Korean, Edit

Mathematical Logic Problems (21-40)

Recommended Article : 【Mathematics】 Mathematics Contents

Problem 21. A, B, C, D, E, F, G, H are employees of a company. They work in one of the three departments of the company: HR, Administration, Marketing. (Assume that no more than four of the eight work in the same department.) They each have different hobbies, which are American football, cricket, volleyball, badminton, tennis, basketball, hockey, and table tennis. (Note: A - F do not necessarily have these hobbies in this order.)

⑴ D works in the Administration department and does not like American football or cricket.

⑵ F works in the HR department with A, who likes table tennis.

⑶ E and H do not work in the same department as D.

⑷ C likes hockey and does not work in the Marketing department.

⑸ G does not work in the Administration department and does not like cricket or badminton.

⑹ One person in the Administration department likes American football.

⑺ The person who likes volleyball works in the HR department.

⑻ No one in the Administration department likes badminton or tennis.

⑼ H does not like cricket.

Who works in the Administration department? And in which department does E work?

Solution 21.

○ First, arrange the information in the order it was discovered.

○ Organize the given information, first inference, second inference, third inference, and fourth inference as follows:

○ A : Table tennis. HR.

○ B : Administration (because of ⑹). Likes American football (because of ⑹).

○ C : Likes hockey. Not Marketing. Administration (because of ⑵, interpreting HR as only two people)

○ D : Administration. Does not like American football, cricket. Not in the same department as E, H. Does not like badminton, tennis. Likes basketball (by process of elimination, ⑺).

○ E : Not Administration (because of ⑶). Marketing (because of ⑵, interpreting HR as only two people). Likes cricket (by process of elimination).

○ F : HR. Likes volleyball (because of ⑺).

○ G : Not Administration. Does not like cricket, badminton. Marketing (because of ⑵, interpreting HR as only two people). Likes tennis (by process of elimination).

○ H : Not Administration (because of ⑶). Does not like cricket. Marketing (because of ⑵, interpreting HR as only two people). Likes badminton (by process of elimination).

○ Answer : People working in the Administration department - B, C, D

○ Answer : E works in the Marketing department

Problem 22. A lion, tiger, and wolf each arrive at a riverbank with their cub. There is a raft at the riverbank that can carry a maximum of two animals. However, the problem is that in the absence of their mother, the cubs are vulnerable to being attacked by other predators. Therefore, cubs cannot be left alone with other predators.

[Rules]

1. Cubs can row the raft across the river.

2. Cubs are safe together without any harm.

3. In the absence of their mother, a cub will be attacked by other predators.

Solution 22.

○ Lion, lion’s cub, tiger, tiger’s cub, wolf, wolf’s cub | ⛵︎ ~ |

○ Lion, tiger, wolf, wolf’s cub | ~ ⛵︎ | lion’s cub, tiger’s cub

○ Lion, lion’s cub, tiger, wolf, wolf’s cub | ⛵︎ ~ | tiger’s cub

○ Lion, tiger, wolf | ~ ⛵︎ | lion’s cub, tiger’s cub, wolf’s cub

○ Lion, lion’s cub, tiger, wolf | ⛵︎ ~ | tiger’s cub, wolf’s cub

○ Lion, lion’s cub | ~ ⛵︎ | tiger, tiger’s cub, wolf, wolf’s cub

○ Lion, lion’s cub, tiger, tiger’s cub | ⛵︎ ~ | wolf, wolf’s cub

○ Lion’s cub, tiger’s cub | ~ ⛵︎ | lion, tiger, wolf, wolf’s cub

○ Lion’s cub, tiger’s cub, wolf’s cub | ⛵︎ ~ | lion, tiger, wolf

○ Wolf’s cub | ~ ⛵︎ | lion, lion’s cub, tiger, tiger’s cub, wolf

○ Wolf, wolf’s cub | ⛵︎ ~ | lion, lion’s cub, tiger, tiger’s cub

○ | ~ ⛵︎ | lion, lion’s cub, tiger, tiger’s cub, wolf, wolf’s cub

Problem 23. Three men and two women need to cross a river. The boat can carry one man or two women at a time. All five people can row, and men and women cannot be in the boat together. What is the way for all five people to cross the river?

Solution 23.

○ Men Men Men Women Women | ⛵︎ ~ |

○ Men Men Men | ~ ⛵︎ | Women Women

○ Men Men Men Women | ⛵︎ ~ | Women

○ Men Men Women | ~ ⛵︎ | Men Women

○ Men Men Women Women | ⛵︎ ~ | Men

○ Men Men | ~ ⛵︎ | Men Women Women

○ Men Men Women | ⛵︎ ~ | Men Women

○ Men Women | ~ ⛵︎ | Men Men Women

○ Men Women Women | ⛵︎ ~ | Men Men

○ Men | ~ ⛵︎ | Men Men Women Women

○ Men Women | ⛵︎ ~ | Men Men Women

○ Women | ~ ⛵︎ | Men Men Men Women

○ Women Women | ⛵︎ ~ | Men Men Men

○ | ~ ⛵︎ | Men Men Men Women Women

Problem 24. “The Genius” Written Exam - Type A

○ Question 1

○ Solution

○ If you look at the empty spaces without black stones, you can see that each time a new figure appears, each empty space moves one space to the right.

○ However, note that when moving one space to the right from the third column, it moves to the first column.

○ Answer : C

○ Question 2

○ Solution

○ Row 1 : 2 × 6 = 12

○ Row 2 : 3 × 7 = 21

○ Row 3 : 4 × 9 = 36

○ Row 4 : 6 × 8 = 48

○ Row 5 : 8 × 7 = 56

○ Answer : 6

○ Question 3

○ Solution

○ From the given information, consider D-F-E, C-A-G as one group.

○ If D is in room 101 or 107, B’s room cannot coexist with C-A-G → Contradiction

○ If D is in room 103, B’s room is 104, G’s room is 107 → Not a contradiction

○ If D is in room 104, B’s room is 103, but there’s no place for C-A-G → Contradiction

○ Conclusion

B

C D

A F

G E

○ Answer : A

○ Question 4

○ Solution

○ Drawing a table helps : The elements marked in the table below represent the winning team.

| | A | B | C | D | E | F | | — | — | — | — | — | — | — | | A | . | . | . | . | . | . | | B | A | . | . | . | . | . | | C | A | B | . | . | . | . | | D | A | draw | D | . | . | . | | E | A | B | C | D | . | . | | F | A | B | F | D | F | . | ​

○ C won against D. (❌)

○ B won against F. (⭕️)

○ B and C drew. (❌)

Problem 25. “The Genius” Written Test - Type (B)

○ Question 1

○ Solution

○ There are a total of 4 segments made up of 3 numbers, and by looking at the segments (11, 8, 3), (8, 5, 9), we can see that the sum of the numbers in each segment is 22.

○ 8 + ? + 8 = 5 + ? + 11 = 22

○ Answer : ? = 6

○ Question 2

○ Solution

○ After considerable thought, I’m not certain, but I would like to suggest two answers.

○ Answer 1. Considering only the numbers 49, 64, 61, ?, which are of equal size, and knowing that 49 and 64 are inverses of each other, ? is 19, which is the inverse of 61.

○ Answer 2. Combining the tens digits of 61 and 33 gives 63, and combining the tens digits of 63 and 49 gives 64, so combining the tens digits of 64 and 57 for ? gives 65.

○ Question 3

○ Solution

○ (Note) Since 242 = 35 - 1, this implies that the bricks should be divided into thirds rather than halves.

○ (Note) When dividing the bricks into n parts and placing only 2 parts on a balance scale, the greater the n, the greater the benefit but also the greater the risk.

○ (Reference) The locations marked with an asterisk (*) indicate defective bricks.

○ Dividing into halves and placing on a balance scale, you can identify the defective brick in 8 uses.

○ 242* → 142* + 142 (1st use)

○ 142* → 81* + 81 (2nd use)

○ 81* → 40* + 40 + 1 (3rd use) : Compare two bundles of 40

○ 40* → 20* + 20 (4th use)

○ 20* → 10* + 10 (5th use)

○ 10* → 5* + 5 (6th use)

○ 5* → 2* + 2 + 1 (7th use) : Compare two bundles of 2

○ 2* → 1* + 1 (8th use)

○ Dividing into thirds and placing on a balance scale, you can identify the defective brick in 5 uses.

○ 242* → 81* + 81 + 80 (1st use) : Compare two bundles of 81

○ 81* → 27* + 27 + 27 (2nd use) : Compare two bundles of 27

○ 27* → 9* + 9 + 9 (3rd use) : Compare two bundles of 9

○ 9* → 3* + 3 + 3 (4th use) : Compare two bundles of 3

○ 3* → 1* + 1 + 1 (5th use) : Compare one by one on the balance scale

○ When divided into quarters and placed on the balance scale, the result is not significantly different from dividing into halves.

○ Dividing into fifths or sixths increases the risk too much, reducing the expected utility in the worst-case scenario.

○ Answer : 5 times

○ Question 4

○ Solution

○ To solve this problem, simply draw a table.

○ C did not choose yellow or black as favorite colors. (✔️)

○ There are more people (3) who chose black than those (2) who chose green. (✔️)

○ D, unlike B, chose black as a favorite color. (✔️)

Problem 26. When arranging coins according to the following rules, there are shapes △ and ▽ between the coins. How many of each shape △ and ▽ are there at position ⑹? Can this be generalized?

Solution 26.

○ △ = 0 + 1 + 2 + ∙∙∙ + 5 = 15 pieces → n pieces at nth position

○ ▽ = 1 + 2 + 3 + ∙∙∙ + 6 = 21 pieces → n pieces at nth position

Problem 27. In a company, there are a president, a department head, and a section chief, each with the surnames Kim, Lee, and Park. However, it is not yet known who has which surname. One day, guests with the surnames Kim, Lee, and Park visited the company. Based on the following conditions, find out the surnames of the president, department head, and section chief!

1) Guest Mr. Park lives in Daejeon.

2) The section chief lives in Suwon.

3) Guest Mr. Lee’s salary is 3,356,511 won.

4) The salary of the guest living in Suwon with the section chief is exactly twice the salary of the section chief. (Note: there are no cents in the salary.)

5) The guest with the same surname as the section chief lives in Seoul.

6) The department head’s surname is not Kim.

Solution 27.

○ First, list the information in the order confirmed.

○ Organize the given information, primary inference, secondary inference, tertiary inference, and quaternary inference as follows.

○ President : Kim ( ∵ By elimination).

○ Department Head : Not Kim. Park ( ∵ By elimination).

○ Section Chief : Suwon. Not Kim ( ∵ 5). Lee ( ∵ 5).

○ Guest Mr. Kim : Suwon ( ∵ By elimination).

○ Guest Mr. Lee : Salary 3,356,511 won. Does not live in Suwon. Seoul ( ∵ 5, By elimination).

○ Guest Mr. Park : Daejeon.

Problem 28. Five students A, B, C, D, E participated in a mathematics competition. Before announcing the results, the teacher asked each of them to predict the rankings of two other students, excluding themselves. Below are their predictions.

A : B will be 4th and C will be 5th.

B : C will be 4th and D will be 3rd.

C : D will be 5th and E will be 4th.

D : E will be 1st and A will be 2nd.

E : A will be 1st and B will be 4th.

According to the teacher, two of them correctly guessed the rankings of both students they predicted, while the other three were all incorrect. What might be the rankings of the five students?

Solution 28.

○ Case 1. If A and B are correct: Contradiction, as C cannot be both 5th and 4th.

○ Case 2. If A and C are correct: Contradiction, as there cannot be two people ranked 5th (C and D).

○ Case 3. If A and D are correct: B = 4, C = 5, E = 1, A = 2. D = 3, but B got that right, so they all can’t be wrong.

○ Case 4. If A and E are correct: B = 4, C = 5, A = 1. If B is all wrong, then D = 2, E = 3.

○ Case 5. If B and C are correct: Contradiction, as there cannot be two people ranked 4th (C and E).

○ Case 6. If B and D are correct: C = 4, D = 3, E = 1, A = 2. B = 5, and the other three are all wrong.

○ Case 7. If B and E are correct: Contradiction, as there cannot be two people ranked 4th (C and B).

○ Case 8. If C and D are correct: Contradiction, as E cannot be both 4th and 1st.

○ Case 9. If C and E are correct: Contradiction, as there cannot be two people ranked 4th (E and B).

○ Case 10. If D and E are correct: Contradiction, as there cannot be two people ranked 1st (D and A).

○ Answer : (A, B, C, D, E) = (1, 4, 5, 2, 3), (2, 5, 4, 3, 1)

Problem 29. Using only a 5-liter barrel and a 3-liter barrel, how can one measure exactly 4 liters of wine?

Solution 29.

○ Step 1. Fill the 5-liter barrel and transfer to the 3-liter barrel.

○ Step 2. Remove 3L of liquor from the barrel, and transfer the remaining 2L of liquor in the 5L barrel to the 3L barrel.

○ Step 3. Fill the 5L barrel to the brim with 5L of liquor, and put 1L in the 3L barrel.

○ This leaves only 4L of liquor in the 5L barrel.

Problem 30. ** On an island, there are 50 people. Each of them has a dog, and they all gather in the square every afternoon before dispersing to their homes. One night, a wizard appears and casts the following three spells:

⑴ Some of the 50 dogs instantly become infected with rabies.

⑵ The owner of an infected dog cannot tell if their dog has rabies (However, they can tell if other people’s dogs are rabid).

⑶ From the moment the spell is cast, the residents can no longer speak.

⑷ Only the owner can kill a dog infected with rabies.

⑸ The residents are aware of the effects of spells ⑴ to ⑷, and know that all the rabies-infected dogs must be killed for the spells to be lifted.

Since a rabies-infected dog could harm others at any time, and the 50 island residents value life highly, they start observing each other’s dogs from the day after the wizard’s visit, to only kill the rabid ones. When an owner is certain their dog has gone mad, they return home and kill it.

On the fourth day after the wizard’s visit, after the residents gather and disperse in the square, all the mad dogs are killed that night. They are then able to speak again and are certain there are no more mad dogs. How many mad dogs were there?

Until the fourth gathering of the residents, no dogs had been killed. All the mad dogs are killed on the same day.

Solution 30.

○ Case 1. If there is 1 rabid dog

○ On the first day after the wizard’s visit, the owner of the rabid dog sees no other mad dogs.

○ Since the residents are still under the spell, the owner quickly realizes their dog has rabies.

○ Therefore, all mad dogs are killed that night.

○ Case 2. If there are 2 rabid dogs

○ On the first day after the wizard’s visit, the owner of each rabid dog sees 1 other mad dog.

○ However, they realize that no mad dogs were killed even on the second day after the wizard’s visit.

○ Thus, the owners quickly realize their own dogs are also rabid.

○ Therefore, all mad dogs are killed that night.

○ Mathematical Induction

○ If there are k rabid dogs, there are no deaths until the kth day after the wizard’s visit, and all die that night (1 ≤ k ≤ n) (※)

○ If there are (n+1) rabid dogs, assume no dogs die until the (n+1)st day after the wizard’s visit.

○ In this case, the owners of the rabid dogs are certain of their dogs’ condition and kill them that night.

○ Thus, (※) still holds true even for k = (n+1).

○ Note, the owner of a rabid dog sees n dogs, so they try to kill their dog after the night of the (n+1)st day, but the owner of a healthy dog sees (n+1) dogs, so they attempt to kill their dog after the (n+2)nd day → There is no case of mistakenly killing a healthy dog.

○ Therefore, (※) always holds true.

○ Answer : There are 4 rabid dogs.

○ More intuitive explanation

Problem 31. A census taker visits a house for a survey and has the following conversation with the parents.

▷ Parents : We have three daughters, and the product of their ages is 36.

▷ Census Taker : I don’t understand.

▷ Parents : The sum of their ages is the same as our house number.

▷ Census Taker : (After checking the house number) I still don’t understand.

▷ Parents : Well, let me call our eldest daughter, then you can guess.

▷ Census Taker : No, now I know their ages.

How did the census taker figure out the ages of the three daughters, and what are their ages?

Solution 31.

○ Possible ages where the product is 36 (sum in parentheses):

○ 1 × 1 × 36 (38)

○ 1 × 2 × 18 (21)

○ 1 × 3 × 12 (16)

○ 1 × 4 × 9 (14)

○ 1 × 6 × 6 (13)

○ 2 × 2 × 9 (13)

○ 2 × 3 × 6 (11)

○ 3 × 3 × 4 (10)

○ There are duplicate sums: (1, 6, 6) and (2, 2, 9).

○ The eldest daughter is unique: Thus, (1, 6, 6) is not the correct solution.

○ Answer : The ages of the daughters are 2, 2, and 9 years.

Problem 32. Here are several pieces of string that appear identical. When one end of a string is lit, it burns unevenly but always takes exactly 1 minute to burn completely. Use these strings to measure exactly 45 seconds.

Solution 32.

○ If both ends of a string are lit, it will burn out in 30 seconds.

. ○ Lay two strings side by side, light one end of one string and both ends of the other.

○ When the string with both ends lit burns out, light the other end of the first string. This will total 30 seconds + 30 seconds / 2 = 45 seconds.

Problem 33. There are five number cards: 0, 1, 2, 3, 4. Picking three at a time, what are the possible sums of these numbers?

Solution 33.

○ Possible sums range from 0 + 1 + 2 = 3 to 2 + 3 + 4 = 9, so there are 7 possible scores.

○ Examples

○ 0 + 1 + 2 = 3

○ 0 + 1 + 3 = 4

○ 0 + 2 + 3 = 5

○ 0 + 2 + 4 = 6

○ 0 + 3 + 4 = 7

○ 1 + 3 + 4 = 8

○ 2 + 3 + 4 = 9

Problem 34. There are 6 marbles of the same size and shape. Three are heavy, and the other three are light. The weights of the three heavy marbles are the same, as are the weights of the three light marbles. Use a balance scale three times to distinguish between the heavy and light marbles.

Solution 34.

⑴ Assign numbers 1 to 6 to the 6 marbles.

⑵ Step 1. Compare (1, 2) with (3, 4).

⑶ Case 1. (1, 2) > (3, 4)

① Possible heavy marble combinations

○ (1, 2, 3)

○ (1, 2, 4)

○ (1, 2, 5)

○ (1, 2, 6)

○ (1, 5, 6)

○ (2, 5, 6)

② Step 2. Compare marble 3 with 6.

③ Step 3. Depending on the result of Step 2, use different strategies.

○ 3 > 6: No further action needed, (1, 2, 3) is determined.

○ 3 = 6: Compare marble 4 with 5 to uniquely determine (1, 2, 4 ), (1, 2, 5 ).

○ 3 < 6: Compare marble 1 with 2 to uniquely determine ( 1 , 2 , 6), ( 1 , 5, 6), (2 , 5, 6).

⑷ Case 2. (1, 2) = (3, 4)

① Possible heavy marble combinations

○ (1, 3, 5)

○ (1, 3, 6)

○ (1, 4, 5)

○ (1, 4, 6)

○ (2, 3, 5)

○ (2, 3, 6)

○ (2, 4, 5)

○ (2, 4, 6)

② Step 2. Compare (1, 3) and (2, 5).

③ Step 3. Depending on the result of Step 2, use different strategies as follows:

○ If (1, 3) > (2, 5) : Compare balls 4 and 5 to uniquely determine among (1, 3, 5), (1, 3, 6), and (1, 4, 6).

○ If (1, 3) = (2, 5) : Compare balls 1 and 2 to uniquely determine between ( 1, 4, 5) and ( 2, 3, 6).

○ If (1, 3) < (2, 5) : Compare balls 3 and 6 to uniquely determine among (2, 3, 5), (2, 4, 5), and (2, 4, 6).

⑸ Detailed Explanation

Problem 35. (Webtoon Kill the King: Flaw Game) Choose one of the flipped cards to replace with the King card. The number of the card I choose becomes the ‘location where the King card is hidden’. The starting player cannot choose cards from the ‘Penalty Zone (26 ~ 33)’. Players take turns asking ‘questions’ to find out where the other player has hidden the King card. The winner is the one who finds out the location first. Questions must be answerable with ‘Yes’ or ‘No’. (Note: ‘Penalty Zone’ refers to a zone exclusive to the player playing second.)

First, it is necessary to properly understand the rules of the game. Let’s assume the starting player is certain that one of two cards contains the opponent’s King card. According to the rules stated above, the starting player would know the location of the opponent’s King card with just one question, which seems to make the starting player the winner. However, according to the content in the webtoon, this is not the case. Therefore, the condition for victory is not whether one knows the location of the King card, but whether one has correctly ‘identified’ the location. Then, it can be understood that there are two types of questions: ‘guessing’ the location of the King card and ‘identifying’ it.

To narrow down the range of the King card with one question, it is ideal to split the combinations into those with and without the King card, which the webtoon introduces as the ‘Half and Half Strategy’.

In the webtoon, the male lead mentions that if the starting player uses 24 cards and the second player uses 32 cards, making a total of 56 cards, both sides have a 50% chance of winning. He inferred that this awkward rule must have some kind of ‘flaw’. Then, where did the male lead get the conclusion that using 56 cards results in a 50% winning rate?

Solution 35. Webtoon Kill the King: Flaw Game Analysis

Entered : 2022.05.20 10:32

Edited : 2023.09.10 10:41