📚 Prime Factors, HCF and LCM | 质因数、最大公因数与最小公倍数
Understanding factors, multiples and prime numbers is a central skill in KS3 Cambridge Mathematics. These ideas build the foundation for fractions, ratios, algebra and problem-solving. In this article, you will learn how to find factor pairs, use divisibility tests, write prime factorisations, and calculate the highest common factor (HCF) and lowest common multiple (LCM) confidently.
理解因数、倍数和质数是 KS3 剑桥数学的一项核心技能。这些概念为分数、比、代数和问题解决打下基础。本文将带你学会如何找因数对、使用整除性检验、写出质因数分解，并自信地计算最大公因数（HCF）和最小公倍数（LCM）。
1. Key Definitions: Factors, Multiples and Primes | 关键定义：因数、倍数与质数
A factor of a whole number divides into it exactly with no remainder. For example, the factors of 12 are 1, 2, 3, 4, 6 and 12.
一个整数的因数是可以整除它而没有余数的数。例如，12 的因数是 1、2、3、4、6 和 12。
A multiple is the product of a number and any whole number. The multiples of 5 include 5, 10, 15, 20, 25 and so on.
倍数是一个数与任意整数的乘积。5 的倍数包括 5、10、15、20、25 等。
A prime number has exactly two distinct factors: 1 and itself. The number 7 is prime because its only factors are 1 and 7.
质数只有两个不同的因数：1 和它本身。7 是质数，因为它的因数只有 1 和 7。
2. Finding All Factor Pairs | 求所有因数对
To find all factors of a number, list factor pairs that multiply to give the number. Start with 1 and the number itself, then test 2, 3, 4 and so on.
要找出一个数的所有因数，列出相乘得到该数的因数对。从 1 和这个数本身开始，然后依次检验 2、3、4 等。
18, the factor pairs are 1 × 18, 2 × 9 and 3 × 6, so the factors are 1, 2, 3, 6, 9 and 18.
按顺序进行，直到因数对开始重复。对于 18，因数对是 1 × 18、2 × 9 和 3 × 6，所以因数是 1、2、3、6、9 和 18。
A common error is stopping too early or missing a middle factor. Always check around the square root of the number.
一个常见错误是过早停止或漏掉中间的因数。一定要在数的平方根附近检查。
3. Divisibility Tests You Must Know | 必须掌握的整除性检验
Divisibility tests help you find factors quickly without long division. The key tests are for 2, 3, 4, 5, 6, 9 and 10.
整除性检验可以帮助你快速找到因数，而无需进行长除法。关键检验针对 2、3、4、5、6、9 和 10。
A number is divisible by 2 if its last digit is even; by 3 if the sum of its digits is divisible by 3; by 5 if the last digit is 0 or 5.
如果一个数的末位是偶数，它能被 2 整除；如果各位数字之和能被 3 整除，它能被 3 整除；如果末位是 0 或 5，它能被 5 整除。
A number is divisible by 4 if the last two digits form a number divisible by 4; by 6 if it is divisible by both 2 and 3; by 9 if the digit sum is divisible by 9.
如果一个数的末两位组成的数能被 4 整除，它能被 4 整除；如果它能同时被 2 和 3 整除，它能被 6 整除；如果各位数字之和能被 9 整除，它能被 9 整除。
A number is divisible by 10 if its last digit is 0. Divisibility by 8 is less common at KS3 but can be tested using the last three digits.
如果一个数的末位是 0，它能被 10 整除。能被 8 整除的检验在 KS3 较少见，但可以用末三位数检验。
| Divisor | Test | Example |
|---|---|---|
| 2 | Last digit is even | 346 ends in 6 |
| 3 | Digit sum divisible by 3 | 411: 4 + 1 + 1 = 6 |
| 4 | Last two digits divisible by 4 | 912: 12 ÷ 4 = 3 |
| 5 | Last digit 0 or 5 | 835 ends in 5 |
| 6 | Divisible by 2 and 3 | 72 is even and digit sum 9 |
| 9 | Digit sum divisible by 9 | 738: 7 + 3 + 8 = 18 |
| 10 | Last digit 0 | 230 ends in 0 |
4. Prime Numbers and Composite Numbers | 质数与合数
Prime numbers have exactly two factors, while composite numbers have more than two factors. The number 1 is neither prime nor composite because it has only one factor.
质数恰好有两个因数，而合数有两个以上的因数。1 既不是质数也不是合数，因为它只有一个因数。
The first ten prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29. Notice that 2 is the only even prime number.
前十个质数是 2、3、5、7、11、13、17、19、23 和 29。注意 2 是唯一的偶质数。
To test whether a number is prime, check divisibility by primes up to its square root. If none divide exactly, the number is prime.
要检验一个数是否为质数，检验它能否被不超过其平方根的质数整除。如果都不能整除，这个数就是质数。
5. Prime Factorisation Using Factor Trees | 用因数树进行质因数分解
Prime factorisation means writing a number as a product of prime numbers. Every whole number greater than 1 has a unique prime factorisation.
质因数分解是指把一个数写成质数的乘积。每个大于 1 的整数都有唯一的质因数分解。
To use a factor tree, split the number into any two factors, then continue splitting composite factors until all branches end in prime numbers.
使用因数树时，把这个数分成任意两个因数，然后继续拆分合数因数，直到所有分支都以质数结束。
nd 10 into 2 × 5. The prime factors are 2, 2, 3 and 5, so 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5.
例如，对 60 进行质因数分解，先分成 6 × 10。然后将 6 分成 2 × 3，将 10 分成 2 × 5。质因数是 2、2、3 和 5，所以 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5。
60 = 2 × 2 × 3 × 5 = 2² × 3 × 5
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