Babylonians used special cuneiform characters. They carved the words on the clay tablet and dried them. After drying, clay pieces are as hard as stones and can be preserved for a long time.
From the unearthed clay tablets, people found the mathematical problems given by the Babylonians more than 3000 years ago:
"10 brothers 100 two silver, more than one person. I only know that the difference of each level is the same, I don't know how much. Now BaDi is divided into six pieces of silver. How much is the difference in the first level? "
If 10 brothers share 100 silver equally, then everyone should share 10 silver. Now the eighth brother only gets six taels, which means that the boss gets the most, and the next one is less than the last one.
According to the conditions given in the title, there should be the following relationship:
The second child gets the difference of the boss minus one time, the third child gets the difference of the boss minus two times, and the fourth child gets the difference of the boss minus three times.
Old ten gets the difference between the boss and nine times.
In this way, the money of the eldest brother and the oldest ten * * = the money of the second child and Laojiu * * * = the money of the third child and the eighth child * * = the money of the fourth child and the seventh child * * = the money of the fifth child and the sixth child * * = 20, the known eighth child gets 6, and the third child can get 20-6 = 14.
Answer: Grade 1 difference 1.6 silver.
Mathematics and astronomy in Babylon developed rapidly. In addition to using hexadecimal system first, they also determined that a month (moon month) has 30 days and a year (moon year) has 12 moon months. In order not to lag behind the solar year, they adopted the method of stipulating leap month for correction in some years.
Babylonians knew the existence of planets. They worship the sun, the moon and Venus, and regard the number 3 as "happiness". Later, they discovered Jupiter, Mars, Mercury and Saturn, and the number 7 was regarded as "happiness".
Babylonians paid special attention to the study of the moon, called the ratio of the bright part of the meniscus to the total area of the moon "moon phase", and recorded the topic about the moon phase on a clay tablet:
"Let the total area of the moon be 240. From the new moon to the full moon, 15 days, the first five days are twice that of the previous day, that is, 5 10, 20, 40, 80, and the following 10 days increase the same value every day. What is the added value? "
The total area of the moon is 240. On the fifth day, the area of the moon is 80. After 10, the increased area of the moon is 240-80 = 160.
So the daily increase value is 160 ÷ 10 = 16.
A: The increased value is 16.
2. The rand papyrus on the papyrus is a math book of ancient Egyptians 4000 years ago, which recorded many interesting math problems in hieroglyphics, such as:
At 7,7× 7,7× 7,7× 7,7× 7,7× 7,7× 7,7× 7,7× 7, ...
There are several hieroglyphs on these numbers: house, cat, mouse, barley and bucket, which translates as:
"There are seven houses. There are seven cats in each family. Each cat eats seven mice. Every mouse eats Three barley. Each ear of barley seeds can grow seven barrels of barley. Please calculate the total number of houses, cats, mice, barley and buckets. "
Strangely, similar arithmetic problems were also circulated among the people in ancient Russia:
"There are seven old people walking on the road. Each old man has seven walking sticks, and each walking stick has seven branches. Seven bamboo baskets are hung on each branch. There are seven bamboo cages in each bamboo basket and seven sparrows in each bamboo cage. How many sparrows are there in all? "
The topic of ancient Russia was relatively simple. The number of old people is 7, the number of walking sticks is 7× 7 = 49, the number of branches is 7× 7× 7 = 49× 7 = 343, the number of bamboo baskets is 7× 7× 7 = 343× 7 = 240 1, and the number of bamboo cages is 7× 7× 7. There are 1 17649 sparrows in total. It's not easy for seven old people to run around with 1 1 more than ten thousand sparrows! If each sparrow weighs 20 grams, these sparrows weigh more than 2 tons.
There is an answer behind the question that cats eat mice and mice eat barley in rand papyrus, saying that it is 280 1 multiplied by 7.
To find the total number of houses, cats, mice, barley and buckets is to sum 7+7× 7+7× 7+7× 7 = 7+49+343+2401+16807. This is the same as the answer of 280 1× 7 = 19607 above. The ancient Egyptians mastered this special summation method more than 4000 years ago.
A similar problem also appeared in an old English nursery rhyme:
"When I went to Iverson Holy Land, I met seven women and children, each with seven bags in his hand. A cat and seven children depended on each other closely. Women and bags, cats and children went to the Holy Land at the same time?"
The Italian mathematician Fibonacci had a similar problem in the abacus book published in 1202:
"There are seven old women on their way to Rome, each with seven mules; Each mule carries seven bags, each movable bag carries seven loaves, each loaf carries seven knives, and each knife has seven sheaths. How many women, mules, bread, knives and scabbard were there on the way to Rome? " The same kind of problems appeared in different forms in different times and countries, but the earliest time was the ancient Egyptian Rand papyrus.
In ancient Egypt, the topic of "someone stole the treasure" also spread:
"One person took 13 from the treasure house, another person took 1 17 from the rest of the treasure house, leaving 150 treasures in the treasure house. How many treasures are there in the treasure house? "
The formulation of this problem is very similar to the topic in the current textbook, and it can be solved like this:
Let the original treasure in the treasure house be 1, then the first person takes 13, the second person takes (1-kloc-0/2) ×117 = 252, and the last treasure house is1.
Therefore, the original treasure in the treasure house is150 ÷ 3251=150× 5132 = 23916.
The comprehensive formula is150 ÷ [1-kloc-0/3-(1-kloc-0/3) ×117 = 2391.
The rand papyrus has such a problem:
"The number of items, two-thirds, one-half, one-seventh, all, ***33 pieces, find the number of items."
Solve problems by arithmetic, let all be 1, and the number of terms is 33 ÷ (23+12+17+1).
=33÷9742=33×4297= 142897 The answer is unique, but the cursive answer is 14, 156, 197, 16544. What's going on here? Are there eight answers to this question?
The original paper cursive script gives the answer in the form of ancient Egyptian score, which means14+14+156+197+194+1388+1679. Might as well work it out and see:
14+ 14+ 156+ 197+ 194+ 1388+ 1679+ 1776 = 14+ 1456+ 156+ 196
3. Greece in poetry is one of the ancient civilizations in the world. It has a splendid ancient culture. There are some math problems written in poetry in the Greek poetry collection.
In The Sorrow of Love, Elos is the goddess of love in ancient Greek mythology, and Ghiblida is the patron saint of the island of Cyprus. Among the nine literary goddesses, Yvette played music, Ella played love poems, Daglia played comedies, Texola played dances, Meribo played tragedies in myna, Cleo played history, Bolunya played carols, Uranija played astronomy and Kaliopa played epics.
The trouble of love "Love Ross is crying by the roadside, tears drop by drop."
Kipling Lida forward and asked:
What makes you so sad?
Can I help you? Love Ross replied:
Nine literary goddesses, I don't know where they came from, almost swept away the apples I picked from Mount Helkang.
Yevterbo quickly caught a twelfth, and Airato caught more-one of the seven apples.
One eighth of the apples were taken away by Dalya, and twice as many apples fell into Texhola's hands.
Meribo myna was the most polite, and only took one twentieth.
But Cleo came again, and she gained more than four times.
There are three goddesses, none of whom are empty-handed:
30 apples belong to Polynesia, 120 apples belong to Uranija and 300 apples belong to Kaliopa.
How many apples do I have in eros, poor eros? There are 50 apples left. "
This 26-line poem gives a math problem with many numbers. The original title is I don't know the number of apples. After being snatched away by nine literary goddesses, Elos has only 50 apples left, which is a mathematical problem of "knowing its parts and seeking all its types".
Let the original number of apples in eros be X.
According to the question, it means112x+17x+120x+15x+30+120+300+50 =
"This is a bronze statue of cyclops. The sculptor is highly skilled. In the bronze statue, there is a clever organ:
The giant's hand, mouth and one eye are all connected with large and small water pipes. Through the water pipe of the hand, Santianchi is full; Through the one-eyed water pipe-it takes a day; The water spit out from the mouth is faster, and two-fifths a day is enough. When will the pool be full when the water is discharged from three places at the same time? "
Let the volume of the pool be 1 and the time required to fill the pool with three pipes at the same time be x days, then13x+x+52x =1∴ x = 623. Here is a jingle by China:
"Li Bai was carrying a pot to buy wine.
Double it when you meet a store, and have a drink when you see flowers.
I met the shop flower three times and drank all the wine in the pot.
How much wine is there in the pot? "
This limerick means that there is wine in Li Bai's pot, and every time he meets a hotel, he doubles the wine in the pot; Li Bai and Shang Huashi will drink and write poems, and drink a barrel of wine at a time (barrels are ancient containers for wine). After that, they will drink all the wine in the pot for three times, and finally ask Li Bai how much wine was left in the pot?
The best way to solve this problem is to use reverse deduction:
When Li Bai saw the flowers for the third time, he drank all the wine in the pot, indicating that there was only one barrel of wine in the pot before he saw the flowers. It is further concluded that before Li Bai met the hotel for the third time, there were 12 fights in the pot. According to this calculation method, it can be calculated that there were 1 12 fights in the pot before he met Hua for the second time, and 1 12÷2=34 fights in the pot before he met the hotel for the second time. The first time I saw a barrel of wine in Qianhuahu 134. Before seeing the hotel for the first time, there was a barrel of wine in the original pot 134÷2=78. There are 78 barrels of wine in the original pot.
There are many interesting mathematical problems in distributing the inheritance according to the will.
Stranoloubowski, a famous Russian mathematician, once raised the question of distributing inheritance: "In his will, the father asked to give 13 of the inheritance to his son and 25 to his daughter; The remaining money, 2500 rubles, is used to pay off debts. 3000 rubles to mother, what a legacy! How many children have been divided! "
Set the total inheritance to x rubles.
Then13x+25x+2500+3000 = x = 20625.
The son's score is 20625× 13 = 6875 (rubles), and the daughter's score is 20625× 25 = 8250 (rubles).
As a result, the daughter got the most, with 8250 rubles, followed by the son, with 6875 rubles, and the mother got the least, with 3000 rubles. It seems that the father loves his daughter.
The following story first spread among Arabs, and later spread to all countries in the world. The story says that an old man raised 17 sheep. After his death, he asked in his will to distribute 17 sheep to his three sons in proportion, the eldest son 12, the second son 13 and the third son 19.
After reading their father's will, the three sons were worried. 17 is a prime number, which can neither be divisible by 2 nor by 3 and 9, and it is not allowed to kill sheep to divide it. What can we do?
After getting the news, the clever neighbor ran to help herd sheep. The neighbor said, "I'll lend you a sheep, so 18 sheep can be divided easily."
The first score is 18× 12 = 9 (only), the second score is 18× 13 = 6 (only), and the third score is18×19.
Together it is 9+6+2 = 17, which is exactly 17 sheep. There was only one sheep left, and the neighbor took it back.
The sheep was divided by the neighbors. If we think deeply about this problem, we will find that there are unreasonable places in the will. If we take the sheep left by the old man as a whole, 1, because12+13+19 =1718, or the three sons can't divide all the sheep, we still leave 65438. Or one more sheep than he left, he can share it. The clever neighbor brought another sheep. According to the score of 17 18, it adds up to 18 18, divided by 17 18, leaving 16544.
Look at another topic about wills:
When someone died, his wife was pregnant. He said to his wife, "If your child is a male, give him one-third of the property;" If it's a woman, give her the property and give you the rest. " Then he died.
Coincidentally, his wife gave birth to a boy and a girl twins. How will the property be divided?
Can be solved in proportion:
The distribution ratio of sons and wives is 23∶ 13=2∶ 1, and that of daughters and wives is 25∶35=2∶3.
It can be seen that the distribution ratio of daughters, wives and sons is 2: 3: 6, which is reasonable.
There are some math problems written in folk songs all over the world.
American folk songs:
"An old drunkard, named Baten, eats pork slices and ribs for 94 cents, and each rib costs only seven cents. Even the ribs and ribs were eaten for ten pieces. Let me ask you:
Our Barton, how many ribs and meat did you eat? "
You can solve it like this:
If Baten ate ten pieces of meat, he spent 70 cents. If he subtracted 70 points from 94 points, the difference was 24 points. What's this 24 cents?
Because Barton doesn't eat all the pieces of meat, there are ribs, and a rib is more expensive than a piece of meat 1 1-7 = 4 points, and this 24 points is the difference between ribs and meat, so we can calculate the number of ribs that Barton eats:
(94-7×10) ÷ (11-7) = 24 ÷ 4 = 6 (block) 10-6 = 4 (slice) Barton ate six steaks and four pieces of meat.
China also has similar folk songs:
"A team of robbers and a team of dogs, two teams together is a team, three hundred and sixty heads, eight hundred and ninety legs. How many robbers and dogs are there? "
This problem is the same as the "chicken and rabbit in the same cage" in Sun Tzu's Art of War, except that the chicken is replaced by a robber and the rabbit is replaced by a dog. The specific algorithm is (360× 4-890) ÷ (4-2) = 275360-275 = 85. There are 275 robbers and 85 dogs.
There is also a China folk song:
"A few old people went to the market and bought a bunch of pears on the way. One person has one more, and one person has two fewer pears.
How many old people and pears are there? "
Let the number of people be x, then the number of pears is X+ 1. According to the meaning of the question, we can get:
2x = (x+ 1)+2, x = 3, x+ 1 = 4 "Western Western jackdaw and the Branches" is a Russian folk song;
"A few western Western jackdaw flew in and landed on the branches.
If a western Western jackdaw falls on every branch, then a western Western jackdaw lacks a branch; If two western Western jackdaw fall on each branch, then a branch will not fall on the western Western jackdaw.
How many western Western jackdaw did you say?
How many branches did you say * * had? "
It can be solved as follows:
If two western Western jackdaw fall on each branch, that is 2+ 1 = 3 western Western jackdaw and one western Western jackdaw fall on each branch. At this time, the difference of the number of Western jackdaw in the west on each branch is 2- 1 = 1.
Dividing the extra western Western jackdaw number by the western Western jackdaw number of each branch is equal to the number of branches.
Therefore, (2+1) present (2-1)
= 3 ÷ 1 = 3 (branch) The number of Western jackdaw in the west is 3+ 1 = 4 (branch).
The answer is that there are three branches and four western Western jackdaw.
The following ballad is also very interesting. This is a China ballad:
"Shepherd boy Wang Xiaoliang, grazing a flock of sheep.
Ask him some sheep, please do something.
Only increase the number of heads, only decrease the number of heads.
Only multiply by the number of heads, only divide by the number of heads.
Four numbers add up to exactly one hundred. "
In fact, the number of heads and the number of heads are the same thing. So only the head count is 0, and only the head count is 1. So there is: simplex × simplex +2 × simplex = 99.
By testing, it can be concluded that the number is equal to 9, because 9× 9+2× 9 = 99, so there are 9 sheep.