An arithmetic problem, math master comes 20

There are 3 results (5, 3, 0) for each question, and there are 20 questions in total.

Score s=5x+3y+0*(20-x-y).

5x+3y (0<=x<=20,0<=y<=20,0<=x+y<=20,x,y as an integer).

The integers from 0 to 100 cannot satisfy the s condition and are decomposed into 5x+3y form only 1 2 4 6 7 11 14 17 87 91 94 97 99

Another idea is that a single digit of a two-digit number is 0, and 5 can be broken down into the form of 5x+3*0.

The single digits of 3, 6, and 9 in the two-digit number can be decomposed into the form of 5*0+3y.

A two-digit number with a single digit of 1 can be decomposed into the form of 5x+3*7 (0<=x<=13), a two-digit number with a single digit of 2 can be decomposed into a form of 5x+3*4 (0<=x<=16), a two-digit number with a single digit of 4 can be decomposed into a form of 5x+3*8 (0<=x<=12), a two-digit number with a single digit of 7 can be decomposed into a form of 5x+3*9 (0<=x<=11), and a two-digit number with a single digit of 8 can be decomposed into a form of 5x+3*6 (0<=x<=14)

y={5x(x=20)

3x(0<=x<=19)

Those who do not meet the above conditions will not be able to get points.

1.1/20 3 and 4/4 20

1/20 20) 3 and 3/4

1 3 and 3 out of 4

3 and 3/4

2.5 and 5/7 81 and 4/9 7/10 = 38/7 733 7/10

38/10 733/9

19/5 733/9

13927/45

309 and 45 molecules 22

3.105 and 97/105 5/8 11 and 2/7 = 11122/105 5/8 79/7 = 5561/105 5/4 79/7 = 2196595 2940/2

747 and 415/2940

The answer is: (too simple, no need to write the process).

and 28,655 3/4

5/7 {32x(.})

5/7 [32 (1 3/8)].

5/7 [32 5/8].

5/7 20

5/7 1/20

13/121/28 x 17+8 21/17

13/21 17 8 17/21

17/21 13 8 17/21

17/21 (13 8).

17/21 21 17

30 = 2×3×5

There is already a common factor 3 in the known conditions, and there is a lack of a 2 and a 5, so there must be a factor 5 in a and a factor 2 in b, so it follows that one of a and b must be 2 and one of 5.

i.e. a = 2*3*2*5 = 60

b = 3*5*2*5 = 150

5x8/9 + 5x2/9

7/10 + 3/4) noisy x5

14 4+15 Summoning infiltration 4

135x8/9 x 3/4

6/11x7/8 - 8/621 of 11 + 2/5) x 5/7

15 and 2 7

2/5 x 15x1/10

Solution: (1) From the meaning of the question:

y=（210-10x）（50+x-40）

10x2+110x+2100 (0 x 15 and x is an integer);

2）y=-10-（

a=-10 0, when x=, y has a maximum.

0 x 15, and x is an integer, when x=5, 50+x=55, y=2400 (yuan), when x=6, 50+x=56, y=2400 (yuan).

When the selling price is set at 55 or 56 yuan per piece, the profit per month is the largest, and the maximum monthly profit is 2,400 yuan

3) When y=2200, -10x2+110x+2100=2200, the solution is: x1=1, x2=10

When x=1, 50+x=51, when x=10, 50+x=60

When the selling price is set at 51 or 60 yuan per piece, the profit per month is 2,200 yuan

When the selling price is not less than 51 or 60 yuan, the profit per month is 2200 yuan

When the selling price is not less than 51 yuan and not higher than 60 yuan and is an integer, the monthly profit shall not be less than 2,200 yuan (or when the selling price is 51, 52, 53, 54, 55, 56, 57, 58, 59, 60 yuan, the monthly profit shall not be less than 2,200 yuan).

It's too late today.

Let's start with one. Clearly say y=2100-310x+10x squared.

(1) y=(10+x)(210-10x) 1=x is n+, so x=5 or 6 is the maximum.

y(max)=2400

3) The profit is 51 when the selling price is 60 or 2200

The monthly profit is not less than 51 when the selling price is between 60 and 2200

(1) y=(50+x-40)(210-10x)(0 can be obtained.

y=-10x²+110x+2100

2) Recipe available.

y=-10(

Because x is a positive integer, when x = 5 or 6, i.e. when the price is 55 or 56, the profit is the largest, which is 2400

The solution yields x = 10 or 1

That is, when the price is 51 yuan or 60 yuan, the profit is exactly 2200, and the monthly profit is not less than 2200, the range of the selling price m is 51 m 60 (m is an integer).

= 14/3 (-7/3) + 9 to the 9th power 8 to the 9th power = -2 + to the 9th power.

2+ to the 9th power.

Four and two-thirds divided by (minus two and one-third) plus to the tenth power multiplied by 8 to the ninth power 14/3 (3/7) (9 to the 9th power 2 1 to the 9th power).

111 and 111 139

11 25+11 11 17 (Is this question written incorrectly) 19 and 1 11

3. 80 and 5/7 x 3/50 + 3x11 and 23/7 50 (80 and 5 7 + 11 and 2 7) 5 and 13 25

8 and 91 100

6. 4568 7890 - 1234 6656 + 4567 + 7890 (see if there is a mistake).

139 + 1) * 111 139 = 111 + 111 139 = 111 and 111 139

Written wrong, right? 3 50 * (5/7 of 80 + 2/7 of 11) = 3 50 * 92 = 138 25

9 100 * (100-1) = 9-9 100 = 8 and 91 100 cross to get 6

There seems to be no easy way ......