Today, it is five feet thick and two mice are wearing it. The mouse has one foot every day, and the mouse has one foot. The daily life of mice is doubled, while the daily life of mice is doubled. Q: When shall we meet? Wear geometry?

There is a wall, the stack is five feet thick (old length unit, 1 foot = 10 inch), and two mice, one large and one small, are digging holes from both sides of the wall at the same time along a straight line. The mouse scored 1 ft on the first day, and the progress every day was twice that of the previous day. The score of the mouse on the first day was also 1 foot, and the progress every day after that was half that of the previous day. How many days can they meet? How many did you fight when you met?

This topic was published in the chapter "Remains and Shortcomings" of Nine Chapters of Arithmetic, a famous classic of classical mathematics in China. Nine Chapters of Arithmetic was written in the first century. Because of its long history, its author and exact writing date have not been confirmed. This book is arranged in the form of a list of mathematical problems. This book * * * collects 246 math problems, which are divided into nine categories, namely nine chapters, so it is called "nine-chapter arithmetic".

It is not difficult to solve this problem. Please have a try.

(2) Han Xin short radius.

Legend has it that Han Xin, a general of the Han Dynasty, used a special method to count the number of soldiers. His method is: let the soldiers line up in three rows (three people in each row), then five rows (five people in each row) and finally seven rows (seven people in each row). As long as he knows the approximate number of soldiers in this group, he can calculate the exact number of soldiers in this group according to the number of soldiers in the last line of these three parades. If Han Xin saw three processions at that time, and the number of soldiers in the last line was 2, 2 and 4 respectively, knowing that the number of soldiers in this team was about 300 to 400, can you quickly calculate the number of soldiers in this team?

(3) Monks divide steamed bread

There is a famous arithmetic problem in Cheng Dawei's masterpiece "Directing at Arithmetic Unity" in Ming Dynasty:

A hundred steamed buns and a hundred monks,

The three monks are even less controversial.

One of the three little monks,

How many monks are there? "

If translated into vernacular, it means: there are 100 monks sharing 100 steamed buns, which is exactly the ending. If the big monk is divided into three parts and the young monk is divided into three parts, how many people are there in each part?

Method one, solve by equation:

Solution: Let the big monk have X people and the little monk have (100-x) people. According to the meaning of the question, the equation is listed:

3x+ 1/3( 100-x)= 100

To solve this equation: x=25.

Little monk: 100-25 = 75 people.

Method two, chickens and rabbits in the same cage:

(1) Suppose 100 people are big monks, how many steamed buns should they eat?

3× 100=300 (pieces).

How much did you eat?

300- 100 = 200 (pieces).

(3) Why did you eat 200 more? This is because the young monk is regarded as a big monk. So how many steamed buns does each young monk count when he is regarded as a big monk?

3- 1/3=8/3

(4) Each young monk bought 8/3 steamed buns, and one * * * bought 200 steamed buns, so the young monk has:

200/8/3 = 75 (person)

Big monk: 100-75 = 25 (person)

Method 3, grouping method:

Because the big monk has three steamed buns and the little monk has three steamed buns. We can group three little monks and 1 big monks, so that four monks in each group are exactly divided into four steamed buns, so that the total number of 100 monks is divided into 100÷(3+ 1)=25 groups, because each group has 1. Because there are three young monks in each group, there are 25× 3 = 75 young monks. This is the solution in "Command Algorithm Unifies Clans". The original words are: "Let 100 monks be the truth, divide by three to get four, and get 25 great monks." The so-called "real" is "dividend" and "law" is "divisor". The formula is:

100÷(3+ 1)=25, 100-25=75。 This shows the wisdom of the ancient working people in China.

(4) You can tell a monk by his bowl.

There is a woman washing dishes by the river. Passers-by asked her why she washed so many dishes. She replied: There are many guests at home. They share a rice bowl for every two people, a soup bowl for every three people and a vegetable bowl for every four people, and * * * uses 65 bowls. Can you infer how many guests have been to her house from the situation of her household bowls?

(5). Hundred yuan problem

There is a chicken Weng today, worth five; One mother hen is worth three; Chicks, chicks are worth one. You can buy a hundred chickens for every hundred dollars. What are the geometric figures of chicken, Weng and chicken?

According to legend, during the Northern and Southern Dynasties (386-589 AD), a "child prodigy" appeared in northern China. He was quick-witted and had excellent computing ability. He solved many problems that even adults could not answer at that time. People from far and near like to ask him to solve math problems.

The fame of "child prodigy" grew bigger and bigger, and it reached the ears of the then Prime Minister. One day, in order to find out whether the "child prodigy" is true or not, the Prime Minister specially called the "child prodigy" father and gave him 100 pence to bring 100 chickens the next day. It is also stipulated that 100 chickens should have roosters, hens and chickens, no more nor less, and it must be 100 yuan for one hundred chickens.

At that time, it took five pence to buy 1 rooster, three pence to buy 1 hen and only 1 penny to buy three chickens. How can I earn 100 yuan and 100 chickens? The "prodigy" thought about it and told his father that all he had to do was send four cocks, 18 hens and 78 chickens.

The next day, the prime minister was surprised to see that the chicken he sent just met the needs of hundreds of dollars and chickens. He thought for a moment, gave 100 pence, and sent 100 chickens tomorrow. It is also stipulated that only four cocks are allowed.

This problem has not stumped the "prodigy". He thought about it and asked his father to send eight cocks, 1 1 hen and 8 1 chicken. He also told his father that if he encountered similar problems, he would just do what he could. The next day, the Prime Minister was amazed to see 100 chickens delivered. He gave 100 pence again, and asked to give 100 chickens next time.

Unexpectedly, just after a while, the "prodigy" father sent 100 chickens. Prime Minister, count: 12 cocks, 4 hens and 84 chickens, just enough to meet the needs of hundreds of dollars and chickens.

This "prodigy" is Zhang Qiujian. He continued to study hard and eventually became a famous mathematician. In his masterpiece "Zhang Qiujian Su 'an Classic", the last topic is this interesting "Hundred Chicken Problem".

"Hundred chickens problem" is an indefinite equation problem. X+y+z= 100

Let the number of roosters, hens and chickens be x, y and z respectively, and the equation can be obtained according to the meaning of the question: 5x+3y+ 1/3z= 100.

In addition, if an integer parameter k is set, it is: x = 4k, y = 25-7k, z = 75+3k.

Because the number of chickens x, y and z can only be positive numbers, the k value satisfying this set of formulas can only be 1, 2, 3. Replace k in the formula with 1, 2, 3 respectively, and the calculated answer is exactly the same as that of Zhang Qiujian.

When Zhang Qiujian lived, people didn't make equations. So, how did he come up with several answers to this question?

It turns out that Zhang Qiujian discovered a secret: four cocks are worth 20 pence and three chickens are worth 1 penny, so the total number of chickens is 7 and the money is 21; As for seven hens, the number of chickens is seven, and the money is also 2 1. If you buy seven fewer hens, you can use the money to buy four more cocks and three more chickens. In this way, a hundred chickens are still a hundred chickens, and a hundred dollars is still a hundred dollars. So as long as only one answer is found, according to this law, other answers can be found immediately.

This is the famous "Hundred Chicken Technique" at home and abroad.

(6). Zhu Shijie, a mathematician in the Yuan Dynasty, has such a topic in "Meeting Philip Burkart" edited by 1303:

999 pence, buy 1000 pears in time,

Eleven pears, nine pears, seven fruits and four pence.

Q: What's the price of pears?

Answer: There are 657 pears, * * * 803p, and 343 pears, * * *196p.

(7). Hundred sheep problem

The Problem of Arithmetic Unity is one of China's ancient mathematical works. There is such a question in the book:

A took a fat sheep and asked the shepherd, "There are about 100 sheep you drove." The shepherd replied, "If you double this flock of sheep, add half of the original flock of sheep, and add 1/4 of the original flock of sheep, even the fat sheep you are leading will only make up a hundred." Please count how many sheep this shepherd has driven.

(8) Li Bai buys wine

Zhang Zhu, an astronomer and mathematician in the Tang Dynasty in China, once wrote a math problem on the topic of "Li Bai drinking": "Li Bai is walking in the street, carrying a pot to buy wine. Double it when you meet a store, and drink a barrel when you see flowers (a barrel is an ancient wine set and can also be used as a unit of measurement). I met the flowers in the shop three times and drank all the wine in the pot. How much wine is there? "

Solution: you need the original amount of alcohol in the pot, and tell the change and final result of the wine in the pot-add (multiply by 2) the amount three times and subtract (lose weight) the light. To solve this problem, it is generally based on the changed results, using the reciprocal relationship of multiplication and division, addition and subtraction, and gradually reverse reduction. "When you meet three flowers in a shop, you will drink all the wine in the pot." As you can see, when you meet three flowers, there will be a bar bucket in the pot, a bar bucket of 1÷2 when you meet three flowers, and a bar bucket of 1÷2+ 1 when you meet two flowers.

[(1÷ 2+1) ÷ 2+1] ÷ 2 = 7/8 (barrel)

So, the pot hit 7/8.

The key point of the above solution lies in inverse reduction, which can also be expressed by schematic diagram or line segment diagram.

Of course, if we use algebra to solve this problem, the quantitative relationship will be more clear. There are x barrels of wine in the pot. List the equations according to the meaning of the question.

2[2(2x- 1)- 1]- 1 = 0

X=7/8 (barrel) is obtained by solving.

(9) Upgrade pagodas

In the Ming Dynasty, there was such a ballad in Cheng Dawei's < < algorithm unification, which was called "Fu Tu Zeng Song Ji".

Seen from a distance, the red light on the seventh floor of the towering tower doubles.

* * * The lights are three hundred and eighty-one. How many lights are sharp?

The pagoda described in this ancient poem was called Pagoda in ancient times. This topic says that it has seven pagodas, and the number of red lights hanging on each floor is twice that of the previous floor. How many lights are there at the top of this tower?

A: There are three pagodas on the top floor. This question means that there is a magnificent pagoda in the distance with many red lights hanging on it. The number of lights on the lower floor is twice that of the upper floor, and the whole tower has 38 1 lights. How many lights are there on the top floor?

1. The ratio of lights on each floor is1:2: 4: 816: 32: 64, and the total number of lights is+2+4+8+16+31+64 =1.

Solution: set a layer of X.

x+2x+4x+8x+ 16x+32x+64x = 38 1

127x=38 1

x=3

8x=24

There is a red light on the fourth floor.

(10) Things are unknown.

China's classic mathematics.

I don't know how many things happened today. Three three numbers leave two, five five numbers leave three, and seven seven numbers leave two. What is the geometry of things?

A number is divided by 3 by 2, by 5 by 3, and by 7 by 2. Find this number.

Can you explain this number?

The solution of Sun Tzu's Art of War goes something like this.

Find the number divisible by 3/2, 5 and 7 at the same time. The minimum value is 140.

The number divisible by 5/3 and by 3 and 7 at the same time, the minimum is 63.

Finally, find the number divisible by 7/2 and 3,5 at the same time, and the minimum value is 30.

So 140+63+30=233 is a required number.

It subtracts or adds a multiple of 105, the least common multiple of 3, 5 and 7, such as 233-2 10=23.

233+ 105=388, ... is also a number that meets the requirements, so there are infinite numbers that meet the requirements. The smallest number is 23.