Who has this year s Olympiad questions? Past editions can also be used, and the answers should be in

Olympiad problems refer to questions in a math contest or math olympiad and are usually very challenging and creative math problems. The purpose of the Olympiad is to cultivate students' mathematical thinking ability, analytical and problem-solving skills, as well as students' logical reasoning and innovative thinking.

Olympiad questions usually require students to solve them independently within a limited amount of time, and the process of solving them needs to be clear, accurate, and detailed. These questions may involve multiple branches of mathematics, such as algebra, geometry, number theory, and combinatorics, or they may be a combination of practical problems.

The difficulty of the Olympiad problems is usually high, which is beyond the scope of general school teaching, and the requirements for students' mathematical foundation are higher, and students need to have the ability to deeply understand and flexibly apply the knowledge of lenient mathematics. Moreover, the Olympiad also focuses on cultivating students' innovative thinking and problem-solving skills, and encourages students to think about different methods and angles to solve problems.

Participating in the Mathematics Olympiad will help students improve their mathematical literacy, broaden their horizons, cultivate self-confidence, stimulate their interest in travel, and may lay a foundation for future academic research and the development of mathematics majors.

The steps to solve the problem are as follows.

Solution 1: C is in the middle, and the number he signs is the sum of the two sides. That is, one side is 22, and the other side is: 38-22=16, then the difference between the two sides is 22-16=6

That is: A and B each signed 22, D and E each signed 16, A signed 6 more than E, and the calculation: 22-16 = 6

Solution 2: Let e sign x, according to the meaning of the question: 2x+6=38 solution: x=16.

Solution: If E is signed x times, then A is signed x + 6 times, and it is known that each student happens to find 3 writers who are next to each other to sign their seats.

So writer c: x+x+6+ (signature with b and d) = 38, so (38-(x+x+6)) multiplied by 2+x+x+6=22x=24

The SOB is signed x+6+38-x-x-x-6=14 times.

1:c is in the middle, and the number of his signatures is the sum of the two sides. That is, one side is 22, and the other side is: 38-22=16, then the difference between the two sides is 22-16=6

That is: A and B each signed 22, D and E each signed 16, A signed 6 more than E, and the calculation: 22-16 = 6

2: Let e sign x, according to the title: 2x+6=38

Solution: x=16.

To arrive at the same time in two shifts, the method is as follows: the car will be sent to Class B first, and Class A will set off on foot at the same time. The car will send Class B to point A, turn back to pick up Class A, and Class B will continue on foot.

The car turns back and picks up Class A at point B, then turns around and drives to the destination, and the last two classes arrive at the same time (you can also send Class A first).

Let the distance from the starting point to point B (i.e. the walking distance of class A) be x, and the distance from point A to the destination (i.e. the walking distance of class B) is y.

Then the walking time of class A is equal to the time when the car sends to class B + the time when the car turns back, and the listed formula is:

x/6=[（

In the same way, the walking time of class B is equal to the time when the car turns back + the time when the car sends to shift A, and the listed formula is:

y/4=[（

Solve the above two equations together to get the answer (note: the units of x and y solved above are kilometers, and the question asks how many meters is, and you need to multiply the resulting number by 1000).

After the father and mother meet, the father and the child begin to meet, and a total of 3 (100+75)=525 are gone

This is the distance between the mother and the child when the father and the mother meet.

525 (90-75) = 35 – It takes 35 minutes for the father and mother to meet.

Full journey: 35 (100+90)=6650