1. The sum of three numbers is 555. These three numbers can be divisible by 3, 5 and 7 respectively, and the same is true for manufacturers. Find these three numbers.

Thinking: Let the quotient be x, then: 3x+5x+7x = 555, and the solution is X = 37.

So the three numbers are: 37× 3 =11,37× 5 = 185, 37× 7 = 259.

2. It is known that A is a natural number, which is a multiple of 15, and there are only two numbers in it: 0 and 8. What is the minimum quantity of a?

Thinking: 15 = 3× 5, so if a natural number is a multiple of 15, it must be divisible by 3 and 5 at the same time.

A number divisible by 5 can only have 5 and 0 at the end, so the end of A is 0.

The sum of digits of a number divisible by 3 is a multiple of 3, so there are at least three eights.

So, a = 8880.

3. Arrange natural numbers into the following array:

1,2,4,7,…

3,5,8,…

6,9,…

10,…

…

Now it is stipulated that the horizontal line is a row and the vertical line is a column. ask

Thinking: The continuous natural numbers from the upper right to the lower left are defined as "stripes", that is, the first stripe is 1, the second stripe is 2,3, and the third stripe is 4,5,6. ...

It is not difficult to find that the number of rows and columns of natural numbers in the same article add up to the same.

What is the number in the fifth column of (1) 10?

The 1 line of the article where the fifth column of 10 is located should be in the 14 column of (10+5- 1 = 14). Therefore, there is+1+2+3+... 1 2+13 = (1+13) ×13 徉

(2) What is the number in column 10 in line 5?

Row 5, column 10 is also in 14, so this number is 92+4 = 96.

(3) In which row and column was it ranked in 2004?

Because (63+1) × 63 ÷ 2 = 2016 >; 2004; (62+ 1)×62÷2 = 1953 & lt； 2004

So in 2004, in article 63. Article 63: 1 behavior (62+1) × 62 ÷ 2+1=1954. In 2004, the line 1 2004- 1954+65438+ was included. The number of columns is 63+1-51=13. So 2004 is in 5 1 row and 13 column.

4. The product of three prime numbers is exactly equal to 1 1 times of their sum. Find these three prime numbers.

Thinking: The product of three prime numbers is 1 1 times of the sum, so 1 of three prime numbers is 1 1.

Let the other two prime numbers be x and y, then XY = X+Y+ 1 1.

Y = (x+11)/(x-1) ≥ 2, and the solution is 13≥x≥2.

Substitute x = 2, 3, 5, 7, 1 1 3 respectively, and the three prime numbers are 2, 1 1, 13 or 3, 7,/kloc-0 respectively.

5. There are two integers, the sum of which happens to be two numbers with the same number, and the product of which happens to be three numbers with the same number. Find these two integers.

Idea: Start with the same three-digit number.

1 1 1 = 37× 3, 37+3 = 40.

222 = 37× 6 = 74× 3, 74+3 = 77.

333 = 37 × 9, 37+9 = 46.

444 = 37× 12 = 74× 6, 37+ 12 = 49, 74+6 = 80.

555 = 37× 15, 37+ 15 = 52.

666 = 37× 18 = 74× 9, 37+ 18 = 55; 74+9 = 83 Give up

777 = 37× 2 1, 37+2 1 = 58.

888 = 37× 24 = 74× 12, 37+24 = 6 1, 74+ 12=86.

999 = 37× 27, 37+27 = 64.

The two integers that match the meaning of the question are 3,74 or18,37.

On the 6.800-meter roundabout, a colorful flag was inserted every 50 meters, and later some colorful flags were added to shorten the interval of colorful flags, and the colorful flags at the starting point did not move. After re-plugging, I found that the colorful flags on all sides did not move. How many meters are the intervals of colored flags now?

Idea: After the distance is shortened, the colored flags at the new interval distance and the common multiple of 50 do not need to be moved.

800 ÷ 4 = 200, and the colorful flags will not move every 200 meters. 200=2×2×2×5×5=50×4

So the spacing can be: 4× 2 = 8m, or 4× 10 = 40m.

7. 135 1 1, 13903, 14589 divided by the natural number m, the remainder is the same. What is the maximum value of m?

Thinking: Let the remainder be a and the quotient be x, y and z respectively.

So:

mx+a= 135 1 1

my+a= 13903

mz+a= 14589

Subtract three formulas:

m(y-x)=392=2×2×2×7×7

m(z-y)=686=2×7×7×7

m(z-x)= 1078 = 2×7×7× 1 1

So m can be 2× 7× 7 = 98 at most.

135 1 1÷98= 137……85

13903÷98= 14 1……85

14589÷98= 148……85

8. 1 to 200 How many natural numbers are not divisible by any of 2, 3 and 5?

Idea: There are 200 divisible by 2 ÷ 2 =100; There are 200 divisible by 3 ÷ 3 = 66...2,66; There are 200 ÷ 5 = 40 divisible by 5.

200 ÷ 6 = 33...2, 33 is divisible by 2 and 3 at the same time; 200 ÷ 10 = 20 is divisible by 2 and 5 at the same time; There are 200 ÷ 15 = 13...5, 13 are divisible by 3 and 5 at the same time.

200 ÷ 30 = 6...20 can be divisible by 2, 3 and 5 at the same time.

So a number with100+66+40-33-20-13+6 =146 can be divisible by 2, 3 or 5.

Then there are 200- 146 = 54 numbers that match the meaning of the question.

9. There is a list of numbers: 1, 999,998, 1, 997,996, 1, … Starting from the third number, each number is the difference of the first two numbers. Find the sum of 999 numbers from number 1 to number 999.

Thinking: every 3 numbers are regarded as a group, then the 999th number is 999 ÷ 3 = 333 groups.

The numbers in each group are 1 and two adjacent natural numbers, so by the 333rd group * * except 1, it is 333× 2 = 666 numbers.

The number 1 is 999, the second is 998, the third is 997, ... 666 is 334.

Their sum: (334+999) × 666 ÷ 2+1× 333 = 444222.

10.200 to 1800 How many natural numbers have odd divisors?

Thinking: Any natural number can be expressed as the product of two natural numbers, including prime numbers, which are the product of itself and 1. In other words, the divisors of a number appear in pairs. There is only one special case, that is, two divisors appearing in pairs are equal, that is, the number is a complete square number.

14× 14 = 196 & lt； 200

15× 15 = 225 & gt； 200

42×42 = 1764 & lt； 1800

43×43 = 1849 & gt； 1800

So it is the square of 15 to 42, and a * * * 42- 15+ 1 = 28.

1 1. In the figure below, there are two isosceles right triangles with the same area, both of which are 100. Cut two small squares along the dotted line in the picture. Please find the area of each square and compare the sizes.

Tu Tu, Tu Tu, Tu Tu, Tu Tu, Tu Tu, Tu Tu, Tu Tu, Tu Tu, Tutu, Tutu, Tutu, Tutu.

12. Party A said, "I have 100 yuan with Party B and Party C." Party B said, "If Party A's money is six times as much as it is now, my money is13, and Party C's money remains unchanged, the three of us still have 100 yuan." C said, "My money is not even 30 yuan." Ask them how much money each of them has.

Idea: This question seems to miss the condition: the money of Party A, Party B and Party C is an integer.

Suppose a has x yuan and b has y yuan.

Then x+y=6x+y/3, x = 2y/ 15, where y is a multiple of 15.

Party A, Party B and Party A each have X+Y yuan, namely 17y/ 15.

So C's money: 0

Solving inequality: 6 1.76

So y = 75 and x = 10.

The amount of money for three people is as follows: A 10 yuan, B 75 yuan, C 15 yuan.

13. Two people are going to explore the desert. They go deep into the desert for 20 kilometers every day. It is known that each person can carry one person's food and water for up to 24 days. If some food is not allowed to be stored on the way, how many kilometers can one of them go deep into the desert (the last two have to return to the starting point)? What if some food can be stored on the way back?

Ideas:

Case 1: Don't store food on the way.

After A and B set off, A will give B as much food as possible to ensure that B goes far.

Suppose A delivers food to B X days after departure, and then returns immediately.

Then A consumes X days of food and needs X days of food to return. The amount that can be given to B is (24-2x) days of food.

For B, it has consumed X days of food, so it can supply X days of food at most.

So there is equation 24-2x = x (what A can give to B is what B consumes the most).

Solve x = 8, that is, 8 days after A and B set out together, A gives B8 days of food and then returns by itself. In this way, B * * * can eat food for 32 days, one way 16 days, and can go deep into the desert for 20× 16 = 320 kilometers.

Situation 2: Food can be stored on the way.

After A and B set off, A will give B as much food as possible to ensure that B goes far.

Suppose a sends food to b x days after departure, and then comes back by himself.

Then A consumes X days of food and needs X days of food to return. The amount that can be given to B is (24-2x) days of food.

For B, food has been consumed for X days. In order to ensure that you go far, you need to bring more food, but you need to ensure that you return to the starting point, so you need to leave enough food in the same place for returning when you get A's supply, that is to say, A can give B up to 2x days of food (X days of food that B has consumed and X days of food that B needs for returning).

So there is equation 24-2x = 2x.

Solve x = 6, that is, after starting together for 6 days, A gives B 12 days of food, and B leaves 6 days of food in the same place waiting to return. In this way, B * * * can eat food for 36 days, one way 18 days, and can go deep into the desert for 20 × 18 = 360 kilometers.

14. Bonuses are divided into first prize, second prize and third prize. The bonus of each first prize is twice that of each second prize, and the bonus of each second prize is twice that of each third prize. There are two people in the first, second and third prizes, and each first prize is 308 yuan; If there are/kloc-0 first prize, 2 second prizes and 3 third prizes, what is the bonus of the first prize?

Idea: If the third prize is X yuan, then the second prize is 2x yuan and the first prize is 4x yuan.

Total amount = 308 ÷×××× (x+2x+4x )× 2 =1078 yuan.

According to the new distribution method, the first prize is1078 ÷ (3x+2× 2x+4x )× 4x = 392 yuan.

15. Divide 1296 into four numbers: A, B, C and D. If A adds 2, B subtracts 2, C multiplies 2, and D divides 2, the four numbers are equal. What are these four numbers?

Thinking: Let four numbers equal X, then A is X-2, B is X+2, C is x/2, and D is 2x.

Meaning: x-2+x+2+x/2+2x = 1296.

9x/2= 1296

x=288

The four numbers are: 286290 144576.