Vinculus Puzzles

Circles are particles and lines joining them are bonds. The objective is to find all the hidden values, following these four rules:

  1. Particle values must be the sum of their bond values.
  2. Particles can have the following values: 0, 1, 2, 4, 8, 12, 16.
  3. Bonds can have the following values: 0, 1, 2, 4.
  4. If two particles have the same value, the bond between them must have value 0.

The solutions are shown below.

Jugs with Water

You have a 3 liter jug, a 5 liter jug, and an infinite amount of tap water. How can you measure exactly 4 liters of water using the jugs?

Call the 3 liter jug “small” the 5 liter jug “large”.

  • Fill the large jug with water.
  • Pour 3 liters of water from the large jug into the small jug.
  • Empty the small jug with water.
  • Pour the remaining 2 liters of water from the large jug into the small jug.
  • Fill the large jug with water.
  • Pour water from the large jug into the small jug, until the small jug is full.

Now there are exactly 4 liters of water in the large jug.

Canals on Mars Maze

Starting with the letter “T” at the bottom, visit all spots exactly once before returning to the beginning, so that the letters you pass through spell a complete sentence.

The path goes through the letters “T”, “H”, “E”, “R”, “E”, “I”, “S”, “N”, “O”, “P”, “O”, “S”, “S”, “I”, “B”, “L”, “E”, “W”, “A”, “Y”, to spell the sentence “There is no possible way”.

Plinks, Plonks, and Plunks

If all plinks are plonks and some plunks are plinks, which of these statements must be true?

  • All plinks are plunks.
  • Some plonks are plunks.
  • Some plinks are not plunks.

Remark: “Some” means more than 0.

The first statement says that the set of plonks contains the set of plinks, and the second statement says that there is at least one plunk-plink. Therefore, that plunk-plink must also be a plonk, and the second statement is true.

The first and the third statements, however, do not need to be true. Indeed, it is possible that there is a plink that is not a plunk, or that all plinks are plunks.

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

Three Cards

There are three playing cards in a row. There is a two to the right of a king. There is a diamond to the left of a spade. There is an ace to the left of a heart. There is a heart to the left of a spade. Identify the three cards.

The cards are an Ace of Diamonds, a King of Hearts, and a Two of Spades.

Cheryl’s Birthday

Cheryl’s birthday is one of 10 possible dates:

May 15
May 16
May 19
June 17
June 18

July 14
July 16
August 14
August 15
August 17

Cheryl tells the month to Albert and the day to Bernard.

Albert: I don’t know the birthday, but I know Bernard doesn’t know either.
Bernard: I didn’t know at first, but now I do know.
Albert: Now I also know Cheryl’s birthday.

When is Cheryl’s birthday?

If Albert knows that Bernard doesn’t know when the birthday is, then the birthday can’t be on May 19 or June 18. Also, Albert must know that the birthday can’t be on these dates, so May and June are completely ruled out.

If Bernard can deduce when the birthday is after Albert’s comment, then the birthday can’t be on 14th. The remaining possibilities are July 16, August 15, and August 17.

Finally, if Albert figures out when the birthday is after Bernard’s comment, then the date must be July 16.

Married Couples

In a small village, there are 100 married couples living. Everyone in the village lives by the following two rules:

  1. If a husband cheats on his wife and she figures it out, the husband gets immediately killed.
  2. The wives gossip about all the infidelities in town, with the only exception that no woman is told whether her husband has cheated on her.

One day a traveler comes to the village and finds out that every man has cheated at least once on his wife. When he leaves, without being specific, he announces in front of everybody that at least one infidelity has occurred. What will happen in the next 100 days in the village?

Let us first see what will happen if there are N married couples in the village and K husbands have cheated, where K=1 or 2.

If K = 1, then on the first day the cheating husband would get killed and nobody else will die. If K = 2, then on the first day nobody will get killed. During the second day, however, both women would think like this: “If my husband didn’t cheat on me, then the other woman would have immediately realized that she is being cheated on and would have killed her husband on the first day. This did not happen and therefore my husband has cheated on me.”. Then both men will get killed on the second day.

Now assume that if there are N couples on the island and K husbands have cheated, then all K cheaters will get killed on day K. Let us examine what will happen if there are N + 1 couples on the island and L husbands have cheated.

Every woman would think like this: “If I assume that my husband didn’t cheat on me, then the behavior of the remaining N couples will not be influenced by my family’s presence on the island.”. Therefore she has to wait and see when and how many men will get killed in the village. After L days pass however and nobody gets killed, every woman who has been cheated on will realize that her assumption is wrong and will kill her husband on the next day. Therefore if there are N + 1 couples on the island, again all L cheating husbands will get killed on day L.

Applying this inductive logic consecutively for 3 couples, 4 couples, 5 couples, etc., we see that when there are 100 married couples on the island, all men will get killed on day 100.