Number Theory Essentials: Primes, Factors and Multiples | 数论基础：质数、因数与倍数

📚 Number Theory Essentials: Primes, Factors and Multiples | 数论基础：质数、因数与倍数

Number theory forms the backbone of arithmetic in the IGCSE Mathematics syllabus. Understanding primes, factors and multiples is not merely about memorising definitions — it is about developing a toolkit for solving problems efficiently across algebra, fractions, ratios and even geometry. This article walks through every essential concept, with worked examples and exam-style insights, to ensure you master this foundational topic with confidence.

数论是 IGCSE 数学课程中算术部分的基石。理解质数、因数和倍数不仅是为了记住定义，更是为了培养一套高效解题的工具，这套工具在代数、分数、比例乃至几何中都有广泛应用。本文将逐一讲解所有核心概念，配合例题和考试视角的剖析，帮助你扎实掌握这一基础主题。

1. What Are Prime Numbers? | 什么是质数？

A prime number is a positive integer greater than 1 that has exactly two distinct positive divisors: 1 and itself. In other words, a prime number cannot be formed by multiplying two smaller positive integers. For example, 7 is prime because its only divisors are 1 and 7. The number 1 is not prime, and 0 is not prime either — 0 has infinitely many divisors.

质数是大于 1 的正整数，且恰好只有两个不同的正因数：1 和它本身。换言之，质数不能由两个更小的正整数相乘得到。例如，7 是质数，因为它的因数只有 1 和 7。数字 1 不是质数，0 也不是质数——因为 0 有无限多个因数。

Here is the list of prime numbers less than 50:

以下是小于 50 的质数列表：

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
Note that 2 is the only even prime number — every other even number has 2 as a divisor.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47
注意 2 是唯一的偶质数——所有其他偶数都有因数 2。

A positive integer n is prime if and only if it has exactly two positive divisors.

正整数 n 是质数当且仅当它恰好有两个正因数。

2. How to Test for Primality | 如何判断质数？

To test whether a number n is prime, you only need to check divisibility by prime numbers up to the square root of n. This is because if n has a divisor greater than √n, then it must also have a divisor smaller than √n. For example, to test whether 97 is prime, check divisibility by 2, 3, 5, and 7 — all primes ≤ √97 ≈ 9.8. Since none divide 97, it is prime.

要判断一个数 n 是否为质数，只需检查不超过 √n 的质数是否能整除 n。原因在于：如果 n 有一个大于 √n 的因数，那么它必然也有一个小于 √n 的因数。例如，要判断 97 是否为质数，只需检查 2、3、5、7——所有不超过 √97 ≈ 9.8 的质数。由于这些数都不能整除 97，因此 97 是质数。

This method is known as trial division and is highly effective for numbers in the IGCSE range (typically less than 200). Below is a quick divisibility checklist:

这种方法称为试除法，对于 IGCSE 范围内的数（通常小于 200）非常有效。以下是一个快速整除性检查清单：

Divisible by 2: the last digit is even.
Divisible by 3: the sum of digits is divisible by 3.
Divisible by 5: the last digit is 0 or 5.
Divisible by 7: double the last digit, subtract from the rest; repeat. Check if result is divisible by 7.

能被 2 整除：末位数字是偶数。
能被 3 整除：各位数字之和能被 3 整除。
能被 5 整除：末位数字是 0 或 5。
能被 7 整除：将末位数字加倍，从其余部分减去；重复此过程。检查结果是否能被 7 整除。

3. Prime Factorisation | 质因数分解

Prime factorisation (also called prime decomposition) is the process of expressing a positive integer as a product of prime numbers. Every integer greater than 1 has a unique prime factorisation, regardless of the order of the factors — this is known as the Fundamental Theorem of Arithmetic.

质因数分解（也称素因子分解）是将一个正整数表示为质数乘积的过程。每个大于 1 的整数都有唯一的质因数分解，无论因数的顺序如何——这被称为算术基本定理。

To perform prime factorisation, use the method of repeated division:

要进行质因数分解，可以使用反复除法的方法：

Example: Express 360 as a product of its prime factors.

示例：将 360 表示为质因数的乘积。

Divide by 2 repeatedly:

反复除以 2：

360 ÷ 2 = 180
180 ÷ 2 = 90
90 ÷ 2 = 45

45 is not divisible by 2, so move to 3:

45 不能被 2 整除，所以换用 3：

45 ÷ 3 = 15
15 ÷ 3 = 5

5 is prime. Therefore:

5 是质数。因此：

360 = 2³ × 3² × 5

An alternative method is the factor tree, where you break the number into any two factors and continue until all branches end at primes. Both methods produce the same result.

另一种方法是因数分解树，将数拆分为任意两个因数，不断分解直到所有分支末端都是质数。两种方法会得到相同的结果。

4. Finding Factors | 求因数

A factor (or divisor) of a positive integer n is any positive integer that divides n without leaving a remainder. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12. Note that factors always come in pairs: (1, 12), (2, 6), and (3, 4).

正整数 n 的因数（或除数）是任何能整除 n 且不留下余数的正整数。例如，12 的因数是 1、2、3、4、6 和 12。注意因数总是成对出现：(1, 12)、(2, 6) 和 (3, 4)。

To find all factors of a number systematically, use prime factorisation to list every combination of prime factors. For example, 12 = 2² × 3. The factors can be generated by choosing exponents 0–2 for 2 and 0–1 for 3:

要系统地找出一个数的所有因数，可以使用质因数分解列出质因数的所有组合。例如，12 = 2² × 3。因数可以通过选择 2 的指数 0–2 和 3 的指数 0–1 来生成：

2⁰ × 3⁰ = 1
2¹ × 3⁰ = 2
2² × 3⁰ = 4
2⁰ × 3¹ = 3
2¹ × 3¹ = 6
2² × 3¹ = 12

This confirms the complete list: 1, 2, 3, 4, 6, 12.

这确认了完整列表：1、2、3、4、6、12。

5. Highest Common Factor (HCF) | 最大公因数（HCF）

The highest common factor (HCF) of two or more numbers is the largest positive integer that divides all of them exactly. It is sometimes called the greatest common divisor (GCD). For example, the HCF of 18 and 24 is 6, because 6 is the largest number that divides both without remainder.

两个或多个数的最大公因数（HCF）是能整除所有这些数的最大正整数。它有时也叫最大公约数（GCD）。例如，18 和 24 的最大公因数是 6，因为 6 是能同时整除这两个数的最大数。

To find the HCF using prime factorisation, follow these steps:

要使用质因数分解求 HCF，可以遵循以下步骤：

Write each number as a product of prime factors.
Identify the common prime factors.
For each common prime, take the smallest exponent.
Multiply the resulting prime powers together.

将每个数写成质因数的乘积。
找出共同的质因数。
对于每个共同质数，取最小指数。
将得到的质数幂相乘。

Example: Find the HCF of 36 and 60.

示例：求 36 和 60 的 HCF。

36 = 2² × 3²
60 = 2² × 3 × 5

Common primes: 2 (smallest exponent = 2) and 3 (smallest exponent = 1). Therefore:

共同质数：2（最小指数 = 2）和 3（最小指数 = 1）。因此：

HCF(36, 60) = 2² × 3 = 4 × 3 = 12

6. Lowest Common Multiple (LCM) | 最小公倍数（LCM）

The lowest common multiple (LCM) of two or more numbers is the smallest positive integer that is a multiple of all of them. For example, the LCM of 4 and 6 is 12, because 12 is the smallest number that is divisible by both 4 and 6.

两个或多个数的最小公倍数（LCM）是所有数的公倍数中最小的正整数。例如，4 和 6 的最小公倍数是 12，因为 12 是同时能被 4 和 6 整除的最小正整数。

To find the LCM using prime factorisation:

要使用质因数分解求 LCM：

Write each number as a product of prime factors.
List every prime factor that appears, with the largest exponent found across all numbers.
Multiply these together.

将每个数写成质因数的乘积。
列出所有出现过的质因数，并取各数中最大的指数。
将这些相乘。

Example: Find the LCM of 36 and 60.

示例：求 36 和 60 的 LCM。

36 = 2² × 3²
60 = 2² × 3 × 5

Primes present: 2 (max exponent = 2), 3 (max exponent = 2), 5 (max exponent = 1). Therefore:

出现的质数：2（最大指数 = 2）、3（最大指数 = 2）、5（最大指数 = 1）。因此：

LCM(36, 60) = 2² × 3² × 5 = 4 × 9 × 5 = 180

7. The Relationship Between HCF and LCM | HCF 与 LCM 的关系

For any two positive integers a and b, there is an important connection between their product, their HCF and their LCM:

对于任意两个正整数 a 和 b，它们的乘积、HCF 和 LCM 之间有一个重要的关系：

a × b = HCF(a, b) × LCM(a, b)

This formula is extremely useful when one of the values is unknown. For example, if HCF(14, n) = 2 and LCM(14, n) = 84, then:

这个公式在其中一个值未知时非常有用。例如，若 HCF(14, n) = 2 且 LCM(14, n) = 84，则：

14 × n = 2 × 84
14n = 168
n = 12

This relationship works only for two numbers. If working with three or more numbers, the simple product relationship does not hold.

这个关系只适用于两个数。对于三个或更多数，简单的乘积关系不再成立。

8. Real-World Applications of HCF and LCM | HCF 与 LCM 的实际应用

HCF and LCM appear frequently in real-world contexts, and IGCSE exams often frame questions around practical scenarios. Understanding when to apply each concept is crucial:

HCF 和 LCM 经常出现在实际情境中，IGCSE 考试也常以实际场景出题。理解何时应用哪个概念至关重要：

HCF in splitting/grouping problems: If you need to divide items into equal groups without leftovers, use the HCF. For example, cutting two ribbons of lengths 18 cm and 24 cm into equal shorter pieces of maximum length — the answer is the HCF, 6 cm.

HCF 用于分配/分组问题：如果需要将物品平均分组且没有剩余，就使用 HCF。例如，将两根长度分别为 18 cm 和 24 cm 的彩带剪成等长的较短段，且每段要最长——答案是 HCF，即 6 cm。

LCM in scheduling/repeating events: If two events repeat at different intervals, the time until both occur together again is the LCM of their intervals. For example, if a person takes medicine every 4 hours and another every 6 hours, both will be taken together again after LCM(4, 6) = 12 hours.

LCM 用于时间表/重复事件：如果两个事件以不同的间隔重复，那么它们再次同时发生的时间是两个间隔的 LCM。例如，一个人每 4 小时吃药，另一个人每 6 小时吃药，两者会再次同时服药的时间是 LCM(4, 6) = 12 小时后。

Scenario	Which to use	Reason
Dividing into equal groups	HCF	Largest common divisor
Repeating events coincide	LCM	Smallest common multiple

场景	使用哪个	原因
分成等组	HCF	最大公共因数
重复事件重合	LCM	最小公共倍数

9. Working with Powers and Roots | 幂与根的处理

Prime factorisation makes working with powers and roots much simpler. To find the square root of a perfect square, divide each exponent in the prime factorisation by 2. For example:

质因数分解使处理幂和根变得简单得多。要求一个完全平方数的平方根，只需将质因数分解中每个指数除以 2。例如：

144 = 2⁴ × 3²
√144 = 2² × 3 = 4 × 3 = 12

Similarly, to find the cube root of a perfect cube, divide each exponent by 3:

类似地，要求一个完全立方数的立方根，只需将每个指数除以 3：

216 = 2³ × 3³
∛216 = 2 × 3 = 6

This technique also helps simplify radicals, such as √72. Since 72 = 2³ × 3², rewrite as √(2² × 2 × 3²) = 2 × 3 × √2 = 6√2.

这种技巧也有助于化简根式，例如 √72。因为 72 = 2³ × 3²，改写为 √(2² × 2 × 3²) = 2 × 3 × √2 = 6√2。

10. Exam Tips and Common Mistakes | 考试技巧与常见错误

Even strong students lose marks on number theory questions due to careless errors. Here are the most common pitfalls and how to avoid them:

即使是优秀的学生也可能因为粗心错误在数论题上失分。以下是最常见的陷阱及如何避开它们：

Forgetting that 1 and 0 are not prime: Always start testing from 2.
Mixing up HCF and LCM signs: HCF uses smallest exponents; LCM uses largest exponents. Write this down before solving.
Incomplete factorisation: Ensure the final product contains only prime numbers.
Using the a × b = HCF × LCM formula for three numbers: It does not apply — solve step by step.
Listing multiples incorrectly: Be systematic; always check that the LCM is indeed divisible by all original numbers.

忘记 1 和 0 不是质数：始终从 2 开始检验。
混淆 HCF 与 LCM 的求法：HCF 取最小指数；LCM 取最大指数。解题前先写下这一点。
分解不完整：确保最终乘积中只包含质数。
对三个数使用 a × b = HCF × LCM 公式：该公式不适用——应逐步求解。
倍数列举错误：要系统化；始终检查 LCM 确实能被所有原始数整除。

It is also wise to verify answers in the exam. For example, after computing the LCM of 36 and 60 as 180, check that 180 ÷ 36 = 5 and 180 ÷ 60 = 3 — both exact.

在考试中检查答案也很明智。例如，算出 36 和 60 的 LCM 为 180 后，检查 180 ÷ 36 = 5 且 180 ÷ 60 = 3——都能整除。

11. Summary of Key Formulas | 核心公式总结

To conclude this revision guide, here is a compact summary of the essential facts and formulas you need for your IGCSE exam:

为总结这本复习指南，以下是 IGCSE 考试所需核心事实与公式的简明汇总：

Prime number: n > 1 with exactly two divisors (1 and itself).
Prime factorisation: express n as a product of prime powers, e.g. 360 = 2³ × 3² × 5.
HCF: product of common primes with smallest exponents.
LCM: product of all primes with largest exponents.
Key formula: a × b = HCF(a, b) × LCM(a, b), valid for two numbers only.

质数：n > 1 且恰好有两个因数（1 和它本身）。
质因数分解：将 n 表示为质数幂的乘积，例如 360 = 2³ × 3² × 5。
HCF：共同质因数取最小指数的乘积。
LCM：所有质因数取最大指数的乘积。
关键公式：a × b = HCF(a, b) × LCM(a, b)，仅适用于两个数。

Mastering these concepts takes practice. Start by working through the examples above without looking at the solutions, then move on to past paper questions. With consistent effort, number theory will become one of the most reliable scoring areas in your mathematics examination.

掌握这些概念需要练习。先不看解答独立完成上述例题，然后转向历年真题。只要持续努力，数论将成为你数学考试中最稳定的得分板块之一。

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Number Theory Essentials: Primes, Factors and Multiples | 数论基础：质数、因数与倍数