Mathematical Olympiad calculations and ingenious calculations in primary school
The following are some common types and methods of quick and clever calculations in primary school Mathematical Olympiad: ** 1. Clever calculations in addition ** 1. ** Rounding Method ** - When adding several numbers, if there were two numbers that could be added together to get a whole ten, a whole hundred, a whole thousand, etc., they could be added first. For example, 24 + 44+56, because 44 + 56 = 100 is a whole hundred, so first calculate 44+56, then add 24, that is, 24+(44 + 56)=24 + 100 = 124. - For an equation like 53+36 + 47, because 53+47 = 100 was a whole hundred,+47 could be moved with the sign to the front of +36, and then (53+47)+36 = 100+36 = 136. 2. ** Split and Rounder Method ** - In the case of 96+15, 15 was split into 15 = 4+11, because 96+4 = 100, which could be rounded up first, that is, 96+15 = 96+(4+11)=(96 + 4)+11 = 100+11 = 111. - For 52+69, since 69+31 = 100, 52 was split into the sum of 21 and 31, and then 31+69 = 100 was rounded up, which was 52+69=(21+31)+69 = 21+(31+69)=21 + 100 = 121. - When calculating 63+18+19, 63 was split into 63 = 60+2+1. Because 2+18 and 1+19 could be rounded up, 63+18+19 = 60+2+1+18+19 = 60+(2+18)+(1+19)=60+20+20 = 100. 3. ** The clever calculation of adding the same number ** - For 28+28+28, you can think of it this way, because 28+2 = 30 can be rounded up, but in the end, you have to subtract the three additional 2s, that is, 28+28+28=(28 + 2)+(28 + 2)+(28+2)-6 = 30+30+30 - 6 = 90 - 6 = 84. ** 2. Clever calculation in the substitution (position swap method)** - In an algorithm, the position of the numbers could be changed, and the order of the calculation of the numbers could be changed. For example, 632 - 156 - 232 = 632 - 232 - 156 = 400 - 156 = 244. ** 3. Clever calculations in multiplication ** 1. **"Tongbu" and "Butong" Quick Calculation Method ** - If the sum of two numbers was equal to 10, then the two numbers were complementary. In the integral multiplication operation, for example, 72×78, where the ten digits of the multiplication and the multiplication were the same, and the single digits were complementary (this kind of formula was called "same head, complementary tail" type), and 26×86, where the ten digits of the multiplication and the multiplication were complementary, and the single digits were the same (this kind of formula was called "complementary head, same tail" type), there were very simple and direct fast calculation methods. They were called the "same-complementing" fast calculation method and the "complementing the same" fast calculation method. ** 4. Other methods ** 1. ** Borrowing Method ** - Borrowing a number from a number, turning it into a whole ten or a whole hundred or a whole thousand, and finally, deducting the borrowed number. For example, 9+99 +999+9999=(10 - 1)+(100 - 1)+(1000 - 1)+(10000 - 1)=10+100 + 1000 + 10000 - 4 = 11106. 2. ** Choose a reference number ** - Among all the numbers, find that everyone is close to this number (this number is the benchmark number), first multiply the number of numbers by this benchmark number, and finally, subtract the excess. For example, 489+487+483+485+484+486+488. First find the base number 490 among these numbers, then use 490×7. Finally, subtract the excess, which is 490×7 - 1 - 3 - 7 - 5 - 6 - 4 - 2 = 3430 - 28 = 3402. Read more exciting novels for free
30 compulsory questions for the primary school Mathematical Olympiad
The following are some of the common types of questions for the primary school Mathematical Olympiad exam: return to one problem, itinerary problem, cow eating grass problem, etc. However, because the reference materials did not completely list the 30 compulsory questions, the contents of all 30 compulsory questions could not be provided. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Two Formula of Substitute for Mathematical Olympiad in Primary School
In elementary school Mathematical Olympiad, equivalent substitution meant that two identical quantities could be replaced with each other. For example, if 1 × = 2 ×, 2 × = 4 ×, then 1 × = 4 ×. However, there was no fixed formula named " two quantity substitution formulas ". It was mainly based on the relationship of equality to replace the quantity. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
An Analysis of the Ten Difficult Questions in Primary School Mathematical Olympiad
The following is an analysis of the ten difficult questions commonly seen in primary school Mathematical Olympiad: ** One, Return to One Problem ** 1. ** Meaning ** - When solving a problem, the first thing to do was to find out how much a portion was (that is, a single amount), and then use the single amount as the standard to find the required amount. 2. ** Number of relationships ** - Total amount/portions = single amount; single amount × portions = the number of portions required; or total amount A/(total amount B/portions B)= portions A. 3. ** Solution train of thought ** - First, find a single quantity. Using a single quantity as the standard, find the required quantity. For example, if you want to buy 5 pens, you need 0.6 yuan. If you want to buy 16 pens, you need to find 0.6 + 5 = 0.12 yuan for one pencil, and 0.12×16 = 1.92 yuan for 16 pens. ** 2. The problem of returning to the main body ** 1. ** Meaning ** - When solving a problem, first find the "total number" and then solve the problem according to the known conditions. The so-called "total quantity" could refer to the total price of the goods, the workload in a few days, the total output of a few acres of land, the total journey in a few hours, and so on. 2. ** Number of relationships ** - 1 serving x number of copies = total amount; total amount/1 serving = number of copies. 3. ** Solution train of thought ** - He would first find the total number before solving the problem. For example, the clothing factory originally used 3.2 meters of cloth to make a set of clothes. Originally, it made 791 sets of clothes, so the total amount of cloth was 3.2×791 = 2531.2 meters. Now, each set of clothes used 2.8 meters of cloth, which could make 2531.2/2.8 = 904 sets. ** 3. Problem of sum and difference ** 1. ** Meaning ** - Given the sum and difference of two quantities, find out what the two quantities are. 2. ** Number of relationships ** - Large numbers =(sum + difference) div2; Decimals =(sum-difference) div2. 3. ** Solution train of thought ** - Simple questions could be solved by applying the formula, while complex questions could be solved by adapting the formula. For example, Class A and Class B had a total of 98 students. Class A had 6 more students than Class B. Class A =(98 + 6) div2 = 52 students, Class B =(98 - 6) div2 = 46 students. ** 4. The Problem of Summing Times ** 1. ** Meaning ** - Given the sum of the two numbers and "how many times the large number is a fraction (or how many fraction of the large number is a fraction)", find out what the two numbers are. 2. ** Number of relationships ** - Sums × (Multiple + 1)= Lesser Number; Sums-Lesser Number = Greater Number; or Lesser Number × Multiple = Greater Number. 3. ** Solution train of thought ** - Simple questions could be solved by applying the formula, while complex questions could be solved by adapting the formula. For example, there were 248 apricot trees and peach trees in the orchard. The peach trees were three times the number of apricot trees. There were 248/(3 + 1)=62 apricot trees and 62×3 = 186 peach trees. ** 5. The problem of the difference ** 1. ** Meaning ** - Given the difference between the two numbers and "how many times the large number is a decimal (or how many times the small number is a large number)", find out what the two numbers are. 2. ** Number of relationships ** - The difference between the two numbers divided by (multiple- 1)= lesser number; lesser number × multiple = greater number. 3. ** Solution train of thought ** - Simple questions could be solved by applying the formula, while complex questions could be solved by adapting the formula. For example, the number of peach trees in the orchard was three times that of apricot trees, and there were 124 more peach trees than apricot trees. There were 124/(3 - 1)=62 apricot trees and 62×3 = 186 peach trees. ** 6. Multiple ratio problem ** 1. ** Meaning ** - There are two known quantities of the same kind, one of which is several times the other. When solving a problem, first find the multiple, and then use the multiple ratio method to calculate the required number. 2. ** Number of relationships ** - Total amount A/quantity A = multiple; quantity B× multiple = total amount B. 3. ** Solution train of thought ** - First, find the multiple, then use the multiple ratio relationship to solve. ** 7. Age problem ** 1. ** Key Points ** - The precession of the equinoxes (the difference in age) never changed, but the sum of years and the multiple of years changed. The precession of the equinoxes, the sum of years, and the difference of sum had to correspond to the multiple and time. ** 8. The problem of cows eating grass ** 1. ** Meaning and Key Points ** - For example, the grassland would grow grass at a constant rate every day. The key to the problem of cows eating grass was to determine how much grass they would grow every day. 2. ** Solution train of thought ** - Usually, a cow was set to eat one unit of grass a day. The original amount of grass and the amount of grass grown per day were calculated by the number of days that different cows ate grass. Then, the number of days that a given number of cows ate grass was calculated. Like a blade of grass that can feed twenty-seven cows for six days (A total of 27×6 = 162 units of grass were eaten, including the original grass and 6-day-old new grass), which could feed 23 cows for 9 days (a total of 23×9 = 207 units of grass were eaten, including the original grass and 9-day-old new grass). The amount of new grass per day could be calculated as (207 - 162)/(9 - 6)=15 units, and the original grass was 162 - 6×15 = 72 units. Then, the number of days that the 21 cows ate grass was calculated. ** 9. Quick and skillful calculations (for lower grades)** 1. ** Meaning ** - For first-year students, calculation was the first problem they encountered in their studies and was the focus of their studies. For second-year students, calculation was also the focus and difficult part of their Mathematical Olympiad studies. ** 10. Enumeration Questions (For Lower Grades)** 1. ** Difficulty analysis ** - For second-year students, orderly and abstract thinking was more difficult. For questions, second-year students were more willing to make up the numbers to try to answer the questions. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
What are the recommendations for innovative thinking in primary school Mathematical Olympiad?
The following are some recommended books for primary school Mathematical Olympiad students: 1. " Mathematical Olympiad Star's Classic Question Bank of innovative thinking," published by China Forest Press on January 1, 2008. The author was Liu Xianguo. 2. Learning and Thinking, Thinking and Creation, also known as the Big White Book, was a set of books with more difficult questions in the learning and thinking system. 3. The 1998 Guangdong-Hong Kong Mathematical Olympiad Invitational Competition for the fourth grade examination paper can also be used as reference material. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
The first grade of primary school had a lobster for the Mathematical Olympiad
Based on the available information, there was a first-grade elementary school exercise about lobsters. It was to use matchsticks to form a lobster with its head up and move three matches so that it was facing down. However, the answer to which three matches to move was not obtained. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Classic Mathematical Olympiad questions and answers for primary school students
The following are some classic Olympiad math questions and answers for primary school students: ** 1. Profits and losses ** 1. [Question: How many children are there in total? Each child has 10 watermelons, 9 less, 8 more than 7.] - ** Answer **: According to the formula,"First add more or less, then divide by the distribution to find the difference." If there are 9 fewer and 7 more, the first sum is [9 + 7=16]; if each person is allocated 10 and each person is allocated 8, the difference is [10 - 8 = 2]. Finally, divide by the number of people. ** 2. Train itinerary ** 1. [Title: There are two trains, one is 102 meters long and travels 20 meters per second, and the other is 120 meters long and travels 17 meters per second.] The two trains were traveling in the same direction. How many seconds would it take for the first train to catch up with the second train and for the two trains to leave? - ** Answer **: From catching up to leaving, the distance of catching up is the sum of the two train leaders, which is <102 + 120 = 222> m. The speed difference is <20 - 17 = 3> m/s. Time = distance difference/speed difference, so <222> div3 = 74> s. 2. [Title: A person's walking speed is 2 meters per second.] A train came from behind and overtook him by 10 seconds. It was known that the train was 90 meters long. Ask for the speed of the train. - [Answer: The distance that a train can travel is 90 meters, the time is 10 seconds, and the speed of a person is 2 meters per second.] Assuming that the train speed is <v> m/s, according to distance = speed x time,<(v - 2)<times10 = 90>, the solution is <v = 11> m/s. 3. [Title: There are two trains moving in the same direction at the same time. After 12 seconds, the fast train overtakes the slow train.] The express train traveled 18 meters per second, while the slow train traveled 10 meters per second. If the two trains were traveling in the same direction, the express train would overtake the slow train in 9 seconds. Find the length of the two trains. - ** Answer **: When moving head to head, the distance that the express train passes over the slow train is the length of the express train, which is [(18 - 10)<times12 = 96] m; when the rear of the train is moving head to head, the distance that the express train passes over the slow train is the length of the slow train, which is [(18 - 10)<times9 = 72] m. 4. [Title: A train needs 40 seconds to cross a 440-meter bridge, and 30 seconds to cross a 310-meter tunnel at the same speed.] What was the speed and length of this train? - ** Answer **: Let the train's length be x meters and its speed be v meters per second. According to the relationship between the distance and time through the bridge and tunnel, the equation system can be obtained.(\begin{cases} 40v = 440 + x30v = 310 + x\end{cases}}), subtract the second equation from the first equation to get\(10v = 130'), the solution is [v = 13') m/s. Substitute [v = 13')] into [30v = 310 + x], the solution is [390 = 310+x]. 5. ** Title **: In order to measure the speed and length of the train, Xiaoying used a watch to record the time it took for the train to pass in front of her was 15 seconds, and Xiaomin used another watch to record the time it took for the train to pass the first telephone pole from the front to the second telephone pole at the back was 20 seconds. It was known that the distance between two utility poles was 100 meters. Find the length and speed of the train. - ** Answer **: Let the train's length be x meters and its speed be v meters per second. The train passes in front of us, the distance it travels is the length of the train (x = 15v\); The distance from the front of the car to the back of the car is the length of the car plus the distance between the poles.(x + 100 = 20v_), substitute_(x = 15v_) into_(x + 100 = 20v_) to get_(15v +100 = 20v_). The solution is_(v = 20_) m/s, the length of the car is_(x = 15, time20 = 300_) m, and the speed is_(20,time3.6 = 72_) km/h. 6. [Title: A train needs 40 seconds to cross a 530-meter bridge, and 30 seconds to cross a 380-meter cave at the same speed.] How many meters is the speed and length of this train? - ** Answer **: Let the train's length be x meters and its speed be v meters per second. We can get the equation set of\(\begin{cases} 40v = 530 + x'30v = 380 + x 'end{cases}). Subtracting the first equation from the second equation gives\(10v = 150), which is solved as\(v = 15) m/s. Substituting\(v = 15) into\(30v = 380 + x) gives\(450 = 380 + x), which is solved as\(x = 70) m. 7. [Title: The two of them walked along the trail beside the railway line, starting from two places and traveling at the same speed.] A train came, and it took 10 seconds for the entire train to pass by A. Three minutes later, B encountered a train. The entire train passed by B in just nine seconds. How long after the train left B did the two meet? - ** Answer **: Let the speed of the train be <v> 1>, the speed of the person be <v> 2>, and the length of the train be <L>. According to the relationship between the train and A, L=(v1v2)<times10>, and the relationship between the train and B, L=(v1v2)<times9>, we can get <(v1v2)<times10>=(v1v2)<times9>, which is simplified to <(v1v2 = 19v2>). 3 minutes after the train leaves A, it meets B. The distance between A and B and the distance between A and B is (180v1 - 180v2). The time after the train leaves B is set to be <t>, then <(v2 + v2)t= 180v1 - 180v2>. Substituting <v1 = 19v2> into <2v2t = 180Times19v2 - 180v2>, the solution is <t = 1710> seconds = 28.5 minutes. 8. [Question: Two trains, one 120 meters long and traveling at 20 meters per second, the other 160 meters long and traveling at 15 meters per second. The two trains are moving in opposite directions. How many seconds does it take for the two trains to meet and leave?] - ** Answer **: The two trains are moving in opposite directions. The distance from the front of the train to the back of the train is the sum of the two train leaders, which is the sum of the speed of the train. The speed is the sum of the speed of the train. The time is the sum of the speed of the train, so the speed of the train is the sum of the speed of the train. 9. [Title: A person's walking speed is 2 meters per second.] A train came from behind and passed him in ten seconds. Given that the train is 90 meters long, find the speed of the train. - ** Answer **: The distance the train can travel over a person is 90 meters, the time is 10 seconds, and the speed of a person is 2 meters per second. Assuming that the train speed is <v> m/s, according to distance = speed x time,<(v - 2)<times10 = 90>, the solution is <v = 11> m/s. 10. [Title: A and B are walking along the railway at the same speed. A train takes 8 seconds to pass by A and meets B 5 minutes later. It only takes 7 seconds to pass by B. How many minutes will it take for B and B to meet each other after B meets the train?] - ** Answer **: Let the speed of the train be <v> 1>, the speed of the person be <v> 2>, and the length of the train be <L>. According to the relationship between the train and A, L=(v1v2)<times8>, and the relationship between the train and B, L=(v1v2)<times7>, we can get <8(v1v2)=7(v1v2)>, which is simplified to <v1v2>. The train will meet B 5 minutes after leaving A. The distance between A and B and the distance between A and B in these 5 minutes is (300v1 - 300v2), from the time B meets the train, the time when A and B meet is set to be (t), then ((v2.1 + v2.1)t = 300v1 - 300v2.1). Substituting <<v1>= 15v2> into <<2v2.1>= 300.1> time15v2 - 300v2>, the solution is <<t = 2100>> seconds = 35 minutes. ** 3. Average problem ** 1. [Question: Cai Chen's average score in the final exam was 89 in politics, Chinese, mathematics, English, and biology.] The average score for politics and mathematics was 91.5. The average score for Chinese and English was 84. The average score for politics and English was 86 points, and English was 10 points higher than Chinese. He asked Cai Chen what his marks for each subject should be for this exam? - ** Answer **: If the Chinese test score is x, then the English test score is x + 10. According to the average score of Chinese and English, the equation could be written as [(x + x + 10)<div2 = 84], and the solution was [x = 79]. The English score was [89]. The average score for politics and English is 86 points, so the political score is 86/89 = 83/. The average score for politics and mathematics is 91.5, so the math score is 91.5'time2 - 83 = 100 '. The biology score was [89 'times5-(79 + 89+83 + 100)=94]. 2. [Title: A and B cotton fields, with an average yield of 185 catties of seed cotton per mu] The A cotton field had 5 mu, with an average yield of 203 catties of seed cotton per mu. The B cotton field had an average yield of 170 catties of seed cotton per mu. How many mu was the B cotton field? - ** Answer **: Suppose B has a cotton field of x mu. According to the total yield equal-countable equation, the solution is x = 6 mu. 3. [Question: Given that the sum of eight consecutive odd numbers is 144, find these eight consecutive odd numbers.] - ** Answer **: If the first odd number is <x>, then the eight consecutive odd numbers are <x,x + 2,x+4,x + 6,x+8,x + 10,x+12,x + 14>. According to the sum of the 144 equations, x+(x + 2)+(x+4)+(x + 6)+(x+8)+(x + 10)+(x+12)+(x + 14)=144, the solution was x = 7. The eight consecutive odd numbers were 7, 9, 11, 13, 15, 17, 19, and 21. 4. ** Question **: A sugar is 8.8 yuan per kilogram, B sugar is 7.2 yuan per kilogram. How much B sugar should be mixed with 5 kilograms of A sugar to make the price of each kilogram of sugar 8.2 yuan? - ** Answer **: Assuming that you need [B] sugar in a kilogram. According to the constant countable equation of the total price before and after mixing, the solution is [8.8'times5 +7.2x=(5 + x)'times8.2]. 5. [Question: Buy five sheep from the canteen. Take out two sheep each time and weigh them. There are ten different weights (kg):47, 50, 51, 52, 53, 54, 55, 57, 58, 59.] How many kilograms did each of the five sheep weigh? - ** Answer **: Let the weight of the five sheep from light to heavy be a, b, c, d, e. Each sheep was called together with the other four sheep once, so each sheep was called four times. Then, 4(a + b + c + d+e)=(47 + 50+51+52+53+54+55+57+58+59)=536, so 4 (a + b + c + d+e = 134). The lightest combination is a + b = 47, and the heaviest combination is d + e = 59, so c = 134- 47 + 59 =28. Because a + c = 50, then a = 50 - 28 = 22, b = 47 - 22 = 25, c+e = 58, then e = 58 - 28 = 30, d = 59 - 30 = 29. ** 4. Arithmetic Sequence Problem ** 1. [Question: Below is a string of numbers arranged according to a pattern: 2, 5, 8, 11, 14…] What was the 1995th item? - ** Answer **: This is an arithmetic sequence. The first term is <a1 = 2>, and the tolerance is <d = 3>. According to the general term formula of arithmetic sequence,<a_n= a1 +(n - 1)d>, the 1995th term is <a_{1995}=2+(1995 - 1)><times3 = 2+ 1994><times3 = 5984>. 2. [Question: Among the natural numbers starting from 1, what is the 100th number that cannot be completely divided by 3?] - ** Answer **: 1, 2, 3, 4, 5, 6, 7,.... Starting from 1, form a group of three numbers. The first two numbers in each group cannot be divided by 3. If there are two numbers in a group, there will be a group of 100 numbers. Each group has three numbers, and there will be a total of 50. Then the 100th number that cannot be divided by 3 is 150 - 1 = 149. ** 5. The problem of cows eating grass ** 1. [Title: A patch of green grass that grows at a constant rate every day.] This patch of grass could feed twenty-seven cows for six days or twenty-three cows for nine days. How many days could this patch of grass feed twenty-one cows? - ** Answer **: Suppose a cow eats one unit of grass a day. First, find the amount of grass that grows every day. Twenty-seven cows ate for six days, eating a total of 27,000 units of grass; Twenty-three cows ate for nine days, eating a total of 20,000 units of grass. (207- 162 = 45 units. This is the amount of grass that grows in a day, so the grass grows in a day. The original amount of grass is in units of [162 - 6'times15 = 72]. Assuming that 21 cows eat for [x] days, the equation [21x = 72+15x] can be solved for [x = 12] days. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Numerology related calculations
Numerology was a theory that used numbers to predict and explain human fate and fortune. In Numerology, people use various numbers to represent human fate and fortune. For example, the numbers 0-9 represent different personality traits and life stages, while the number 10 represents success and wealth. Numerology calculations usually required combining and calculating numbers to obtain a prediction result. These calculations included: The calculation of the five elements of numbers: The five elements of numbers can affect the fate and fortune of humans. For example, the number wood, fire, earth, metal, and water could determine a person's character and class choice. Combination of numbers: Different combinations of numbers can produce different results. For example, the combination of the numbers 5 and 3 means that there may be some challenges and difficulties, while the combination of the numbers 9 and 2 means that there may be some success and wealth. 3. Calculating the relationship between numbers and astrological signs and dates of birth: Numerology also involves calculating the relationship between numbers and human fate by calculating the relationship between astrological signs, dates of birth, and other factors. It should be noted that Numerology is only a theory, and its predictions are not absolutely accurate. It is only for reference and entertainment.
Elementary School Students 'Mathematical Olympiad
There were many ways for elementary school students to calculate Olympiad math: 1. ** Borrowing Number Method **: Borrowing a number from a number, turning it into a whole ten, a whole hundred, or a whole thousand, and finally deducting the borrowed number. For example, if you calculate [9 + 99+999 + 9999], you can convert it to [(10 - 1)+(100 - 1)+(1000 - 1)+(10000 - 1)=10+100 + 1000 + 10000 - 4 = 11106]. 2. ** Choose Base Number **: Find a number that everyone is close to as the base number. For example, if you want to calculate [489+487+483+485+484+486+488], first determine the base number [490], then use [490×7 - 1 - 3 - 7 - 5 - 6 - 4 - 2 = 3430 - 28 = 3402]. 3. ** Transposition Method **: Transposition the position of the numbers in the calculation and change the order of the numbers. For example,[632 - 156 - 232 = 632 - 232 - 156 = 400 - 156 = 244];[128+186+72 - 86 = 128+72+186 - 86=(128 + 72)+(186 - 86)=200 + 100 = 300]. 4. ** Common factor extraction method **: Use the multiplication distribution law to extract the same factor and add and subtract the remaining terms to obtain an integral number. For example,[0.92×1.41 + 0.92×8.59 = 0.92×(1.41 + 8.59)]. 5. ** Borrowing and borrowing method **: Use it when you observe that there are numbers close to 10 or 100 in the formula that are easy to calculate. Pay attention to borrowing and returning. For example, 9999+999+99+9 = 9999+ 1+999 + 1+99+1+9+1 - 4. 6. ** Splitting Method **: In order to split a number into several numbers, you need to master some "good friend" numbers, such as <2> and <5>,<4> and <5>, and do not change the size of the numbers when splitting. For example, 3.2×12.5×25 = 8×0.4×12.5×25 = 8×12.5×0.4×25. 7. ** Association Law of Additions **: Use the rule of "(a + b)+c = a+(b + c)" to change the position of the addend to obtain a simpler operation. For example,[5.76+13.67+4.24+6.33=(5.76 + 4.24)+(13.67 + 6.33)]. 8. ** Combination of splitting method and multiplication distribution law **: When you see numbers that are close to an integral number such as <99>,<101>,<9.8>, you must first consider splitting. For example, 34×9.9 = 34×(10 - 0.1). 9. ** Using the formula method **: - The law of addition: Commutative law (a + b = b + a), Association law ((a + b)+c = a+(b + c)). - Subtraction operation properties: <a-(b + c)= a-b- c>,<a-(b-c)= a-b + c>,<a-b- c = a-c- b>,<(a + b)-c = a-c + b = b-c + a>. - Multiplication (similar to addition): Commutativity, Association, Assignment Rate, A + B, C = ac + Bc, A - B, C = ac -Bc. - The properties of division (similar to substitution) were: <a Mum (b×c)=<a Mum (b× c)>,<a Mum (b × c)=<a Mum (b×c)>,<a Mum (b × c)=> a Mum (b ×c)>,<a Mum (b × c)=> a Mum (b × c)>,<a Mum (a Mum (b)= a Mum (b × c)> c & 10. ** Split term method **: Split term method is to split the terms in the score formula so that the split terms can be offset. The common split term method is to split the number into the sum or difference of two or more numerical units. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Fan Sizhe's calculations
Fan Sizhe had shown his outstanding ability to calculate in the TV series Joy of Life. He was good at doing business and was very sensitive to numbers. He could easily calculate complicated numbers. Under Fan Xian's guidance, he formulated a detailed plan, opened the Danbo Bookstore, and sold " Story of the Stone " and " Banxianzhai Poetry Collection," achieving great commercial success. Therefore, Fan Sizhe was very good at calculation.
E-book on macro calculations
" Calculating the Big Balance: An Essay on Economics " was published by the Economics and Sciences Press in 2011. No e-book version was found. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>