47208 關於“百雞問題”術文的理解及其他

### 二、對「百雞術」的一種解讀

$$\tag{1}$$ $$\left\{\begin{array}{l} x+y+z=100,\\ 5x+3y+\frac 13 z=100.\end{array}\right.$$ $$\tag{2}$$

$$14x+8y=200, \tag{3}$$

$$7x+4y=100, \tag{4}$$

$$\left\{\begin{array}{l} x+y+z=100\\ 5x+4y+\dfrac 14z=100 \end{array}\right.\quad \hbox{和}\quad \left\{\begin{array}{l} x+y+z=100\\ 4x+3y+\dfrac 13z=100 \end{array}\right. \hbox{。}$$

$$\left\{\begin{array}{l} x+y+z=100\\ 7x+3y+\dfrac 13z=100 \end{array}\right.\quad \hbox{和}\quad \left\{\begin{array}{l} x+y+z=10\\ 7x+3y+\dfrac 13z=10 \end{array}\right. \hbox{。}$$

$$\left\{\begin{array}{l} x+y+z=100\\ \dfrac 7{10}x+\dfrac 3{10}y+\dfrac 1{30}z=10 \end{array}\right.\quad \Leftrightarrow\quad \left\{\begin{array}{l} x+y+z=100\\ 7x+3y+\dfrac 1{3}z=100 \end{array}\right. \hbox{。}$$

$$200=28\times 6+32=(20+8)\times 6+8\times 4=20\times 6+8\times (6+4)=20\times \fbox{6}+8\times \fbox{10}\hbox{。}$$

$$50=7\times 6+8=(5+2)\times 6+2\times 4=5\times 6 +2\times (6+4)=5\times \fbox{6}+2\times \fbox{10}\hbox{。}$$

$$x=\frac 25 (25-y),$$

$$\left\{\begin{array}{l} x=2\\[-5pt] y=20\\[-5pt] z=78\end{array}\right.;\quad \left\{\begin{array}{l} x=4\\[-5pt] y=15\\[-5pt] z=81\end{array}\right.;\quad \left\{\begin{array}{l} x=6\\[-5pt] y=10\\[-5pt] z=84\end{array}\right.;\quad \left\{\begin{array}{l} x=8\\[-5pt] y=5\\[-5pt] z=87\end{array}\right.\hbox{。}$$

$$\left\{\begin{array}{l} x+y+z=100\\[-5pt] 9x+2y+\dfrac 1{6}z=100 \end{array}\right. \hbox{。}$$

\begin{align*} x=\,&\frac 1{53}(500-11y)=\frac 1{53}(11\times 45-11y+5)=\frac 1{53}(11\times (31+14)-11y+5)\\ =\,&\frac 1{53}(11\times 31-11y+11\times 14+5)=\frac 1{53}(11\times 31-11y+159)=\frac {11}{53}(31-y)+3. \end{align*}

\begin{align*} ax+by=c\tag{*} \end{align*}

$$\left\{\begin{array}{l} x=\dfrac{bq_0+r}{a}+bt\\[-5pt] y=q-q_0-at \end{array}\right., \qquad \Big(0\le t\le \Big[\frac{q-q_0}a\Big]\Big)\hbox{。}$$

$$ax_0+by_0=c=bq+r\lt bq+b=b(q+1)\hbox{。}$$

$$x=\frac 1a(c-by)=\frac 1a(bq-by+r)\hbox{。}$$

(I) 若 $r=0$, 取 $q_0=0$, 則由 $x=\frac ba(q-y)$, 令 $q-y=at$, 可得 (*) 的非負整數解為

$$\left\{\begin{array}{l} x=bt\\[-5pt] y=q-at \end{array}\right., \qquad \Big(0\le t\le \Big[\frac{q}a\Big]\Big);$$

(II) 若 $0\lt r\lt b$, 取 $q_0=q-y_0$, 因 $0\le y_0\le q$, 故有 $0\le q_0\le q$。 在此情形, 有

$$x=\frac 1a(bq-by+r)=\frac 1a(b(y_0+q_0)-by+r)=\frac ba(y_0-y)+\frac{bq_0+r}{a},$$

$$\left\{\begin{array}{l} x=\dfrac{bq_0+r}{a}+bt\\ y=q-q_0-at \end{array}\right., \qquad \Big(0\le t\le \Big[\frac{q-q_0}a\Big]\Big)\hbox{。}{\Box}$$

$$\left\{\begin{array}{l} x+y+z=20\\ 3x+\dfrac 32 y+\dfrac 12 z=20 \end{array}\right.; %\qquad\hbox{證畢。}\eqno{\bm\Box}$$

### 三、一個與之有關的問題

$$\left\{\begin{array}{l} 2x+y=3y+z\\ 3y+z=4z+u\\ 4z+u=5u+v\\ 5u+v=6v+x\\ 6v+x=2x+y \end{array}\right.\qquad\Leftrightarrow\qquad \left\{\begin{array}{l} 2x-2y-z=0\\ 3y-3z-u=0\\ 4z-4u-v=0\\ x-5u+5v=0\\ x+y-6v=0 \end{array}\right.$$