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[Algebra] Hundred Fowls Problem

2021-07-04 17:26 作者:AoiSTZ23  | 我要投稿

By: Tao Steven Zheng (鄭濤)

【Problem】

Zhang Qiujian Suanjing Chapter 3 Problem 38

Suppose roosters costs?5?coins each, hens?3?coins each and three chickens cost?1?coin. If?100?fowls are bought for?100?coins, how many roosters, hens and chickens are there?

【Solution】

Let x%2Cy%2Cz?represent the number of roosters, hens, chickens respectively. The problem is a system of two linear equations in three unknowns:

x%2By%2Bz%3D100%20......%20(1)

5x%2B3y%2B%5Cfrac%7Bz%7D%7B3%7D%3D100%20......(2)

?

Multiply equation (2) by 3 to get

x%2By%2Bz%3D100%20......%20(1)

15x%2B9y%2Bz%3D300%20......(2)

Subtract equation (1) from equation (2) to obtain

14x%2B8y%3D200

Hence,

y%3D25-%5Cfrac%7B14%7D%7B8%7D%20x

Substitute y?into equation (1) to obtain

z%3D75%2B%5Cfrac%7B3%7D%7B4%7D%20x

?Therefore,

(x%2Cy%2Cz)%3D(x%2C%2025-7x%2F4%2C75%2B3x%2F4)

By the nature of the problem, the solutions for x%2Cy%2Cz must be non-negative integers; hence, x?must be divisible by 4. Substitute x%3D4t? into the above solution set to obtain

(x%2Cy%2Cz)%3D(4t%2C%2025-7t%2C75%2B3t)

where t%3D0%2C1%2C2%2C3.

?

Solution 1: 0 roosters, 25 hens, and 75 chickens. [1]

Solution 2: 4 roosters, 18 hens, and 78 chickens.

Solution 3: 8 roosters, 11 hens, and 81 chickens.

Solution 4: 12 roosters, 4 hens, and 84 chickens.

?

[1] Zhang Qiujian only gave positive integer solutions, so the first solution (0 roosters, 25 hens, and 75 chickens) was not included in the “Zhang Qiujian Suanjing”.


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