A Dream Last Night

The director of a company wakes up early for a morning flight. He realizes that he forgot some papers at work and goes there. In the office, he meets the night watchman who is leaving for home. The watchman stops the director and tells him that he shouldn’t fly. “I had a dream last night,” the watchman says, “I saw you crash. I saw you die sir, please do not fly today.”

The director listens to the advice of the watchman and decides not to travel. On the next day, the director returns, gives the watchman a generous bonus, and then he fires him. Why?

The plane indeed crashed, so the watchman saved the director’s live. However, apparently, he has been sleeping during his shift, which is the reason the director fires him.

Trips in Bulmenia

In the country of Bulmenia there are 40 big cities. Each of them is connected with 4 other big cities via paths, and you can get from any city to any other via these paths.

  1. Show that you can create a trip passing through every path exactly once that ends in the city it starts from.
  2. Show that you can create one or multiple trips, such that every trip passes through different cities, ends in the city it starts from, and also every city is part of exactly one trip.

Remark: The paths can intersect each other, but you cannot switch from one path to another midway.

  1. Let us call a trip that ends in the city it starts from a “loop”. Start from any city and keep traveling without using any path twice. If at some point you can’t continue, stop, creating a loop, and modify your trip as follows. Pick any city you have visited from which there are unused paths going out, and once again start traveling along the unused paths until you can’t continue further. Add the newly formed loop to the original trip and continue this procedure until there are no unused paths left, thus completing a loop passing through every path exactly once. This method works because there is an even number of paths going out from every city and you can get from any city to any other.
  2. Use the loop from 1. and color every second path on it in black. Then, notice that there are 2 black paths going out from every city. Therefore, these black paths create one or multiple disjoint loops passing through every city in Bulmenia exactly once.
Source:

IMO 2020

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

Beautiful Tapestry

A piece of a beautiful tapestry is missing. Can you figure out what its colors are?

The tapestry represents the factorizations of the numbers from 2 to 26.

Each 12×12 square on the tapestry represents a number between 2 and 26, such that all squares representing prime numbers are painted in single colors. The colors of the squares representing composite numbers are determined by the factors of these numbers.

The number 2 is represented by orange color (top left corner). The number 3 is represented by green color. The number 4 = 2×2 is represented once again by orange (2, 2) color. The number 5 is represented by red color. The number 6 = 2×3 is represented by orange (2) and green (3) colors. The number 7 is represented by blue color. The number 8 = 2×2×2 is represented once again by orange (2, 2, 2) color. The number 9 =3×3 is represented once again by green (3, 3) color. The number 10 = 2×5 is represented by orange (2) and red (5) colors, and so on.

FEATURED

Does Not Divide Another

How many numbers between 1 and 100 can you pick at most, so that none of them divide another?

You can choose fifty numbers at most: 51, 52, 53, … , 100.

In order to see that you cannot choose more than fifty, express each number in the form 2ⁿ×m, where m is an odd number. Since no two numbers can have the same m in their expressions, and there are only fifty odd numbers between 1 and 100, the statement of the problem follows.