There is a math problem that asks the master to give me a specific solution to the problem

Let's buy x oranges, y apples, and z melons.

x+y+z=100

x/11 + 3y + 4z = 100

x must be a multiple of 11: 11 22 33 44 55 66 77 88 99

They are: 1 2 3 4 5 6 7 8 9 yuan.

Carry-in: when x=11.

Get: 368+z=100

Not true. When x=22.

Get: 236+z=100

Not true. When x=33.

Get: 204+z=100

Not true. When x=44.

Get: 172+z=100

Not true. When x=55.

Get: 140+z=100

Not true. When x=66.

Get: 108+z=100

Not true. When x=77.

Get: 76+z=100

z=24 calculates: x=77

y=-1z=24

Not true. When x=88.

Get: 44+z=100

z=56 calculates: x=88

y=-44z=56

Not true. When x=99.

Get: 12+z=100

z=88 calculates: x=99

y=-87z=88

Not true. There is no solution to this question.

x oranges, y apples, z melons.

x+y+z=100

x 11 + 3y + 4z = 100, when x = 0, 11, 22, 33, 44, 55, 66, 77, 88, 99, all of which are not in line with the topic, so there is no solution to this problem.

a/11+3b+4c=100

a+b+c=100

A is a multiple of 11 and can be considered in this way.

x/11+3y+4z=100

x+y+z=100

z=3*(100-x 11) (100-x)xcan =11,22,33,44,55,..99, bring in z without an integer less than 100, there is no solution?

x oranges, y apples, z melons.

x+y+z=100

x 11 + 3y + 4z = 100, when x = 0, 11, 22, 33, 44, 55, 66, 77, 88, 99, it is not suitable for the topic, and it is discarded.

Let the number of oranges be x, the number of apples be y, and the number of melons be z

Then 1 11x+3y+4z=100(1).

and x+y+z=100(2).

1)*11 x+33y+44z=1100(3)2)*33 33x+33y+33z=3300(4)4)-(3) 32x-11z=2200(5) One condition is missing.

Column the ternary equation and try it algebraically.

Polynomials formed into 1 a, 1 b, 1 c:

ab (a+b)=1 (1 a+1 b)=1 2,1 a+1 b=2bc (rough b+c)=1 bend (1 b+1 c)=1 3,1 b+1 c=3

Similarly: 1 c + 1 a = 4

Three-formula addition: 1 a+1 b+1 c=(2+3+4) 2=9 2abc (ab+bc+ca)=1 (bury stool scale 1 a+1 b+1 c)=2 9

s trapezoid: (2+10)*12 2=72(cm2) Find the area of the entire trapezoid.

s: small triangle: 3*4 2=6(cm2) Then calculate the area of the triangle in a.

S: Small trapezoid: (4+6)*4 2=20(cm2) Then calculate the area of the trapezoid in a.

The remaining area: 72-20=52 (cm2) Subtract the area of the triangle from the area of the trapezoid.

In total: s:a: 52-6=64 (cm2) minus the area of the trapezoid is equal to the area of a.

It's clearer to see it this way.

If there are x coins of 2 cents, the value of 2 cents is 2x;

Then there are (2x-13) coins of 1 cent, and the value is also (2x-13);

A coin of 5 cents is the remaining [100-x-(2x-13)], which is worth 5 [100-x-(2x-13)];

The equation can be listed as:

2x+(2x-13)+5[100-x-(2x-13)]=200 solution, x=32

So there are 32 2 coins; 1 cent coin has (2*32-13) = 51 coins; There are 100-32-51 = 17 coins for 5 cents.

an=a1+(n-1)d, where d=2

When n=50, a50=-50+(50-1)*2=48, the 50th digit is: 48

sn=n(a1+an)/2

Not the upstairs 50,0

Sequences of numbers are calculated by formulas.

Solution: Set up x boys and y girls.

y=x=30，y=15

There are 30 + 15 = 45 students in the class.

Re-columnar sub.

0·4x+12 - the number of boys.

0·4x+12) 50% – the number of female students.

0·4x+12) 50%+0·4x+12=x - just solve this, answer x=45

The number of boys, the number of girls is 50% of the number of boys, i.e., then, so x=45

You should set the girl as x, that is easy to understand, so the class is 3x.

The column equation 2x=40%x3x+12 is solved to get x=15, and then x3 to get 45

It should be multiplied by 50% instead of 25%.

There are no problems with the rest of the place!!

The first person does y hours The second person does y-t, the third person does y-2t , and the last person does 1 4 y

The time that these people work is the first term y last term 1 4 y difference -t total x terms.

y -t(x-1)=1 4 y (t is known) then 3y=4t(x-1) (Eq. 1).

Suppose there are x stevedores with a speed of 1 (10x) per stevedore

Sum of work time (y+1 4y) x 2 speed 1 (10x).

Then: [(y+1 4y)x 2][1 (10x)]=1 y 16=1 y=16 (the time when the first person worked is the total time).

Bring in (Eq. 1 ) 3 *16=4t(x-1) x=12 t+1 t and x are positive integers, so x can be ,

t can be taken, and the corresponding x is (since the weight says that the last stevedore is the first stevedore's working time 1 4), so x=(rounded).

Answer: (1) The time required is 16 hours (2) There are a total of stevedores who can be individuals.

If the number of stevedores is m, and the time of the last person to do it is r hours, then the total amount of work can be expressed as 10*m, and the time from the last person to work can also be r, t r, 2t r, 3t r, (m-1)t + r....

According to the total activity column, 10m = m(m-1)t2 + mr, i.e., (m-1)t2 + r = 10

From the last person time is a quarter of the first person: 4r = (m-1)t + r, i.e. r = (m-1)t 3

Solving the binary equation yields: r=4, m=12 t +1

Since m can only be an integer, all t can only be taken by , and the corresponding m values are respectively