📚 Mastering Prime Factors, HCF and LCM | 掌握质因数、HCF 与 LCM
Welcome to a focused revision guide on prime factorisation, highest common factor (HCF) and lowest common multiple (LCM). These number skills lie at the heart of the Cambridge KS3 Mathematics curriculum and appear regularly in problem-solving contexts. Whether you are building a solid foundation for future algebra work or simply aiming to boost your mental arithmetic, a clear understanding of factors, multiples and primes will make a real difference to your confidence and accuracy.
欢迎阅读这篇关于质因数分解、最高公因数（HCF）和最小公倍数（LCM）的重点复习文章。这些数论技能是剑桥 KS3 数学课程的核心，经常在解决问题中出现。无论你是在为未来的代数学习打基础，还是只是想提高心算能力，清楚地理解因数、倍数与质数都会大大增强你的信心与准确性。
1. Factors and Multiples | 因数与倍数
A factor of a whole number is any whole number that divides into it exactly, leaving no remainder. For example, the factors of 12 are 1, 2, 3, 4, 6 and 12 because 12 can be divided evenly by each of these. A multiple of a number is the result of multiplying that number by any integer. So the multiples of 3 include 3, 6, 9, 12, 15 and so on. Notice that a number can be both a factor and a multiple in different contexts.
一个整数的因数是能够整除它、没有余数的任何整数。例如，12 的因数有 1、2、3、4、6 和 12，因为 12 可以被这些数整除。倍数是一个数与任何整数相乘得到的结果。因此 3 的倍数有 3、6、9、12、15 等。请注意，在有的情境下一个数既可以是因数也可以是倍数。
- Factor – divides a number exactly without leaving a remainder
- 因数 – 能整除一个数，且没有余数的整数
- Multiple – the product of the number and any whole number
- 倍数 – 该数与任意整数相乘的积
2. Prime and Composite Numbers | 质数与合数
A prime number is a whole number greater than 1 that has exactly two distinct factors: 1 and itself. Examples of prime numbers up to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29. Remember that 2 is the only even prime. A composite number has more than two factors. For instance, 8 is composite because it can be divided by 1, 2, 4 and 8. The number 1 is neither prime nor composite – it has only one factor.
质数是大于 1 且只有两个不同因数的整数：1 和它本身。30 以内的质数有 2、3、5、7、11、13、17、19、23 和 29。记住 2 是唯一的偶质数。合数有多于两个因数。例如 8 是合数，因为它可以被 1、2、4、8 整除。数字 1 既不是质数也不是合数——它只有一个因数。
Prime numbers are the building blocks of all whole numbers because every composite number can be written as a product of primes. This fact is called the fundamental theorem of arithmetic and it underpins prime factorisation.
质数是所有整数的“积木块”，因为每一个合数都可以写成质数的乘积。这一事实被称为算术基本定理，也是质因数分解的基础。
3. Prime Factorisation Using Factor Trees | 使用因子树进行质因数分解
To perform prime factorisation, we break a composite number down into factors until every branch ends in a prime number. One of the most visual methods is the factor tree. Start with the number, split it into any factor pair, and continue splitting each composite factor. Keep going until you only have prime numbers at the ends. Finally, write the original number as a product of these primes.
进行质因数分解时，我们把合数分解成因数，直到每个分支末端都是质数。最直观的方法之一是因子树。从该数开始，将其拆分成任意一对因数，继续拆分每一个合数因数，直至末端全剩下质数。最后，将原数写成这些质数的乘积。
ll primes gives 24 = 2 × 2 × 2 × 3. We can write this more compactly using index notation.
例如：将 24 写成质数的乘积。一棵因子树可以先将 24 拆成 4 × 6，再将 4 拆成 2 × 2，6 拆成 2 × 3。收集全部的质数得到 24 = 2 × 2 × 2 × 3。我们还可以用指数形式更简洁地表示。
24 = 2 × 2 × 2 × 3 = 2³ × 3
4. Expressing Prime Factors in Index Form | 用指数形式表示质因数
Once you have listed all the prime factors, it is good practice to group them in ascending order and use powers (indices) s you compare factorisations quickly and is essential for finding HCF and LCM efficiently.
当你列出所有质因数后，最好按从小到大的顺序排列，并用指数简化重复的乘法。对于 120，因子树得到 2 × 2 × 2 × 3 × 5。将三个 2 组合写成 2³。于是我们得到 120 = 2³ × 3 × 5。指数形式有助于快速比较分解式，是高效求 HCF 和 LCM 的关键。
= 120, which matches the original number. This step confirms your factor tree is correct.
一定要把质因数乘回去检验。对于 2³ × 3 × 5，2³ = 8，8 × 3 = 24，24 × 5 = 120，与原数相符。这一步骤可以确认你的因子树是正确的。
5. Highest Common Factor (HCF) | 最高公因数
The highest common factor (HCF) of two or more numbers is the largest whole number that divides all of them exactly. A simple method is to list all factors of each number and then pick the biggest one they share. For example, to find the HCF of 18 and 24: factors of 18 are 1, 2, 3, 6, 9, 18; factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. The common factors are 1, 2, 3, 6, so the HCF is 6.
两个或多个数的最高公因数是能整除所有给数的最大整数。一个简单方法是列出每个数的全部因数，然后选出它们共有的最大一个。例如，求 18 和 24 的 HCF：18 的因数为 1、2、3、6、9、18；24 的因数为 1、2、3、4、6、8、12、24。公因数是 1、2、3、6，因此 HCF 是 6。
While the listing method works well for small numbers, it becomes slow for larger numbers. That is why the prime factor method (introduced next) is more powerful.
虽然列表法对较小的数很管用，但当数变大时会很慢。因此接下来介绍的质因数法会更为强大。
6. Finding HCF Using Prime Factorisation | 用质因数分解求 HCF
To find the HCF with prime factorisation, write each number as a product of prime factors in index form. Then identify the common prime bases and, for each base, take the lowest power 2², and the lowest power of 3 is 3 (to the power 1). Therefore HCF = 2² × 3 = 4 × 3 = 12.
使用质因数分解求 HCF 时，先将每个数写成质因数的指数形式。接着找出公共的质因数底数，并对每个底数取出现过的最小指数，再把它们乘起来。以 36 和 60 为例，分解得 36 = 2² × 3²，60 = 2² × 3 × 5。公共底数是 2 和 3。两者中 2 的最小指数是 2²，3 的最小指数是 3（指数 1）。因此 HCF = 2² × 3 = 4 × 3 = 12。
36 = 2² × 3² | 60 = 2² × 3 × 5 | HCF = 2² × 3 = 12
The rule is simple: common primes × smallest indices. Always write the repeated multiplication clearly to avoid dropping a factor.
规则很简单：公共质因数 × 最小指数。始终把乘法写清楚，以免漏掉因数。
7. Lowest Common Multiple (LCM) | 最小公倍数
The lowest common multiple (LCM) ultiples of 5 are 5, 10, 15, 20, 25 ... The first shared multiple is 20, so LCM(4,5) = 20.
两个或多个数的最小公倍数是同时为它们倍数的最小正数。一种求法是依次列出每个数的倍数，直到找到第一个公共倍数。对于 4 和 5，4 的倍数为 4、8、12、16、20、24…… 5 的倍数为 5、10、15、20、25…… 第一个公共倍数是 20，所以 LCM(4,5) = 20。
Listing multiples can be quick for small numbers but becomes impractical for larger ones or when you need a precise answer under time pressure. The prime factor method is far more reliable and will be covered now.
列表法对小数字比较快，但对于大数字或在时间紧迫需要精确答案时就不太适用。更可靠的方法是质因数法，下面将详细说明。
8. Finding LCM Using Prime Factorisation | 用质因数分解求 LCM
With prime factorisation, the LCM is found by writing each number in index form, then taking every prime base that appears and using the highest power ghest power of 3 is 3², and the highest of 5 is 5. So LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180.
使用质因数分解求 LCM 时，先把每个数写成指数形式，然后取出所有出现过的质数底数，并对每个底数使用最高的指数，再把它们乘起来。继续用 36 和 60 为例。质因数为 36 = 2² × 3²，60 = 2² × 3 × 5。出现的底数有 2、3 和 5。2 的最高指数是 2²，3 的最高指数是 3²，5 的最高指数是 5。因此 LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180。
36 = 2² × 3² | 60 = 2² × 3 × 5 | LCM = 2² × 3² × 5 = 180
Contrast this with the HCF calculation: for HCF we used the smallest powers of common primes, whereas for LCM we use the largest powers of all primes present. This distinction is critical.
将此与 HCF 计算相对比：求 HCF 时，我们对公共质因数取最小的指数；而求 LCM 时，我们对所有出现的质因数取最大的指数。理清这一区别至关重要。
9. Word Problems Involving HCF and LCM | 涉及 HCF 和 LCM 的应用题
Many real-life problems rely on HCF or LCM. A HCF problem l often involves splitting items into equal groups or finding the largest possible size that fits measurements exactly. For example: “ cm by 180 cm. What is the side length of the largest square tile that can cover it exactly?” s, so we need HCF(240,180). 240 = 2⁴ × 3 × 5, 180 = 2² × 3² × 5. HCF = 2² × 3 × 5 = 60 cm.
许多现实问题依赖 HCF 或 LCM。HCF 题通常涉及将物品等分或找出能够恰好适合尺寸的最大规格。例如：“一块长方形地板长 240 厘米，宽 180 厘米。最大正方形地砖的边长是多少才能正好铺满？”边长必须能整除两个维度，所以我们要求 HCF(240,180)。分解得 240 = 2⁴ × 3 × 5, 180 = 2² × 3² × 5。HCF = 2² × 3 × 5 = 60 cm。
An LCM problem often deals with events that repeat at different intervals. For instance: “ e departs every 12 minutes, the other every 18 minutes. When will they next leave together?” . They leave together every 36 minutes, so the next simultaneous departure is at 8:36 a.m.
LCM 题常涉及不同重复间隔的事件。例如：“两辆公交车在上午 8:00 同时从总站出发。一辆每 12 分钟一班，另一辆每 18 分钟一班。它们下一次同时发车是什么时候？”我们需要求 12 和 18 的 LCM。12 = 2² × 3, 18 = 2 × 3²。LCM = 2² × 3² = 36。它们每 36 分钟同时发车，因此下一次同时发车是上午 8:36。
Read the question carefully – if keywords suggest ‘greatest size’ or ‘largest number of groups’, think HCF. If they point to ‘next time together’ or ‘smallest common length’, think LCM.
仔细读题——如果关键词提示“最大尺寸”或“最多组数”，就考虑 HCF；如果指向“下一次同时发生”或“最小公共长度”，就考虑 LCM。
10. Common Mistakes and How to Avoid Them | 常见错误及避免方法
One common mistake is confusing HCF with LCM. A quick check: HCF will always be less than or equal to the smallest number, whereas LCM will be greater than or equal to the largest number (except when numbers are multiples of each other). Another frequent error is forgetting that 1 is not a prime. Writing a factor tree incorrectly, such as missing a factor pair, can lead to wrong prime factorisation. Always multiply back to verify.
一个常见错误是将 HCF 和 LCM 混淆。快速检验：HCF 总是小于或等于最小的那个数，而 LCM 总是大于或等于最大的那个数（除非它们互为倍数）。另一个常见错误是忘记 1 不是质数。画因子树出错，例如漏掉一对因数，会导致错误的质因数分解。一定要乘回去检验。
Some learners use the ‘listing’ method on very large numbers and run out of time. For numbers above 50, using prime factorisation is far safer. Finally, when writing index form, always arrange primes in ascending order and use dot or cross symbols clearly; avoid missing a prime base when comparing HCF and LCM.
有些学生用列表法处理很大的数字，结果浪费时间。对于大于 50 的数字，使用质因数分解要稳妥得多。最后，书写指数形式时，一定要按从小到大的顺序排列质数，清楚地使用点或乘号；在比较 HCF 和 LCM 时，不要漏掉质数底数。
Practice with a range of problems, and you will soon spot the patterns that make these topics predictable and rewarding.
通过练习各种题目，你很快就会发现其中的规律，让这些主题变得可预测并且充满成就感。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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