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。

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。

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。

3. Method 1: Factor Tree | 方法一：因子树

A factor tree splits a number into factors, then splits any composite factor until 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。

Since 535 is divisible 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

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。

5. Prime Factorisation of 1070 | 分解 1070

Let us verify: 1070 is even, so 2 is a factor. 1070 ÷ 2 = 535. The digit sum of 535 is 5 + 3 + 5 = 13, so 3 is not a factor; however, 535 ends in 5, so 5 is a factor. 535 ÷ 5 = 107. Now test 107: 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

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

Given two numbers, list their prime factorisation. The Highest Common Factor is found by multiplying the common prime factors with the smallest power. For example, take 1070 and 2100. We have 2100 = 2² × 3 × 5² × 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。

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

The Lowest Common Multiple is found by multiplying all prime factors that appear in either number, using the largest power. For 1070 = 2 × 5 × 107 and 2100 = 2² × 3 × 5² × 7, the LCM is

Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）