Throughout history and across the world, thinking ability is the most obvious symbol of intelligence. Do you want to know how high your thinking ability is? Let's take a simple test with the editor of Huayi Fortune-telling Network.

【Questions】

1. A rich man hires someone to work for 7 days, and his salary is 7 connected gold bars (1 per day). To ensure that the worker gets the correct daily wage (not more, not less), the rich man can only break the gold bars twice. How should he break them to meet the requirements?

2. There are 100 balls in total. A and B take turns taking up to 5 balls each time. A goes first. How should A take the balls to ensure that the last one is his?

3. Ten bags of gold coins, each containing ten coins. Nine of the bags have coins weighing 10 grams each, and one bag has coins weighing 9 grams each. There is a scale. How can you find out which bag contains the 9-gram coins by weighing it once?

4. There are 12 balls of the same appearance, and one of them has a different weight. You are given a balance scale. How can you find the different ball within three weighings?

5. There are 13 balls of the same appearance, and only one has a different weight. How can you find the different ball using a balance scale within three weighings? Please explain your process.

【Answers】

1. Break the bars into 1, 2, and 4. On the first day, give 1; on the second day, give 2 and take back 1; on the third day, give 2+1; on the fourth day, give 4 and take back 2+1; on the fifth day, give 4+1; on the sixth day, give 4+2 and take back 1; on the seventh day, give all.

2. A takes 4 balls first, then for every n balls B takes, A takes 6-n (n is any number from 1 to 5). So the order is A, B, A, B, A... A, B, A, B, A. When it's A's turn, he has already taken 4 + (5×18) = 94 balls. No matter how many N (any number from 1 to 5) balls B takes, the remaining (6-N) will be taken by A.

3. Number the bags from 1 to 10, then take 1 ball from bag 1, 2 balls from bag 2, ..., 9 balls from bag 9, and 10 balls from bag 10. Weigh these 55 balls. If the total weight is n grams less than 550 grams, then the bag numbered n is the one with 9-gram coins.

4. Number the 12 balls and divide them into 3 groups randomly. Without loss of generality, they are: (1, 2, 3, 4)…①; (5, 6, 7, 8)…②; (9, 10, 11, 12)…③.

First weighing: Place group ① and group ② on both sides of the balance. There are two possible outcomes: balanced or unbalanced. Suppose group ① is heavier than group ②. First consider the balanced case. Then balls 1-8 are all normal, and the defective ball must be in group ③, i.e., balls 9-12. Take 3 balls from group ③, say (9, 10, 11)…④, and leave out ball 12. Take 3 normal balls from group 1-8, say (1, 2, 3)…⑤. Weigh group ④ against group ⑤ in the second weighing. There are three possible outcomes: ④ = ⑤; ④ > ⑤; ④ < ⑤. If ④ = ⑤, the defective ball is ball 12. In the third weighing, compare ball 12 with any normal ball to determine whether it is heavier or lighter. If ④ > ⑤, the defective ball must be among 9, 10, and 11, and it is heavier than normal balls. Take any two of them, say 9 and 10, and weigh them against each other in the third weighing. There are three possibilities: 9 = 10; 9 > 10; 9 < 10. If 9 = 10, then the defective ball is 11, which is heavier. If 9 > 10, then the defective ball is 9, which is heavier. If 9 < 10, then the defective ball is 10, which is heavier. The same logic applies if ④ < ⑤. Now consider the unbalanced case. If group ① > group ②, meaning (1, 2, 3, 4) is heavier than (5, 6, 7, 8). Rearrange the balls from groups ① and ② and form new groups: keep ball 3 from group ①, remove ball 4, move balls 1 and 2 to group ②, and add a normal ball, say ball 9. Keep ball 7 from group ②, remove balls 6 and 8, and move ball 5 to group ①. Form new groups: (5, 3, 9)…③; (1, 2, 7)…④. Weigh group ③ against group ④ in the second weighing.

The results can be three: ③ = ④; ③ > ④; ③ < ④. If ③ = ④, the defective ball must be among the removed ones, i.e., balls 4, 6, and 8, and ball 4 is at least heavier than one of 6 or 8. Compare 6 and 8 in the third weighing. If 6 = 8, then ball 4 is the defective one, which is heavier. If 6 > 8, then ball 8 is the defective one, which is lighter. If 6 < 8, then ball 6 is the defective one, which is lighter.

If ③ > ④, this indicates that the original weight difference remains, due to the unchanged balls in the groups. Therefore, the defective ball must be between ball 3 and ball 7, and ball 3 is definitely heavier than ball 7. In the third weighing, compare ball 3 with a normal ball, say ball 9. The results are: 3 = 9; 3 > 9; 3 < 9. If 3 = 9, then ball 7 is the defective one, which is lighter. If 3 > 9, then ball 3 is the defective one, which is heavier. If 3 < 9, then ball 3 is lighter, but since 3 > 7, ball 7 would also be lighter, which contradicts the condition that there is only one defective ball. If ③ < ④, this is caused by the swapped balls, so the defective ball must be among balls 1, 2, and 5, and ball 5 is at least lighter than one of 1 or 2. Compare balls 1 and 2 in the third weighing. The results are: 1 = 2; 1 > 2; 1 < 2. If 1 = 2, then ball 5 is the defective one, which is lighter. If 1 > 2, then ball 1 is the defective one, which is heavier. If 1 < 2, then ball 2 is the defective one, which is heavier.