题目

Problem

Given that $a$ , $b$ and $c$ are positive integers such that

prove, by exhaustion, that the product $abc$ is always a multiple of $6$

You may use the table below to illustrate your answer.

(3)

You may not need to use all rows of this table.

\begin{array}{c|c|c|c} a&b&c&abc\\ \hline &&& \end{array}

解答

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因为 $a,b,c$ 都是正整数，而且 $c=a^2-1$ ，所以 $a$ 不能太大。逐个找出所有可能的 $a$ ，这就是 exhaustion。

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Since $b$ is a positive integer,

\begin{align*} b=20-a-c>0. \end{align*}

Also $c=a^2-1$ , so

\begin{align*} b=20-a-(a^2-1)=21-a-a^2. \end{align*}

Now test the possible positive integer values of $a$ .

If $a=1$ , then $c=0$ , so this is not allowed. If $a=5$ , then $b<0$ , so no larger value of $a$ is possible.

\begin{array}{c|c|c|c} a&b&c&abc\\ \hline 2&15&3&90\\ 3&9&8&216\\ 4&1&15&60 \end{array}

Since $b$ and $c$ must be positive, the possible cases are

\begin{align*} (a,b,c)=(2,15,3),\ (3,9,8),\ (4,1,15). \end{align*}

Their products are

\begin{align*} 90,\quad 216,\quad 60. \end{align*}

Each of these is divisible by $6$ .

Therefore, $abc$ is always a multiple of $6$ .