已知 $a,b,c$ 为大于 0 的整数，且 $c=b+2$、$a+b+c=10$。

用穷举法证明 $a,b,c$ 的乘积总是偶数。你可以使用下表说明答案，不一定需要用完所有行。

（3 分）

先把 $c=b+2$ 代入 $a+b+c=10$，得到 $a=8-2b$。由于 $a,b,c$ 均为正整数，$1\leqslant b\leqslant3.5$，所以 $b$ 只有 $1,2,3$ 三种可能。穷举相应的 $(a,b,c)$，计算乘积 $abc$，逐个验证为偶数。

Substitute $c = b + 2$ into $a + b + c = 10$:

Since $a$, $b$, and $c$ are integers greater than 0, we must have $b \geqslant 1$. Also:

Since $b$ must be an integer, the only possible values for $b$ are $1$, $2$, and $3$. We list the cases:

Case 1: $b = 1$

The product is:

Case 2: $b = 2$

The product is:

Case 3: $b = 3$

The product is:

The combinations can be summarized in the table below:

Since all possible combinations have been exhausted and in every case the product $abc$ is even, the product of $a$, $b$ and $c$ is always even.

我们可以通过代数结构分析变量的奇偶性： 由 $a = 8 - 2b = 2(4 - b)$ 可知，由于 $4-b$ 是整数，所以 $a$ 必定是 2 的倍数，即 $a$ 必定是一个偶数。 因为任意整数与偶数的乘积必然是偶数（$\text{even} \times \text{integer} = \text{even}$），所以只要 $a$ 是偶数，不管 $b$ 和 $c$ 的奇偶性如何，$abc$ 的乘积必然是偶数。

From the equations $c = b + 2$ and $a + b + c = 10$, we obtain:

Since $b$ is an integer, $4 - b$ is also an integer. Therefore, $a$ must be an even integer.

Since the product of any integers with at least one even factor is always even, the product $abc$ must be even for any valid integer combinations.

Hence, the product of $a$, $b$ and $c$ is always even.

将乘积 $abc$ 完全用单一变量 $b$ 表示出来：

提取公因子 2，写成 $2 \times \text{integer}$ 的形式，从而直接证明其整除性。

Using the equations $a = 8 - 2b$ and $c = b + 2$, we can express the product $abc$ in terms of $b$ only:

Since $b$ is an integer, $(4 - b)b(b + 2)$ is also an integer.

Since $abc$ has a factor of $2$, $abc$ is divisible by $2$, which means the product $abc$ is always even.