Olympiad math classic questions. What are the classic math Olympiad problems? Please let me know

1. There are 19 coins of one yuan, five jiao, one dime, five cents, two cents and one cent, totaling yuan. Q: How many coins are there?

2. Prove: (cos) 2+(cos) 2+(cos) 2-2cos cos cos -1<0 and + can be pushed each other.

3. If you are told that the two farms of A and B are about the same size, do you agree with Xiaoyu's statement at this time? Let's talk about the reason A farm....Grain accounted for 65. Vegetables account for 20.

Cotton accounts for 15 percent. B Farm. Cotton 57.

Grain accounts for 25. Vegetables account for 18. Also, Xiaoyu said that the grain crop planting area of Farm A must be much larger than that of Farm B.

Do you agree with him? Why?

4. Find (1+4*2 4)(1+4*4 4)(1+4*6 4)....(1+4*20^4)/(1+4*1^4)(1+4*3^4)(1+4*5^4)…(1+4*19^4)

5. Seek: 1+3 2 +5 2 2 +7 2 3+...201/2^100+205/2^100

A piece of engineering. A completed in 30 days, completed in 40 days, and completed in 50 days.

A dried for 15 days, and the remaining 3 people and a few days of dry were completed.

In terms of modules: Calculation Module, Equation Module, Number Theory Module, Application Problem Module, Geometry Module, Counting Module, Special Topic Module, and Miscellaneous Problem Module.

Answering process:

Method 1: The number of participants in all three types of training is x.

29+21+25）-（17+15+10）+x=4575-42+x=45

33+x=45

x=45-33

x = 12 A: The number of participants in all three types of training is 12. Method 2:

12 people. A: The number of participants in all three types of training is 12.

Participate in computer training + participate in composition training: 29 + 25-17 = 37 people only participate in Olympiad training: 45-37 = 8 people.

Participate in all three types of training: (15+10)-(21-8)=12 peopleThe result is greater than the number of people who participated in both Olympiad training and computer training 10 people, and you have a problem with your topic.

There are x number of people who participate in all three types of training.

then participate in both computer training and composition training (do not participate in Olympiad training): 17-x participate in both composition training and Olympiad training (do not participate in computer training): 15-x participate in both Olympiad training and computer training (do not participate in composition training):

10-x only participate in computer training: 29-x-(17-x)-(10-x)=2+x only participate in composition training: 25-x-(17-x)-(15-x)=x-7 only participate in Olympiad training:

21-x-(10-x)-(15-x)=x-4x+17-x+15-x+10-x+2+x+x-7+x-4=45x=12

So you have a problem with the topic.

29＋21＋25＝75

Detailed explanation: One person who reports one class counts as 1 person, two classes count as 2 people, and three classes as 3 people.

29 21 25 75 (The total number of students in the three classes is 75.) 17 2 15 2 10 2 84 (One person registers for two classes, which counts as 2 people, and 15 people apply for Olympiad mathematics and composition classes, so there are 15 2 people.) In the same way, the number of people who apply for the Olympiad and computer classes, and the computer and composition classes must be calculated by using the number of 2.

In total, there were 84 visits. ）

84 75) (2 3 3) 3 (The total number of students who applied for the Olympiad and composition class, the Olympiad and the computer class, and the computer and composition class was 84 75 9 more than the actual number of students, because if a student applied for three classes at the same time, it should actually be counted as 3 people, but when he counted the number of people who applied for the Olympiad and composition class, the Olympiad and the computer class, and the computer and composition class, he was counted 2 3 6 times, which was 6 3 3 times more than the actual number. Finally, because the students who applied for three classes at the same time were overcounted by a total of 9 people, and each person was overcounted by 3 times, it can be found that there are 3 students who applied for three classes at the same time by dividing 9 by 3. Agreed.

I was wrong just now, there is no definite solution to this problem, 10, 9, 8, 7 are OK, note that there may be people who have not participated in the training.

The 10 people who participated in the Olympiad and the computer, all three of them obviously could not be greater than 10, so 12 was obviously wrong, and I calculated that it should be 6

This question is solved in the sixth-grade Olympiad math course, using the above formula, that is: (1) 29 plus 21 plus 25 minus 17 minus 15 minus 10 equals 33 (2) 45 minus 33 equals 12 There are 12 people who participated in all three types of training.

The answer is 3 people. It's ridiculous to say that the topic is problematic when you can't do it yourself. It's ridiculous that those admins also list it as a recommended answer.

29 21 25=75

48—45=3 (person).

However, there were no participants, so there were more than three people. (Just add those who don't participate.) )

Why 46It's not 45 - it's one more person.

65 (Jian Feng kilograms) - there are 65 kilograms of fruit candy to know the difficulty.

The second method:

Featuring x kg of fruit candy.

35+x)=

Solution x=65.

Middle section (2+1) (1-1 2) = 3 1 2 = 6 kg.

Rear section 6 1 2 + 2 = 3 + 2 = 5 kg.

The carp weighs 6 + 5 + 2 = 13 kg.

Analysis:1It is known that the weight of the middle section is exactly 1 kg less than the sum of the weight of the front and rear sections, 2From the sum of the weight of the rear section equal to half of the weight of the middle section and the weight of the front section, we know that the weight of the middle section = 2x (weight of the front section - weight of the rear end).

3.It is known that the weight of the front segment = 2 kg.

To sum up, the weight of the last section = 2 = 5 kg is calculated first.

Middle section weight = } = 6 kg.

The total weight of the carp is 2 + 6 + 5 = 13 kg.

The weight of the middle section is 1kg more than the back section, and the weight of the rear section is equal to half of the middle section plus 2kg decisively guess that the middle section is 6kg, then the back section is 5kg Meets the conditions, no equation, the total amount is 13kg

The posterior weight is equal to half of the weight of the middle section and the sum of the weight of the anterior segment. That is, the weight of the rear section of 2 times is equal to the sum of the weight of the middle section and the weight of the front section 2 times, and because the weight of the middle section is exactly 1 kg less than the sum of the weight of the front and rear sections, that is, the weight of the rear section of 2 times is equal to the sum of the weight of the rear section and the weight of the front section 3 times less than 1 kg, that is, the weight of the rear section is equal to the weight of the front section 3 times less than 1 kg, that is, the rear section is 5 kg, and then the middle section is 6 kg, a total of 13 kg.

The text description is actually the process of solving the equation!

There will be more solutions to this kind of problem, but the principle is almost the same, the essence is listed in the equation, here I will say a relatively simple topic, as follows: let's count 4 fish, it is equal to the sum of the 4 sections of the front, middle and back, and the weight of the back section is equal to half of the weight of the middle section and the sum of the weight of the front section, 4 back ends are equal to 2 middle sections plus 4 front sections, so 4 fish are equal to 8 front sections and 6 middle sections, and then 3 middle sections are equal to 3 front sections plus 3 rear ends less than 3, So 4 fish is equal to 11 front segments plus 3 middle segments plus 3 rear segments less 3, put 3 front segments, 3 middle segments, 3 rear segments, it is 3 fish, so 4 fish is equal to 3 fish plus 8 front segments less 3, and 8 front segments are equal to 16, so 4 fish are equal to 3 fish plus 16-3 = 13, of course, each fish is 13 kilograms, I hope it can help you.

This is a better way to solve the problem with line segments.

Draw a long line first, divide it into 3 sections, and mark the front, middle, and back.

Then write the data on: medium +1 = after 2+. Medium = 2+post-1 = 1+post. After = back + 2 [half of the first section].

1 2) after =

According to the first condition, it is known that after 2+-1=, it is medium.

According to the second condition, it is known that after 2+2=.

Replace the first equation with the second equation.

2+Medium 2+2-1=Medium.

Then after medium = 6 = 52 + 6 + 5 = 13 (kg).

Known: Middle segment = 2 + posterior segment - 1

Rear segment = 1 2 middle segment + 2

Then: middle section = (2+2-1)*2

6 (kg) rear section = 6 * 1 2 + 2

5 (kg) then the total weight; 6+5+2=13（kg）

Rear weight: [(1+1 2) 2-1] (2 1-1) = 5kg

Interrupt weight: 2 + 5 + 1 = 6kg

Total weight: 2 + 6 + 5 = 13kg

It's easy.

Weight: 2 (1 2-1 3) = 12 kg.

121, 169

From the title, it can be seen that the number has only one factor except 1 and itself, and the factors appear in pairs, so this number is a square number, and this factor is a prime number, 100-200 is the square of 10 to the square of 14, and only 11 and 13 meet the above conditions, so the result is 11*11 = 121; 13*13 = 169。

Do the fastest first, and then decide by the length of time, from short to long to do, the order of state width is b, c, a, e, d, at this time, the sum of the delay is the least, this time is the ruler track 0 + 3 + 3 + 4 + 3 + 4 + 5 + 3 + 4 + 5 + 6 = 37 minutes.

Hello! d e b c a time is the assembly time of parts 3 + 4 + 5 + 6 + 7 + 5.

First of all, the longest and second longest ones don't have to waste any more time. Then choose the second and third short. Finally, the fastest solo one. Therefore, the equation is: 7 + 6 + 5 + 4 + 3 = 25 (divide and take the clock).