for the person who smokes Pall Mall and raising birds (6) is the red house. Then the Norwegian must keep cats (10) and the German is left with the pet fish in the green house.

Close the Loop

Alex and Bob are playing a game. They are taking turns drawing arrows over the segments of an infinite grid. Alex wins if he manages to create a closed loop, Bob wins if Alex does not win within the first 1000 moves. Who has a winning strategy if:

a) Alex starts first (easy)
b) Bob starts first (hard)

Remark: The loop can include arrows drawn both by Alex and Bob.

In both cases, Bob wins. An easy strategy for part a) is the following:

Every time Alex draws an arrow, Bob draws an arrow in such a way, that the two arrows form an L-shaped piece and either point towards or away from each other. Since every closed loop must contain a bottom left corner, Alex cannot win.

For part b), Bob should use a modification of his strategy in part a). First, he draws a horizontal arrow. Then, he splits the remaining edges into pairs, as shown on the image below. If Alex draws one arrow on the grid, then Bob draws its paired arrow, such that the two arrows point either towards or away from each other. The only place where a loop can have a bottom left corner is where Bob drew the first arrow. However, if a loop has a bottom left corner in this positio, then it must have at least one more bottom left corner, which is impossible.