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1. A rich man hires a person to do 7 days of work for him. The wages he gives him are 7 gold bars (1 yuan per day) that are linked together. It is necessary to ensure that the hired workers receive the wages they should receive each day (no more and no less). The rich can only judge the gold bars that are linked together twice, and ask: What kind of method can they pay the employees as required? 2. There are a total of 100 balls. A and B take rounds. The maximum is 5 at a time. A takes first. How can he take to ensure that the last one is his? 3. Ten bags of gold coins, ten in each bag, nine bags of gold coins each weighing 10 grams, one bag of gold coins each weighing 9 grams, there is a scale, and the scale picks out a bag of 9 grams. How to weigh? 4, 12 balls have the same appearance, only one has a difference in weight with other balls. Give you a balance scale. How to find out this difference ball within three times? 5, 13 are exactly the same, only 1 Balls with different masses, how to find this ball with different masses three times with a balance scale? Speak your process.
1. Make three copies of 1, 2, and 4: take the first day 1, take the next day 2 and return 1, take the third day 2 + 1, take the fourth day 4 and 2 + 1, and take the fifth day 4+ 1. Take 4 + 2 on the sixth day and 1. Take everything on the seventh day.
2. A takes 4 for the first time and then 6-n for n for A (n is any number of 1, 2, 3, 4, 5), so the order is A, B, A, B ... A, B, A, and A have already taken 4 + (5 × 18) = 94 last B no matter how much N (N is any of 1, 2, 3, 4, 5), the remaining (6-N) I removed everything
3. Number the bags 1, 2 ... 10 and then take out 1 ball from the No. 1 bag and 2 balls from the No. 2 bag ............... 9 balls from the No. 9 bag Take out 10 balls in the No. 10 bag and take these 55 balls to weigh n grams less than 550g. Then the bag with the number n is 9 grams.
4. Number 12 balls separately and divide them into 3 groups at will. Without loss of generality, they are: (1,2,3,4) .. ①; (5,6,7,8) .. ②; (9,10,11,12) .. ③.
The first title: put the groups ① and ② on both ends of the balance. There are two kinds of results: one is flat; the other is uneven. It may be assumed that group ① is heavier than group ②. Let's look at the situation first. The 1-8 balls are all normal. The defective product must be in the group ③, that is, in the 9-12 ball. Choose 3 out of 9-12 balls, you may choose (9, 10, 11) ... ④, save the 12th ball: you can also choose 3 in normal balls 1-8, you may choose ( 1, 2, 3) ... ⑤. Weigh ④ and ⑤ a second time. There are three results: ④ = ⑤; ④ ＞ ⑤; ④ ＜ ⑤. If ④ = ⑤, the defective ball is the 12th ball. The third time with the 12th ball and any normal ball, you can immediately judge whether the 12th ball is too heavy or light. If ④ ＞ ⑤, the defective ball must be within the 3 balls of group ④ and heavier than the normal ball. At this time, choose two of the three balls of 9-11 (may be set to 9 and 10 balls), and then put it on the balance for the third time.
There are three cases: 9 = 10, 9> 10, 9 <10. When 9 = 10, the defective ball must be the 11th ball, which is heavier than the normal ball; when 9> 10, the weighted 9 ball is a defective product; when 9 <10, the weighted 10 ball is a defective ball Product. The same can be proved for ④ ＜ ⑤. For another uneven situation, change the proof again. Continue to prove. When there are irregularities, there are two cases, namely group ①> group ②; group ① <group ②. Now let's discuss the situation when group ①> group ②. That is, (1, 2, 3, 4) is heavier than (5, 6, 7, 8). Adjust the balls in group ① and group ② and regroup them: leave the number 3 ball in group ①, take out the number 4 ball, and put the 1 and 2 balls into group ② and add the normal ball One, it may be set as the No. 9 ball; leave the No. 7 ball in the group ②, take out the No. 6 and No. 8 balls, and change the No. 5 ball into the group ①, and make up a new group: ) ... ③; (1,2,7) ... ④. Now proceed to the second one, which is to put group ③ and group ④ on the balance.
There are three results: ③ = ④; ③ ＞ ④; ③ ＜ ④. When ③ = ④. Then the defective ball must be in the several balls taken out, that is, in the 3 balls of No. 4, 6, and No. 8, and the No. 4 ball is heavier than at least one of No. 6 and No. 8 balls. At this time, the 6th ball and the 8th ball are used for the third weighing, and the result is 6th = 8th; 6th> 8th; 6th <8th. When No. 6 = No. 8, then the No. 4 ball is a defective ball, and it is heavier than a normal ball; when No. 6> No. 8, the defective ball is a No. 8 ball, which is lighter than a normal ball; When the number is less than 8, the defective ball is the 6th ball, which is lighter than the normal ball.
When ③ ＞ ④. Explanation: The changed group still maintains the essence of the original group. This is caused by the ball that remains unchanged in the group. The defective ball must be between the 3rd and 7th balls, and the 3rd ball is known. Must be heavier than the 7th ball. At this time, the third call is made: choose one of the 3rd and 7th balls to call the normal ball, and you may choose the 3rd ball and the normal ball to call the 9th ball. The results are: No. 3 = No. 9; No. 3> No. 9; No. 3 <9 No. When No. 3 = No. 9, the defective product is a No. 7 ball, which is lighter than a normal ball; when No. 3> No. 9, the defective product is a No. 3 ball, which is heavier than a normal ball; When No. 3 < On the 9th, from the 3rd to the 7th, both the 3rd and the 7th are defective products, which is impossible because there is only one contradiction with the defective products specified in the conditions. When ③ <④.
This is caused by the exchange of the group balls, so the defective ball must be between 1, 2, and 5, and the 5 ball is at least lighter than one of the 1 and 2 balls. At this time, we used the No. 1 and No. 2 balls for the third time. The results are: No. 1 = No. 2; No. 1> No. 2; No. 1 <No. 2. When No. 1 = No. 2, the defective product is No. 5 and it is lighter than the normal ball; when No. 1> No. 2, then the defective product is No. 1 and it is heavier than the normal ball; , And No. 5 is smaller than No. 2, the defective product is No. 2, which is heavier than the normal ball.
Huayi Fortune Telling Network reminds you: No matter whether you are lucky or bad, do n’t have to be happy or discouraged. Fortunately, good fortune is bad luck, bad luck is good luck in the past, and doing good things is the root of good luck. (Reprinted, please indicate from -Huayi.com: allblogsite.com)