Prime Factors, HCF and LCM | 质因数、最大公因数与最小公倍数

📚 Prime Factors, HCF and LCM | 质因数、最大公因数与最小公倍数

Prime factors, highest common factor (HCF) and lowest common multiple (LCM) are core number skills in IGCSE Mathematics. They appear in direct questions and as hidden steps in fractions, ratios, algebra and real-world time problems. Building a clear method here saves marks across the whole paper.

质因数、最大公因数（HCF）和最小公倍数（LCM）是 IGCSE 数学的核心数论技能。它们既会直接考查，也会隐藏在分数、比、代数以及现实生活中的时间问题里。在这一部分建立清晰的方法，可以在整张试卷中稳定得分。

1. Why This Topic Matters | 为什么要掌握这个主题

IGCSE exam papers often combine HCF and LCM with word problems, fraction simplification and ratios. A reliable prime factor method prevents careless arithmetic errors and makes your working clear to examiners.

IGCSE 考试常把 HCF 和 LCM 与文字题、分数约分和比率结合考查。一套可靠的质因数方法可以避免粗心计算错误，也能让阅卷人更清楚地看到你的解题过程。

The same skills are also needed when adding or subtracting fractions with different denominators, simplifying surds and solving problems that involve equal groups or repeated events.

这些技能还用于异分母分数的加减、根式化简，以及涉及等分组或重复事件的问题。

2. Factors and Multiples Recap | 因数与倍数的回顾

A factor of a number divides exactly into that number, while a multiple is the result of multiplying the number by an integer. For example, the factors of 12 are 1, 2, 3, 4, 6 and 12; the first few multiples of 12 are 12, 24, 36, 48 and 60.

因数是能整除一个数的数，倍数则是该数乘以整数得到的结果。例如，12 的因数有 1、2、3、4、6、12；12 的前几个倍数是 12、24、36、48、60。

Factor: a number that divides exactly into another number with no remainder.
因数：能整除另一个数且没有余数的数。
Multiple: the product of a number and any integer.
倍数：一个数与任意整数相乘得到的积。
Common factor: a factor shared by two or more numbers.
公因数：两个或更多数共同拥有的因数。
Common multiple: a multiple shared by two or more numbers.
公倍数：两个或更多数共同拥有的倍数。

3. Prime Numbers and Composite Numbers | 质数与合数

A prime number has exactly two distinct factors: 1 and itself. For example, 7 is prime because its only factors are 1 and 7. The number 1 is not prime because it has only one factor. A composite number has more than two factors, such as 12.

质数正好有两个不同的因数：1 和它本身。例如，7 是质数，因为它的因数只有 1 和 7。数字 1 不是质数，因为它只有一个因数。合数则有超过两个因数，例如 12。

The prime numbers below 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29. Notice that 2 is the only even prime number, so every even number greater than 2 is composite.

30 以内的质数是 2、3、5、7、11、13、17、19、23 和 29。注意 2 是唯一的偶质数，所以任何大于 2 的偶数都是合数。

4. Prime Factor Decomposition | 质因数分解

Every integer greater than 1 can be written as a unique product of prime factors. This is called the fundamental theorem of arithmetic. Writing a number in prime factor form makes HCF and LCM questions much easier.

每个大于 1 的整数都可以唯一地写成质因数的乘积，这称为算术基本定理。把数字写成质因数形式，会让 HCF 和 LCM 问题简单很多。

For example, 60 can be written as:

例如，60 可以写成：

60 = 2² × 3 × 5

The prime factors are usually written in ascending order. This helps you compare two numbers quickly when finding HCF or LCM.

质因数通常按从小到大排列。这样在求 HCF 或 LCM 时，可以快速比较两个数。

5. Using Factor Trees | 使用因数树

A factor tree splits a number into factor pairs until all end branches are prime. For 72, split it into 8 and 9, then 8 = 2 × 4, 4 = 2 × 2, 9 = 3 × 3, giving:

因数树把一个数逐层拆成因数对，直到所有末端都是质数。对于 72，可以先拆为 8 和 9，再把 8 拆为 2 × 4，4 拆为 2 × 2，9 拆为 3 × 3，得到：

72 = 2³ × 3²

Step 1: Write the number at the top of the tree.
步骤 1：把数字写在因数树的顶部。
Step 2: Split it into any factor pair that is not 1 and the number itself.
步骤 2：把它拆成任意一对因数，不要使用 1 和它本身。
Step 3: Circle prime factors as soon as they appear.
步骤 3：一旦出现质数就圈起来。
Step 4: Continue until every branch ends in a prime number.
步骤 4：继续拆分，直到每个分叉都结束于质数。
Step 5: Write the prime factors as powers, arranged in ascending order.
步骤 5：把相同质因数写成乘方，并按从小到大排列。

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

The highest common factor (HCF) is the largest number that divides exactly into two or more given numbers. You can find it by listing factors, but prime factor form is faster and less likely to cause mistakes.

最大公因数（HCF）是能同时整除两个或多个给定数的最大数。你可以通过列出因数来寻找，但使用质因数形式更快，也更不容易出错。

For 48 and 60, write each number as a product of prime factors:

对于 48 和 60，先把每个数写成质因数乘积：

48 = 2⁴ × 3

60 = 2² × 3 × 5

Multiply only the common prime factors, using the lowest power each time. The common prime factors are 2 and 3, so:

只计算共同的质因数，每次使用较低次幂。共同的质因数是 2 和 3，因此：

HCF = 2² × 3 = 12

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

The lowest common multiple (LCM) is the smallest number that is a multiple of two or more given numbers. Listing multiples can work for small numbers, but prime factor form is much more efficient for larger values.

最小公倍数（LCM）是两个或多个给定数的倍数中最小的一个。对于较小的数字，可以列出倍数来寻找；但对于较大的数值，使用质因数形式更高效。

For 8 and 12, write the prime factor forms:

对于 8 和 12，写出质因数形式：

8 = 2³

12 = 2² × 3

Use every prime factor that appears, but take the highest power of each. So the LCM is:

使用所有出现过的质因数，但取每个质因数的最高次幂。因此：

LCM = 2³ × 3 = 24

8. HCF and LCM from Prime Factor Form | 由质因数形式求 HCF 和 LCM

Once you have prime factor forms, use these two rules:

得到质因数形式后，可以使用以下两条规则：

HCF: multiply common prime factors with their lowest powers.
HCF：把共同质因数按较低次幂相乘。
LCM: multiply all prime factors with their highest powers.
LCM：把全部质因数按最高次幂相乘。

Worked example for 48 and 180:

48 和 180 的示例：

48 = 2⁴ × 3

180 = 2² × 3² × 5

For HCF, take common prime factors 2 and 3 with the lowest powers:

求 HCF 时，取共同质因数 2 和 3，并使用较低次幂：

HCF = 2² × 3 = 12

For LCM, take all prime factors 2, 3 and 5 with the highest powers:

求 LCM 时，取全部质因数 2、3 和 5，并使用最高次幂：

LCM = 2⁴ × 3² × 5 = 720

9. Working with Indices | 指数运算入门

Prime factor form uses indices, so you should be confident with the basic rules. When multiplying powers of the same base, add the exponents. When dividing powers of the same base, subtract the exponents. When raising a power to another power, multiply the exponents.

质因数形式会使用指数，因此你需要熟练基本指数规则。同底数幂相乘时，指数相加；同底数幂相除时，指数相减；幂的乘方时，指数相乘。

2² × 2³ = 2⁵

2⁵ ÷ 2² = 2³

(2²)³ = 2⁶

These rules help you compare powers quickly when several prime factors are involved. Always write the final answer in index form unless the question asks for a numerical value.

这些规则可以帮助你在涉及多个质因数时快速比较幂。除非题目要求具体数值，否则最终答案应保留指数形式。

10. Common Word Problems | 常见应用题

Typical IGCSE scenarios include buses leaving a station at different intervals, lights flashing together, and cutting lengths into equal pieces. The key is deciding whether the situation asks for a largest common size or a next simultaneous time.

IGCSE 常见情境包括公交车按不同时间间隔发车、灯光同时闪烁，以及把不同长度剪成相等小段。关键是判断题目要求的是最大共同尺寸，还是下一次同时发生的时间。

If two bells ring every 6 minutes and 8 minutes, the LCM tells when they will next ring together:

如果两个铃铛每 6 分钟和 8 分钟响一次，LCM 可以给出下一次同时响铃的时间：

LCM(6, 8) = 24 minutes

If you cut ribbons of lengths 24 cm and 36 cm into equal pieces with no waste, the HCF tells the largest possible piece length:

如果把 24 cm 和 36 cm 的丝带剪成相等长度且没有剩余，HCF 给出每条的最大长度：

HCF(24, 36) = 12 cm

11. Exam Tips | 考试技巧

Show your method clearly by writing the prime factor form for every number before finding the HCF or LCM. This earns method marks even if you make a small arithmetic slip later.

在求 HCF 或 LCM 之前，先把每个数写成质因数形式，清晰展示方法。这样即使之后出现小的计算错误，也能获得步骤分。

A useful check is:

一个有效的检验方法是：

HCF × LCM = product of the two original numbers

For 48 and 180, this gives 12 × 720 = 8640 and 48 × 180 = 8640, so the answers are consistent. Remember that multiplying the two original numbers directly gives a common multiple, but not always the lowest common multiple.

对于 48 和 180，12 × 720 = 8640，而 48 × 180 也等于 8640，因此答案一致。要记住，直接相乘得到的是一个公倍数，但不一定是最小公倍数。

12. Quick Practice Check | 快速自测

Try these three IGCSE-style questions to test your understanding before moving on.

在继续学习之前，试着完成以下三道 IGCSE 风格的问题来检验理解。

Find the HCF and LCM of 72 and 90.
求 72 和 90 的 HCF 和 LCM。
Write 126 as a product of prime factors.
把 126 写成质因数乘积。
Two traffic lights cycle every 45 seconds and 60 seconds. After how many seconds do they next change together?
两个交通信号灯分别每 45 秒和 60 秒变换一次。它们下一次同时变换是在多少秒之后？

Answers:

答案：

72 = 2³ × 3², 90 = 2 × 3² × 5, HCF = 18, LCM = 360

126 = 2 × 3² × 7

LCM(45, 60) = 180 seconds = 3 minutes

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Prime Factors, HCF and LCM | 质因数、最大公因数与最小公倍数