2005 AMC 8 第 8 题

Question

设

m

和

n

是正奇数。下列哪一个一定也是奇数？

Suppose

m

and

n

are positive odd integers. Which of the following must also be an odd integer?

m + 3n

3m - n

3m^2 + 3n^2

(nm + 3)^2

3mn

答案：E

知识点：奇偶性

难度评级：1000

小提示：

用奇数的奇偶性规则检验各选项

Test the choices using parity rules for odd numbers.

大提示：

奇数的乘积是奇数

The product of odd integers is odd.

解答：

先回顾以下四条规则：

• 奇数加奇数、偶数加偶数，结果都是偶数

• 偶数加奇数，结果是奇数

• 偶数乘任何整数，结果都是偶数

• 奇数乘奇数，结果是奇数

任意奇数可写成

2m+1

，任意偶数可写成

2n

，其中

m

、

n

为整数，因此很容易验证这些规则。

下面逐一检查各选项：

A:

注意

3

是奇数，因此对模

2

而言，该式为

1 + 1 \cdot 1 \equiv 0 \pmod 2\text{。}

根据上述规则，结果是偶数。

B:

1 \cdot 1 - 1 \equiv 0 \pmod 2\text{。}

结果仍是偶数。

C:

1 \cdot 1^2 + 1 \cdot 1^2 \equiv 0 \pmod 2\text{。}

结果也是偶数。

D:

(1 \cdot 1 + 1)^2 \equiv 0 \pmod 2\text{。}

结果同样是偶数。

E:

1 \cdot 1 \cdot 1 \equiv 1 \pmod 2\text{。}

结果是奇数。

因此只有 E 一定是奇数。

所以正确答案是 E。

Recall the four following rules:

• odd plus odd and even plus even is even

• even plus odd is odd

• even times anything is even

• odd times odd is odd

These rules can be easily verified by representing arbitrary odd numbers as

2m+1

and arbitrary even numbers as

2n

respectively, for integers

m,

n.

With this in mind, let’s examine each answer choice individually:

A:

Note that

3

is odd. Modulo

2

, this gives us

1 + 1 \cdot 1 \equiv 0 \pmod 2.

From our above rules, we know that this is even.

B:

1 \cdot 1 - 1 \equiv 0 \pmod 2.

Once again, this is even.

C:

1 \cdot 1^2 + 1 \cdot 1^2 \equiv 0 \pmod 2.

This is also even.

D:

(1 \cdot 1 + 1)^2 \equiv 0 \pmod 2.

Unfortunately, this is also even.

E:

1 \cdot 1 \cdot 1 \equiv 1 \pmod 2.

This is odd.

Therefore, E is the only answer choice that is an odd integer.

Thus, E is the correct answer.

2005 AMC 8 第 8 题

答案：E

难度评级：1000

小提示：

大提示：

解答：

其他年份的第 8 题