Elementary Math OlympiadThe primary school mathematics Olympiad could also be called the "primary school mathematics Olympiad" or "primary school mathematics competition." In 2003, he published the Primary School Mathematical Olympiad (A Collection of Past Questions of the National Primary School Mathematical Olympiad). In 2010, the Hubei Changjiang publishing group and the Hubei Education Press published the Primary School Mathematical Olympiad Quanzhen Test Questions, which aimed to provide primary school students with a collection of competition Quanzhen test questions to improve their ability to solve problems. Its name originated from "Mathematics is the gymnastics of the mind", reflecting the commonality between mathematics and the Olympic spirit (faster, higher, stronger). The scope of the Mathematical Olympiad questions exceeded the compulsory education level, and some of the questions were more difficult than the university entrance exam, even making it difficult for mathematicians. After the abolition of the unified examination from primary school to junior high school, key middle schools favored students with good results in Mathematical Olympiad. Excellent results in Mathematical Olympiad could help students enter famous middle schools smoothly. In the class placement examination of famous schools, students with good Mathematical Olympiad skills had an advantage when they were admitted to the mathematics experimental class. While watching the Olympics, you can also read the wonderful novels related to the Olympics!
The difference between primary school Olympiad math innovative thinking and Olympiad math tutorialMathematical Olympiad courses usually focused on the systematic construction of the Mathematical Olympiad knowledge system and the teaching of solving skills. They used classic Mathematical Olympiad questions as a carrier to help students understand Mathematical Olympiad knowledge in depth and improve their ability to solve problems. The content mainly revolved around mathematical knowledge, and the difficulty exceeded the level of compulsory education. It focused more on the logical thinking exercise in mathematical solving.
The primary school's Mathematical Olympiad innovative thinking emphasized the cultivation of innovative thinking. It focused on developing new ways of thinking based on the knowledge of the primary school's Mathematical Olympiad. It was not only limited to conventional solution ideas, but also might involve the innovative application of traditional Mathematical Olympiad knowledge and methods, as well as combining new teaching concepts and teaching methods to inspire students to think about Mathematical Olympiad problems from different angles. It had a higher requirement for the expansion of students 'thinking.
<a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
From textbooks to Olympiad math a edition e-booksHe only mentioned that there were all the textbooks from grade 1 to grade 6 to the e-textbook sample of Mathematical Olympiad. The file size was 379.07M, but the specific way to obtain it was not specified. It was impossible to provide an effective way to obtain the e-book from the textbook to the Mathematical Olympiad A edition.
<a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
The concentration of primary school Olympiad math is slightly difficultThe following is a primary school Olympiad math concentration problem and its solution:
For example, the bottle contains 1000 grams of 15% alcohol solution. Now, he poured 100 grams and 400 grams of alcohol solution A and B into the bottle, and the concentration in the bottle became 14%. It was known that the concentration of alcohol solution A was twice that of alcohol solution B. Find the concentration of alcohol solution A.
1. First, find the total weight of the mixed solution:
- The original solution weighed 1000 grams. After adding 100 grams and 400 grams of solution, the total weight of the solution after mixing was [1000 + 100+400 = 1500] grams.
2. Then find the total weight of the mixed alcohol:
- After mixing, the concentration of the solution was 14%. According to the weight of the solution x concentration = the weight of the solute, the total weight of the alcohol after mixing was 1500 x 14%= 210%.
3. Then, he asked for the original weight of the alcohol.
- The original concentration of the solution was 15%, and the original weight of the solution was 1000 grams, so the original weight of the alcohol was (1000× 15%= 150) grams.
4. Find the total weight of alcohol in solution A and B:
- Because the total weight of the alcohol after mixing was 210 grams, and the original weight of the alcohol was 150 grams, the total weight of the alcohol in the two solutions A and B was [210 - 150 = 60] grams.
5. Since the concentration of alcohol solution A is twice the concentration of alcohol solution B, if the concentration of alcohol solution B is <x>, then the concentration of alcohol solution A is <2x>:
- According to the weight of solute = solution weight x concentration, the equation for the weight of alcohol in solution A and B could be obtained.
- The weight of alcohol in 400 grams of solution B is 400x, and the weight of alcohol in 100 grams of solution A is 100×2x = 200x.
- Then, the solution to the problem is: 400x +200x = 60, which is 600x = 60. The solution is x = 0.1.
6. Finally, find the concentration of alcohol solution A:
- The concentration of alcohol solution A is [2x], so the concentration of alcohol solution A is [2×0.1 = 20%].
<a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
What are the solutions to the Olympiad math problems for primary school students?There were several ways to solve the Mathematical Olympiad problems for primary school students:
1. ** Visualization Drawing Method **: By reasonably using points, lines, surfaces, graphs, and tables, the Mathematical Olympiad problems can be visualized, visualizing the abstract quantitative relationship. This will allow students to easily understand the quantitative relationship and grasp the essence of the problem to solve it quickly.
2. ** Inverse Method **: Starting from the final result of the question, you can work backwards according to the known conditions until the problem is solved.
3. ** Enumeration Method **: For questions with special numerical relationships that are difficult to answer using ordinary methods, you can list the data that basically meets the requirements according to the requirements of the question and then choose the correct answer.
4. If there is a problem from the positive side of the condition, you can think from the negative side of the result or the problem to solve the problem.
5. ** Ingenious Transformation Method **: When solving a problem, remind yourself if you can transform a new problem into an old problem. The transformation types include condition transformation, problem transformation, relationship transformation, graph transformation, etc. You can solve the problem by transforming it into a familiar problem.
6. [Holistic approach: For some Olympiad math questions that are complicated and unnecessary from the details, you can grasp them from the overall perspective. You can solve the problem by studying the overall form, structure, and the internal relationship between the part and the whole.]
<a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Primary school Olympiad math questions vertical style complete bookThe following are some examples of the types and solutions of the primary school Mathematical Olympiad vertical questions:
** 1. The vertical addition puzzle **
1. ** When the same number is added up, the same number will still be obtained **
- For example, if a three-digit number plus a three-digit number is equal to a three-digit number, if a triangle plus a triangle is equal to a triangle, it is only satisfied when the triangle is 0, because 0 + 0=0. Then, according to the numerical relationship between the 10th and 100th digits, the circle plus the square on the 10th digit was equal to 0 (in fact, 10 to 100th digit was 1), and the square plus the square plus the carry on the 100th digit was equal to the circle. By transforming the vertical form into the horizontal form, the values of the square and the circle were solved by equivalent substitution.
2. ** Normal addition, find the relationship between numbers vertically **
- For example, when calculating the addition of two numbers, one had to consider the rules of digit alignment and addition, starting from the single digit. If the addition of the single digit had a carry, it had to be carried to the tenth digit, and when the tenth digit was added, the number of the single digit carry had to be added. For example, if two two-digit numbers were added together, the addition of the one-digit numbers was equal to 11, and the addition of the ten-digit numbers plus the one-digit carry was equal to 13. According to this relationship, the one-digit and ten-digit numbers of the two numbers could be obtained respectively.
** 2. Multiplication vertical riddle (Take a five-digit number multiplied by a one-digit number as an example)**
1. ** Confirm the range of key numbers **
- First, determine the possible value of one of the multiplying factors (one digit). For example, in a five-digit number multiplied by a one-digit number equals a five-digit number, the value of the one-digit multiplication factor is determined according to the highest digit without carry, the single-digit calculation condition, the highest digit calculation condition, and so on. For example, this one-digit number could not be 0 or 1. If this one-digit number was 9, there might be conflicts when deducing other digits according to the calculation rules (such as obtaining the same number does not meet the requirements of each Chinese character representing different numbers, etc.). The value of this one-digit number was determined by gradual elimination.
2. ** Derives other numbers based on the determined number **
- After determining this one-digit number, the numbers on the other digits were derived according to the multiplication calculation rules. For example, according to the one digit of the product of the one digit numbers and the carry situation to determine the number on the other five-digit number, then according to the calculation on the ten digit (including the one digit carry) to determine the number on the ten digit, and so on to calculate the number of other digits.
These vertical questions were mainly solved by analyzing the operational relationship of the numbers, the carry situation, and the range of the numbers. They were solved by mathematical methods such as elimination and equivalent substitution.
<a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
How to do the two square questions in the elementary school Olympiad mathThere were many ways to solve the question of two squares in the elementary school Mathematical Olympiad. The following were some common situations:
** 1. Area related **
1. ** Area difference problem **
- If you know the difference in the perimeter of two squares, you can first find the difference in the length of the sides. For example, the difference between the circumference of two squares is 32cm. Since the circumference of a square = 4 x the length of a side, the difference in the length of a side is 32 × 4 = 8cm. If the area difference was known, it could be solved by making an auxiliary line. For example, the shadow part (the area difference part) was cut into several parts, and the area of the special figure (such as a small square) constructed from it was used to calculate the area of the rectangular shape, and then other unknown quantities such as the length of the side were calculated.
2. ** Overlapping Area Problem **
- When two squares overlapped, the area of the shadow (usually the non-overlapping part) needed to be calculated according to the known conditions (such as the length of the side, the length of the line segment of the overlapping part, etc.) through the combination and decomposition of the figures. He might use the area formula of a triangle, a echelon, and other shapes, as well as the idea of equal-area transformation to convert the area of an irregular shape into the area of a regular shape for calculation.
** 2. Edge length related **
1. ** Proportional relationship of side length **
- If the proportional relationship between the sides of the two squares and other related conditions (such as the sum of the area, the sum of the perimeter, etc.) were given, one could set the side of one of the squares as an unknown number and solve the equation according to the conditions. For example, let the side length of a small square be x, and the side length of a large square be kx (k is a known proportional coefficient). Then, according to the perimeter sum or area sum conditions, the equation is listed to solve the side length.
2. ** The relationship between length of side and line segment **
- When two squares have a specific position relationship and the length of some line segments is given (such as the line segment connecting the two square vertexes, the distance between the two squares, etc.), the Pythagorean theorem or the similarity of geometric figures can be used to establish the relationship between the length of the line and these line segments, thereby solving the length of the line or other related quantities.
** 3. Combination Diagram **
1. ** Construct a special pattern **
- For example, when calculating the area of the shadow part in the combination of two squares, one could construct a echelon by connecting the diagonal lines, and use the properties of the echelon (such as the equal area of the wings, that is, the area of the two non-adjacent triangle that is divided into four by the diagonal of the echelon is equal) to convert and calculate the area.
2. ** Relationship between the whole and the parts **
- Take the graph composed of two squares as a whole, analyze the relationship between the various parts (such as squares, triangle, echelon, etc.), and calculate the unknown variables such as the area and side length of the whole or part according to the known conditions.
<a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Primary school students 'Olympiad math problems explanation video tutorialYou can refer to the "2020 New Primary School Mathematical Olympiad Video Course for Grades 1 - 6 + 300 Questions + Formulas" published on March 14,2020, which contains the content of the Mathematical Olympiad video course for each grade. For example, 15 lessons for Grade 1, including I know how to compare, folding of graphs, etc.; 15 lessons for Grade 2, including counting and calculating, hands-on operation of cutting, etc.; 15 lessons for Grade 3, including I know how to calculate addition and substitution, etc. There were 15 lectures in the fourth grade, covering the definition of new operations, 15 lectures in the fifth grade, such as the basic calculation of scores, and 15 lectures in the sixth grade, such as the key examination content of the calculation of primary school, which could be used as a video tutorial to explain the difficult problems of the Olympiad.
<a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>