📚 Edexcel IGCSE Maths Topic 1.1: Integers, Factors and Multiples | Edexcel IGCSE 数学主题1.1：整数、因数与倍数
In the Edexcel IGCSE Mathematics specification, Topic 1.1 lays the groundwork for all number work. It covers the properties of integers, how to perform operations with them, and the related concepts of factors, multiples, primes, HCF and LCM. This revision guide will walk you through the theory and typical problem-solving approaches to help you secure full marks.
在Edexcel IGCSE数学考试大纲中，主题1.1是整个数字运算的基础。它涵盖了整数的性质、运算法则，以及因数、倍数、质数、最大公因数和最小公倍数等重要概念。本篇复习指南将带你系统梳理理论和常见题型，帮助你稳稳拿下这部分分数。
1. Understanding Integers | 理解整数
An integer is any whole number that can be positive, negative or zero. This includes numbers like −3, 0, 7, and 2024, but not fractions or decimals such as ½ or 3.14. The set of integers is sometimes denoted by the symbol ℤ. On a number line, integers are equally spaced, and the further right you go, the larger the number. Negative integers are written with a minus sign, while positive integers may be written with or without a plus sign.
整数是指任何正整数、负整数或零，比如−3、0、7和2024，但不包括像½或3.14这样的分数或小数。整数集有时用符号ℤ表示。在数轴上，整数等距分布，越往右数值越大。负整数前面写负号，正整数前面可写也可不写正号。
2. Operations with Integers | 整数的运算
sion, the sign rules are straightforward: the product or quotient of two integers with the same sign is positive, while that of two integers with different signs is negative.
加上一个负整数等同于减去其相反数：5 + (−3) = 5 − 3 = 2。减去一个负整数则变为加法：5 − (−3) = 5 + 3 = 8。对于乘法和除法，符号规则很简单：同号得正，异号得负。
The table below summarises the sign rules for multiplication and division:
下表总结了乘除运算的符号法则：
| Sign of first integer | Sign of second integer | Sign of result (× or ÷) |
|---|---|---|
| + | + | + |
| + | − | − |
| − | + | − |
| − | − | + |
When several operations are mixed, always follow the order of operations – by BIDMAS or BODMAS. This prevents errors like calculating 2 + 3 × 4 as 20 instead of 14.
当多种运算混合时，一定要遵循运算次序——通常用BIDMAS或BODMAS记忆法则。这样才能避免诸如把2 + 3 × 4错算成20而非14的错误。
3. Order of Operations (BIDMAS/BODMAS) | 运算次序 (BIDMAS/BODMAS)
te the bracket: 3² = 9, so 9 − 1 = 8. Then multiply: 2 × 8 = 16. Finally add: 8 + 16 = 24.
BIDMAS代表括号、指数（幂）、除法、乘法、加法和减法。除法和乘法同级，从左到右运算；加法和减法同理。例如，对于表达式8 + 2 × (3² − 1)，先算括号：3² = 9，9 − 1 = 8；再算乘法：2 × 8 = 16；最后算加法：8 + 16 = 24。
Always show your working step by step to avoid losing marks on multi-step problems. Even one misstep can lead to an incorrect final answer, and with negative numbers extra care is needed when deciding the sign.
一定要逐步写出运算过程，以免在多步计算中丢分。一步算错就可能导致最终答案错误，尤其是涉及负数时更要反复确认符号。
4. Factors and Multiples | 因数与倍数
A factor of an integer is a whole number that divides exactly into it, leaving no remainder. For example, the factors of 12 are 1, 2, 3, 4, 6 and 12. Every integer is a factor of itself, and 1 is a factor of every integer. A multiple of an integer is the product of that integer and any whole number; multiples of 5 include 5, 10, 15, 20, and so on. The first multiple of any number is the number itself.
一个整数的因数是指能整除该数且无余数的整数。例如，12的因数有1, 2, 3, 4, 6和12。每个整数都是自身的因数，而1是所有整数的因数。一个整数的倍数是指该整数与任意整数的乘积；5的倍数包括5, 10, 15, 20等。任何数的第一个倍数都是它本身。
Understanding factors and multiples is essential for simplifying fractions, solving ratio problems and finding common denominators in algebra. Make sure you can list them systematically.
理解因数和倍数对于约分、解决比例问题以及代数中寻找公分母至关重要。请确保自己能系统地列出一个数的所有因数或前几个倍数。
5. Prime Numbers and Prime Factorisation | 质数与质因数分解
xactly two distinct factors: 1 and itself. Examples are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 … 2 × 4 and 3 × 3, continuing until all factors are prime: 72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3².
质数是大于1且只有两个不同因数（1和它本身）的整数。例如2, 3, 5, 7, 11, 13, 17, 19, 23, 29…… 注意1不是质数。数字2是唯一的偶质数；其他所有偶数都有因数2，因而都是合数。将一个数写成质因数乘积的形式叫做质因数分解。可以利用“因数树”来分解：例如72可以拆成8 × 9，再继续拆成2 × 4和3 × 3，直到所有因数都是质数为止：72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²。
Always write the prime factors in ascending order and use index notation where possible. This makes it much easier to find the HCF and LCM later.
一定要按从小到大的顺序书写质因数，并尽可能用指数形式表示。这会让你在后续求最大公因数和最小公倍数时轻松很多。
6. Highest Common Factor (HCF) | 最大公因数
The HCF of two or more integers is the largest integer that divides all of them exactly. One method is to list all factors and pick the greatest common one, but for larger numbers prime factorisation is far more efficient. Write each number as a product of prime factors in index form, then multiply the common prime factors, each raised to the lowest power that appears in any of the numbers.
两个或多个整数的最大公因数（HCF）是指能同时整除这些数的最大整数。一种方法是将所有因数罗列出来，选出最大的公共因数；但对于较大的数，质因数分解法高效得多。先将每个数写成质因数指数形式，然后把公共质因数按照各数中出现的最低指数相乘便得到HCF。
Worked example: Find the HCF of 36 and 48.
36 = 2² × 3²
48 = 2⁴ × 3
prime factors: 2 (lowest power 2²) and 3 (lowest power 3¹). So HCF = 2² × 3 = 4 × 3 = 12.
例题：求36和48的HCF。
36 = 2² × 3²
48 = 2⁴ × 3
公共质因数：2（最低指数为2²）和3（最低指数为3¹）。因此HCF = 2² × 3 = 4 × 3 = 12。
An alternative method is the Euclidean algorithm, but at IGCSE level the prime factor method is perfectly acceptable and often required.
另一种方法是欧几里得算法，但在IGCSE阶段，质因数分解法是完全可以接受且经常被要求的。
7. Lowest Common Multiple (LCM) | 最小公倍数
The LCM is the smallest positive integer that is a multiple of all given numbers. To find it using prime factors, write each number in index form. Then, for every prime that appears, take the highest power present in any of the numbers, and multiply them together.
最小公倍数（LCM）是指能同时被若干个给定数整除的最小的正整数。用质因数法求LCM时，先把每个数写成指数形式；然后找出每个质数在所有数中出现的最高指数，将它们相乘即可。
2⁴ × 3² = 16 × 9 = 144. Notice that HCF × LCM = 12 × 144 = 1728, which also equals 36 × 48 – a useful check.
沿用上面的数字：36 = 2² × 3²，48 = 2⁴ × 3。质数2的最高指数是2⁴，质数3的最高指数是3²。因此LCM = 2⁴ × 3² = 16 × 9 = 144。注意HCF × LCM = 12 × 144 = 1728，恰好等于36 × 48——这是一条有用的检验方法。
When dealing with more than two numbers, simply extend the method: include every prime factor that appears and take the highest power for the LCM, and the lowest power for the HCF.
当涉及多于两个数字时，只需扩展此法：求LCM时取每个质因数的最高指数，求HCF时取最低指数。
8. Word Problems and Real‑Life Applications | 应用题与实际应用
HCF and LCM frequently appear in contextual problems. For example, ' equal pieces of maximum possible length without any waste. Find the length of each piece.' Prime factorisation gives 120 = 2³ × 3 × 5, 150 = 2 × 3 × 5², so HCF = 2 × 3 × 5 = 30 cm.
HCF和LCM经常出现在实际情境题中。例如：“两条丝带长度分别为120 cm和150 cm，需剪成尽可能长的等长小段且无剩余，求每小段的长度。”这里每段的最大长度就是120和150的HCF。质因数分解得120 = 2³ × 3 × 5，150 = 2 × 3 × 5²，因此HCF = 2 × 3 × 5 = 30 cm。
In LCM problems, you might be asked: ' the other every 18 minutes. After how many minutes will they next be at the depot together?' e answer is the LCM of 12 and 18. 12 = 2² × 3, 18 = 2 × 3², so LCM = 2² × 3² = 36 minutes.
而在LCM问题中，可能会这样问：“两辆公交车同时从车站出发，一辆每12分钟返回一次，另一辆每18分钟返回一次。请问多少分钟后两车会再次同时在车站？”答案就是12和18的LCM。12 = 2² × 3，18 = 2 × 3²，故LCM = 2² × 3² = 36分钟。
Read the wording carefully – look for keywords like 'greatest', 'maximum', 'longest' (HCF), or 'least', 'smallest', 'next time together' (LCM) to decide which concept to use.
仔细审题——注意诸如“最大”、“最长”（HCF）或“最小”、“下一次同时”（LCM）等关键词，以决定使用哪个概念。
9. Exam Tips and Common Pitfalls | 考试技巧与常见误区
ponent only applies to the 3 unless brackets are used. Misreading this is a very common mistake.
IGCSE考官经常会在大型题目中嵌套负数运算，比如将负值代入代数式。一定要记住(−3)² = 9，但−3² = −9，因为指数仅作用于3，除非有括号。看错这一点是非常典型的错误。
form using × for multiplication. Do not forget to include prime factors that appear only in one of the numbers when finding the LCM – that is another frequent error.
求HCF和LCM时，除非数字很小，否则尽量避免单纯列举因数的方法。使用质因数分解更可靠，且能展示出清晰的解题步骤，即便犯个小错也能获得方法分。请务必用相乘的形式（如 × ）清晰地写出质因数指数表达式。在求LCM时，别忘了纳入仅在某数中出现的质因数——这也是常见的失分点。
he relationship HCF × LCM = product of the numbers (for two numbers). If the product doesn't match, you know an error has been made.
最后，永远用HCF × LCM = 两数之积这一关系检验答案（限于两个数）。若乘积不符，说明计算过程有误。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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