已知 $a$、$b$、$c$ 是正整数，且满足：

• $c = 3a + 1$
• $a + b + c = 15$

用穷举法证明 $abc$ 一定是4的倍数。

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（3分）

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由两条条件求出 $b=14-4a$。正整数条件把 $a$ 限定为 1、2、3，逐一列出全部组合。

Given $c = 3a + 1$, substitute into $a + b + c = 15$:

Since $a, b, c$ are positive integers:

• $a \geqslant 1$
• $b = 14 - 4a \geqslant 1 \Rightarrow a \leqslant 3$

Therefore $a$ can only take the values $1, 2, 3$. Check each case:

All values of $abc$ are multiples of 4. $\quad \blacksquare$

也可以穷尽 $a$ 的奇偶两种情形。$b$ 必为偶数；$a$ 为偶数时 $a,b$ 均偶，为奇数时 $b,c$ 均偶，所以乘积总含因子 4。

From the conditions,

Thus $b$ is even. Exhaust the two possible parities of $a$:

• If $a$ is even, both $a$ and $b$ are even.
• If $a$ is odd, $c=3a+1$ is even, so both $b$ and $c$ are even.

In either case there are two even factors. Their product has the form $(2m)(2n)=4mn$, so multiplying by the third integer still gives a multiple of 4.

Therefore $abc$ is always a multiple of 4. $\blacksquare$