12 Olympiad math questions, write the process clearly

1.81. Find the surface area of the large cylinder, and add the side area of each small cylinder.

2.15 3 (15-10) - 3 6 cm.

4.The solution is that motorcycles should run non-stop, and those who walk should also keep walking.

The fastest way is to take B (or C) to the finish line on a motorcycle, and then return to bring C (or B) to the finish line, which is already halfway there.

The second leg is 52-(52 50)*5, A and C go in opposite directions, and the time is [52-(52 50)*5] (50+5).

During this time period, C travels a distance of 5[52-(52 50)*5] (50+5).

The third section of the journey is the second section of the journey - the distance of C, which is the whole journey of motorcycles, with a total time of 52 50 + (52-52 50*5) (50 + 5) + [52-52 50*5-5 (52-52 50*5) (50 + 5)] 50

How to simplify the intermediate process, please deal with it yourself and calculate the results yourself.

The idea is this, the equation is too long to guarantee the correctness, and I recalculate the equation according to this idea.

5.The speed ratio of A, B, C, D is 6 9 12 15 = 2 3 4 5, that is, when A runs 2 laps, B runs 3 laps, C runs 4 laps, and D runs 5 laps, and the four people meet again at this time, so it takes (1 6) 2 6 = 1 18 hours = 200 seconds from the start to the four people meeting again.

The focus of this question is divisibility and the common multiple is high, firstly, the common multiple is 6;

The common multiple is 30;

Let's start with divisible:

From 1 to 2006, 1003 are divisible by 2;

Of the numbers from 1 to 2006, 668 are divisible by 3;

From 1 to 2006, 401 numbers are divisible by 5;

From 1 to 2006, 334 are divisible by 6;

Of the numbers from 1 to 2006, 66 are divisible by 30;

The first time the lights are pulled: 1,003 lights are turned off and another 1,003 lights are on;

The second time the lights are pulled: the lights numbered as 3 multiples are all extinguished, that is, there will be 668 lights off, but 334 of them are numbered as multiples of 6, and the lights have been turned off in the first pull, so there are still 1003-668 + 334 = 669 lights on for the second pull;

The third time the lights are pulled: then the lights numbered as 5 multiples are all extinguished, that is, there will be 401 lights out, but 66 of them are numbered as multiples of 30, and they have been extinguished after the first and second lights, so the third time the lights are still 669-401 + 66 = 334 lights on;

So in the end there will be 334 lights on!

1 squared 1 2 squared 2 3 squared 3 ....The square of 99 and 99 can be used again.

Write homework at 2:10 and 10 11 at 3:16 and 4 11.

1.The largest number is 35

The smallest composite number is 4, 4, 4, 4, 20, so even numbers with large rents slipped equal to 20 have this property.

Odd numbers greater than 20 can be considered in 4 categories according to the remainder divided by 8, and the smallest composite number divided by 8 by the remainder 1 is 9, so 25 4 4 9, 33 4 6 9....It can be seen that all odd numbers greater than 17 and divided by 8 by 1 are OK.

The smallest composite number divided by 8 by 3 is 27, so 43 4 4 27, 51 4 6 27....It can be seen that all odd numbers greater than 35 and divided by 8 by 3 are OK. Disadvantages of wax.

The minimum composite number divided by 8 and the remainder of 5 is 21, so 37 4 4 21, 45 4 6 21....It can be seen that all odd numbers greater than 29 and divided by 8 by 5 are OK.

The least composite number divided by 8 is 15, so 31 4 4 15, 39 4 6 15....It can be seen that all odd numbers greater than 23 and divided by 8 by 7 are OK.

All can still be expressed in the form of cosams of 4 even and odd composite numbers.

From the above analysis, it can be seen that odd numbers larger than 35 can be represented.

2.No, e.g. 99 99, and a 98, the sum of these numbers is not divisible by 100, no matter how many there are.

1. Natural numbers are divided into remainders divided by 8.

8m+8 8m+9 8m+10 8m+11 8m+12 8m+13 8m+14 8m+15

m>=2, only 8m+11 8m+13 does not have this property.

When m>=3, only 8(m-1)+19 does not have this property.

When m>=4, the natural number can be written as 8 (m-2) + 27, which has such a property.

So the maximum natural number is 8*2+19=35

2. Let the remainders of the 100 natural numbers r1, r2 and so on respectivelyr100.

r1}, where there must be two sets of the same remainder, set to the manuscript search bend.

Therefore, the sum is divisible by 100.

So, such a number can always be found.

Question 2: Can the chain be missing. For example, 100 is divisible by 100, 99 is divisible by 100, and their sum is 100, which is divisible by 100, many, many.

What is the significance of this kind of education, as the knowledge involved in the questions about numbers in the primary school math Olympiad is nothing more than elementary number theory in college?

Solution: There are x students, each of whom has bought Y books. 3x horizontal line books, x field character books, and 5x exercise books were issued. According to the title, it can be obtained:

y -3x = 24, the total number of Tian character books and exercise books left is: 2y - x - 5x ) = 48 books.

Solution: Let the reverse water velocity be x kilometers per hour, and the distance from A to B is y, then take 2 more (y-x) = 6, and the total time is y x + y (x+8) = 2

Lianlid: distance y=12

Hope it helps!

If there is a certain point where the car is being greeted, the car is chased, and the person meets at the same time, the next car and car chase are considered.

The distance between the car and the person in 6 minutes is 2 times the distance of the person in 6 minutes, and the distance between 6 minutes and 9 minutes is 2 times the distance of 6 minutes + (9-6) minutes, so that the ratio of the speed of the person to the speed of the car is 3 [2*6 + (9-6)] = 1 5

Therefore, the departure interval is 6*(1+1 5) or 9*(1-1 5)=minutes.

The speed of the car is x, the speed of the person is y, and the interval between the departure is t and the distance between the second car and the person is xt when it meets the first car

After meeting the first car, continue walking, and the distance between the person and the car when meeting the second car is 6(x+y)=xt

When the first car catches up with him, the distance between the second car and the person is also xt, and the distance that the second car catches up with him minus the distance traveled by the person is the distance of the original difference xt=9(x-y).

So we get: 6(x+y)=9(x-y) to get x=5y, and substitute x=5y into 6(x+y)=xt

The solution gives t = 36 5 (minutes), which is 7 minutes and 12 seconds.

The following units are kilograms, and only the weight of the container is considered to be a whole kilogram

19630 = 2 * 5 * 13 * 151, the product of the multiplication of the factors is less than 350, that is to say, the weight of each box of goods can only be one of the above weights. When loading, try to load as many boxes as possible into the car, so that the car is fully loaded, that is, the weight of the cargo box is multiplied by a certain integer to make it closest to 1500, respectively. When the weight of the goods on each car is the smallest, the most cars are needed, that is, 1208 is taken, each box of goods weighs 302 kg, a total of 65 boxes, each car is loaded with 4 boxes, a total of 17 vehicles are needed.

I don't know if I have a clear expression, and the paintings that are interested in me can hi me.

Actually decompose the prime factor and unit agreement problem.

Tons = 19,630 kg = 151 x 13 x 5 x 2 The maximum weight is 151 x 2 = 302 kg (i.e. 302 kg x 65 = 19,630) because each box does not exceed 350 kg (and the same weight).

The largest number of cars is needed at this time, 302*4=1208 kg= tons less than tons.

Cars required: 65 = 4*16+1 i.e.: 15 cars.

Answer: When each box weighs 302 kg, you need the most cars, and you need up to 15 cars to be happy: haha This question is a little bit of a problem If each box of goods is very light and lighter than sand It's okay I'm a big money I have nothing to do One box and one car What's wrong.

Solution: 19630 kg = 151 x 13 x 5 x 2 Since each box does not exceed 350 kg (and the same weight), the maximum weight is 302kg (i.e. 302kg x 65 pieces).

At this time, the largest number of cars is needed, 302*4=1208 kg= tons.

Cars required: 65 = 4*16+1 i.e.: 15 cars.

A: When each box weighs 302kg, you need the most cars, up to 15 cars.

Tons = 1500 kg.

Tons = 19,630 kg.

Find an approximate of 1500 and less than 350; The last 325, i.e. 325 kilograms per box of goods, 19630 divided by 1500 to get 13 more than 130

That is, 14 cars are needed.

The weight of each box may be kilograms, a total of four types.

The answer should be the kind of 151.

Each car is loaded with 9 boxes, and a total of 130 9= cars are needed

Equal to 15 cars.

It depends on how big the box is and how big the volume of the deadweight ton car is, 1kg per box, but the volume of 1 cubic meter per box is definitely more than the volume of 10kg per box and 2 cubic meters per box.