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There should be a hidden condition in the question, and there should be exactly one tree in each corner.

To plant trees at the maximum distance, then the maximum distance is 532 and 308 greatest common divisors.

That is, 28, then the circumference of the land is (532+308) 2=1680, and because the total distance is sealed.

The total number of trees that can be planted is direct.

1680 28 = 60 trees.

When there are 208 students in the dormitory, when there are 350 students in the dormitory with a total of 4 2 8 vacant bunks, and if there are 2 vacant berths in the dormitory, the greatest common divisor of 208 + 8 216 and 350 + 2 352 is 8, so each dormitory has a maximum of 8 bunks.

1.You can find the greatest common divisor of 532 and 308, and the distance is 4 meters. Count how many trees you have.

2.If there is a maximum of x bunks per dormitory, (208+2 4) x is an integer, and (350+2) x is also an integer.

That is, find the greatest common divisor of 216 and 352 and calculate it yourself.

1. Find the greatest common divisor of 532 and 308 is 28 (the distance between two trees) 2. Calculate how many trees can be planted on each side of the length and width 532 28-1=18 308 28-1=10

3. How many trees are planted in total (18+10)*2 56

Question 2. 1. The number of bunks that have been accommodated this semester: 2*4 2 62, the number of students who have stayed in this semester is 350 208 1423, and the number of students in each bunk is 142 6=24

4. There are a total of 208 24 = 8 more than 16 16 people occupy two bunks.

5. Each berth 8 2+1+4=9

Solution: Let the teacher have x computers and students have (50-x) computers.

5000x+3000 (slag 50-x) = 1580005000x + 150000-3000x = 158000150000 + 2000x = 1580000

2000x=158000-150000

2000x=8000

x=4, so lead field 50-x=46

Set up the purchase of student computer x

3000*x+(50-x)*5000=158000 to get x=46

That is, there are 46 sets of students' electric suspicion and 4 sets of teachers.

Suppose the teacher has x computers, and the students use 50-x computers, 5000*x+3000(50-x)=158000

Set up the teaching hall to dress up and buy the A platform, and the Qingyu students (50-A) platform.

5000a+（50-a）3000=158000

A = 46 teachers: 50-46 = 4 (sets).

1.Only when one odd and one even, the sum is an odd number, so from 1 to 99 of the 99 numbers, 50 odd numbers, 49 even numbers, choose two different numbers, so that the sum is an even number of the total number of selections is c50 2 + c49 2 = 2401 (or c100 2-[c50 1] [c49 1]).

2.With the baffle method, it is all eaten up in one day, C9 0 = 1 kind.

One plate (2 days) when there is C9 1

2 pcs., C9 2

.A total of c9 0 + c9 1 + c9 2 +...c9 9 = 2 9 = 512 species.

Odd + Odd = Even.

Even + Even = Even.

Odd + Even = Odd.

Even plus odd = odd.

There are 50 odd numbers in 1-99 and 49 even numbers.

So a total of (50 49 + 49 48) 2 = 49 2 = 2401 species.

A method for solving indefinite equations (interpolation).

c99+c98+c97+c96+……c91 + c90 = 2 9 = 512 species.

Hello: There are 6 possibilities for randomly drawing two shoes, 3 of which are a pair of shoes.

So the probability of randomly drawing one pair of shoes from three identical shoes is 50%.

1 will be randomly selected, regardless of left or right, taking the left as an example.

Then there are 5 left, of which 3 are on the right and 2 are on the left.

Then the probability of drawing to the right is 3 5

So the probability of randomly selecting one pair of shoes from three identical shoes is 60%.

There is a 100% chance that one pair of shoes will be randomly selected from three identical pairs

The probability is 3 out of 5, and after taking a shoe out, there is only a 3 out of 5 chance of taking another one out to match.

First of all, take any one and then leave 2 out of 5 so the probability is 40%, if you use the formula to calculate 2 . c = 15 and then the correct even number is 6 (there are two ways to take a pair) and the same result is 40%.6

If the sum of two numbers is known to be negative, then (c).

a.Both numbers must be positive. b.Both numbers are negative. c.At least one of the two numbers is negative. d.Two numbers must be one positive and one negative.

If 丨x丨+(-1(2 3))=1, then x is equal to (b).

or -2 3. or -2 ()2 3. or -1 3. or -1 (1 3).

If 丨m丨=3 and 丨n丨=5, then the value of m+n must be (d).

c.-8。 d.None of the above conclusions are correct.

The sum of the three numbers -12, -2, +7 and is smaller than the sum of their absolute values (d).

a.-4。 。c.-28。 。

In the numbers 1, -1, and -2, the maximum value of the sum of any two numbers is (b).

c.-1。 d.-2。

If a+b = 0, then the value of the real numbers a,b must be (b).

a.It's all 0. b.are inverse numbers to each other. c.There is at least one 0. d.Each other is the reciprocal.

In a football round-robin match, the red team beats the yellow team 4:1, the yellow team beats the blue team 1:0, and the blue team beats the red team 1:0, then the goal difference of the red, yellow, and blue teams is [A] respectively.

If the sum of the two numbers is less than each of the added numbers, then it must be [b].

a.Both additions are positive. b.Both additions are negative. c.The symbols for the two additions are different. d.One of the two additions is zero.

There are the following sayings: When the sum of two numbers is positive, both numbers are positive; The sum of two numbers is negative, and both numbers are negative; The sum of two rational numbers may be equal to one of the additives; The sum of two rational numbers may be equal to 0, where the correct one is [ b ].

Piece. Piece. Piece. Piece.

The absolute value of the sum of two rational numbers is equal to the sum of their absolute values, then [a].

a.Both of these rational numbers are positive. b.Both of these rational numbers are negative. c.These two rational numbers have the same name. d.These two rational numbers have the same sign or at least one of them is zero.

The following statement is true.

a.When two numbers of the same sign are added, their sum is greater than the addition number. b.The sum of the two numbers is equal to the addition of their absolute values. c.The sum of the two numbers is 0. d.The sum of two positive numbers is positive, and the sum of two negative numbers is negative.

Solution: Set: There are x books of science and technology, then there are 3x storybooks, and there are x-80 books.

According to the meaning of the topic and known, there are:

x+3x+x-80=620

5x=700

x=140 (book).

3 140 = 420 (Ben).

140-80 = 60 (Ben).

A: There are 140 science and technology books, 420 story books, and 60 books.

If there are x science and technology books, then the storybook is 3x, ***(620-x) x-(620-x)=80

2x-620=80

2x=700

x=3503x=350*3=1050 copies.

620-350 = 270 copies.

There are 350 science and technology books, 1050 story books, and 270 books.

Solution: There are x science and technology books.

x+x - 80 = 620

2x - 80 = 620

2x = 700

x = 350

: 350 - 80 = 270 (books) Storybook: 350 3 = 1050 (books).

If there are x science and technology books, then it can be obtained from the title: x+3x+x-80=620, and the solution is x=140, then 3x=420, x-80=60, so there are 420 science and technology books, 420 story books, and 60 books.

The science book is x, y***, and the storybook v x+y=620 v=3x y=x-80 x=350 y=270 z=1050, I don't know if it's right.

Set up x science and technology books.

3x+x+x-80=620

then x=140

So 420 storybooks, 140 science and technology books, ***60 books.

620 plus 80 divided by 4 calculates how many science and technology books there are, multiply the number by 3 is the book of the storybook, and add 80 to the number of the first one