Olympiad problem solving requires a detailed process, the first and second problems need to be solv

1.Solution: Let the speed of car A be 7x, then the speed of car B is 11x, and the time t will elapse for the two cars to meet for the second time.

The equation is: 11xt-18x) + (7xt-18x) = 18x7xt-18x = 80

Solution: x=80 3 So the distance between the two places is 18x=18 (80 3)=480 km.

2.Solution: There are already x people queuing up before the ticket check, the number of people coming every minute is y, and the ticket checking speed of each ticket gate is z.

The equation is: x+30y=5*30z; x+20y=6*20z: x=60z, y=3z

If you want the queue to disappear in 10 minutes, you need to open n ticket gates.

then x+10y=n*10z

Substituting the relationship between x, y and z above, we can eliminate z, and then solve the value of n as 9

The third question, assuming that the commodity ** is X yuan, then the buyer gives X 7 yuan, and the seller finds him X 6 yuan, so you can always find it!!

1.If AB and the two places are separated by x km, then A has taken x+80 km. B walked 2x-80 km.

x+80）/7=（2x-80）/11

x=480

Grasp the center of the problem and sort out all the known conditions yourself, preferably with equations.

Olympiad mathematics requires a flexible mind, but in fact, there is still a certain mode of thinking, but it is not a formula. You try to look at the questions in each category.

1) You may want to set z>y>x

then x>=1

y>=2

z>=3

So a<=1 1+1 2+1 3<2

So a=1 then x>=2

If x>=3, then 1 x+1 y+1 z<=1 3+1 4+1 5<1 so x=2

And then apparently there is.

y=3 z=6

2) Solution: 1 x+1 y+1 z<=1+1 2+1 3<2, i.e., k=1, if x>=3, 1 x+1 y+1 z<1, i.e., x=2, i.e., 1 y+1 z=1 2

When y>=4, 1 y+1 z<1 4+1 4=1 2, i.e., y=3, is obtained as z=6

Actually, I don't know, hehe.

1. All settings are 100000+x, (x is a five-digit number).

Then move the number 1 to the far right, and the resulting six-digit number is 10x+110x+1=3 (100000+x).

10x+1=300000+3x

7x=299999

x=42857

The original number 142857

Equation solution: Let the number of the last 5 digits be x

Get it according to the title.

100000 + x ) 3 = x * 10 + 1 so 300000 + 3x = 10x + 17x = 299999

x=42857

So the original number is 142857

Let the other five digits be x, then 3 (100000+x) = 10x+1

x=42857

Let : x=1*10**5+y x,y is an integer according to the title: 3x=3*10**5+3y=y*10+1So: 3*10**5-1=10*y-3*y=7yy=42857

x=1*10**6+42857=142857

1.Teacher Zhang walked from home to school for 3 minutes at a speed of 80 meters per minute. At this rate, you will be 3 minutes late, but if you change to 110 meters per minute from now on, you can arrive 3 minutes early. How far is Ms. Zhang's house from the school?

Answer: If you walk at different speeds twice, you will walk 80*3=240 meters less in the specified time in the previous time.

After the specified time, walk 110 * 3 = 330 meters, and the difference between the two actual walks is 570 meters.

The speed difference between the two is 110-80-30 meters, so the specified time is 570 30 = 19 minutes.

The distance between schools is 80 * (19 + 3) or 110 * (19-3) = 1760 meters.

22.Use a rope to measure the depth of the well platform to the water surface, fold the rope in half once and hang it to the water surface, the rope exceeds the well platform by 4 meters, fold the rope in half twice and hang down to the water surface, the upper end of the rope is located 3 meters below the well platform, so how many meters is the rope long? What is the distance from the well platform to the water surface?

Answer: Consider the first time I take two ropes and fold them in half once, the question becomes 2 ropes folded in half at a time to measure 4 wells, more than 4 meters.

The second time I took a rope and folded it in half twice, which means that 1 rope was folded in half twice to measure 4 wells, less than 3 meters.

The difference between the first and second time is 4*4+3*4=28 meters, in fact, the difference between the first and second time is the length of a rope, that is, the length of the rope is 28 meters, so the distance from the well platform to the water surface is (28-4*2) 2=10 meters.

x@3=x(x+1)(x+2), let (x@3)=y, (x@3)@2=(x@3)[(x@3)+1]=(x@3) +x@3)=y+y=3660

y²+y-3660=0

y²+y-60×61=0

y+61)(y-60)=0

y+61=0, get y1=-61, not a natural number, can't do a new operation y-60=0, get y2=60, yes.

x(x+1)(x+2)=60

The multiplication of the three adjacent natural numbers equals 60, and the three numbers are 3, 4, and 5 respectively, so x=3

If the number of students in School B is x, then the number of students in School A is 2x 5

School A girls are: 2x 5 3 10=3x 25 students.

School B has boys: 21x 50.

Then the girls in school B have x-21x 50=29x 50, and the girls in the two schools have 3x 25+29x 50=35x 50, and the girls in the two schools have x+2x 5=7x 5

The total number of female students in the two schools is 35 50 (7 5) = 50%.

Solution: The number of students in school B is x

Then the number of students in school A is 2x 5, and the number of girls in school A is 2x 5 3 10, because the number of boys in school B is 21x 50, so the number of girls is 29x 50, and the total number of students in the two schools is x+2x 5

The ratio of the total number of girls in the two schools is (2x 5 3 10 + 29x 50) (x + 2x 5).

Solution 1 2

The relationship between the distance of the snail from the origin s1 and the number n is: s1 = (n + 1) 2 (unit length) (n is an odd number).

Or: s1=-n 2 (unit length) (n is an even number).

The relationship between the crawling distance of the snail s2 and the number of times n is: s2=(1+n)n2 (unit length).

1. Because 100 is an even number, s1=-100 2=-50 (unit length), that is, 50 unit lengths to the left of the origin.

2. According to 1, the snail passes 100 times to point C, and the crawling distance of 100 times is (1+100)*100 2=5050 unit length, 5050 2=2525 minutes.

2 = 120 unit length, 120 = (1 + n) n 2, solution n = 15

Because 15 is an odd number, s1=(15+1) 2=8, i.e., after 1 hour, the snail is 8 units of length to the right of the origin.

-100 2 = -50 units of length.

100 + 1] * 50 = 505 units of length = 5050 2 = 2525 minutes.

60 * 2 = 120 unit length = positive 8 [the 15th positive walk to the positive 8 length unit length, the cumulative stroke is 120 unit length, and it is at the positive 8 punctuation point at this time].

I'm sorry to change it many times.

The total number of the third monkey = 2 * 3 = 6;

The total number of the second monkey = (6 + 2) * 3 2 = 12, and the total number of the first monkey = 12 (3 2) = 18;

The number of the first little monkey = 18-12 = 6;

There were 18 peaches;

1.The third monkey divided the rest of the peaches into 3 piles, took 2 of them, and finally left 2 of them.

Explain that the second monkey has 6 more after taking it.

2.The second monkey divided the remaining peaches into 3 piles, and took away 2 more peaches;

Illustrate that the second monkey is before the score is 12 more.

3.The first monkey divides the pile of peaches into 3 piles and takes one of the piles, indicating that there are 18 peaches at the beginning.

There are x peaches in total, the first monkey takes x 3 peaches, and there are 2x 3 left, and the second monkey is divided into 3 piles, each pile has 2x 3*1 3=2x 9, and (2x 9+2) peaches are taken, and there are 2x 3-(2x 9+2)=4x 9-2 left;

The third monkey divided into 3 piles and took 2 piles, leaving 1 pile, and 1 pile is 2 peaches, so 1 3 * (4 x 9-2) = 2

4x/9-2=6

4x/9=8

4x=72x=18

So there were a total of 18 peaches, and the first little monkey took 1 3 piles, that is, 6 peaches.

(2+2) 3 2 3 2 3 = 27.

Problem solving ideas: Thinking backwards, starting from the third monkey, a bunch of peaches take away 2 and there are 2 left, which should be (2+2)=4, three piles of 12 in total; That is, after the second monkey takes away one pile, the remaining two piles, one pile is 6, and the three piles are 18 in total; That is, after the first monkey took away one pile, the remaining two piles, one pile was 9, and the three piles were 27 for a total of 27.

Suppose there are a peach, then.

After the first monkey has taken it, there is a 2 3 left, divided into three parts, each part is (a*2 3) 3 After the second monkey has taken it, a 2 3-a*2 3*1 3-2=a*4 9-2, divided into 3 parts, each part is (a*4 9-2) 3

After the third monkey took it, there was (a*4 9-2) 3=2 so a=18, and the first monkey took 6.

It turns out that there are x peaches. The first little monkey divided the pile of peaches into 3 piles and took one of them, leaving 2 3*x pieces. The second little monkey divided the remaining peaches into 3 piles, and took away a pile of more than 2, leaving 2 3*2 3*x-2.

The third monkey divided the rest of the peaches into 3 piles, took 2 of them, and left 1 3* (2 3*2 3*x-2). 1/3*（2/3*2/3*x-2）=2。The solution is x=18

The first monkey walked 1 3 * 18 = 6 pcs.

2*3=6 (6+2)*3 2=12 12(3 2)=18 18-12=6.

There are eighteen howls

Analysis: We know that if the number obtained by adding all the numbers of a number can be divisible by 3, then this number can also be divisible by 3, and the operation in the problem is exactly like this, and 6 can be divisible by 3, so all "good numbers" can be divisible by 3, but not by 6 at the same time, so the greatest common divisor of these "good numbers" is 3

In other words, these "good numbers" arise from multiples of 3, and now let's see what are the rules of the operation in the multiples of 3 execution problem, from 3, 6, 9, 12, 15, 18, 21 ,......, and what you get from the operation in the problem to the end is that 3, 6, and 9 appear alternately, and the 6 is the required "good number", that is, there is only one "good number" in every 9 consecutive natural numbers.

So we can get that all "good numbers" can be expressed as 6+9n, n=0, 1, 2,......The maximum value of n obtained by 6+9n<2012 is 222, so there are 223 "good numbers" that do not exceed 2012.

So the answer to the two blanks in the question is 223, 3