Ten Lanterns

You have ten lanterns, five of which are working, and five of which are broken. You are allowed to choose any two lanterns and make a test that tells you whether there is a broken lantern among them or not. How many tests do you need until you find a lantern you know for sure is working?

Remark: If the test detects that there are broken lanterns, it does not tell you which ones and how many (one or two) they are.

You need 6 tests:

(1, 2) → (3, 4) → (5, 6) → (7, 8) → (7, 9) → (8, 9)

If at least one of these tests is positive, then you have found two working lanterns.

It all of these tests are negative, then lantern #10 must be working. Indeed, since at least one lantern in each of the pairs (1, 2), (3, 4), (5, 6) is not working. Therefore, there are at least 2 working lanterns among #7, #8, #9, #10. If #10 is not working, then at least one of the pairs (7, 8), (7, 9), or (8, 9) must yield a positive test, which is a contradiction.

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.

King Arthur and the Knights

King Arthur and his eleven honorable knights must sit on a round-table. In how many ways can you arrange the group, if no honorable knight can sit between two older honorable knights?

The answer is 1024 ways, up to rotation around the table. To see this, notice that the youngest honorable knight must sit right next to King Arthur – there are two possible places for him. Then, the second youngest knight must sit right next to this group of two. Once again, there are two possible places for him. Continuing like this, we see that for all honorable knights, except for the oldest one, there are two possible spots on the table. Multiplying two to the power of ten out, we get 1024.

Non-Transitive Dice

This is a non-transitive dice set, i.e. every dice in it is weaker than some other dice. Can you design a non-transitive set with only 3 dice?

Remark: “Weaker” means that it loses more often than it wins.

The simplest solution is given by:

2, 2, 4, 4, 9, 9;
1, 1, 6, 6, 8, 8;
3, 3, 5, 5, 7, 7.

Another solution is given by the so-called “Miwin’s dice”. They are as follows:

1, 1, 3, 5, 5, 6;
2, 3, 3, 4, 4, 5;
1, 2, 2, 4, 6, 6.