Prime Factorisation: Breaking Down Numbers | 质因数分解（素因数分解）

📚 Prime Factorisation: Breaking Down Numbers | 质因数分解（素因数分解）

Every integer can be written as a product of prime numbers in exactly one way (ignoring order). This is called the Fundamental Theorem of Arithmetic, and it is the cornerstone of many IGCSE questions on factors, HCF and LCM.

每个整数都可以唯一地表示为若干个质数的乘积（忽略顺序）。这就是算术基本定理，也是 IGCSE 考试中因数、最大公因数与最小公倍数相关题目的基石。

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

A prime number has exactly two distinct positive factors: 1 and itself. The first few primes are 2, 3, 5, 7, 11 and 13. A composite number has more than two positive factors. The number 1 is neither prime nor composite.

质数恰好有两个不同的正因数：1 和它本身。最小的几个质数是 2、3、5、7、11、13。合数则多于两个正因数。数字 1 既不是质数也不是合数。

• Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19 …
• Composite numbers: 4, 6, 8, 9, 10, 12, 14, 15 …
• Remember: 2 is the only even prime number.

• 质数：2、3、5、7、11、13、17、19 ……
• 合数：4、6、8、9、10、12、14、15 ……
• 注意：2 是唯一的偶质数。

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2. Divisibility Tests | 整除判断法则

Before factorising large numbers, quickly test divisibility by small primes. This saves time in an exam and reduces errors.

在分解大数之前，先用小质数测试整除性，这样既能节省考试时间，又能减少错误。

Divisor	Rule	Example with 154
2	Last digit even	154 ends in 4 → divisible by 2
3	Sum of digits divisible by 3	1+5+4=10 → not divisible by 3
5	Ends in 0 or 5	154 does not end in 0/5 → not divisible
7	Double last digit, subtract from rest, repeat	15 − 2×4 = 7 → divisible by 7
11	Alternating sum of digits divisible by 11	1 − 5 + 4 = 0 → divisible by 11

154 ÷ 2 = 77, 77 ÷ 7 = 11, 11 ÷ 11 = 1

So 154 = 2 × 7 × 11. All three factors are prime.

因此 154 = 2 × 7 × 11。这三个因数均为质数。

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3. Factor Tree Method | 因数分解树法

A factor tree helps you visualise breaking down a number by splitting it into two factors, then breaking each composite factor further until only primes remain.

因数树通过将一个数字分成两个因数，再继续分解每个合数因数，直到只剩下质数为止，从而帮助直观展示分解过程。

154
  /  \
 2  77
    /  \
   7  11

Write the final answer in ascending order: 154 = 2 × 7 × 11.

最终答案按升序写出：154 = 2 × 7 × 11。

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4. Division Method | 短除法

An alternative is to divide by the smallest prime factors one by one until you reach 1.

另一种方法是按从小到大的质因数逐步相除，直到商为 1 为止。

2)154
7)77
11)11
 1

lt: 154 = 2 × 7 × 11. The division method is compact and often preferred in written exams.

短除法得到相同结果：154 = 2 × 7 × 11。它书写紧凑，在笔试中经常被优先采用。

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5. Writing a Number as a Product of Primes | 将数写成质数的乘积

Any composite number can be written as a product of primes. This is sometimes called prime factor decomposition.

任何合数都可以写成质数的乘积，这种形式也称为质因数分解。

• Check if the number is divisible by the smallest prime first.
• Continue dividing by the same prime until it no longer divides evenly.
• Then move to the next prime.
• Finish when the quotient is 1.

• 先用最小的质数检验是否能整除。
• 持续用同一个质数除，直到不能整除为止。
• 然后换下一个质数继续除。
• 当商为 1 时停止。

154 = 2 × 7 × 11

ctors, so it remains as a product of three distinct primes. Wait, what about 2¹ × 7¹ × 11¹? Yes, the index form is allowed but optional.

若可能则使用指数形式：154 没有重复的质因数，因此可以写成三个不同质数的乘积。实际上也可以写作 2¹ × 7¹ × 11¹，但通常指数 1 省略。

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6. Listing All Factors of a Number | 列出一个数的所有因数

Once you have the prime factorisation, you can systematically list every factor of the number by combining the prime factors.

有了质因数分解后，就可以通过组合质因数来系统地列出该数的所有正因数。

For 154 = 2 × 7 × 11, list all products of subsets of {2, 7, 11}:

对于 154 = 2 × 7 × 11，列出 {2, 7, 11} 的所有子集乘积：

Subset	Product
{} (empty)	1
{2}	2
{7}	7
{11}	11
{2,7}	14
{2,11}	22
{7,11}	77
{2,7,11}	154

Thus the factors of 154 are 1, 2, 7, 11, 14, 22, 77 and 154.

因此 154 的因数有 1、2、7、11、14、22、77、154。

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7. HCF and LCM Using Prime Factors | 利用质因数求最大公因数和最小公倍数

The highest common factor (HCF) and lowest common multiple (LCM) of two numbers can be found from their prime factorisations.

两个数的最大公因数（HCF）和最小公倍数（LCM）可以从它们的质因数分解中求出。

Example: Find the HCF and LCM of 154 and 210.

例：求 154 和 210 的最大公因数与最小公倍数。

154 = 2 × 7 × 11

210 = 2 × 3 × 5 × 7

• HCF: multiply the lowest power of each common prime: 2 × 7 = 14
• LCM: multiply the highest power of every prime that appears: 2 × 3 × 5 × 7 × 11 = 2310

• HCF：把所有共有质因数的最低次幂相乘：2 × 7 = 14
• LCM：把所有出现的质因数的最高次幂相乘：2 × 3 × 5 × 7 × 11 = 2310

Check: 154 × 210 = HCF × LCM? 154 × 210 = 32340, 14 × 2310 = 32340 ✓.

检验：154 × 210 = HCF × LCM？154 × 210 = 32340，14 × 2310 = 32340 ✓。

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8. Real-Life Applications | 实际应用

Prime factorisation helps solve problems involving equal splitting, repeating cycles and shared arrangements.

质因数分解常用于解决平均分配、重复周期和共享排列等问题。

Example: A shop has 154 pens and 198 pencils. Both are to be packed into identical gift packs, with each pack containing the same number of pens and the same number of pencils. What is the largest possible number of packs?

例：一家商店有 154 支笔和 198 支铅笔。将两者分别装入相同的礼品袋，每袋中笔的数量和铅笔的数量都相同。最多能装多少袋？

f 154 and 198. 198 = 2 × 3² × 11, so HCF = 2 × 11 = 22. The largest number of packs is 22.

解：求 154 和 198 的最大公因数。198 = 2 × 3² × 11，所以 HCF = 2 × 11 = 22。最多能装 22 袋。

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9. Common Mistakes and Tips | 常见错误与提示

• Mistake 1: Thinking 1 is a prime number. 1 is not prime.
• Mistake 2: Forgetting that 2 is prime; always test 2 first.
• Mistake 3: Stopping the factor tree too early when a factor is still composite.
• Mistake 4: Missing a prime factor when using the division method.
• Tip: Use your divisibility tests to avoid missing primes.
• Tip: Write the final product in ascending order to keep answers tidy.

• 错误 1：认为 1 是质数。1 不是质数。
• 错误 2：忘记 2 是质数；务必先试验 2。
• 错误 3：因数树只分解到一半，仍有合数因数时就停止。
• 错误 4：使用短除法时漏掉某个质因数。
• 提示：利用整除判断法则，避免漏掉质因数。
• 提示：最终乘积按升序书写，让答案更整洁。

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10. Exam-Style Practice | 考试题型练习

Practise these typical IGCSE questions after revisiting the method.

复习完方法后，请练习这些典型的 IGCSE 题目。

Question 1: Write 154 as a product of its prime factors.

题 1：将 154 写成质因数的乘积。

Question 2: Find the Highest Common Factor (HCF) of 154 and 110.

题 2：求 154 与 110 的最大公因数（HCF）。

Question 3: Find the Lowest Common Multiple (LCM) of 154 and 175.

题 3：求 154 与 175 的最小公倍数（LCM）。

Answers: 1) 2 × 7 × 11 2) 2 × 11 = 22 3) 175 = 5² × 7, so LCM = 2 × 5² × 7 × 11 = 3850.

答案：1) 2 × 7 × 11 2) 2 × 11 = 22 3) 175 = 5² × 7，所以 LCM = 2 × 5² × 7 × 11 = 3850。

Remember to always check whether your final factors are prime. A good check: multiply your primes back together — you should get the original number.

请牢记：最后确认每个因数都是质数。验算方法：将各个质数重新相乘，应当得到原始数字。

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Published by TutorHao | Mathematics Revision Series | aleveler.com

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