Prime Factorization & Number Theory | 质因数分解与数论基础

📚 Prime Factorization & Number Theory | 质因数分解与数论基础

In this revision article, we will explore prime factorization, a core topic in the Edexcel IGCSE Mathematics syllabus. We will use the number 202 as a running example to illustrate how to break down numbers, find highest common factors and lowest common multiples, and apply these skills to exam-style problems.

在本篇复习文章中，我们将深入探讨质因数分解——这是 Edexcel IGCSE 数学大纲中的核心考点。我们以数字 202 作为贯穿示例，逐步讲解如何对一个数进行质因数分解、如何求最大公约数与最小公倍数，以及如何将这些技能应用到考试题型中。

1. What Are Prime and Composite Numbers? | 什么是质数与合数？

A prime number has exactly two distinct positive factors: 1 and itself. Examples include 2, 3, 5, 7, 11, 13, 17, 19, and 101. The number 2 is the smallest and the only even prime number.

质数恰好只有两个不同的正因数：1 和它本身。例如 2、3、5、7、11、13、17、19 和 101。其中 2 是最小的质数，也是唯一的偶质数。

A composite number has more than two factors. For example, 4, 6, 8, 9, 10, 12, and 202 are composite numbers because each has at least one factor other than 1 and itself.

合数拥有两个以上的因数。例如 4、6、8、9、10、12 和 202 都是合数，因为除了 1 和自身之外，它们还至少含有其他因数。

The number 1 is neither prime nor composite, because it has only one positive factor.

数字 1 既不是质数也不是合数，因为它只有一个正因数。

2. The Fundamental Theorem of Arithmetic | 算术基本定理

The Fundamental Theorem of Arithmetic states that every integer greater than 1 can be written as a product of prime numbers in exactly one way, ignoring the order of the factors.

算术基本定理指出：每一个大于 1 的整数都可以唯一地表示成质数的乘积，这里不考虑各质因子的排列顺序。

For example, 202 = 2 × 101, and there is no other way to write 202 as a product of primes other than rearranging these two numbers.

例如，202 = 2 × 101，除重新排列这两个质数之外，202 不存在其他质数乘积的写法。

This theorem is the reason why prime factorization is so powerful: it gives every number a unique “fingerprint” made of primes.

这一定理使质因数分解非常强大：它为每个数字提供了一组由质数构成的独特“指纹”。

3. How to Factorise 202: First Steps | 如何分解 202：初步尝试

Let us apply prime factorization to 202. First, test whether 202 is divisible by the smallest prime, 2. Since 202 is even, 202 ÷ 2 = 101.

让我们对 202 进行质因数分解。首先，检验 202 是否能被最小的质数 2 整除。由于 202 是偶数，202 ÷ 2 = 101。

Now we ask: is 101 a prime number? We only need to test prime divisors up to the square root of 101, which is approximately 10.05.

现在我们要问：101 是质数吗？我们只需要用不超过 101 平方根（约为 10.05）的质数来检验。

The primes below 10.05 are 2, 3, 5, and 7. None of them divides 101 exactly (101 ÷ 7 ≈ 14.43, and 101 = 7 × 14 + 3). Therefore, 101 is prime.

小于 10.05 的质数有 2、3、5 和 7。它们都不能整除 101（101 ÷ 7 ≈ 14.43，即 101 = 7 × 14 + 3）。因此，101 是质数。

So the prime factorization of 202 is:

因此，202 的质因数分解为：

202 = 2 × 101

4. Using Factor Trees | 使用因数树

A factor tree is a visual method for prime factorization. Write the number at the top, then split it into two factors, and continue splitting until all branches end at primes.

因数树是一种直观进行质因数分解的方法。把原数写在上方，然后将其分成两个因数，再继续拆分，直到所有分支末端都是质数。

For 202, a simple factor tree looks like this:

对于 202，简单因数树如下：

202 → 2 × 101

Since both 2 and 101 are prime, the tree stops immediately. If you encounter a larger number, keep branching until every leaf is a prime number.

因为 2 和 101 都是质数，因数树立即结束。如果遇到更大的数，请继续分支，直到每个叶节点都是质数。

When writing the final answer, collect the primes in ascending order and use index notation if a prime repeats.

写出最终答案时，应按升序列出质数，若某一质数重复出现，则使用指数记号。

5. Index Notation and Powers | 指数记号与幂

If a prime factor appears more than once, we express it with a power. For instance, 100 = 2 × 2 × 5 × 5 = 2² × 5².

如果某个质因数出现不止一次，我们用幂来表示。例如，100 = 2 × 2 × 5 × 5 = 2² × 5²。

In the case of 202, the prime factors are 2 and 101, each appearing once, so we write 202 = 2¹ × 101¹. Usually the exponent 1 is omitted.

对于 202，其质因子为 2 和 101，各自仅出现一次，因此我们写 202 = 2¹ × 101¹。通常指数 1 可以省略。

In exams, you may be asked to give your answer in index form, especially for numbers with repeated factors.

考试中，你可能会被要求以指数形式给出答案，尤其是当数字含有重复因子时。

6. Highest Common Factor (HCF) | 最大公约数 (HCF)

The highest common factor (HCF) of two or more numbers is the largest number that divides all of them without leaving a remainder.

两个或多个数的最大公约数是指能同时整除这些数且不产生余数的最大整数。

To find the HCF using prime factorization, list the prime factors common to all numbers, each raised to the smallest exponent that appears in any of the factorizations, and multiply them together.

使用质因数分解求最大公约数时，先找出所有数中共同的质因子，取每个共同质因子在所有分解中出现的最小指数，然后相乘。

Let us find the HCF of 202 and 106. We have 202 = 2 × 101 and 106 = 2 × 53. The only common prime factor is 2, with minimum exponent 1.

让我们求 202 和 106 的最大公约数。已知 202 = 2 × 101，106 = 2 × 53。它们唯一的共同质因子是 2，最小指数为 1。

HCF(202, 106) = 2

7. 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 the numbers.

两个或多个数的最小公倍数是指能被所有这些数整除的最小正整数。

To find the LCM using prime factorization, take every prime factor that appears in any of the numbers, each raised to the highest exponent that appears in the factorizations, and multiply them together.

使用质因数分解求最小公倍数时，取出任意一个数中出现过的所有质因子，每个质因子取在所有分解中的最大指数，然后相乘。

Let us find the LCM of 202 and 50. We have 202 = 2 × 101 and 50 = 2 × 5². The primes present are 2, 5, and 101. The maximum exponents are 2¹, 5², and 101¹.

让我们求 202 和 50 的最小公倍数。已知 202 = 2 × 101，50 = 2 × 5²。涉及到的质因子有 2、5、101。最高指数分别是 2¹、5²、101¹。

LCM(202, 50) = 2¹ × 5² × 101¹ = 2 × 25 × 101 = 5050

Notice that 202 × 50 = 10100, which is 2 times the LCM because the HCF is 2. In general, for two numbers a and b, we have a × b = HCF(a, b) × LCM(a, b).

注意 202 × 50 = 10100，它是最小公倍数的 2 倍，因为最大公约数是 2。一般而言，对于两个数 a 和 b，有 a × b = HCF(a, b) × LCM(a, b)。

8. Using Division Ladders for HCF and LCM | 用短除法求 HCF 与 LCM

Another useful method is the division ladder. Write both numbers side by side, then repeatedly divide by a common prime until no common prime factor remains.

另一个实用方法是短除法。将两个数并排写出，反复除以一个共同的质数，直到不再有共同质因子为止。

For example, to find HCF and LCM of 202 and 106, first divide both by 2: 202 ÷ 2 = 101, and 106 ÷ 2 = 53. No further common prime factor exists.

例如，求 202 和 106 的 HCF 与 LCM 时，先将两者同时除以 2：202 ÷ 2 = 101，106 ÷ 2 = 53。此时不再有共同的质因子。

The HCF is the product of the numbers on the left: 2. The LCM is the product of the left column and the bottom row: 2 × 101 × 53 = 10706.

HCF 就是左边所有除数的乘积：2。LCM 是左边除数乘积再乘以下方的商，即 2 × 101 × 53 = 10706。

This method is fast and helps avoid missing factors in exam conditions.

这种方法速度快，在考试中能帮助你避免遗漏因数。

9. Real-Life Applications | 实际应用

Prime factorization is not only an abstract exercise. It has real applications in scheduling, packaging, and music.

质因数分解不仅是一种抽象练习。它在排班、包装和音乐等领域都有实际应用。

Suppose two buses arrive at a station every 202 minutes and every 50 minutes respectively. The next time both buses arrive together is given by the LCM of 202 and 50, which we found to be 5050 minutes.

假设两辆公交车分别每隔 202 分钟和每隔 50 分钟到达同一个站点。它们下一次同时到达的时间间隔就是 202 和 50 的最小公倍数，即我们求出的 5050 分钟。

Similarly, if you want to cut two pieces of ribbon of lengths 202 cm and 106 cm into equal smaller pieces with no waste, the maximum piece length is the HCF, which is 2 cm.

类似地，若想把长度为 202 厘米和 106 厘米的两段彩带剪成等长且无剩余的小段，每段的最大长度就是它们的 HCF，即 2 厘米。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

A common mistake is stopping factorization too early. For example, if you write 202 = 2 × 101 and do not check that 101 is prime, you might wrongly continue or stop without justification. Always test prime divisors up to the square root.

常见错误是过早停止分解。例如，如果你写 202 = 2 × 101 却不检查 101 是否为质数，你可能会错误地继续拆分或没有根据地停止。始终要用不超过平方根的质数去检验。

Another mistake is confusing HCF and LCM. Remember: HCF uses the smallest exponent, LCM uses the largest exponent.

另一个错误是混淆 HCF 和 LCM。记住：HCF 取最小指数，LCM 取最大指数。

In exams, always present your prime factorization with a clear method such as a factor tree, and show all steps. Marks are often awarded for correct method even if your final answer has a small slip.

考试中，务必使用清晰的方法（如因数树）展示质因数分解，并写出所有步骤。即使最终答案有细小失误，正确的方法也常常能获得步骤分。

If the product of two numbers is given and one HCF is known, use the relation a × b = HCF × LCM to find the missing value quickly.

如果已知两个数的乘积和其中一个 HCF，可以利用 a × b = HCF × LCM 的关系快速求出未知值。

11. Practice Questions | 练习题目

Try these questions yourself before checking the answers.

请先尝试解决以下问题，再核对答案。

Question 1: Write 202 as a product of prime factors.

第 1 题：把 202 写成质因数的乘积。
Question 2: Find the HCF of 202 and 106.

第 2 题：求 202 和 106 的最大公约数。
Question 3: Find the LCM of 202, 50, and 30.

第 3 题：求 202、50 和 30 的最小公倍数。
Question 4: Given that HCF(202, 50) = 2, use the formula to find LCM(202, 50).

第 4 题：已知 HCF(202, 50) = 2，使用公式求 LCM(202, 50)。

Answers: 1. 202 = 2 × 101. 2. HCF(202, 106) = 2. 3. 202 = 2 × 101, 50 = 2 × 5², 30 = 2 × 3 × 5, so LCM = 2 × 3 × 5² × 101 = 15150. 4. LCM = 202 × 50 ÷ 2 = 5050.

答案：1. 202 = 2 × 101。2. HCF(202, 106) = 2。3. 202 = 2 × 101，50 = 2 × 5²，30 = 2 × 3 × 5，因此 LCM = 2 × 3 × 5² × 101 = 15150。4. LCM = 202 × 50 ÷ 2 = 5050。

12. Summary | 总结

Prime factorization is a vital skill in IGCSE Mathematics. Every composite number can be uniquely expressed as a product of primes, as we demonstrated with 202 = 2 × 101.

质因数分解是 IGCSE 数学中至关重要的技能。每个合数都能唯一地表示成质数的乘积，正如我们用 202 = 2 × 101 所示。

From this factorization, we can easily find HCF and LCM by comparing exponents. Remember: HCF uses minimum exponents, LCM uses maximum exponents.

基于这种分解，我们可以通过比较指数轻松求出 HCF 和 LCM。记住：HCF 使用最小指数，LCM 使用最大指数。

Keep practising with different numbers, draw factor trees carefully, and check whether each factor is truly prime. With consistent revision, you will master this topic and gain confidence in your exams.

请继续用不同数字进行练习，仔细绘制因数树，并检查每个因数是否确为质数。通过持续复习，你将掌握这一主题并在考试中更加自信。

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Prime Factorization & Number Theory | 质因数分解与数论基础