Target Audience: Junior High School Mathematics Enrichment / Competition Introduction
Recommended Time: 15 minutes
I. Knowledge Card
1. Definition
If two integers \(a\) and \(b\) have the same remainder when divided by \(m\), then \(a\) and \(b\) are said to be congruent modulo \(m\), denoted as:
Equivalent Statement: \(a\) and \(b\) are congruent modulo \(m\) \(\Longleftrightarrow\) \(a-b\) (or \(b-a\)) is an integer multiple of \(m\), i.e., \(m \mid (a-b)\).
2. Seven Properties of Congruence
| Property Name | Expression | Explanation |
|---|---|---|
| Reflexivity | \(a \equiv a \pmod{m}\) | Every number is congruent to itself |
| Symmetry | If \(a \equiv b \pmod{m}\), then \(b \equiv a \pmod{m}\) | Left and right sides can be interchanged |
| Transitivity | If \(a \equiv b \pmod{m}\) and \(b \equiv c \pmod{m}\), then \(a \equiv c \pmod{m}\) | Transmits like the "equals" sign |
| Additivity | If \(a \equiv b \pmod{m}\) and \(c \equiv d \pmod{m}\), then \(a \pm c \equiv b \pm d \pmod{m}\) | Congruences can be added/subtracted |
| Multiplicativity | If \(a \equiv b \pmod{m}\) and \(c \equiv d \pmod{m}\), then \(ac \equiv bd \pmod{m}\) | Congruences can be multiplied |
| Power Property | If \(a \equiv b \pmod{m}\), then \(a^n \equiv b^n \pmod{m}\) (for any natural number \(n\)) | Congruences can be raised to the \(n\)-th power |
II. Basic Exercises with Answers (5 questions, 20 points each, total 100 points)
Question 1 (Basic Definition)
Determine whether \(17\) and \(7\) are congruent modulo \(5\).
【Answer】
\(17 \div 5\) leaves a remainder of \(2\), and \(7 \div 5\) leaves a remainder of \(2\). The remainders are the same.
\(\therefore 17 \equiv 7 \pmod{5}\) holds true.
Question 2 (Reflexivity and Symmetry)
Write any two different integers that are congruent to \(8\) modulo \(3\).
【Answer】
\(8 \div 3\) leaves a remainder of \(2\), so we just need to find numbers that leave a remainder of \(2\) when divided by \(3\).
For example: \(5\) (\(5 \div 3\) leaves remainder \(2\)), and \(11\) (\(11 \div 3\) leaves remainder \(2\)).
\(\therefore 5 \equiv 8 \pmod{3}\) and \(11 \equiv 8 \pmod{3}\).
Question 3 (Transitivity)
Given that \(13 \equiv 3 \pmod{5}\) and \(3 \equiv 8 \pmod{5}\), determine whether \(13 \equiv 8 \pmod{5}\) holds true, and explain your reasoning.
【Answer】
It holds true.
By transitivity, if \(13 \equiv 3 \pmod{5}\) and \(3 \equiv 8 \pmod{5}\), then \(13 \equiv 8 \pmod{5}\).
Verification: \(13-8=5\), which is an integer multiple of \(5\). \(\therefore\) it holds true.
Question 4 (Additivity)
Given that \(7 \equiv 2 \pmod{5}\) and \(4 \equiv 9 \pmod{5}\), determine whether \(7+4\) and \(2+9\) are congruent modulo \(5\). Verify by calculating their difference.
【Answer】
\(7+4 = 11\), \(2+9 = 11\). Clearly \(11 \equiv 11 \pmod{5}\), so it holds true.
Difference verification: \((7+4) - (2+9) = 11-11 = 0\), which is an integer multiple of \(5\).
(Alternatively, this can be directly concluded using additivity.)
Question 5 (Multiplicativity)
Given that \(3 \equiv 8 \pmod{5}\) and \(2 \equiv 7 \pmod{5}\), determine whether \(3 \times 2\) and \(8 \times 7\) are congruent modulo \(5\). Verify your conclusion.
【Answer】
\(3 \times 2 = 6\), \(8 \times 7 = 56\). \(56-6=50\), which is an integer multiple of \(5\) (\(50=5\times10\)).
\(\therefore 6 \equiv 56 \pmod{5}\), i.e., \(3 \times 2 \equiv 8 \times 7 \pmod{5}\) holds true.
III. Notes
- The definition and properties of congruence are the foundation for solving divisibility problems, remainder problems, and number theory competition problems. When reviewing, it is recommended to emphasize the core criterion: "\(a-b\) is an integer multiple of \(m\)."
- This exercise can be used as an independent introductory worksheet on congruence or as a pre-assessment diagnostic for a number theory unit.
同余（Modular Congruence）基础练习与答案
适用年级：初中数学拓展/竞赛入门
建议用时：15分钟
一、知识卡片
1. 定义
如果两个整数 \(a\) 和 \(b\) 除以 \(m\) 的余数相同，则称 \(a\) 和 \(b\) 对于 \(m\) 同余，记作：
等价说法：\(a\) 和 \(b\) 对 \(m\) 同余 \(\Longleftrightarrow\) \(a-b\)（或 \(b-a\)）是 \(m\) 的整数倍，即 \(m \mid (a-b)\)。
2. 同余的七大性质
| 性质名称 | 表达式 | 说明 |
|---|---|---|
| 自反性 | \(a \equiv a \pmod{m}\) | 任何数与自己同余 |
| 对称性 | 若 \(a \equiv b \pmod{m}\)，则 \(b \equiv a \pmod{m}\) | 左右可互换 |
| 传递性 | 若 \(a \equiv b \pmod{m}\)，\(b \equiv c \pmod{m}\)，则 \(a \equiv c \pmod{m}\) | 像“等号”一样传递 |
| 可加性 | 若 \(a \equiv b \pmod{m}\)，\(c \equiv d \pmod{m}\)，则 \(a \pm c \equiv b \pm d \pmod{m}\) | 同余式可相加/相减 |
| 可乘性 | 若 \(a \equiv b \pmod{m}\)，\(c \equiv d \pmod{m}\)，则 \(ac \equiv bd \pmod{m}\) | 同余式可相乘 |
| 乘方性 | 若 \(a \equiv b \pmod{m}\)，则 \(a^n \equiv b^n \pmod{m}\)（\(n\) 为任意自然数） | 同余式可同时取 \(n\) 次幂 |
二、基础练习与答案（共5题，每题20分，总分100分）
第1题（基础定义）
判断 \(17\) 和 \(7\) 是否对 \(5\) 同余。
【答案】
\(17 \div 5\) 余 \(2\)，\(7 \div 5\) 余 \(2\)，余数相同。
\(\therefore 17 \equiv 7 \pmod{5}\) 成立。
第2题（自反性与对称性）
写出与 \(8\) 对 \(3\) 同余的任意两个不同的整数。
【答案】
\(8 \div 3\) 余 \(2\)，所以只要找除以3余2的数即可。
例如：\(5\)（\(5 \div 3\) 余2），\(11\)（\(11 \div 3\) 余2）。
\(\therefore 5 \equiv 8 \pmod{3}\)，\(11 \equiv 8 \pmod{3}\)。
第3题（传递性）
已知 \(13 \equiv 3 \pmod{5}\)，\(3 \equiv 8 \pmod{5}\)，请判断 \(13 \equiv 8 \pmod{5}\) 是否成立，并说明理由。
【答案】
成立。
由传递性可知，若 \(13 \equiv 3 \pmod{5}\) 且 \(3 \equiv 8 \pmod{5}\)，则 \(13 \equiv 8 \pmod{5}\)。
验证：\(13-8=5\)，是 \(5\) 的整数倍，\(\therefore\) 成立。
第4题（可加性）
已知 \(7 \equiv 2 \pmod{5}\)，\(4 \equiv 9 \pmod{5}\)，求 \(7+4\) 与 \(2+9\) 是否对 \(5\) 同余？并计算两者的差值验证。
【答案】
\(7+4 = 11\)，\(2+9 = 11\)，显然 \(11 \equiv 11 \pmod{5}\)，成立。
差值验证：\((7+4) - (2+9) = 11-11 = 0\)，是 \(5\) 的整数倍。
（也可用可加性直接判断。）
第5题（可乘性）
已知 \(3 \equiv 8 \pmod{5}\)，\(2 \equiv 7 \pmod{5}\)，判断 \(3 \times 2\) 与 \(8 \times 7\) 是否对 \(5\) 同余？验证你的结论。
【答案】
\(3 \times 2 = 6\)，\(8 \times 7 = 56\)，\(56-6=50\)，是 \(5\) 的整数倍（\(50=5\times10\)）。
\(\therefore 6 \equiv 56 \pmod{5}\)，即 \(3 \times 2 \equiv 8 \times 7 \pmod{5}\) 成立。
三、建议
- 同余的定义和性质是后续解决整除问题、余数问题、数论竞赛题的基础，建议讲评时重点强调“\(a-b\) 是 \(m\) 的整数倍”这一核心判断标准。
- 本练习可作为“同余入门”独立练习，也可作为数论专题的课前诊断。