FEATURED

Prisoners and Boxes

There are 100 inmates living in solitary cells in a prison. In a room inside the prison there are 100 boxes and in each box there is a paper with some prisoner’s name (all different). One day the warden tells the prisoners that he has aligned next to the wall in a special room 100 closed boxes, each of them containing some prisoner’s name (all different). He will let every prisoner go to the room, open 50 of the boxes, then close them and leave the room the way it was, without communicating with anybody. If all prisoners find their names in the boxes they open, they will be set free, otherwise they will be executed. The prisoners are allowed to come up with a quick plan before the challenge begins. Can you find a strategy which will ensure a success rate of more than 30%?

The prisoners can assign numbers to their names – 1, 2, 3, … , 100. When prisoner X enters the room, he should open first the X-th box in the line. If he sees inside prisoner’s Y name, he should open next the Y-th box in the line. If he sees in it prisoner’s Z name, he should open next the Z-th box in the line and so on.

The only way which will prevent all prisoners from finding their names is if there is a long cycle of boxes (length 51 or more), such that the first box in the cycle directs to the second box in it, the second box to the third box, the third box to the fourth box and so on.

It is not hard to compute that the probability of having a cycle of length K>50 is exactly 1/K. Then the probability for failure will be equal to the sum 1/51 + 1/52 + … + 1/100, which is very close to ln(100) – ln(50) = ln(2) ~ 69%. Therefore, this strategy ensures a success rate of more than 30%.

FEATURED

Gun Duel

Mick, Nick, and Rick arrange a three-person gun duel. Mick hits his target 1 out of every 3 times, Nick hits his target 2 out of every 3 times, and Rick hits his target every time. If the three are taking turns shooting at each other, with Mick starting first and Nick second, what should be Mick’s strategy?

Clearly, Mick should not aim for Nick, because if he kills him, then he will be killed by Rick. Similarly, Nick should not aim for Mick, because if he kills him, then he also will be killed by Rick. Therefore, if Rick ends up against alive Mick and Nick, he will aim at Nick, because he would prefer to face off a weaker opponent afterward. This means that if Nick is alive after Mick shoots, he will shoot at Rick.

Thus, if Mick shoots at Rick and kills him, then he will have to face off Nick with chance of survival less than 1/3. Instead, if he decides to shoot in the air, then he will face off Nick or Rick with chance of survival at least 1/3. Therefore, Mick’s strategy is to keep shooting in the air, until he ends up alone against one of his opponents.

Scoring penalties

At some point in Leonel Messi’s career, the football player had less than 80% success when performing penalty kicks. Later in his career, he had more than 80% success when performing penalty kicks. Show that there was a moment in Leonel Messi’s career when he had exactly 80% success when performing penalty kicks.

Let us see that it is impossible for Messi to jump from under 80% success rate to over 80% success rate in just one attempt. Indeed, if Messi’s success rate was below 80% after N attempts, then he scored at most 4N/5 – 1/5 = (4N-1)/5 times. If his success rate was above 80% after N+1 attempts, then he scored at least 4(N+1)/5 + 1/5 = (4N-1)/5 + 6/5 times. However, Messi can not score more than one goal in a single attempt, which completes the proof.

Envelopes with Numbers

You are given 2 sealed envelopes with numbers inside. You are told that one of the numbers is twice as much as the other one. You grab one of the envelopes and right before you open it, you make the following calculation:

“If this envelope contains X inside, then the other envelope contains either X/2 or 2X inside. Since the chance that the other envelope contains a larger number is exactly 50%, the expected money I will get after switching is X/4 + X = 1.25X > X. Therefore, I should switch!”

Clearly, this reasoning is wrong, since you can’t possibly deduce which envelope of the two contains a larger number. Where is the mistake?

The trick is that conditionally on the fact that your envelope contains X, it is not true that the other envelope has 50% chance of containing either X/2 or 2X. The reason is that it is impossible that all amounts of dollars appear in the envelopes with the same probabilities (densities). Thus, for example, if it is very unlikely that an envelope contains more than 1000, and you open an envelope with 800 inside, you will not think that the other envelope has 50% chance of containing 1600.

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

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.

Self-Describing Number

Find all 10-digit numbers with the following property:

  • the first digit shows the number of 0s in the number
  • the second digit shows the number of 1s in the number
  • the third digit shows the number of 2s in the number, and so on

Let the number be ABCDEFGHIJ. The number of all digits is 10:

A + B + C + D + E + F + G + H + I + J = 10

Therefore, the sum of all digits is 10. Then:

0×A + 1×B + 2×C + 3×D + 4×E + 5×F + 6×G + 7×H + 8×I + 9×J = 10

We see that F, G, H, I, J < 2. If H = 1, I = 1, or J = 1, then the number contains at least 7 identical digits, clearly 0s. We find A > 6 and E = F = G = 0. It is easy to see that this does not lead to solutions, and then H = I = J = 0.

If G = 1, we get E = F = 0. There is a 6 in the number, so it must be A. We get 6BCD001000, and easily find the solution 6210001000.

If G = 0, F can be 0 or 1. If F = 1, then there must be a 5 in the number, so it must be A. We get 5BCDE10000. We don’t find any solutions.

If F = 0, then the number has at least five 0s, and therefore A > 4. However, since F = G = H = I = J = 0, the number does not have any digits larger than 4, and we get a contradiction.

Thus, the only solution is 6210001000.

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.