April Online Questions:

The following are two Remainder Theorem related questions that appear in the Chinese Literature/Novel.
Please contribute ONE POSSIBLE method to find their solution(s).
Remainder to state your class and name at the end of it.
(I) 韓信點兵又稱為中國剩餘定理,相傳漢高祖劉邦問大將軍韓信統御兵士多少,韓信答說,每3人一列餘1人、5人一列餘2人、7人一列餘4人、13人一列餘6人……。劉邦茫然而不知其數。
ANS:

Let the number of the soldiers be x

We could see that 15 is the multiple of 3,5, and the remainder will be 1 if it is divided by 7
70 is the multiple of 7,5, and the remainder will be 1 if it is divided by 3
21 is the multiple of 3,7, and the remainder will be 1 if it is divided by 5
So the remainder will be 2 if 21*2 is divided by 5, and the remainder will be 4 if 15*4 is divided by 7
21*2+15*4+70=172
172/105=1……67
So x=105a+67=13b+6
Because a,b are non-negetive integers
(105a+61)is a multiple of 13
So 4 is the smallest number of a
So x=481+1365k(k is a non-negetive integer)
(3s1 Cao Wei)


(II) 射雕英雄傳裡寫道,有一次黃蓉因瑛姑而感到氣急敗壞,正欲反唇相譏之時,
念頭一轉,便用竹杖在㆞㆘細沙㆖寫了三道算題,裡面便有一道題目與餘數問題相關。這題目
是這麼出的:今有物,不知其数,三三数之,剩二,五五数之,剩三,七七数之,剩二,问物几何?」

ANS:
Let the number be x
Because the remainder is 2 when x is divided by 3 and 7,the remainder is 2 when x is divided by 21
x=21a+2=5b+3(a,b are non-negetive integers)
21a-1=5b
So the smallest value of a is 1
So x=23+105k(k is a non-negetive integer)
(3s1 Cao Wei)


韩信点兵

淮安民间传说着一则故事——“韩信点兵”,其次有成语“韩信点兵,多多益善”。韩信带1500名兵士打仗,战死四五百人,站3人一排,多出2人;站5人一排,多出3人;站7人一排,多出2人。韩信马上说出人数:1073。


The method
1) Give a list of the multiple of 3 with a remainder of 2
2,5,8,11,14,17……
2) Give a list of the multiple of 5 with a remainder of 3
8,13,18,23……
3) So 8 will be the fist number and f(x)=8+15x(x>=0)
4) The list of f(x) is
8,23,38,43,……
5) Give a list of the multiple of 7 with a remainder of 2
2,9,16,23,30.......
6) So the second number is 23 and g(x)=23+105x(x>=0)
because the total number is between 1000 and 1100, when x = 10, g(x)=1073

(Lu Zhen 3S1 14)

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