Solve the math problem Hundred Chicken Problem algorithm problem 1 Hundred Chicken Problem

This question is "the problem of buying 100 chickens for 100 dollars". Generally, indefinite equations are used to solve, which is difficult for primary school students and even junior high school students to understand.

Analysis and Solution Because 100 yuan buys 100 chickens, an average of 1 yuan buys 1 chicken. 4 chickens per group: 1 hen and 3 chicks, worth a total of 4 cents.

Because 1 hen costs 3 yuan, 3 chicks 1 yuan), it happens to be an average of 1 yuan to buy 1 chicken.

7 chickens per large group: 1 rooster and 6 chicks. It is worth a total of 7 cents. (Because 1 rooster costs 5 yuan, 3 chicks cost 1 yuan, and 6 chicks cost 2 yuan), which happens to be an average of 1 yuan to buy 1 chicken.

No matter how many large groups and how many groups 100 chickens can be divided into, they buy an average of 1 chicken per penny. How many groups and how many groups can 100 chickens be divided into?

Through analysis and exploration, the following situations can be found.

Divided into 4 large groups, 18 groups.

There are 4 large groups of roosters: 1 4 = 4 (only).

There are 4 large groups of chicks: 6 4 = 24 (birds).

In the 18 groups, there are 1 18 = 18 hens

In the 18 groups, there were chicks: 3 18 = 54 (birds).

In this case, there are 4 roosters, 18 hens, and 78 chicks (24 + 54 =).

Divided into 8 large groups, 11 groups.

There are 8 roosters in the 8 large groups: 1 8 = 8 (only).

There are 8 large groups of chicks: 6 8 = 48 (only).

In the 11 groups, there are 1 11 = 11 hens

There are 11 chicks in the group: 3 11 = 33 (birds).

In this case, there are 8 roosters, 11 hens, and 81 chicks (48 + 33 =).

Divided into 12 large groups, 4 groups.

There are 12 roosters in the 12 large groups: 1 12 = 12 (only).

There are 12 large groups of chicks: 6 12 = 72 (birds).

There are 4 hens in the group: 1 4 = 4 (only).

There are 4 chicks in the group: 3 4 = 12 (chickens).

In this case, there are 12 roosters, 4 hens, and 84 chicks (72 + 12 =). So there are three possibilities for this question: buy 4 roosters, 18 hens, and 78 chicks; or 8 roosters, 11 hens, and 81 chicks; Or buy 12 roosters, 4 hens, and 84 chicks.

Absolutely correct if you want another way to solve the madness

Set up rooster x hen y chick z

x+y+z=100① 5x+3y+z/3=100③→15x+9y+z=300②

The two equations want to subtract 7x+4y=100

Because the chicken is whole, xy must be an integer, solved to be x=4, y=18, z=78 or x=8, y=11, z=81 or x=12, y=4, z=84

The upstairs seems to have misread the title, which is 5 yuan for roosters, 3 yuan for hens, and 1 yuan for 3 chicks.

Categories: Education Academic Exams >> Study Help.

Problem description: Buy 100 chickens with 100 yuan, a rooster for 5 yuan, a hen for 3 yuan, and three chickens for 1 yuan, how many roosters, hens, and chicks do you buy? I saw some of my friends listed the content of history, I don't need it, I just need a specific solution or an explanation for this problem, thank you.!!

Analysis: Solution: Suppose there are x roosters, y hens, and (100-x-y) chicks.

Then: 5x+3y+(1 3)(100-x-y)=100

15x+9y+100-x-y=300

14x+8y=200

7x+4y=100

At this point, the problem becomes finding a non-negative positive integer solution of 7x+4y=100.

4y=100-7x

y=(100-7x)/4

Since y is a natural number, (100-7x) 4 is a natural number.

And since 100 is divisible by 4, 7x must be divisible by 4

And because 7 is coprime with 4, x is a multiple of 4.

So if x is 4, then y = (100-4 7) 4 = 18 and (100-x-y) = 78

If x 8, then y = (100-8 7) 4 = 11, (100-x-y) = 81

If x 12, then y = (100-12 7) 4 = 4, (100-x-y) = 84

So there are several options:

1. Buy 4 roosters, 18 hens, and 78 chicks.

2. Buy 8 roosters, 11 hens, and 81 chicks.

3. Buy 12 roosters, 4 hens, and 84 chicks.

Topic: 5 yuan for each rooster, 3 yuan for each hen, 1 yuan for every 3 chickens, how many ways to buy 100 chickens for 100 yuan?

Problem solving idea: Let Qi Hao x, y, z be the number of three kinds of chickens, and the algebraic equation can be obtained: x+y+z=100, 5x+3y+z3=100. For this indefinite equation, it is more convenient to use the enumeration method.

The specific implementation is as follows:

Output: Rooster x=4; hen y=18; Chick Z=78VM410:7 Rooster x=8; hen y=11; Chick Z=81VM410:7 Rooster x=12; hen y=4; Chick z=84

Earned -2 yuan There are 3 transactions in the whole event, the first transaction: 8 yuan to buy, 9 yuan to sell, profit 1 yuan; The second transaction: sell for 9 yuan, buy for 10 yuan, profit -1 yuan; Third Transaction:

10 yuan to buy, 11 yuan to sell and profit 1 yuan; The whole process: 1-1+1=1 yuan So the analysis learned that this person is a fool, because the last two transactions are equal to nothing, he could have earned 3 yuan directly, and after his 3 transactions, the total profit became 1 yuan, so 1-3=-2

Suppose you have 10 yuan to buy chicken, and there are 2 yuan, after selling it, you have 2 + 9 = 11, and then you spend 10 yuan after selling, and you have 1 yuan after you sell it, and you have 12 yuan after selling, so you earn 2 yuan!

This is a question that has been asked by others, and the answer is 2 yuan as everyone says. I don't know if the landlord has paid attention to the problem that has appeared here.

Let the hen x, the rooster y, and the chick 100-x-y, so 5y+3x+(100-x-y) 3=100 and x,y are integers, so the correct answer can be obtained, and there are three cases.

1.4 roosters, 18 hens, 78 chicks.

2.8 roosters, 11 hens, 81 chicks.

3.12 roosters, 4 hens and 84 chicks.

If the small is x, then the female is 9x the male is 15x

15x+9x+x=100

x=4 4 (1 3)=12.

9x=36 36 3=12.

15x=60 60 5=12.

Copy someone else's, I don't know if it's right or not, you can use it as a reference.

Solution: Suppose there are x roosters, y hens, and (100-x-y) chicks, then: 5x+3y+(1 3)(100-x-y)=10015x+9y+100-x-y=300

14x+8y=200

7x+4y=100

At this point, the problem becomes finding a non-negative positive integer solution of 7x+4y=100.

4y=100-7x

y=(100-7x)/4

Because y is a natural number, (100-7x) 4 is a natural number and since 100 is divisible by 4, 7x must be divisible by 4 and because 7 and 4 are coprime, x is a multiple of 4.

So if x is 4, then y = (100-4 7) 4 = 18 and (100-x-y) = 78

If x 8, then y = (100-8 7) 4 = 11, (100-x-y) = 81

If x 12, then y = (100-12 7) 4 = 4, (100-x-y) = 84

So there are several options:

1. Buy 4 roosters, 18 hens, and 78 chicks.

2. Buy 8 roosters, 11 hens, and 81 chicks.

3. Buy 12 roosters, 4 hens, and 84 chicks.

Buy a total of 10 roosters 10 8=80 yuan 5 hens 3 5=15 yuan 15 chicks 15 3=5 yuan.

80 + 15 + 5 = 100 yuan.

8 roosters 10 8 = 80 yuan.

5 hens 3 5 = 15 yuan.

15 chicks 15 3=5 yuan.

80 + 15 + 5 = 100 yuan.