1． The location of the ancient Babylonian kingdom on the clay tablet is in the middle and lower reaches of the Tigris and Euphrates rivers in Western Asia, in what is now Iraq. The Babylonian country was established in the 19th century BC and is one of the four major ancient civilizations in the world. .
The Babylonians used special cuneiform writing. They carved the writing on clay tablets and then dried them in the sun. After the clay tablets were dried, they were as hard as stone and could be preserved for a long time.
From the excavated clay tablets, people discovered a mathematical problem posed by the Babylonians more than 3,000 years ago:
“10 brothers are divided into 100 taels of silver, and each person has more than one person. , I only know that the amount of difference at each level is the same, but I don’t know how much the difference is. Now the eighth brother is divided into 6 taels of silver. How much is the difference at the first level? ”
If the 10 brothers are divided equally? For 100 taels of silver, each person should get 10 taels. Now the eighth brother only got 6 taels, which means that the eldest brother got the most, and each one below received less.
According to the conditions given in the question, there should be the following relationship:
The second child gets the difference of the eldest child minus twice,
The third child gets What the fourth child gets is the difference between the oldest child minus two times.
What the fourth child gets is the difference between the oldest child minus three times.
......
The tenth child gets The difference is the boss minus nine times.
In this way, the eldest brother and the tenth old man get a silver tael
=The second child and the ninth old man*** get a silver tael
=The third child and the eighth child* **gets a silver tael
=The fourth and seventh ***get a silver tael
=The fifth and sixth ***get a silver tael
=20 taels
It is known that Lao Ba gets 6 taels, and it can be found that Lao San gets 20-6=14 taels. Lao San has more than Lao Ba and gets 14-6=8. On the other hand, Lao San and Lao Ba get 14-6=8. The difference is 7-2=5 times the difference. Therefore,
Difference=8÷5=1.6 (taels)
Answer: The difference of one level is 1.6 taels of silver.
Babylonian mathematics and astronomy developed rapidly. In addition to the first use of the base 60 system, they also determined that a month (lunar month) has 30 days and a year (lunar year) has 12 lunar months. In order Not to lag behind the solar year, in some years the method of specifying leap months is used to correct it.
The Babylonians understood the existence of planets. They worshiped the sun, moon, and Venus, and regarded the number 3 as "happiness." Later, they discovered Jupiter, Mars, Mercury, and Saturn. The hour number 7 is considered "happy".
The Babylonians paid special attention to the study of the moon. They called the ratio of the bright part of the crescent moon to the full area of ??the moon "lunar phase" and recorded the topic about the moon phase on a clay tablet:
"Suppose the total area of ??the moon is 240. In the 15 days from new moon to full moon, the first 5 days are twice the size of the previous day, that is, 5, 10, 20, 40, 80, and the next 10 days are double the size of the previous day. The same value increases, what is the increased value? "
The total area of ??the moon is 240, the area of ??the moon on the fifth day is 80, and the maximum area of ??the moon in the next 10 days is 240-80=160.
Therefore, the daily increase in value is 160÷10=16.
Answer: The increased value is 16.
2. On papyrus
The "Lant Papyrus" is a mathematics book written by the ancient Egyptians 4,000 years ago. It records many interesting mathematical problems in hieroglyphs, such as:
In 7, 7×7, 7×7×7, 7×7×7×7, 7×7×7×7×7,…
There are several pictographic symbols on these numbers: House, cat, mouse, barley, bucket, translated as:
"There are 7 houses, and there are 7 cats in each house. Each cat eats 7 mice, and each mouse eats There are 7 ears of barley. Each ear of barley seeds can grow 7 bushels of barley. Please calculate the total number of houses, cats, mice, barley and bushels.”
Strangely, similar arithmetic problems were circulated among ancient Russian folk. :
"Seven old men were walking on the road,
Each old man held seven canes,
Each cane had seven branches,< /p>
There are seven bamboo baskets hanging on each tree branch.
There are seven bamboo cages in each bamboo basket.
There are seven bamboo cages in each bamboo cage. Sparrows,
How many sparrows are there in total?”
The ancient Russian question is relatively simple. The number of old men is 7, the number of walking sticks is 7×7=49, and the number of branches is 7. It is 7×7×7=49×7=343, the number of bamboo baskets is 7×7×7×7=343×7=2401, the number of bamboo cages is 7×7×7×7×7=2401×7=16807 , the number of sparrows is 7×7×7×7×7×7=16807×7=117649. There are 117,649 sparrows in total. It is not easy for seven old men to carry more than 110,000 sparrows around! If each sparrow is calculated as 20 grams, these sparrows weigh 2 tons.
The "Lant Papyrus" has an answer to the question of cat eating mouse and mouse eating barley, which is to multiply 2801 by 7.
To find the total number of houses, cats, mice, barley and buckets, the sum is 7+7×7+7×7×7+7×7×7×7+7×7×7×7×7=7+49+343+2401+16807=19607. This is the same as the answer of 2801×7=19607 above. The ancient Egyptians mastered this special summation method more than 4,000 years ago.
Similar questions also appeared in an old British nursery rhyme:
"I went to the holy land of Ephesus,
I met many women on the way Seven,
One person carries seven bags in his hand,
One cat and seven children are closely dependent on each other,
The woman and the bag, the cat and the children,
Geometry Going to the Holy Land at the same time?"
The Italian mathematician Fibonacci also had a similar question in his "Abacus" published in 1202:
"There are seven old women going to Rome. On the way, each person had 7 mules; each mule carried 7 bags, each bag carried 7 large loaves of bread, each loaf carried 7 knives, and each knives had seven sheaths. On the way to Rome, How many women, mules, bread, knives and scabbards are there in one bag? "The same type of question appears in different forms in different times and countries, but the earliest one is the ancient Egyptian "Lan" Special Papyrus".
There is also a popular topic in ancient Egypt about "someone stealing treasure":
"Someone took treasure 13 from the treasure house, and another person took away 117 from the remaining treasure. There are 150 treasures left in the treasure house. How many treasures are there originally in the treasure house? "
The formulation of this question is very similar to the question in the current textbook. It can be solved like this:
Suppose The original treasure in the treasure house is 1, then the first person takes 13, and the second person takes (1-12) × 117 = 252
The last treasure in the treasure house is 1-13- (1-13)×117=1-13-251=3251.
Therefore, the original treasure in the treasure house
150÷3251=150×5132=23916.
The comprehensive calculation formula is
150÷\[1-13-(1-13)×117=239116.
The "Lant Papyrus" also has this question:
"There are several items, two-thirds of them, half of them, one-seventh of them and all of them, ***33 pieces, find the number of items. ”
Use arithmetic to solve, you can set all to 1, then the number of items is
33÷(23+12+ 17+1)
=33÷9742=33×4297
=142897
The answer is unique, but the answer on the papyrus is
=142897
p>
14, 14, 156, 197, 1194, 1388, 1679, 1776. What's going on? Are there eight answers to this question?
It turns out that the papyrus gave the answer in the form of ancient Egyptian fractions, which means 14+14+156+197+1194+1388+1679+1776. Let’s work it out:
14+14+156+197+1194+1388+1679+1776
=14+1456+156+197+197×2+197× 4+197×7+197×8
=14+1456+8+4+2+197×8+197×7
=14+1456+1597×8+ 197×7
=14+1456+11397×56
=14+156897×56=142897
This is the same answer we calculated.
3. In Poetry
Greece is one of the ancient civilizations in the world. It has a splendid ancient culture. There are some mathematical problems written in poetry in the "Greek Collection".
In "The Troubles of Cupid", Eros is the goddess of love in ancient Greek mythology, Giborida is the patron saint of the island of Cyprus, and among the nine goddesses of literature and art, Ephtera plays music. , Ella hosted love poetry, Dalia cameo, Tehihora danced, Melipomena tragedied, Creole history, Polynian ode, Urania astronomy, Callio Papipe epic.
The Troubles of Cupid
“Eros wept by the roadside,
tears came one after another.
Giborida moved forward Asked:
'What makes you so sad?
Maybe I can help you?'
Eros replied:
< p>'Nine goddesses of literature and art,I don’t know where they came from,
almost wiped away all the apples I picked from Mount Helkang.
.
Ephterpo quickly snatched away one-twelfth of the apples,
Erato grabbed even more——
Seven apples. Take one.
One-eighth of it was snatched away by Dalia, and twice as many apples fell into the hands of Tehihora.
Melibomena was the most polite,
only took one-twentieth.
But here comes Cleo again,
Her harvest is four times more than this.
There are also three goddesses,
none of them is empty-handed:
30 apples return to Polynia,
120 The apples belong to Urania, and 300 apples to Calliopa.
I, poor Eros,
How many apples did Eros have? There are 50 apples left. '"
This 26-line poem presents a math problem with a lot of numbers. The original number of apples in the problem is unknown. After being robbed by nine goddesses of literature and art, only 50 Eros are left. Apple is a mathematical problem of "finding the type of the part and finding the whole"
Suppose the original number of apples in Eros is x
According to the meaning of the question, we get
112x+17x+18x+14x+120x+15x
+3123050=x
After sorting, we get 143168x+500=x
∴x=33600 (pieces)
Another type of math problem is given in "Cyclops" below:
"This is a bronze statue of a Cyclops,< /p>
The sculptor is highly skilled,
The bronze statue has ingenious mechanisms:
The giant’s hands, mouth and one eye are all connected.
Large and small water pipes,
Through the water pipe of the hand,
Three days to fill the pool;
Through the one-eyed water pipe - it takes one day;
Water spits out of the mouth faster,
Two-fifths of a day is enough,
Water is released from three places at the same time,
When will the pool be full? ”
Suppose the volume of the pool is 1, and the time required for the three pipes to open to fill the pool is x days,
then 13x+x+52x=1
∴x=623
The following is a limerick from our country:
"Li Bai went to buy wine with a pot:
When you encounter a shop, double the price,
p>Drink a bucket of flowers.
Encountered shops and flowers three times,
Drank up the wine in the pot.
How much wine was originally in the pot? ”
The meaning of this limerick is that Li Bai originally had wine in his pot, and he would double the amount of wine in the pot every time he met a tavern; Li Bai would drink and compose poems when admiring flowers. Drink a dou of wine at a time (a dou is an ancient vessel for holding wine), repeat this process three times, and finally drink up all the wine in the pot. Ask Li Bai how much wine was in the pot?
Solution to this question? It is best to use the inference method to solve:
The third time Li Bai saw the flowers, he drank all the wine in the pot, which means that before he saw the flowers, there was only one bucket of wine in the pot. Before Li Bai met the hotel for the third time, there were 12 buckets of wine in the pot. According to this calculation method, it can be calculated that before he met Hua Hua for the second time, there were 112 buckets of wine in the pot, and before he met the hotel for the second time, there were 112÷ 2=34 Doujiu; the first time I saw Huaqian, there was Doujiu in the pot 134. The first time I met in front of the hotel, there was a Doujiu in the pot. It turned out that there was a Doujiu in the pot. 134÷2=78
It turned out that there was a Doujiu in the pot. 78 fights of wine.
4. In the will, there are many interesting mathematical problems in the distribution of inheritance according to the will.
The famous Russian mathematician Si. Tranolyubovsky once raised the issue of inheritance distribution: "The father requested in his will that 13 points of the inheritance should be given to his son and 25 points to his daughter; 2,500 rubles of the remaining money should be used to repay debts. 3,000 rubles are left to mother, how much is the inheritance! How much should each child get? ”
Suppose the total inheritance is x rubles.
Then 13x+25x+2503000=x
The solution is: 20625×13=6875 (rubles),
The daughter gets 20625×25=8250 (rubles).
The result is that the daughter gets the most, 8250 rubles, followed by the son, who gets 8250 rubles. 6875 rubles, the mother got the least, 3000 rubles. It seems that the father loves his daughter.
The following story was first circulated among Arab people, and later spread to countries around the world. The story says that an old man raised his daughter. After his death, the old man requested that the 17 sheep be divided among his three sons in proportion. The eldest son would be given 12, the second son would be given 13, and the third son would be given 19. The sheep were not allowed to be slaughtered when dividing the sheep. .
After reading their father’s will, the three sons were worried. 17 is a prime number. It is not divisible by 2, nor by 3 and 9, and it is not allowed to kill a sheep to divide it. What’s the matter? What to do?
After the clever neighbor got the news, he ran to help with a sheep. The neighbor said: "I will lend you a sheep, so that the 18 sheep can be divided easily." ”
The eldest child is divided into 18×12=9 (only),
The second child is divided into 18×13=6 (only),
The third child is divided into 18×19 = 2 (only).
The total is 9+6+2=17, exactly 17 sheep, and there is one sheep left.
The neighbor took it back. Finished.
Thinking about this issue further, we will find that there is something unreasonable in the will. If the sheep left by the old man are taken as a whole 1, because
12+13+19=1718
< p>So either the three sons could not divide all the sheep and left 118, and none of the sons gave 1817; or they could only divide the sheep when there was one more sheep than what he had left, smart neighbor Based on the score of 1718, he brought in another sheep to make up 1818. After dividing 1718, there were still 118 sheep left, which was his own sheep.Look at another question about wills:
When a man was dying, his wife was pregnant. He said to his wife: "If the child you give birth to is a boy, Give him 23 of the property, and if it’s a woman, give her the 25, and give the rest to you.” After that, he died.
Coincidentally, his wife gave birth to twins, a boy and a girl. How will the property be divided?
It can be solved in proportion:
The distribution ratio between son and wife is 23:13=2:1
The distribution ratio between daughter and wife is 25 ∶35=2∶3.
It can be seen that the distribution ratio of daughters, wives, and sons is 2:3:6, and it is reasonable to distribute according to this ratio.
5. In ballads
There are some mathematical problems written in the form of ballads circulating around the world. < /p>
Each rib is a dime, and each piece of meat is only 7 cents.
I ate a full ten yuan with the ribs and meat slices.
Ask you:< /p>
How many pieces of ribs and slices of meat did our Batten eat? ”
It can be solved like this:
Suppose that Batten ate ten slices of meat. If so, he spends 70 cents a day, subtract 70 cents from 94 cents, and the difference is 24 cents. What is this 24 cents!
Since what Buten eats is not just sliced ??meat, but also ribs, and a piece of ribs is more expensive than a piece of meat. 11-7 = 4 points. These 24 points are obtained from the price difference between the ribs and the sliced ??meat. We can calculate the price of Buten. Number of ribs eaten:
(94-7×10)÷(11-7)
＝24÷4＝6 (pieces)
10- 6=4 (slices)
Barton ate six steaks and four slices of meat.
There is also a similar folk song in China:
"One team of robbers and one team of dogs,
The two teams walk together as one team,
Counting the head, one*** three hundred and six,
Counting the legs, one***89,
How many robbers and dogs are there?”
This? The question is the same type of question as the "chicken and rabbit in the same cage" in "Sun Zi Suan Jing", 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)=275
360-275=85
There are 275 robbers and 85 dogs.
There is also a Chinese folk song:
"A few old men went to the market,
and bought a bunch of pears on the way,
each of them had one One more person, two less pears per person
How many old men and pears are there? ”
Suppose the number of people is x, then the number of pears is x+1, according to the meaning of the question, we get:
2x=(x+1)+2,
x=3,
x+1=4
"Jackdaws and Branches" is a Russian folk song:
"A few jackdaws flew over,
and landed on the branches to rest.
If only
A jackdaw falls on each branch,
then there is a jackdaw
One branch is missing;
If
Two jackdaws fell on each branch.
There is one branch that
cannot catch.
You. How many jackdaws does *** have?
How many branches do you think *** has? ”
It can be explained like this:
If each branch has Two jackdaws fall on the branches, which is 2+1=3 jackdaws more than one jackdaw falls on each branch. At this time, the difference in the number of jackdaws falling on each branch is 2-1=1 Only.
Divide the number of extra jackdaws by the number of jackdaws on each branch, which equals the number of branches.
Therefore,
(2+1)÷(2-1)
=3÷1=3 (branch)
Jackdaw The number is 3+1=4 (only).
The answer is that there are 3 branches and 4 jackdaws.
The following folk song is also very interesting. It is a Chinese folk song:
"The shepherd boy Wang Xiaoliang is herding a group of sheep.
Ask him how many sheep he has. Please tell me how many sheep he has. Think about it.
The number of heads is added to the number of heads.
The number of heads is multiplied by the number of heads.
The number of heads is divided by the number of heads. Adding them up, we get exactly one hundred.”
In fact, the number of heads and the number of heads are the same thing. Therefore, the number of heads minus the number of heads is 0, and the number of heads divided by the number of heads is 1.
In this way, we have: only number × only number + 2 × only number = 99.
Using the experimental method, the number of sheep can be found to be 9, because
9×9+2×9=99, so there are 9 sheep.