Mathematics 5 - Puzzle Prime

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Soccer Ball

Section: Brain Teasers Category: Mathematics

Difficulty: Hard Tags: Featured

Almost everyone knows what a soccer ball looks like – there are several black regular pentagons on it and around each of them – five white regular hexagons. Around each hexagon there are three pentagons and three more hexagons. However, can you figure out how many pentagons and hexagons are there in total on the surface of the ball?

SOLUTION

Let the number of pentagons is equal to P and the number of hexagons is equal to H. Then the number of edges is equal to (5P + 6H)/2 – that’s because every pentagon has five edges, every hexagon has 6 edges and every edge belongs to 2 sides. Also, the number of vertices is equal to (5P + 6H)/3 – that’s because every pentagon has five vertices, every hexagon has 6 vertices and every vertex belongs to 3 sides. Now using Euler’s Theorem we get P + H + (5P + 6H)/3 – (5P + 6H)/2 = 2, or equivalently P/6=2 and therefore P = 12. Since around every pentagon there are exactly 5 hexagons and around every hexagon there are exactly 3 pentagons, we get H = 5P/3 = 20. Therefore, there are 12 pentagons and 20 hexagons on a soccer ball.

Unknown Author March 1, 2022

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Prisoners and Boxes

Section: Brain Teasers Category: Mathematics

Difficulty: Expert Tags: Featured

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%?

SOLUTION

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%.

Unknown Author January 22, 2022

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Gun Duel

Section: Brain Teasers Category: Mathematics

Difficulty: Advanced Tags: Featured, Probability

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?

SOLUTION

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.

Unknown Author January 16, 2022

4 Comments

Scoring penalties

Section: Brain Teasers Category: Mathematics

Difficulty: Advanced

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.

SOLUTION

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.

Unknown Author December 13, 2021

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Envelopes with Numbers

Section: Brain Teasers Category: Mathematics

Difficulty: Hard Tags: Probability

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?

SOLUTION

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.

Unknown Author October 10, 2021

5 Comments

Trips in Bulmenia

Section: Brain Teasers Category: Mathematics

Difficulty: Expert Tags: Graph Theory

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.

Show that you can create a trip passing through every path exactly once that ends in the city it starts from.
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.

SOLUTION

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.
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

Unknown Author September 28, 2021

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Beautiful Tapestry

Section: Brain Teasers Category: Mathematics

Difficulty: Advanced

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

SOLUTION

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.

Unknown Author September 19, 2021

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Does Not Divide Another

Section: Brain Teasers Category: Mathematics

Difficulty: Hard Tags: Featured

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

SOLUTION

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.

Unknown Author September 13, 2021

1 Comment

Self-Describing Number

Section: Brain Teasers Category: Mathematics

Difficulty: Advanced

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

SOLUTION

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.

Unknown Author August 19, 2021

3 Comments

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Five Points in a Square

Section: Brain Teasers Category: Mathematics

Difficulty: Medium Tags: Featured

There are 5 points in a square 1×1. Show that 2 of the points are within distance 0.75.

SOLUTION

Split the unit square into 4 small squares with side lengths 0.5. At least one of these squares will contain 2 of the points. Since the diagonals of the small squares have lengths less than 0.75, these 2 points must be within such distance.

Unknown Author August 16, 2021

6 Comments