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。
English
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.