Need an Olympiad math problem!!!!!!!!!!!! To be specific, the number can not be written .

According to the problem, you can make the following table, in which you can see the time when and the two pipes are open and the corresponding amount of water (note that the total amount of irrigation is reduced to the amount of irrigation per irrigation).

The first time, the second time, the third time, the fourth time.

9 minutes, 4 minutes, 10 minutes, 0 minutes.

0 minutes 4 minutes 10 minutes ?Minute.

Water: -1 3 1 3 1 3 (filled).

Supplicate: The amount of irrigation per minute is x, is y, turn off the tube and only turn on the tube to fill up in z minutes.

Column equation: 13x+4y=1 3

10x+10y=1/3

From the above equations, y=1 90 is solved

And because: z*1 90=1 3

Solution: z=30 minutes.

30 + 4 + 9 + 10 = 53 (points).

A: Fill the pool after 53 minutes from the start.

Refined. Solution: Set the x of the irrigation basin per minute, and the y of the irrigation basin per minute.

9x+4y+4x=1/3

10x+10y=2/3-1/3

Solution: x=1 45

y=1/90

1-2 3) (1 90) = 30 (points).

9 + 4 + 10 + 30 = 53 (points).

53 minutes.

Let a system of binary equations and calculate it !!

Solution: Set the x of the irrigation basin per minute, and the y of the irrigation basin per minute.

9x+4y+4x=1/3

10x+10y=2/3-1/3

Solution: x=1 45

y=1/90

1-2 3) (1 90) = 30 (points).

9 + 4 + 10 + 30 = 53 (points).

Yes?! Want to do arithmetic? I'm dizzy!

53 minutes.

Let this number be abc, abc = 100 * a + b * 10 + c, abc is a natural number from 1 to 9.

If you don't learn negative numbers, a>c

abc-cba=a*100+c-c*100-a=(a-c)*100-(a-c)

99*(a-c)

So a-c = 4 or 8

a-c=8, a=9, c=1, b=0 to 9 for a total of 10 digits a-c=4, b equals 0 to 9, a total of 10 digits a=9, c=5

a=8,c=4,a=7,c=3

a=6,c=2

a=5,c=1

So a-c=4 has a total of 5*10=50

So there are a total of 50 + 10 = 60 such three-digit numbers.

c 60.

1) Solution: Let the hundredth digit of this three-digit number be a (selected from 1-9), the ten-digit digit is b, and the single digit number is c; Because there is an inverse ordinal number, the single digit c is not 0; Then: (100a+10b+c)-(100c+10b+a)=99a-99c=99(a-c) can be a multiple of 4, so a-c=4 or 8, then it is possible that a=5, c=1;a=6,c=2；a=7,c=3；a=8,c=4；a=9,c=5；a=9,c=1；i.e. there are 6 options; B can choose any number from 0 to 9, so there are 6*10=60 options.

The answer is c.

The second class comes and the third class plants trees and trees account for the whole class: 1-1 5 = 4 5 The second class plants trees and trees accounts for the whole class.

4/5÷（3+5）×zhi3=3/10

Class 3 DAO planted trees in the class: 4 5 (3+5) 5=5 10 Class 2 planted fewer trees than Class 3 : 5 10-3 10=2 10 and because: Class 2 planted 40 fewer trees than Class 3.

So the total number of trees planted is: 40 (2 10)=200 trees planted in the first shift: 200 1 5=40 trees planted in the second shift:

200 3 10=60 Trees planted in the third shift: 200 5 10=100.

Let's say a total of x trees are planted

1 class accounts for 1 5

So the 2+3 class accounts for 4 5

2 shifts: 3 shifts = 3:5

Then 2 classes: 2+3 classes = 3:8

So 2 classes account for the total = 4 5

Version 3 8=3 10

2 classes account for the total = 4 5 5 8 = 5 10

The difference between class 2 and shift 3 weighs 2x 10=40

x=200, so 1 shift = 200 1 5=40

2 shifts = 200 3 10 = 60

3 shifts = 200 5 10 = 100

Set up 1 class x trees, 2 classes y trees, and 3 classes z trees.

From the question x=1 5(x+y+z).

y/z=3/5

y+40=z

Solve the system of equations.

x=40y=60

z=100: 40 trees planted in the first shift, 60 trees in the second shift, and 100 trees in the third shift.

It should be very simple, the first class accounts for one-fifth, which means that the second and third classes account for a total of four-fifths. And because the ratio of the second and third classes is 3:5, it means that the second class occupies eight tolerances.

3 of the parts, the third class accounts for 5 of the eight parts, then the third class has more than the second class 2 out of 8 parts, and 2 out of 8 parts is 40, then 1 part is 20So the second class accounts for 3 parts is 60, and the third class accounts for 5 parts is 100And because the total of these eight copies is 160

That's four-fifths of the original. Then find the original 5 parts, divide 160 by 4 5 to get 200then a fifth is a fifth of 200.

That's a class of trees. This question uses the method of the fifth grade, although you have not learned the comparison, but you should know it when I talk about it like this.

The age difference between A and B remains unchanged, and the ratio of the age difference between A and B and the sum of the ages of A and B 10 years ago was (3-2) (3+2)=1 5, and now it is 1 7, (4-3) (4+3)=1 7The values are the same, so the ratio of the sum of ages 10 years ago to the sum of ages now is 5:7, and the difference in the middle is 10*2 years, because both people are increasing in age.

It is found that the current is 30 for A and 40 for B, and after 10 years, A is 40 and B is 50, and the ratio is 4:5

Find a pattern! 2:3 3:4 should be 4:5 after that! Guess.

Zhao Qiansun Li has set a total of 2 + 2 + 4 + 3 = 11 copies.

a, b, c, d are set in total: 1 + 2 + 2 + 2 = 7 copies According to the title, at least 1 copy is ordered per week.

A total of 5 people have a minimum of 11 + 1 = 12 copies.

Then there are 12-7=5 households that set e.

If there is more than 1 copy per week, it is assumed that at least 2 copies are set per week.

Then the total number of subscriptions for 5 people is at least 11 + 2 = 13 copies.

Then there are at least 13-7 = 6 households, which contradicts a total of 5 households, so only 1 type can be fixed every week, and there are 5 households that are fixed e.

Observe that every 6 numbers are divided into 1 group, and the result of each group is calculated as:

2007 6 = 334 remainder 3

A total of 2007 numbers, which can be divided into 334 groups and 3 numbers remaining, which are 3,2,1 so:

One less 2 and 1 12, right?

1 1 6 (with fractions) + 2 and 1 12 + 3 1 20 (with fractions) +4 1 30 (with fractions) +5 1 42 (with fractions).

15 and 5 14

If it is true that there are no 2 and 1 12

Then the original = 15 and 5 14-2 and 1 12

13 and 23 84

1. A bus is more than a passenger car to sit 20 people, 6 buses and 8 passenger cars are equal to the number of 6 buses than six small cars 20 * 6 = 120 people, these 120 people just sit in two small cars, so a small car sit 120 2 = 60 people If they all sit in a small car, they need 720 60 = 12 cars!

2. 6th grade + other grades = 16 students (other grades are except 5th grade and 6th grade) 5th grade + other grades = 12

5th grade + 6th grade = 20 students.

The total number of winners of the school's mathematics competition is:

16 + 12 + 20) 2 = 24 (person).

If you don't understand, please ask! Have fun!

(1) The number of people sitting in each passenger car is 20*6 (8-6)=60

So passenger cars 720 60=12 are needed.

A bus seats 20 more people than a minibus, the number of 6 buses and 8 minibuses is equal, the number of 6 buses and 6 minibuses + 120 people is equal, and the number of 2 minibuses = 120

720 120 = 6 6 * 2 = 12 cars.

A total of 20 students from Year 5 and Year 6 received the award.

There are 16 people who are not in 5th grade, =1 2 3 4 6 There are 12 people who are not in 6th grade =1 2 3 4 51 2 3 4 =(16+12-20) 2=44+20=24 people.

Question 1:

There are x buses and small y buses, with 6x=8y and x=y+20

Substituting x=y+20, we get y=60

12 passenger cars are required.

Question 2: Set the number of children in grades 1 to 4 as x.

Fifth grade has: 16-x people.

Sixth grade has: 12-x people.

So (16-x) + (12-x) = 20

Calculated as x=4, so the school has 20 + 4 = 24 people awarded.

……Speechless.........

I didn't expect so many well-wishers to ......... in front of me when I was struggling to calculate and type

A square of 256 is a perfectly square number of 225 256 289

The last time Team A planted 10 trees, and Team B planted less than 10 trees, it means that it is an odd number in the 10th place, so it can only be 256

It should be 16x16=256 trees planted for 13 days, and on the 13th day, team A planted 10 trees and team B planted 8 trees.

The square of 16 is 256 and the square of 17 is 289

256 10 = 25 surplus 6, the last time B species, eligible 289 10 = 28 surplus 9, the last time A species, not eligible so it is 256 trees.