The master of the Olympiad is coming

If there were x tons of grain in warehouse A, there would be 2,600-x tons in warehouse B.

1-3/5)(2600-x)-(1-3/4)x=130x=1400

2600-x = 2600-1400 = 1200 x hours.

2(1/12+1/8-1/36)+(1/8-1/36)x=2/3x=22/7

A x, B y

x+y=2600

x/4+130=2y/5

x=1400

y=1200

A speed 1 12, B speed 1 8, C speed 1 36

2 3-2 (1 12 + 1 8-1 36) (1 8-1 36) xx 22 7 hours.

1 If you think: after transporting away, A and B have the same amount of 2600 + 130 = 2730 (tons), then A's 5 20 is the same as B's 8 20, knowing the total amount and number of copies, I think six years should be calculated, you should pay attention to the end B subtract 130.

2 The efficiency of A is 1 12, the efficiency of B is 1 8, the efficiency of C is 1 36, the efficiency of all at the same time is 1 12 + 1 8-1 36 = 13 72, the efficiency of 2 hours is 2 13 72 = 13 36, 2 3-13 36 = 11 36, the efficiency of B and C is 7 72, 11 36 7 72

This kind of question also needs an Olympiad master? Dizzy.

1. Because Xiaodong is only short of 1 cent.

So this book can only be 2 yuan and 7 jiao.

Xiaodong has money, 2 yuan, 6 jiao, and 9 points.

Fang Fang has no money.

Because there is only one identical one for each location.

It can be found that only the hundred digit is not repeated, and the others are repeated, so the hundred digit is 7

2 out of 10 6

Because the one with 6 has the same hundred place with 765.

So it's 2 on the tenth

The next single digit can only be 4.

The article number is 724

Xiaodong lacks 1 penny, and the money of the two is not enough to buy together, so Fangfang doesn't have 1 penny. The book ** is 2 yuan and 7 corners.

724。Note that there are no duplicates in the hundreds, 2 6s in the 100s, and 2 5s in the 100s

1. Set up the book x corner.

Fangfang Xiaodong X-27

Add x to the two

The price of the book is at least a corner.

Thank you.

1. Through the question, we can know that the total walking distance and time of the two classes are the same. So let's assume that A takes the car first and then walks, and B walks first and then takes the car.

Suppose that A's distance in the car is 1, then B's distance on foot should be 1 15 (because the ratio of their speeds is 4 60), and in the remaining distance of 14 15, B and the car become the meeting problem. When the car returns, B is still on foot until they meet, and B can get into the car. So how much did B go on the 14-15 journey?

Then the total distance traveled by B is 7 120 + 1 15 = 1 8

Note here that we assume that the distance of A is 1, that is, the distance of B is also 1, and the distance traveled by B is 1 8In other words, the ratio of walking distance to riding is 1:8

Then the total time spent: 27 9 4 + 27 9 * 8 60 = hours = 1 hour and 9 minutes. That is, it can arrive at 9:09.

2. It is easy to see that the remaining 5 12 (1-1 3-1 4) of the original 1 and 2 classes are 30 people, so the total number of 1 class + 2 class is 30 5 * 12 = 72 people, the number of people in the new class is (72-20) (11 (11 + 10)) = 22 people, and the number of people in the new second class is 20 people. The number of people in the new 1st class - the new 2nd class is 2, which is equivalent to 1 12 of the original 1st class - 1 12 of the original 2nd classThen the original 1 class - the original 2 class = 24 people.

The number of students in the original class is (72+24) 2=48.

3. Because the pond is round, the distance between Xiao Ming and Xiao Ying should remain the same. 20-7 = 13, 94-7 = 87, the reason why the two numbers are different is that the proof is a matter of direction. One is the difference of the small semicircle, and the other is the difference of the large semicircle, so it adds up to the number of trees in the entire pond:

4. According to the report, the number of children at the beginning can be divisible by 11, and when the number of children is added by 1, it can be divisible by 9. The title "Adding a box of gummies to a box happens to have the same composite number of two pieces of sugar" means that the quotient of this number divisible by 11 is smaller than the quotient divisible by 9 after adding 1 to this numberIt can be estimated that the number is 44

Then there were two more children. So the final number of children is 44 + 2 = 46

Let the initial gummy x box be 11x+1=9 (x+1), and the solution will be x=4, and finally the number of children will be 4*11+1+1=46

3-2=1

All small cubes have a total of 1 + 3 +19 without red paint on all six sides, 1 +3 +17 with red paint on at least one side: 19 = 6859.

It's a simple question, but it's a lot of calculations, and I say that at least one side of the idea = all of them - no red on one side.

First of all, calculate how many small cubes with an edge length of 1 can be sawn into the total number of small cubes with an edge length of 1 = 1*1*1+3*3*3+5*5*5+7*7*7+.19*19*19

Then there is no red 1*1*1 on one side, there is no 3*3*3 in 1 (i.e., 1*1*1), 5*5*5 has 27 in it (i.e., 3*3*3), I don't know if you can understand it.

Then the total number of colors at all = 1*1*1+3*3*3+5*5*5+7*7*7+.17*17*17 (because 19*19*19 is 17*17*17 colorless).

Subtract the two, and in the end, at least one of the answers is 19*19*19=6859, and then ask me if you don't understand, ask me to adopt!

From the beginning of the edge length of 3, there is 1 central cube without paint on one side, ·· 5··· There are 27 of them...

And so on, the nth has (n-2).

There is no lacquer. At least a cube with red paint on one side = 1 +3 +·19 - (1 +3 +·17 )

There is a calculation formula, but I am too lazy to calculate, and the formula is a bit forgotten--

Topic.

Either find a number, or 4 or more even numbers!

The number of 2010 certainly meets the requirements.

One must be satisfied!

Suppose 4 consecutive satisfies from a, i.e., 4a+6=2010 a=501 must be satisfied.

Let sn=a1+a2+. an, then in the value of sn mode 2010 (take 0 to 2009), if one of them is 0, then that si is the set of numbers that meet the conditions. If there is no 0, then there must be the same remainder (because the value of 2010 numbers is from 1 to 2009), you may wish to set si, sk congruence, and k>i.

Then SK-Si is divisible by 2010, AI+1, and the number between AK satisfies the condition.

First of all, after the original simplification.

1x2x3 3x2x5 5x2x7 7x2x9 9x2x11 11x2x13

1/2（1/3+1/3x5+1/5x7+1/7x9+1/9x11+1/11x13）

1/4x(12/13)

You can go and search for the formula of the split term and you will understand that I am almost off work and I don't have time to answer.

You can't do the math with a calculator.

The y power of x + the x power of y is expressed as x y + y x If x, y has a greater than or equal to 1, then obviously, otherwise, x < 1, y<1 first regards x as an unknown and y as a constant, we find the extreme value of x y + y x, if the extreme value is greater than 1, then it must be greater than 1

Derivative, the extreme point satisfies y*x (y-1)+y x*lny=0 and substitutes x y+y x to obtain -x*y (x-1)*lny+y x, that is, y x-y (x-1)*ln(y x).

To prove that this equation is greater than 1, note that from the assumption of x<1, y<1, y (x-1)>1 and then assume y x<1 (otherwise there is no need to prove it), then ln(y x) < 0, and after detracting y (x-1), only prove y x-ln(y x)>=1.

Let t=y x, 0=1

That is, y x+x y>1

Is your x-power y+y, or is it the y's behind alone?

Well, it's very troublesome to verify, and it's even more troublesome to enter, so you'd better give some reward points

If that's the case, then x y is an integer.

And because it's positive, then the smallest x y is 1.

Don't you understand that?