A summary of the theoretical knowledge of Olympic mathematics in primary schools
1, sum difference times problem
2. Three basic characteristics of the age problem:
(1) The age difference between two people is constant;
② The age of two people increases or decreases at the same time;
(3) the multiple changes of two people's ages;
3. The basic characteristics of standardization.
There is a constant quantity in the problem, usually that "single quantity", and the topic is generally expressed by words such as "at this speed".
Key problems: determine and find a single quantity according to the conditions in the topic;
4. Planting trees
5. The problem that chickens and rabbits are in the same cage
Basic concepts: the problem of chicken and rabbit in the same cage, also known as replacement problem and hypothesis problem, is the wrong part of replacement hypothesis;
Basic idea:
(1) hypothesis, that is, assume that some phenomenon exists (like a and b or b and a):
(2) Assuming that there are differences different from the subject conditions, find out what the differences are;
(3) The difference caused by everything is fixed, so as to find out the reasons for this difference;
(4) According to these two differences, make appropriate adjustments to eliminate the differences.
Basic formula:
① Assume that all chickens are rabbits: the number of chickens = (number of rabbit feet × number of total heads-number of total feet) ÷ (number of rabbit feet-number of chicken feet)
② Suppose all rabbits are chickens: the number of rabbits = (total number of feet-chicken feet × total number of heads) ÷ (rabbit feet-chicken feet)
Key problem: find out the difference between total quantity and unit quantity.
6. Profit and loss issues
Basic concept: a certain number of objects are grouped according to a certain standard to produce one result; grouping according to another standard to produce another result. Because different grouping standards have different results, we can find the number of groups or the total number of objects grouped from their relationship.
Basic idea: first, compare the two distribution schemes, analyze the change of results caused by different standards, calculate the total number of copies participating in distribution according to this relationship, and then calculate the total number of objects according to the meaning of the question.
Basic question:
(1) One remainder and the other deficiency;
Basic formula: total number of copies = (remainder+shortage) ÷ twice the difference per copy.
(2) When there is a remainder twice;
Basic formula: total number of copies = (larger remainder-smaller remainder) ÷ twice the difference of each copy.
(3) when it is insufficient twice;
Basic formula: total number of copies = (larger shortage-smaller shortage) ÷ twice the difference per copy.
Basic characteristics: The total number of objects and groups remains unchanged.
Key problem: Determine the total number of objects and groups.
7, cattle grazing problem
Basic idea: Assuming that the grazing speed of each cow is "1", the difference of total grass quantity can be found according to two different eating methods; Then find out the reason of this difference, and we can determine the growth rate and total grass quantity of grass.
Basic characteristics: the amount of original grass and the growth rate of new grass remain unchanged;
Key question: Determine two invariants.
Basic formula:
Growth = (long time x long time-short time x short time) ÷ (long time-short time);
Total grass amount = long time × number of bulls for long time-long time × growth amount;
8, cycle and table rules
Periodicity: In the process of movement change, some features appear regularly and periodically.
Period: We call the time elapsed between two consecutive appearances as a period.
Key problem: determine the cycle.
Leap year: there are 366 days in a year;
(1) year is divisible by 4; ② If the year is divisible by 100, the year must be divisible by 400;
Average year: There are 365 days in a year.
① The year cannot be divisible by 4; ② If the year is divisible by 100, but not by 400;
9. Average
Basic formula: ① Average value = total quantity ÷ total number of copies.
Total quantity = average x total number of copies
Total number of copies = total quantity/average value
② Average value = benchmark number+sum of differences between each number and benchmark number ÷ total number of copies.
Basic algorithm:
① Find out the total quantity and total number of copies, and calculate with the basic formula ①.
② Benchmark number method: according to the relationship between given numbers, determine a benchmark number; Generally, the number or intermediate number close to all numbers is selected as the reference number; Taking the reference number as the standard, find the difference between all given numbers and the reference number; Then find the sum of all differences; Then find the average of these differences; Finally, the sum of this difference and the average value of the reference number is the average value, and the specific relationship is shown in the basic formula ②.
10, pigeon coop principle
Drawer principle 1: If (n+ 1) objects are put in n drawers, there must be at least 2 objects in one drawer.
Example: put four objects in three drawers, that is, decompose four into the sum of three integers, then there are the following four situations:
①4=4+0+0 ②4=3+ 1+0 ③4=2+2+0 ④4=2+ 1+ 1
Observing the arrangement of the above four items, we will find a common feature: there are always two or more items in a drawer, which means there must be at least two items in a drawer.
Pigeonhole principle II: If you put n objects in m drawers, where n >；; M, then there must be at least:
①k=[n/m ]+ 1 object: when n is not divisible by m.
②k=n/m objects: when n is divisible by m.
Understanding knowledge points: [x] refers to the largest integer that does not exceed X.
Example [4.351] = 4; [0.32 1]=0; [2.9999]=2;
Key issues: constructing objects and drawers. That is, find the quantities representing objects and drawers, and then calculate them according to pigeonhole principle.