Prime Numbers and Prime Factorization | 质数与质因数分解

📚 Prime Numbers and Prime Factorization | 质数与质因数分解

In this article, we explore the concept of prime numbers, how to test whether a number is prime, and how to use prime factorisation in problems. As a running example, we will use the number 157 — a prime number that often appears in IGCSE questions.

在本文中，我们将探索质数的概念、如何判断一个数是否为质数，以及如何在解题中使用质因数分解。我们以数字 157 作为贯穿示例——它是在 IGCSE 考题中经常出现的质数。

1. What Is a Prime Number? | 什么是质数？

A prime number is a positive integer greater than 1 that has exactly two distinct positive factors: 1 and itself. For example, 2, 3, 5, 7, 11 and 13 are prime numbers.

质数是大于 1 的正整数，它恰好只有两个不同的正因数：1 和它本身。例如 2、3、5、7、11 和 13 都是质数。

Notice that 1 is not a prime number because it has only one factor. Also, 2 is the smallest and the only even prime number.

注意 1 不是质数，因为它只有一个因数。此外，2 是最小的质数，也是唯一的偶质数。

Now consider 157. To be prime, it must have no factors other than 1 and 157. We will soon confirm this.

现在看 157。若它是质数，就必须除了 1 和 157 之外没有其他因数。我们很快会验证这一点。

2. Composite Numbers and Unique Factorisation | 合数与唯一分解

A composite number is a positive integer greater than 1 that has more than two factors. For instance, 12 is composite because its factors are 1, 2, 3, 4, 6 and 12.

合数是大于 1 且拥有多于两个因数的正整数。例如 12 是合数，因为它的因数是 1、2、3、4、6 和 12。

Every composite number can be written as a product of prime numbers. This is called prime factorisation. The order of the factors does not matter; the list of primes is unique.

每个合数都可以写成若干质数的乘积，这称为质因数分解。因子的顺序不重要；质数的列表是唯一的。

For example, 157 is prime, so its prime factorisation is simply 157 itself. No further breaking down is possible.

例如，157 是质数，所以它的质因数分解就是 157 本身，无法再分解。

3. How to Test for Primality | 如何判断一个数是否为质数

To test whether a number n is prime, we only need to check divisibility by prime numbers up to √n. If none of these primes divide n, then n is prime.

要判断一个数 n 是否为质数，我们只需要检查不超过 √n 的质数能否整除 n。如果这些质数都不能整除 n，那么 n 就是质数。

For 157, we compute √157. Since 12² = 144 and 13² = 169, we know that 12 < √157 < 13. Therefore we only need to test the primes 2, 3, 5, 7 and 11.

对于 157，我们计算 √157。因为 12² = 144，13² = 169，所以 12 < √157 < 13。因此我们只需要检验质数 2、3、5、7 和 11。

157 is odd, so it is not divisible by 2.
1 + 5 + 7 = 13, which is not divisible by 3, so 157 is not divisible by 3.
157 does not end in 0 or 5, so it is not divisible by 5.
7 × 22 = 154, and 157 − 154 = 3, so 7 does not divide 157.
11 × 14 = 154, and 157 − 154 = 3, so 11 does not divide 157.

None of the required primes divide 157, so 157 is prime. This method is the standard “trial division” method used in IGCSE.

以上需要检验的质数都不能整除 157，所以 157 是质数。这种方法就是 IGCSE 中使用的标准“试除法”。

4. Prime Factorisation Using Factor Trees | 使用因数树进行质因数分解

A factor tree splits a number into two factors, then continues splitting until all factors are prime. This is a common exam technique.

因数树把一个数拆成两个因数，然后继续拆分直到所有因数都是质数。这是一种常见的考试技巧。

For a prime number like 157, the factor tree is just a single branch: 157 itself.

对于像 157 这样的质数，因数树只有一个分支：157 本身。

Let us demonstrate with a composite number: 72. Split 72 as 8 × 9. Then split 8 as 2 × 4, and 4 as 2 × 2. Split 9 as 3 × 3. This gives:

我们用一个合数来演示：72。将 72 拆成 8 × 9。再把 8 拆成 2 × 4，4 拆成 2 × 2。9 拆成 3 × 3。于是得到：

72 = 2³ × 3²

Always write the final answer with exponents arranged in ascending order of the prime bases.

最终答案应按质数底数从小到大的顺序，用指数形式书写。

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

The Fundamental Theorem of Arithmetic states that every integer greater than 1 is either prime or can be expressed as a product of primes in exactly one way, apart from the order of the factors.

算术基本定理指出：每个大于 1 的整数，要么是质数，要么可以表示为质数的乘积，且不考虑顺序时，表示方式是唯一的。

This uniqueness is why prime factorisation is so powerful in mathematics. For example, if you factor 157, you get only 157. No other product of primes equals 157.

这种唯一性正是质因数分解在数学中如此强大的原因。例如，分解 157 只能得到 157。没有任何其他质数乘积等于 157。

Another way to write the same idea is:

同样地，我们可以这样表述：

n = p₁^a × p₂^b × p₃^c × …

where p₁, p₂, p₃, … are distinct primes and a, b, c, … are positive integers.

其中 p₁、p₂、p₃……是不同的质数，a、b、c……是正整数。

6. Using Prime Factors: HCF and LCM | 利用质因数计算最大公因数与最小公倍数

Prime factorisation helps us find the Highest Common Factor (HCF) and Lowest Common Multiple (LCM) of two or more numbers.

质因数分解帮助我们求两个或多个数的最大公因数（HCF）和最小公倍数（LCM）。

Take 157 and another number, say 168. First factorise 168: 168 = 2³ × 3 × 7. Since 157 is prime and not among 2, 3 or 7, the two numbers share no common prime factors.

取 157 和另一个数，比如 168。先分解 168：168 = 2³ × 3 × 7。由于 157 是质数且不是 2、3 或 7，所以这两个数没有共同的质因数。

Hence HCF(157, 168) = 1, meaning they are coprime. The LCM is their product:

因此 HCF(157, 168) = 1，即它们互质。LCM 就是它们的乘积：

LCM = 157 × 168 = 26376

When numbers share prime factors, for HCF take the lowest power of each common prime; for LCM take the highest power of every prime that appears.

当数字有共同的质因数时，求 HCF 取每个共同质数的最低次幂；求 LCM 取出现的每个质数的最高次幂。

7. Divisibility Rules You Should Know | 你应该掌握的整除规则

Quick divisibility tests save time in primality testing and factorisation. Here are the most useful ones for IGCSE.

快速的整除判法能在判断质数和分解因数时节省时间。以下是 IGCSE 中最常用的一些规则。

Divisible by	Rule	Example
2	Last digit is even	158 is divisible by 2
3	Sum of digits is divisible by 3	156 → 1+5+6=12 → divisible
5	Last digit is 0 or 5	155 is divisible by 5
7	Double last digit, subtract from the rest; repeat	157 → 15 − 14 = 1 → not divisible
11	Alternating sum of digits is divisible by 11	157 → 1 − 5 + 7 = 3 → not divisible

For 157, the rules for 2, 3, 5, 7 and 11 all fail, which again confirms that 157 is prime.

对于 157，2、3、5、7 和 11 的整除规则全部不成立，这也再次确认 157 是质数。

8. Common Mistakes and Examination Tips | 常见错误与考试提示

One common mistake is forgetting that 1 is not prime. Another is checking divisibility by every integer up to n instead of √n, which wastes time.

一个常见错误是忘记 1 不是质数。另一个错误是检查到 n 而不是 √n，这浪费了时间。

When using a factor tree, always continue until all branches end in prime numbers. Check your final product by expanding it back to the original number.

使用因数树时，一定要继续拆分直到所有分支都以质数结束。通过把最终乘积展开回来，检查是否等于原数。

In multiple-choice questions, a quick way to rule out a candidate is to see if it is divisible by 2, 3, 5 or 11. For larger primes like 157, remember the square root shortcut.

在选择题中，快速排除候选数的方法是看它能否被 2、3、5 或 11 整除。对于像 157 这样较大的质数，记住平方根这个捷径。

Finally, when giving the prime factorisation of a prime number, do not write it as 1 × itself as the final answer; simply write the number itself.

最后，给出质数的质因数分解时，不要把最终答案写成 1 × 它本身；直接写这个数即可。

9. Practice Questions | 练习题

Test your understanding with these short questions.

用下面这些小题来检验你的理解。

Determine whether 91 is prime or composite.
Write 240 as a product of prime factors in index form.
Find the HCF and LCM of 157 and 157.
Explain why 157 is prime using trial division.

判断 91 是质数还是合数。
用指数形式写出 240 的质因数分解。
求 157 和 157 的 HCF 与 LCM。
用试除法解释为什么 157 是质数。

Answers: (1) 91 = 7 × 13, composite. (2) 240 = 2⁴ × 3 × 5. (3) HCF = 157, LCM = 157. (4) √157 lies between 12 and 13; testing primes 2, 3, 5, 7, 11 leads to a non-zero remainder each time.

答案：（1）91 = 7 × 13，是合数。（2）240 = 2⁴ × 3 × 5。（3）HCF = 157，LCM = 157。（4）√157 在 12 和 13 之间；分别试除质数 2、3、5、7、11，每次余数都不为 0。

10. Summary | 总结

A prime number has exactly two factors: 1 and itself. The number 157 is prime because no prime ≤ √157 divides it. Prime factorisation expresses every integer as a unique product of primes, and this idea underpins HCF, LCM and many number-theory problems.

质数恰好有两个因数：1 和它本身。157 是质数，因为没有不超过 √157 的质数能整除它。质因数分解将每个整数表示为质数的唯一乘积，这一思想支撑着 HCF、LCM 以及许多数论问题。

Mastering prime numbers and factorisation will help you solve problems quickly and avoid careless errors. The next time you see 157, you will know it is prime beyond any doubt.

掌握质数和因式分解能帮助你快速解题并避免粗心错误。下次再看到 157，你会确信它是一个质数。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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Prime Numbers and Prime Factorization | 质数与质因数分解