Mr. White: "It's urgent! All medicine bottles must be inspected before they are sold. Due to mistakes, one of the pills was overweight 10 mg each. Please return the wrong number of medicine bottles immediately. Mr. White is very angry.
Mr. White: "Unfortunately, I have to take one from each bottle and weigh it. What nonsense.
Mr. White was about to start work when Miss Black stopped him. Miss Black: "Wait a minute, there is no need to weigh it ten times, just once." How is that possible?
Miss Black's brilliant idea is to take 1 capsule from the first bottle, 2 capsules from the second bottle, 3 capsules from the third bottle, and so on until1capsule is taken from the tenth bottle. Put these 55 pills on the scale and write down the total weight. If its weight is 55 10 mg, that is, it exceeds the specification of 10 mg, she will immediately understand that only one is overweight and was taken out of the first bottle.
If the total weight exceeds the specification of 20 mg, two bottles are overweight, take them out from the second bottle, and so on. So Miss Black only needs to be weighed once, right?
Six months later, the drugstore received ten more bottles of this medicine. An urgent telegram followed, pointing out that a more serious error had occurred.
This time, there is no comment on the number of bottles of overweight pills. Mr. White is very angry. Mr. White: "What shall we do, Miss Black?" Our last method is useless. Miss Black didn't answer immediately; She is thinking about this problem.
Miss Blake: "That's right. But if we change this method, we can still only identify the wrong drug once. What's Miss Black's good idea this time?
In the first question of weighing pills, we know that only one bottle of pills is overweight. Take out a different number of pills from each bottle (the simplest way is to use counting sequence), and we can make a set of numbers have a one-to-one relationship with a set of bottles.
In order to solve the second problem, we must use a numerical sequence to mark each bottle of medicine, and each subset in this sequence must have a separate sum. Is there such an order? Yes, the simplest is the following double sequences: 1, 2, 4, 8, 16. . . These numbers are continuous powers of 2, and this series lays the foundation of binary notation.
In this problem, the solution is to arrange the medicine bottles, take out/kloc-0 tablets from the first bottle, 2 tablets from the second bottle, 4 tablets from the third bottle, and so on. Take out the pills and weigh them on the scale. Suppose the total weight is 270 mg overweight, because each pill with wrong weight is overweight 10 mg, so we divide 270 by 10 to get 27, which is the number of overweight pills. Convert 27 into a binary number:11011. In 1 1 01,from right to left, "1"in the first, second, fourth and fifth digits indicates that the weights are1,2, 8 and/kloc-respectively. So the wrong number of bottles are the first bottle, the second bottle, the fourth bottle and the fifth bottle.
In a set of powers of 2, each positive integer is the sum of elements in a single different combination. In view of this fact, binary notation is extremely useful. In the fields of computer science and a large number of applied mathematics, binary notation is essential. In interesting mathematics, there are also countless applications.
Here is a simple poker trick that can confuse your friends. This trick seems to have nothing to do with the problem of medicine bottles, but they are all based on the same, and they are all binary principles.
Ask someone to wash a deck of cards, then put them in your pocket, and then ask someone to say a number between 1 and 15. Then you put your hand into your pocket, and as soon as you reach out, you take out a set of cards, and the sum is exactly equal to the number he said.
The secret is simple. Before you do magic, take out A, 2, 4 and 8 in advance and put them in your pocket. Only four cards in this deck are missing, so it is unlikely to be noticed. After the washed cards are put in the pocket, they are secretly put behind the four cards that have been put in the pocket. Ask someone to name a number, and you will express it as the sum of the powers of 2. If it is 10, then you should think of: 8+2= 10, then reach into your pocket and take out the cards of 2 and 8 for public display.
The basis of counting cards is also the binary principle. Prepare six cards, labeled A, B, C, D, E and F respectively. Then fill in some numbers on the card, and the rules for determining the numbers set on each card are as follows: in the binary representation of a number, if the first digit from the right is "1", then this number is on card A, and the numbers set on the card start from 1, and all the numbers are odd numbers ranging from 1 to 63; Card B includes all numbers from 1 to 63 in binary notation, and the second number on the right is "1"; Card C includes all numbers from 1 to 63 in binary notation, and the third number from the right is "1"; Cards d, e, f and so on. Note: The binary notation of the number 63 is "1111",so every card has this number.
These six cards can be used to determine any number in the range of 1 to 63. Ask an audience to think of a number (such as someone's age) in this range, and then ask him to give you all the cards with this number on them. You say the numbers in his mind at once. The secret is to add up the first number of the power of 2 on each card. For example, if cards C and F are given to you, you only need to add the first numbers 4 and 32 above to know that the number in others' minds is 36.
Sometimes, in order to make this trick more mysterious, the magician will deliberately paint each card with a different color. He just needs to remember the power of 2 represented by each color. For example, red card stands for 1, orange card stands for 2, yellowcard stands for 4, green card stands for 8, blue card stands for 16, and purple card stands for 32 (according to the color order of the rainbow). So, the magician stood at one end of the big room and asked people to come up with a number. He put the card with this number next to him and asked him to tell others' hearts casually according to the color of the card next to him.
Three people go for 30 yuan a night. They each paid 10 yuan to make a 30 yuan and gave it to the boss. Later, the boss said that 25 yuan was enough for today's discount, so he took out the 5 yuan and asked the waiter to return it to them. The waiter secretly hid 2 yuan's money, and then distributed the rest of 3 yuan's money to three people, each 10 yuan. In this way, everyone handed in 10 yuan at the beginning. Now back to 1 yuan, that is, 10- 1=9, each person only spent 9 yuan money, three people each 9 yuan, 3×9 = 27 yuan+2 yuan hidden by the waiter =29 yuan, where did the one yuan go?
This is a typical misleading question. The cost of staying in a hotel for three people is 27 yuan. This 27 yuan includes 25 yuan for accommodation (in the hands of the boss) +2 yuan for waiter corruption, 3 yuan for meeting with whom, and 30 yuan.
Xiaoming and Xiao Qiang are both students of Mr. Zhang. Mr. Zhang's birthday is M, and both of them know Mr. Zhang's birthday.
It's a day in the 10 group below. Teacher Zhang tells Xiaoming the value of m and the value of Xiao Qiang n. Teacher Zhang asked them if they knew when his birthday was.
March 4, March 5, March 8
June 4, June 7
September1September 5
65438+February165438+February 2nd 65438+February 8th
Xiao Ming said: If I don't know, Xiao Qiang certainly doesn't know.
Xiao Qiang said: I didn't know at first, but now I know.
Xiao Ming said: Oh, I know that, too.
Please infer when teacher Zhang's birthday is from the above conversation.
The answer is: September 1
Related reasoning:
1. Xiaoming said, "If I don't know, Xiao Qiang certainly doesn't know."
The subtext of this sentence is actually: "I should guess right. If I guess wrong, Xiao Qiang will definitely not know." But Xiao Ming is still not sure whether he guessed correctly. He needs Xiao Qiang to prove it. What value does m take for Xiao Ming to say so? Obviously, neither 6 nor 12 is desirable. If m is 6 or 12, n may be 2 or 7-Xiao Qiang can know Mr. Zhang's birthday by the number 2 or 7. Then m can only be 3 or 9, and n can only take the values in 1, 4, 5 and 8.
If m is 3, n can take three values, and the result becomes "If Xiao Ming doesn't know, Xiao Qiang may or may not know (2-4, 3-8)." In this case, Xiao Ming's statement that "if I don't know, Xiao Qiang certainly doesn't know" is not valid, and Xiao Ming is not confident enough to say so.
If m is 9, Xiao Ming knows that n can only be 1 or 5. At this point, Xiao Ming's guess is N= 1, and Xiao Ming is not sure whether n is 1. If n is not 1 but 5, Xiao Ming said, "If I don't know, Xiao Qiang certainly doesn't know." At this point, in fact, Xiao Ming already knows that there are only two situations, only waiting for Xiao Qiang to confirm whether n is 5.
Xiao Qiang said, "I didn't know at first, but now I know."
Xiao Qiang said, "I didn't know", and verified that N is really neither 2 nor 7. At the same time, Xiao Qiang also knew that "m is not 6 or 12, only 3 and 9 are left in m". If n is 5, Xiao Qiang should say, "I didn't know at first, but I still don't know." According to the inference in the first section, N= 1, so Xiao Qiang can say, "I didn't know it at first, but now I know it."
Xiao Ming said, "Then I know."
Xiao Ming is waiting for Xiao Qiang. No matter how Xiao Qiang answers, Xiao Ming will know the correct answer. If Xiao Qiang says "I don't know yet", then Xiao Mingcan still knows that "only N=5 will make Xiao Qiang feel at a loss", so the answer is September 5; If Xiao Qiang said "I know", it must be 1 September.
In fact, from beginning to end, Xiao Ming knew that he only needed Xiao Qiang to say something to verify his guess. For Xiao Ming, this is a multiple-choice question that is not A or B, so according to the story development clue of the topic itself, Xiao Ming's third sentence is unnecessary, but many people will use this condition when deducing-it's a bit like doing a math problem.
One day, a customer went to Lao Zhang's toy store and took a fancy to a toy frog. The retail price was 23 yuan (cost 16 yuan), so he took out a bill of 100 yuan for Lao Zhang. Since Lao Zhang had no change, he went to the neighborhood to change the bill of 100 yuan and returned it to 77 yuan to give it to the customer.
Later, the neighborhood said that Lao Zhang's 100 yuan was counterfeit, and Lao Zhang wanted to return 100 yuan to the neighborhood.
How much did Lao Zhang lose in this transaction?
93
12 balls, one ball is broken, light and heavy. Now that you have a balance, how can you find the broken ball only by weighing it three times?
Number the twelve balls as 1- 12.
For the first time, put 1-4 on the left and 5-8 on the right.
1. If the weight is right, the bad ball is in 1-8.
Remove No.2-4 for the second time, move No.6-8 from right to left, and put No.9-11.
It's on the right. That is to say, put 1, 6, 7, 8 on the left and 5, 9, 10,1/on the right.
1. If the weight, the bad ball in the untouched 1, 5. If it is 1,
It is lighter than the standard ball; If it's a No.5 ball, it's heavier than the standard ball.
Put 1 on the left and No.2 on the right for the third time.
1. If the weight is appropriate, 1 is a bad ball, which is lighter than the standard ball;
2. If it is balanced, the 5th ball is a bad ball, which is heavier than the standard ball;
It can't be left-handed this time.
2. If it is balanced, the bad ball is lighter than the standard ball in the number 2-4 that has been removed.
Put No.2 on the left and No.3 on the right for the third time.
1. If the weight is appropriate, the No.2 ball is a bad ball, which is lighter than the standard ball;
2. If it is balanced, the No.4 ball is a bad ball and lighter than the standard ball;
3. If the left is heavy, No.3 is a bad ball, which is lighter than the standard ball.
3. If the left ball is heavy, the bad ball will be taken to the left 6-8, which is heavier than the standard ball.
For the third time, put No.6 on the left and No.7 on the right.
1. If the weight is appropriate, No.7 is a bad ball, which is heavier than the standard ball;
2. If it is balanced, the No.8 ball is a bad ball and heavier than the standard ball;
3. If the left ball is heavy, No.6 is a bad ball, which is heavier than the standard ball.
2. If the balance is balanced, the bad ball is 9- 12.
The second time, put 1-3 on the left and 9- 1 1 on the right.
1. If the weight is appropriate, the bad ball is 9- 1 1, and the bad ball is heavier.
Put No.9 on the left and 10 on the right for the third time.
1. If the weight is appropriate, 10 is a bad ball, which is heavier than the standard ball;
2. If it is balanced, the number 1 1 is a bad ball, which is heavier than the standard ball;
3. If the left ball is heavy, No.9 is a bad ball, which is heavier than the standard ball.
2. If it is balanced, the bad ball is 12.
For the third time, put 1 on the left and 12 on the right.
1. If the weight is appropriate, 12 is a bad ball, which is heavier than the standard ball;
2. It is impossible to balance at this time;
3. If the left ball is heavy, the number 12 is a bad ball, which is lighter than the standard ball.
3. If the left is heavy, the bad ball is 9- 1 1, and the bad ball is lighter.
Put No.9 on the left and 10 on the right for the third time.
1. If the weight is appropriate, the No.9 ball is a bad ball, which is lighter than the standard ball;
2. If it is a balance ball, the number 1 1 is a bad ball, which is lighter than the standard ball;
3. If the left ball is heavy, 10 is a bad ball, which is lighter than the standard ball.
3. If the left is heavy, the bad ball is in 1-8.
Remove No.2-4 for the second time, move No.6-8 from right to left, and put No.9-11.
It's on the right. That is to say, put 1, 6, 7, 8 on the left and 5, 9, 10,1/on the right.
1. If the right ball is heavy, the bad ball will be taken to the left 6-8, which is lighter than the standard ball.
For the third time, put No.6 on the left and No.7 on the right.
1. If the weight is appropriate, No.6 is a bad ball, which is lighter than the standard ball;
2. If it is a balance ball, the No.8 ball is a bad ball, which is lighter than the standard ball;
3. If the left is heavy, No.7 is a bad ball, which is lighter than the standard ball.
2. If it is balanced, the bad ball is heavier than the standard ball in the number 2-4 that has been removed.
Put No.2 on the left and No.3 on the right for the third time.
1. If the weight is appropriate, the No.3 ball is a bad ball, which is heavier than the standard ball;
2. If it is balanced, No.4 is a bad ball, which is heavier than the standard ball;
3. If the left ball is heavy, No.2 is a bad ball, which is heavier than the standard ball.
3. If the left is heavy, the bad ball is at the untouched number 1, 5. If it is 1,
It is heavier than the standard ball; If it's a No.5 ball, it's lighter than the standard ball.
Put 1 on the left and No.2 on the right for the third time.
1. It can't be right this time.
2. If it is balanced, the No.5 ball is a bad ball and lighter than the standard ball;
3. If the left ball is heavy, the number 1 is a bad ball, which is heavier than the standard ball;
It's enough trouble. In fact, there are many situations that are symmetrical, such as the right weight when weighing for the first time and the right light weight. Only one can be considered, and the other can be implemented by comparison. I wrote down the whole process just to scare everyone.
Try it a little, and you will know that it is impossible to find a bad ball only by weighing it twice. If we give thirteen balls, the above solution is basically effective, except for a small change, that is, in this case, when the first ball and the second ball are balanced, the third ball may still be balanced (that is, step 2.2.2 above), then we can determine that the bad ball is 13, but we can't know whether it is lighter than the standard ball or heavier than the standard ball. If we give fourteen balls, we will find that it is impossible to find the bad ball only by weighing it three times.
A natural question is: For a given natural number n, how do we solve the problem of weighing balls with n balls?
In the following discussion, given any natural number n, we must solve the following problems:
(1) Find out the minimum number of times required for the N-ball weighing problem, and prove that the minimum number given above is indeed the smallest;
⑵ Give the specific method of minimum ball weighing times;
(3) If only the bad ball is required to be found, but the weight of the bad ball is not known, solve the above two problems for the weighing problem of N balls;
There is another problem that we are not interested in, but as a by-product:
(4) If a standard ball is given in addition to the given n balls, the above three problems can be solved.