Prime Factorisation | 质因数分解

📚 Prime Factorisation | 质因数分解

In IGCSE Mathematics, prime factorisation is a core skill that supports many other topics, from fractions to number theory. In this article, we use the number 1070 as our running example.

在 IGCSE 数学中，质因数分解是一项核心技能，它支撑着从分数到数论等多个主题。本文以数字 1070 作为贯穿全篇的例子。

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

A prime number is a positive integer greater than 1 that has exactly two factors: 1 and itself. For example, 2, 3, 5, 7, 11 and 13 are primes. A composite number has more than two factors, such as 4, 6, 8, 9 and 10. The number 1 is neither prime nor composite.

质数是大于 1 且恰好只有两个因数的正整数：1 和它本身。例如，2、3、5、7、11 和 13 都是质数。合数有超过两个因数，例如 4、6、8、9 和 10。数字 1 既不是质数也不是合数。

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2. What is Prime Factorisation? | 什么是质因数分解？

Prime factorisation is the process of writing a positive integer as a product of prime numbers. The Fundamental Theorem of Arithmetic states that every integer greater than 1 has a unique prime factorisation, apart from the order of the factors.

质因数分解是将一个正整数写成质数乘积的过程。算术基本定理指出，每个大于 1 的整数都有唯一的质因数分解，不考虑因数的顺序。

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3. Method 1: Factor Tree | 方法一：因子树

l only primes remain. For example, to factor 1070, we can start by writing 1070 = 2 × 535.

因子树将一个数分解为因数，再继续分解其中的合数因数，直到只剩质数。例如，分解 1070 时，可以先写 1070 = 2 × 535。

sible by 5, write 535 = 5 × 107. Now 107 is prime, so the tree ends. The final product is:

由于 535 能被 5 整除，写 535 = 5 × 107。现在 107 是质数，因此分解完成。最终乘积为：

1070 = 2 × 5 × 107

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4. Method 2: Repeated Division (Short Division) | 方法二：连续除法（短除法）

Alternatively, divide by prime numbers in order until the result is 1. For 1070, divide by 2 to get 535, divide by 5 to get 107, and since 107 is prime, stop. The divisors give the prime factors.

另一种方法是按顺序用质数连续除，直到结果为 1。对于 1070，先除以 2 得到 535，再除以 5 得到 107，由于 107 是质数，停止。除数就是质因数。

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5. Prime Factorisation of 1070 | 分解 1070

07: it is not divisible by 2, 3, 5, 7, or 11, and 13² = 169 > 107. Therefore 107 is prime.

让我们验证：1070 是偶数，因此 2 是因数。1070 ÷ 2 = 535。535 的数位和为 5 + 3 + 5 = 13，所以 3 不是因数；但 535 以 5 结尾，所以 5 是因数。535 ÷ 5 = 107。现在检验 107：它不能被 2、3、5、7 或 11 整除，而且 13² = 169 > 107，因此 107 是质数。

So the complete factorisation is:

所以完整的分解是：

2 × 5 × 107

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6. Using Prime Factors to Find HCF | 用质因数求最大公因数

7. The common factors are 2 and 5 with powers min(1,2) and min(1,2), so HCF = 2 × 5 = 10.

给定两个数，列出它们的质因数分解。最大公因数通过取公共质因数中次幂最小的相乘得到。例如，取 1070 和 2100。2100 = 2² × 3 × 5² × 7。公共因数有 2 和 5，指数分别为 min(1,2) 和 min(1,2)，所以 HCF = 2 × 5 = 10。

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7. Using Prime Factors to Find LCM | 用质因数求最小公倍数

er, using the largest power. For 1070 = 2 × 5 × 107 and 2100 = 2² × 3 × 5² × 7, the LCM is

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