Because the number of digits is 9, the added digits do not carry digits.

That is, the sum of single digits Y+W=9 instead of19,29,39. ....

So the sum of ten digits X+Z= 13.

So: x+y+z+w=22.

There is a 500-meter-long circular runway. Both parties start from a certain point on the runway at the same time. If they run in the opposite direction, they will meet in 1 minute. If you run in the same direction, 10 minutes later. To know that A runs faster than him, ask: How many meters does A run per minute?

In the opposite direction, the sum of their velocities is 500/ 1=500.

In the same direction, the speed difference between them is 500/ 10=50.

The speed of A is: (500+50)/2 = 275m/min.

The speed of B is: (500-50)/2 = 225m/min.

On a circular runway, at 1:00 in the afternoon, Xiaoming set off from point A and Xiao Qiang set off from point B at the same time. They met at 1:06 pm, at 1: 18 pm, Xiaoming arrived at point B, at 1: 18pm.

For example, Xiao Qiang walked for 6 minutes before meeting him for the first time, and Xiaoming walked for 4 minutes, so Xiaoming's speed is Xiao Qiang's: 6/4= 1. Five times.

It took 18-6= 12 minutes from the first meeting to the second meeting.

So the speed of Xiao Qiang is: (112)/(1+1. 5)= 1/30

That is, Xiao Ming's speed is: 1/30* 1. 5= 1/20

Then Xiao Ming's travel time is: 1/( 1/20)=20 minutes.

4.A, B and C are natural numbers with two digits, the single digits of A and B are 7 and 5 respectively, and the ten digits of C are 1. If the equation ab+c=2005 is satisfied, then a+b+c=?

First of all, we can judge that the unit of C should be 0 by the unit of B is 5.

In this way, we can know that the unit and ten digits of c are 10.

Then AB should be 2005- 10= 1995,

1995, only the single digits of 57 and 35 are 7 and 5 respectively, so it is determined that.

a+b+c=57+35+ 10= 102

5- 1 1

1, 22 ...2 [2000 2] divided by 13, what is the remainder?

What is the remainder of the square of 2.65438+the square of 0+2+the square of 3 ...+the square of 20065438+the square of 2002 divided by 4?

3. What is the remainder of the product of the number 1998 * 1998 * ... *1998 [2000 times of multiplication] divided by 7?

4. The remainder of an integer divided by 84 is 46, so what is the sum of the three remainders obtained by dividing by 3, 4 and 7 respectively?

There are 69, 85, 93 and 97 tourists in the four tour groups A, B, C and D respectively. Now, the four tour groups should be divided into groups, and each group should be tourists, so that we can go to go on road trip. It is known that after the three groups A, B and C are divided into several groups of A people in each group, the remaining number of people is the same. How many people are left after the tour group is divided into several groups of A people in each group?

6. Four players with numbers 37, 57, 77 and 97 play table tennis. It is stipulated that the number of sets played by every two players is the remainder of the sum of their numbers divided by 3. What is the number of the player who plays the most CDs? How many sets did he play?

1 and 222222 are divisible by 13, so 2000 twos contain 333 cycles, leaving the last 22, so the remainder is 9.

Because every even term is divisible by 4, there are only odd terms left. We can know that the square of 1 +3 +5 +7 can also be divisible by 4, and there is also the square of11+15.

3. The remainder of1998 divided by 7 is 3, so we can divide by 1998=7*n+3.

A total of 2000 * * 1998 = 7 * n+3, so it was three times that of 2000, that is, 3 2000 = 91000 = (7+2)1000, so it became 2 65438 again. 2 1000 =1024100 = (146 * 7+2)100, which becomes the remainder of 2100 divided by 7. Similarly, it finally became 65438+.

4. If it is set to 84a+46, then 84a can be divisible by 3, 4 and 7, and the answer is the sum of three remainders obtained by dividing 46 by 3, 4 and 7. 1+2+4=7.

This question means 69=n 1*A+a, 85=n2*A+a, 93 = n3 * a+a.

16 =(N2-n 1)* A 8 =(n3-N2)* A 24 =(n3-n 1)* A

So we can know that A=8 or 4, or 2. If it is 8, the number of people left is 1. If a is 4, the remainder is 1, so whether A is 8, 4 or 2, the remainder is 1.

6. Because the sum of digits of No.37 is 10, 57 is 12, 77 is 14, and 97 is 16, we know that the remainder when 10+ 12 is divided by 3 is. The remainder of 12+ 14 is 2, the remainder of 12+ 16 is 1, and the remainder of 14+ 16 is 0, so we know that there will be three games on the 37th, 50.

12—— 16T

1. It takes two typists A and B 10 days to finish a book. We played together for eight days, and the rest was played by B alone. If A typed this book alone, it would take 28 days to finish. How many days have you been asking B?

2. A batch of goods, two cars A and B, can transport 5/6 of this batch of goods in 6 days. If shipped separately, shipment A 1/3 and shipment B 1/2. If they are shipped separately, how many days will it take for A and B respectively?

There are some machine parts. It takes 17 days for A to complete alone, which is more than that for B 1 day. After 8 days of cooperation, the remaining 420 parts were made by A alone. How many parts is a * * *? A * * * for a few days?

4. The swimming pool is equipped with two water pipes, A and B. Both water pipes are opened at the same time, and the swimming pool is filled 12 hours. If tube A is on for 5 hours and tube B is on for 6 hours, only 9/20 of the pool can be filled. How many hours does it take to fill the first tube and the second tube respectively?

1. It takes 28 days for Party A to fight alone, so Party A can complete 1/28 of the task every day, and both parties can complete 10 of the task together, so both parties can complete1165438 of the task every day.

2. It takes 6 days to complete 5/6 of the combined transportation of two vehicles, so 5/36 can be completed in one day. When A has finished 1/3 and B can finish 1/2, then the speed of B is 1.5 times that of A, then A can finish 2/36 of this shipment and B can finish 3/36 every day.

3. Party A can complete117 and Party B can complete116 every day. It takes 8 days to complete 33/34, leaving 420 pieces 1/34, so there are 420 * * pieces in this part. A * * * did14280 * 8/17+420 = 7140, a * * * did 1/34 divided by1/7.

4. Party A and Party B start at the same time 12 hours, and Party A and Party B inject112 every hour. Let Party A inject X every hour and Party B inject Y, 5X+6Y=9/20. The above formula is combined into 5(x+y)+y=9/20, and x+y is on at the same time. So x =112-1/30 =1/20, so it takes 20 hours for single filling and 30 hours for single filling.

17. On the 300-meter-long circular runway, Party A and Party B set off side by side in the same direction at the same time. Party A runs 5 meters per second on average, and Party B runs 4.4 meters per second on average. How many meters before the starting line did they meet for the first time after starting? (List the formulas and work out the answers (comprehensive formulas can be written)

300/(5-4.4)=500 seconds

500 * 4.4 = 2200m

2200 divided by 300 equals 7 laps 100.

So their first meeting after departure was before the starting line 100 meters.

18——20

1. If Xiaohong walks from Zhang Cun to Licun at the speed of15km per hour, she will arrive 24 minutes earlier than originally planned. If she walks 12 kilometers per hour, she will arrive 15 minutes later than originally planned. How far is it from Zhang Cun to Licun?

Suppose it takes x hours from Zhang Cun to Lizhuang.

24 points = 0.4, 15 points = 0.25.

Because the distance is fixed, the speed is inversely proportional to the time.

15×(X-0.4)= 12×(X+0.25)

X=3

The distance from Zhangzhuang to Lizhuang is: 15× (3-0.4) = 39 (km).

2. A bookcase is 88 cm wide, and a certain floor is full of math books and Chinese books, with 90 books. A math book is 0.8 cm thick, and the language is 1.2 cm. How many Chinese books and math books are there?

Set up X math books and 90 Chinese books.

0.8x+ 1.2(90-x)=88

x=50

90 x = 40

50 math books

40 Chinese books

A football match was held in the seventh grade of a middle school. Provisions: win a game with 3 points, draw a game 1 point, and lose a game with 0 points. Seven-year class 1 field, * * * scored 8 points, in which the number of winners was the same, and the negative number was more than the number of winners 1. How many wins, draws and losses are there?

Solution: let the number of winning games be X.

3x+ 1x+0*(x+ 1)=8

4x=8

x=2

Win two games

Ping 2 field

Lose three games