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[Number Theory] Objects of Unknown Number

2021-09-04 19:45 作者:AoiSTZ23  | 我要投稿

By: Tao Steven Zheng (郑涛)

【Problem】

The following problem is from the Sunzi Suanjing (孙子算经), a text written by an obscure mathematician with the surname Sun (name unknown) sometime around the 3rd to 5th centuries AD.

Suppose we have an unknown number of objects. When counted in threes, 2 are left over, when counted in fives, 3 are left over, and when counted in sevens, 2 are left over. How many objects are there?

[Assume the lowest positive integer solution]


【Solution】

This problem is a system of indeterminate equations with infinitely many solutions. According to the problem, we get


%5Cbegin%7Balign%7D%0A%20N%20%26%5Cequiv%202%20%5Cpmod%7B3%7D%20%5C%5C%0A%20N%20%26%5Cequiv%203%20%5Cpmod%7B5%7D%20%5C%5C%0A%20N%20%26%5Cequiv%202%20%5Cpmod%7B7%7D%0A%5Cend%7Balign%7D

Calculate the product of the moduli

M%20%3D%203%20%5Ctimes%205%20%5Ctimes%207%20%3D%20105

The solution of the Chinese remainder theorem prescribes that

N%20%3D%20%5Cleft%5Br_1%20M_1%20s_1%20%2B%20r_2%20M_2%20s_2%20%2B%20r_3%20M_3%20s_3%20%5Cright%5D%20%5Cpmod%20M

For this problem

M_1%20%3D%20%5Cfrac%7B105%7D%7B3%7D%20%3D%2035%2C%20%5Cquad%20M_2%20%3D%20%5Cfrac%7B105%7D%7B5%7D%20%3D%2021%2C%20%5Cquad%20M_3%20%3D%20%5Cfrac%7B105%7D%7B7%7D%20%3D%2015


N%20%3D%20%5Cleft%5B2(35)s_1%20%2B%203(21)s_2%20%2B%202(15)s_3%20%5Cright%5D%20%5Cpmod%7B105%7D

where

%5Cbegin%7Balign%7D%0A%2035s_1%20%26%5Cequiv%201%20%5Cpmod%7B3%7D%20%5C%5C%0A%2021s_2%20%26%5Cequiv%201%20%5Cpmod%7B5%7D%20%5C%5C%0A%2015s_3%20%26%5Cequiv%201%20%5Cpmod%7B7%7D%0A%5Cend%7Balign%7D


and s_1%2C%20s_2%2C%20s_3 represent the modular inverses of each respective remainder.  The modular inverses can be solved systematically using Qin Jiushao's (大衍求一术) ; however, the numbers involved in this problem are small enough to be obtained by guessing and checking.


%5Cbegin%7Balign%7D%0A%2035(2)%20%26%5Cequiv%201%20%5Cpmod%7B3%7D%20%5C%5C%0A%2021(1)%20%26%5Cequiv%201%20%5Cpmod%7B5%7D%20%5C%5C%0A%2015(1)%20%26%5Cequiv%201%20%5Cpmod%7B7%7D%0A%5Cend%7Balign%7D

Final Calculation

N%20%3D%20%5Cleft%5B2(35)(2)%20%2B%203(21)(1)%20%2B%202(15)(1)%20%5Cright%5D%20%5Cpmod%7B105%7D

N%20%5Cequiv%20%5Cleft%5B140%20%2B%2063%20%2B%2030%20%5Cright%5D%20%5Cpmod%7B105%7D

N%20%3D%20233%20%5Cpmod%7B105%7D

Here, 233 is a solution, but this is not the lowest positive integer solution. The lowest positive integer solution is


233%20-%202(105)%20%3D%2023

So there are 23 objects.


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