Can you give me some math Olympiad puzzles?

Many parents are facing the prospect of going to school, and they only start to learn Olympiad mathematics in the 5th or 6th grade, which is definitely difficult to learn at this time. First of all, it is a problem for you to start learning the Olympiad from the first grade. It is equivalent to learning Olympiad mathematics for six years in one year.

Not to mention children, even adults are difficult to do. Therefore, children are tired of school as soon as they come into contact with it. It is best to learn the Olympiad correctly at the same time as your own school year, or even about a year earlier.

The first grade Olympiad is basically about common sense. After the third grade, it started to get harder and harder. It's hard to see what you measure, and it's still difficult to get a prize if you go to the competition.

If you just want to score in the top 3 of your class in math and the top 15 in your grade. Then the Olympiad is not difficult. The main thing is to learn early and step by step.

Some topics are difficult and some topics are simple.

Single. It is characterized by:

1. Apply the knowledge of the upper grades to the topics of the lower grades. This part of the topic is simple. Most people will soon.

2. Introduce some knowledge that is never tested in ordinary exams. This part of the questions is of medium difficulty.

3. Special problem-solving skills. This part of the question is super difficult, and most people can't figure it out.

It is not recommended to study Olympiad mathematics. Because, 1, the knowledge of the senior students will be learned sooner or later.

2. The knowledge that is never tested in the ordinary examination is not helpful for the entrance examination and the college entrance examination.

3. In the recruitment and college entrance examination, special problem-solving skills cannot effectively improve the accuracy, and then cannot effectively improve the score.

4. If you want to practice your math skills through the Olympiad, you might as well study the basic concepts and definitions in books. Don't think you understand all these concepts and definitions.

5. If you want to exercise your brain through the Olympiad, it is better to exercise more and read more history and literature books.

6. If you want to help you go to junior high school and get into college through the help of the Olympiad, then the Olympiad basically can't help much.

7. Don't equate good math grades in elementary school, junior high and high school with math talent, and don't easily think that you can become a mathematician in the future.

In the end, the Olympiad and mathematicians are only a small group of people who can play it.

Olympiad is a game for smart people, and those with insufficient intelligence will find it difficult. It's really good to play, only the first 5%.

, where x is the raw gasoline, x = 140

x is a large rectangle and y is small, so x y=4 3

3.Let x days be complete, 32 100x=16, find x as 50, and the rest is 50-16=34 days.

x y = x divided by 10 11x = 11 10, 11 10-1 = 1 10, so add 1 10

x = 120 seconds, 24 remaining, so B walked 24 meters more after 4 laps, so he still has to walk 30-24 = 6 meters.

x = 240, 600-240 = 360, evaporation 360g

Find out x.

x=200 is actually written directly according to what he said, set to the total number to subtract what he said, = what he said at the end, the same, think about it, it's very simple.

The efficiency of A and B is 1 7 and 1 5 respectively (where is the score line, the same below), the efficiency and 1 7 + 1 5 = 12 35, A has done one day, that is, 1 7 of the project, there is 1-1 7 = 6 7 of the project, 6 7 12 35 = 5 2 = days.

A and B complete all 1 12 every day

A completed all 1-3 10 = 7 10 in 14 days, so A completed all 7 10 14 = 1 20 every day, and B completed all = 1 12-1 20=1 30 every day, so B actually worked 3 10 (1 30) = 9 days, so B took 14-9 = 5 days off.

As shown in the attached diagram, the area of the shaded part corresponds to the area of the square, so the area of a shadow nucleus = the area of a square is defined by the title, and the diagonal of the square = 2cm, and the side length = 2cm of the first digging

The area of the square can be counted as 2 cm 2

There are 4 shaded sections in total, so the total shadow area = 4*2cm2=8cm2

Solution: Set the camel to fall after a day.

23000+5y-500y+｛（y-1）²÷2+（y-1）÷2｝×5=0

The positive number y is found to be equal to 74

y-1) is because camels don't eat turnips on the first day on the road, and if you understand that they eat turnips on the first day on the road, they die after 73 days.

The physical strength at the beginning of day nth is a(n).

The stamina starting on day n+1 is a(n+1).

At the beginning of the nth day, the number of turnips that the camel has eaten is 5 (n-1).

The number of radishes left is 500-5(n-1) = 5(101-n).

Then there is. a(n+1) = a(n) -5(101 - n) +5

Simplify. a(n+1) -a(n) = 5n - 500

a(n) -a(n-1) = 5(n-1) -500

a(n-1) -a(n-2) = 5(n-2) -500

.a(3) -a(2) = 5×2 - 500

a(2) -a(1) = 5×1 - 500

Add all the above equations to get it.

a(n+1) -a(1) = 5[1+2+3+4+5+..n] -500n = [5n(n+1)]/2 - 500n

a(n+1) = - 23000

This is the remaining physical strength of the camel when it has not started to walk every day, you can bring it in and calculate, when n=0, a(1)=23000, and the physical strength left at the beginning of the first day is 23000.

Let f(x+1) = - 23000 = 0

23000 = 0 divided by both sides at the same time.

x^2 - 199x + 9200 = 0

x^2 - 2( += ( = -9200 =

The solution is x = - the positive root is rounded off, because the camel dies in the number of days corresponding to the previous smaller root.

It shows that the camel still had physical strength at the beginning of the 74th day, but it ran out by the middle of the 74th day, so the camel died on the 74th day.

Calculation: a(74) = - 23000 = - 23000 = 5

And the number of turnips left on this day is 500 - 73 5 = 135, so the camel died halfway through the 74th day.

Remaining stamina on the first day 23000-500+5=22505 on the second day. 22505-495+5=22015 Day 3... 22015-490+5=21530 Let the physical strength of the camel be A1 on the first day, A2 on the second day, and so on, and the physical strength on the nth day is ANA1=23000-500+5

a2=a1-（500-5）+5

an=a（n-1）-【500-5（n-1）】+5an-an-1=5n-500 ..1

a(n-1)-a(n-2)=5(n-1)-500 ..2

Add all of the above together.

an=22505+5（1+2+3+..n)-500nIf you want to know which day it fell, you just need to find the value of n when an=o.

Mathematical models such as: 500 + 495 + 490 + 485 + 480....=23000+5*(n-1) (0n*(1000-5*n+5) 2=23000+5*n-5 (if you don't understand, check the sum formula of the difference series).

n*n-197n+9198=0

Solve n approximately equal to or.

The result was 77 days.

23000 (500-5) 46 days, will fall around the 45th day.

All meta-assumptions are 1 yuan.

28 is 28 yuan.

1 piece of 1 corner is less than 1 piece of 1 yuan.

Now the yuan is less than the 28 yuan.

So there is 1 corner.

Answer: There are 3 pieces of 1 yuan, 25 pieces of 1 corner.

80 divided by 120 percent = approximately equal to 67

80-67=13 means 13 earned

80 divided. 100-80 equals 20, which means that you lose 20

20-13=7 She lost 7 dollars.

Three pieces of one dollar and twenty-five pieces of one dime.

Solution: Let the number of unary sheets be x

28-x+10x=55

28-9x=55

9x=55-28

9x=27x=3, then there are 3 unary cards.

There are 25 in one corner.

3 pieces of 1 yuan, 25 pieces of a corner will do.

Suppose there are n sheets of one yuan and l of one corner.

Which n*1+l*

Again: n+l=28 sheets.

n=28-l, put it in the previous formula, and it becomes (28-l)*1+l*, and after combination, it is calculated that l=25 sheets.

n=28-25=3 sheets.

Solution: If all 28 sheets are 1 yuan, the total amount is 28 yuan.

There are more yuan than before.

And a piece of 1 yuan is more than a piece of 1 dime.

There are 25 pieces of 1 corner of explanations.

So 3 cards for 1 yuan and 25 cards for 1 corner.

Let 1 yuan be x and 1 corner be (28 x) 5 10 5 10 x (28 x) and get x 3, so there are 3 pieces of 1 yuan and 25 pieces of 1 corner.

Does the landlord understand?

Suppose it is changed to x sheets of one yuan, 28-x sheets of one corner.

Then x+ gets x=3, so 3 pieces of 1 yuan, 25 pieces of 1 corner.

Set 1 yuan x sheet, 1 corner is (28-x) sheet.

x+x+ =

x = 3 because x = 3 so 28-x = 25

So 1 yuan 3 sheets 1 corner 25 sheets.

Set x for one yuan, and y for one corner.

10x+y=55 1 formula.

x+y=28 2 formula.

Formula 1-2 yields 9x=27

x=3 so y=25

Set 1 yuan has x sheets, 1 corner has y sheets, according to the title.

x+y=28

10x+y=5*10+5=55

Solving the above equations together yields x=3, y=25

There are a total of 55 angles, set one yuan x zhang, one corner 28-x zhang, then 10x+1(28-x)=55, the solution is x=3, so one yuan 3 zhang, one corner 25 zhang.

3 pieces of 1 dollar, 25 pieces of 1 corner.

You can think that 28 pieces are all a dime, and the result is less than two yuan and seven dimes, so there should be three more pieces of one dollar and less three pieces of a dime The deformation of the chicken and rabbit in the same cage

There are x sheets for one yuan, and there are y sheets for one corner.

then x+y=28

10x+y=55

The solution: x=3, y=25

Solution: The total amount of money is 55 jiao, let one yuan have x zhang, and one dime has y zhang by x+y=28

10x+y=55

x=3 y=25

Answer: In exchange for 3 pieces of 1 yuan, 25 pieces of 1 corner.

Solution: Set 1 yuan x sheet, 1 corner y sheet.

10x+y=55

x+y=28

x=3 y=25

Answer: 3 sheets of 1 yuan, 25 sheets of 1 corner.

3 pieces of one dollar, 25 pieces of one corner.

Let the length of a and b be x

According to the known results, when the countercurrent speed is 8-2 = 6 km, the downstream speed is 8 + 2 = 10 km.

This question should be discussed and analyzed on a case-by-case basis

When C is downstream of A: (x 6) + (x + 2) 10=3 is solved: x=

When C is upstream of A: (x 6) + (x-2) 10=3 The solution is: x = 12

A: The distance of a and b is one kilometer or 12 kilometers.

1) Solution: Let the per capita donation of this school be X yuan.

10x+20=12x-30

x=25 yuan.

2) Solution: The per capita donation amount of this school is X yuan.

10x+20) 50-(12x-30) 54=x=yuan.

There is no topic!! It's too hard to ask.