Vectors -1, 0, 1

Take 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.

Odd Rectangle

The sides of a rectangle have lengths which are odd numbers. The rectangle is split into smaller rectangles with sides which have integer lengths. Show that there is a small rectangle, such that all distances between its sides and the sides of the large rectangle have the same parity, i.e. they are all even or they are all odd.

Source: Shortlist IMO 2017

Split the large rectangle into small 1×1 squares and color it in black and white, chessboard-style, such that the four corner squares are black. Since the large rectangle has more black squares than white squares, one of the smaller rectangles also must have more black squares than white squares. Therefore, the four corners of that smaller rectangle are all black. Then, it is easy to see that all distances between its sides and the sides of the large rectangle have the same parity.

Larger or Smaller

Alice secretly picks two different integers by an unknown process and puts them in two envelopes. Bob chooses one of the two envelopes randomly (with a fair coin toss) and shows you the number in that envelope. Now you must guess whether the number in the other, closed envelope is larger or smaller than the one you have seen.

Is there a strategy which gives you a better than 50% chance of guessing correctly, no matter what procedure Alice used to pick her numbers?

Choose any strictly decreasing function F on the set of all integers which takes values between 0 and 1. Now, if you see the number X in Bob’s envelope, guess with probability F(X) that this number is smaller. If the two numbers in the envelopes are A and B, then your probability of guessing correctly is equal to:

F(A) * 0.5 + (1 – F(B)) * 0.5 = 0.5 + 0.5 * (F(A) – F(B)) > 50%.

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.

Moms’ Talk

Two moms, Sarah and Courtney, are talking to each other.

Sarah: I have two children.
What is the probability that both of Sarah’s children are boys?

Courtney: Me too! Do you have any boys?
What is the probability that both of Courtney’s children are boys?

Sarah: Yes, I do! What is your younger child?
What is the probability that both of Sarah’s children are boys?

Courtney: It is a boy. He is so mischievous!
What is the probability that both of Courtney’s children are boys?

Sarah: Is he Sagittarius? Sagittarius boys are known to drive their mothers crazy. I can testify from personal experience.
What is the probability that both of Sarah’s children are boys?

Courtney: No, but actually I have the opposite personal experience to yours.
What is the probability that both of Courtney’s children are boys?

Sarah: Well, I guess astrology does not always get it right.

Courtney: I assume it does about half of the time.

The answers are: ~1/4, ~1/4, ~1/3, ~1/2, ~23/47, 1.

Explanation:

Initially, we do not have any information about the children and therefore the chance that both of them boys is 1/2 × 1/2. This applies to the first and the second question.

After Sarah says that she has at least one boy, there are equal possibilities that she has Boy + Boy, Boy + Girl, or Girl + Boy. Therefore, the chance that both children are boys is 1/3.

After Courtney says that her younger child is a boy, the only remaining question is what is the gender of her older child, and therefore the chance is 1/2.

The fifth exchange implies that Sarah has a Sagittarius boy. There are 23 combinations such that both children are boys and at least one of them is Sagittarius. There are 47 combinations such that at least one of the children is a Sagittarius boy. Therefore, the chance that both children are boys is 23/47.

Finally, Courtney says that her younger child, which is a boy, is not Sagittarius, but her personal experience with Sagittarius boys is positive. Therefore, her older child is a Sagittarius boy and the chance is 1.