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}

题目中文翻译

已知 $a$ 、 $b$ 、 $c$ 都是正整数，且

用穷举法证明乘积 $abc$ 总是 $6$ 的倍数。

可以使用上面的表格说明答案。

(3)

不一定需要使用表格中的所有行。

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 $a$ .

Since $c=a^2-1>0$ , the positive integer $a$ must satisfy $a\geq2$ . If $a\geq5$ , then

b=21-a-a^2\leq21-5-25=-9,

which contradicts $b>0$ . Therefore the only possible values are $a=2,3,4$ , giving the following exhaustive table.

\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}

The three products are divisible by $6$ , since

\begin{align*} 90=&\,6\cdot15,\\ 216=&\,6\cdot36,\\ 60=&\,6\cdot10. \end{align*}

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