📚 HCF and LCM | 最大公因数与最小公倍数
Understanding the Highest Common Factor (HCF) and Lowest Common Multiple (LCM) is a fundamental skill in IGCSE Mathematics. These concepts appear in many topics, from fractions to algebra, and are frequently tested in Edexcel papers.
理解最大公因数（HCF）和最小公倍数（LCM）是 IGCSE 数学的基本技能。这些概念出现在许多主题中，从分数到代数，并在 Edexcel 试卷中经常出现。
1. Prime Numbers and Prime Factorisation | 质数与质因数分解
A prime number is a positive integer greater than 1 that has exactly two distinct positive divisors: 1 and itself. For example, 2, 3, 5, 7, 11, and 13 are prime numbers.
质数是大于 1 且只有两个不同正因数（1 和本身）的正整数。例如，2、3、5、7、11、13 都是质数。
Prime factorisation (or prime factor decomposition) means writing a number as a product of its prime factors. A common method is the factor tree.
质因数分解（或质因数拆分）是指将一个数写成其质因数的乘积。常用的方法是因数树。
48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3
Notice how we repeatedly divide by the smallest prime until the result is 1. This gives the unique prime factorisation of any positive integer.
注意如何反复用最小的质数去除，直到结果为 1。这样可以得到任意正整数的唯一质因数分解。
2. Finding the HCF | 求最大公因数
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因数（HCF）是能够整除它们且不留余数的最大数。
Method using prime factorisation: List all common prime factors, then multiply them together.
使用质因数分解的方法：列出所有共同的质因数，然后将它们相乘。
Example: Find the HCF of 48 and 36.
例：求 48 和 36 的 HCF。
48 = 2⁴ × 3, 36 = 2² × 3²
rs at least twice in both) and 3¹ (3 appears at least once in both). So HCF = 2² × 3 = 12.
共同的质因数：2²（因为 2 在两个数中都至少出现两次）和 3¹（3 在两个数中都至少出现一次）。因此 HCF = 2² × 3 = 12。
3. Finding the LCM | 求最小公倍数
The Lowest Common Multiple (LCM) of two or more numbers is the smallest positive number that is a multiple of all of them.
两个或多个数的最小公倍数（LCM）是它们的共同倍数中最小的正数。
Method using prime factorisation: For each prime factor, take the highest power that appears in any of the numbers, then multiply.
使用质因数分解的方法：对于每个质因数，取它在任意一个数中出现的最高次幂，然后相乘。
Example: Find the LCM of 48 and 36.
例：求 48 和 36 的 LCM。
48 = 2⁴ × 3⁰? has 3². Highest power of 2 is 2⁴, highest power of 3 is 3². LCM = 2⁴ × 3² = 16 × 9 = 144.
So the LCM is 144.
所以 LCM 是 144。
4. HCF and LCM for Three or More Numbers | 三个或更多数的 HCF 与 LCM
For HCF, take the minimum power of each common prime across all numbers. For LCM, take the maximum power of each prime that appears in any number.
对于 HCF，取每个共同质因数在所有数中的最小次幂。对于 LCM，取每个出现过的质因数的最大次幂。
Example: Find HCF and LCM of 24, 36, and 60.
例：求 24、36 和 60 的 HCF 和 LCM。
| Number | Prime factorisation |
| 24 | 2³ × 3 |
| 36 | 2² × 3² |
| 60 | 2² × 3 × 5 |
ors are 2 and 3. Minimum power of 2 is 2², minimum power of 3 is 3¹. So HCF = 2² × 3 = 12.
HCF：共同质因数为 2 和 3。2 的最小次幂为 2²，3 的最小次幂为 3¹。所以 HCF = 2² × 3 = 12。
take all primes appearing: 2, 3, 5. Maximum powers: 2³, 3², 5¹. So LCM = 8 × 9 × 5 = 360.
LCM：取所有出现的质因数：2、3、5。最大次幂：2³、3²、5¹。所以 LCM = 8 × 9 × 5 = 360。
5. Relationship Between HCF and LCM | HCF 与 LCM 的关系
For two positive integers a and b, there is a simple relationship:
对于两个正整数 a 和 b，有一个简单的关系：
a × b = HCF(a, b) × LCM(a, b)
This formula can be used to find either the HCF or the LCM if the other is known. It only works for two numbers, not for three or more.
这个公式可用于在已知其中一个时求 HCF 或 LCM。它只适用于两个数，不适用于三个或更多。
36, we found HCF = 12 and LCM = 144. Check: 48 × 36 = 1728, and 12 × 144 = 1728. Correct.
例：对于 48 和 36，我们求出 HCF = 12，LCM = 144。验证：48 × 36 = 1728，12 × 144 = 1728。正确。
6. Word Problems Involving HCF | 涉及 HCF 的应用题
HCF is often used when dividing items into equal groups with no remainder. The HCF tells you the largest possible group size.
HCF 常用于将物品无剩余地分成相等的组。HCF 告诉你最大可能的组大小。
Example: A teacher has 48 pencils and 36 erasers. She wants to create identical gift packages using all items. What is the maximum number of packages she can make?
例：一位老师有 48 支铅笔和 36 块橡皮。她想用所有物品制作相同的礼品袋。她最多能制作多少个袋子？
e can make 12 packages. Each package contains 48 ÷ 12 = 4 pencils and 36 ÷ 12 = 3 erasers.
我们需要 48 和 36 的 HCF，即 12。所以她可以制作 12 个袋子。每个袋子包含 48 ÷ 12 = 4 支铅笔和 36 ÷ 12 = 3 块橡皮。
7. Word Problems Involving LCM | 涉及 LCM 的应用题
LCM is used when two events repeat at different intervals and we want to know when they coincide next.
LCM 用于两个事件以不同间隔重复时，求它们下一次同时发生的时间。
Example: Two lighthouses flash their lights every 12 and 18 minutes respectively. If they flash together at noon, when will they flash together again?
例：两座灯塔分别每 12 分钟和 18 分钟闪一次光。如果它们在中午同时闪光，下一次同时闪光是什么时候？
, 18 = 2 × 3². LCM = 2² × 3² = 36. So they will flash together after 36 minutes, at 12:36.
求 12 和 18 的 LCM。质因数分解：12 = 2² × 3，18 = 2 × 3²。LCM = 2² × 3² = 36。所以 36 分钟后它们会再次同时闪光，即 12:36。
8. Common Mistakes and How to Avoid Them | 常见错误及避免方法
- Confusing HCF and LCM: Remember HCF is always less than or equal to the smallest number; LCM is always greater than or equal to the largest number (if one does not divide the other).
- 错误混淆 HCF 和 LCM：记住 HCF 总是小于或等于最小的数；LCM 总是大于或等于最大的数（除非一个数整除另一个数）。
- Using the relationship for three numbers: HCF × LCM only works for two numbers. For three numbers, you must use prime factorisation.
- 对三个数使用关系式：公式 a × b = HCF × LCM 只适用于两个数。对于三个数，必须使用质因数分解。
- Incorrect powers: When finding HCF, use the smaller exponent; for LCM, use the larger exponent. Mixing these up leads to errors.
- 次幂错误：求 HCF 时用较小的指数；求 LCM 时用较大的指数。混淆会导致错误。
9. Exam-Style Worked Example | 考试风格例题
⁴ × 3 × 7, and 540 = 2² × 3³ × 5. Work out the Highest Common Factor (HCF) of 336 and 540.
题目：336 和 540 都写成质因数的乘积：336 = 2⁴ × 3 × 7，540 = 2² × 3³ × 5。求 336 和 540 的最大公因数（HCF）。
s 2 (from 540). For 3, the smaller exponent is 1 (from 336). So HCF = 2² × 3 = 4 × 3 = 12.
解答：共同质因数为 2 和 3。对于 2，较小的指数是 2（来自 540）。对于 3，较小的指数是 1（来自 336）。所以 HCF = 2² × 3 = 4 × 3 = 12。
Also find the LCM of 336 and 540.
再求 336 和 540 的 LCM。
ake the maximum exponent for each prime: 2⁴, 3³, 5¹, 7¹. So LCM = 16 × 27 × 5 × 7 = 15120.
解答：求 LCM 时取每个质因数的最大指数：2⁴、3³、5¹、7¹。所以 LCM = 16 × 27 × 5 × 7 = 15120。
10. Practice Questions | 练习题
1. Find the HCF of 72 and 108.
1. 求 72 和 108 的 HCF。
2. Find the LCM of 14 and 22.
2. 求 14 和 22 的 LCM。
3. Two buses leave a station at the same time. One returns every 45 minutes, the other every 60 minutes. In how many minutes will they both be at the station again?
3. 两辆公交车同时离开车站。一辆每 45 分钟返回，另一辆每 60 分钟返回。多少分钟后它们会再次同时回到车站？
Answers: 1. HCF = 36. 2. LCM = 154. 3. LCM of 45 and 60 = 180 minutes.
答案：1. HCF = 36。2. LCM = 154。3. 45 和 60 的 LCM = 180 分钟。
11. Quick Check: HCF vs LCM | 快速检查：HCF 与 LCM
| Situation | Use | Reason |
| Splitting items into equal groups with nothing left | HCF | Find largest group size |
| Finding when events occur together again | LCM | Find smallest common time period |
| Comparing ratios of quantities | HCF | Simplify ratio |
| Adding or subtracting fractions | LCM | Find common denominator |
12. Conclusion | 总结
Mastering HCF and LCM helps you solve a range of problems in arithmetic, algebra, and real-life contexts. Always start with prime factorisation, then apply the correct rule: smaller powers for HCF, larger powers for LCM.
掌握 HCF 和 LCM 有助于你解决算术、代数和实际生活中的各种问题。始终从质因数分解开始，然后应用正确的规则：HCF 取较小次幂，LCM 取较大次幂。
Practice these methods until they become second nature. You will encounter them in many examination questions, so being confident with HCF and LCM is essential for your IGCSE success.
练习这些方法直到熟练为止。你会在许多考题中遇到它们，因此对 HCF 和 LCM 有信心是 IGCSE 成功的关键。
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