📚 Prime Numbers and Factorisation: Exploring 179 | 质数与因数分解：探索179
Prime numbers are the building blocks of all whole numbers. In this article, we take a closer look at 179, a prime number, and use it to review the core ideas of factors, multiples, prime factorisation, HCF and LCM — all essential for Edexcel IGCSE Mathematics.
质数是所有整数的构建基石。在以质数179为引子的基础上，本文将系统复习因数、倍数、质因数分解、最大公因数（HCF）和最小公倍数（LCM）等核心概念——这些正是Edexcel IGCSE数学的重要考点。
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 factors: 1 and itself. For example, 2, 3, 5, 7 and 11 are the first five primes.
质数是大于1的正整数，且只有两个不同的因数：1和它本身。例如，2、3、5、7、11是最小的五个质数。
Numbers that have more than two factors are called composite numbers. The number 1 is neither prime nor composite because it has only one factor.
拥有两个以上因数的整数称为合数。数字1既不是质数也不是合数，因为它只有一个因数。
- Prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, …
- 质数：2、3、5、7、11、13、17、19、23、29……
Notice that 2 is the only even prime number. Every other even number has 2 as a factor, so it cannot be prime.
注意，2是唯一的偶质数。所有其他偶数都以2为因数，因此不可能是质数。
2. Is 179 Prime? Checking Divisibility | 179是质数吗？检验整除性
e numbers up to the square root of 179. Since √179 ≈ 13.38, we test 2, 3, 5, 7, 11 and 13.
要检验179是否为质数，我们只需用不大于√179的所有质数来检验。由于√179≈13.38，我们需要测试2、3、5、7、11和13。
- 179 is not even, so it is not divisible by 2.
- 179不是偶数，因此不能被2整除。
- 1 + 7 + 9 = 17, which is not a multiple of 3, so 179 is not divisible by 3.
- 1+7+9=17，不是3的倍数，因此179不能被3整除。
- 179 does not end in 0 or 5, so it is not divisible by 5.
- 179的末位不是0或5，因此不能被5整除。
- 7 × 25 = 175 and 7 × 26 = 182, so 179 is not divisible by 7.
- 7×25=175，7×26=182，所以179不能被7整除。
- 11 × 16 = 176 and 11 × 17 = 187, so 179 is not divisible by 11.
- 11×16=176，11×17=187，所以179不能被11整除。
- 13 × 13 = 169 and 13 × 14 = 182, so 179 is not divisible by 13.
- 13×13=169，13×14=182，所以179不能被13整除。
Since none of these primes divides 179, we conclude that 179 is a prime number.
由于这些质数都不能整除179，我们可以得出结论：179是质数。
3. Factors and Multiples | 因数与倍数
The factors of a number are positive integers that divide the number exactly. Because 179 is prime, its only positive factors are 1 and 179.
一个数的因数是能整除该数的正整数。由于179是质数，它的正因数只有1和179。
Multiples of a number are obtained by multiplying it by positive integers. The first few multiples of 179 are 179, 358, 537, 716, 895, and so on.
一个数的倍数由它与正整数相乘得到。179的前几个倍数是179、358、537、716、895，依此类推。
For any prime number p, the factor list is always {1, p}. This property makes primes especially useful for constructing unique products.
对于任何质数p，它的因数列表始终是{1, p}。这个性质使得质数在构造唯一乘积时特别有用。
4. Prime Factorisation | 质因数分解
Every integer greater than 1 can be written uniquely as a product of prime factors. This is called prime factorisation or the fundamental theorem of arithmetic.
每个大于1的整数都可以唯一地写成质因数的乘积，这称为质因数分解，也称算术基本定理。
For 179, since it is already prime, its prime factorisation is simply 179.
对于179，由于它本身是质数，它的质因数分解就是179本身。
For a composite number, we use a factor tree. For example, 60 = 2² × 3 × 5.
对于合数，我们可以用分解质因数的“树图”。例如，60 = 2² × 3 × 5。
60 = 2² × 3 × 5
When writing prime factorisation, always use the smallest primes first and arrange them in ascending order of base. Use index notation for repeated factors.
书写质因数分解时，通常从小到大排列质数，并用指数记号表示重复出现的因数。
5. Highest Common Factor (HCF) | 最大公因数
The Highest Common Factor (HCF) of two or more numbers is the largest positive integer that divides all of them exactly. It is also called the Greatest Common Divisor (GCD).
两个或多个数的最大公因数（HCF）是能同时整除所有这些数的最大正整数，也叫最大公约数（GCD）。
To find the HCF using prime factorisation, first write each number as a product of its prime factors. Then, for each common prime, take the smallest exponent that appears in all the factorisations. Multiply these together.
利用质因数分解求HCF的方法：先将每个数写成质因数乘积的形式；然后，对每个共同质数，取所有分解中出现的最小指数；最后将这些结果相乘。
Example: Find the HCF of 60 and 84.
例如：求60和84的HCF。
- 60 = 2² × 3 × 5
- 60 = 2² × 3 × 5
- 84 = 2² × 3 × 7
- 84 = 2² × 3 × 7
Common primes are 2 and 3. The lowest powers are 2² and 3, so HCF = 2² × 3 = 12.
共同质数是2和3。最低次幂分别为2²和3，因此HCF = 2² × 3 = 12。
HCF(60, 84) = 12
6. Lowest Common Multiple (LCM) | 最小公倍数
The Lowest Common Multiple (LCM) of two or more numbers is the smallest positive integer that is a multiple of all of them. Another name is Least Common Multiple.
两个或多个数的最小公倍数（LCM）是能同时被所有这些数整除的最小正整数，也叫最小公倍数。
To find the LCM using prime factorisation, write each number as a product of primes. Then take the highest power of every prime that appears in any of the factorisations, and multiply these together.
利用质因数分解求LCM的方法：将每个数写成质因数乘积；然后对出现的每一个质数，取其最高次幂；最后相乘。
s are 2, 3, 5 and 7. The highest powers are 2², 3, 5 and 7, so LCM = 2² × 3 × 5 × 7 = 420.
用同一个例子：60 = 2² × 3 × 5，84 = 2² × 3 × 7。出现的不同质数是2、3、5、7，最高次幂分别为2²、3、5、7，因此LCM = 2² × 3 × 5 × 7 = 420。
LCM(60, 84) = 420
7. Using Prime Factorisation for HCF and LCM | 用质因数分解求HCF和LCM
Prime factorisation provides a systematic method for finding both HCF and LCM. Let us compare the two rules side by side.
质因数分解为求HCF和LCM提供了一种系统方法。下面我们把两种规则放在一起对比。
| HCF | Take common primes only, using the lowest exponent for each. |
| HCF（最大公因数） | 只取共同质的质数，每个质数取最低指数。 |
| LCM | Take all primes that appear in any factorisation, using the highest exponent for each. |
| LCM（最小公倍数） | 取所有分解中出现的质数，每个质数取最高指数。 |
factor with 60. Also LCM(179, 60) = 179 × 60 = 10740, because the two numbers are coprime.
对于179而言，由于它是质数，HCF(179, 60) = 1，因为179与60没有共同的质因数。同时LCM(179, 60) = 179 × 60 = 10740，因为这两个数互质。
For coprime numbers a and b: HCF = 1 and LCM = a × b
对于互质数a和b：HCF = 1，LCM = a × b
8. Real-Life Applications | 实际应用
HCF and LCM are not just abstract ideas; they appear in many everyday situations.
HCF和LCM不仅仅是抽象概念，它们出现在许多日常情境中。
- HCF is used when dividing items into equal groups with no leftovers. For example, packing 60 apples and 84 oranges into identical gift bags.
- HCF用于把物品分成均等的组且没有剩余。例如，把60个苹果和84个橙子装入相同的礼品袋。
- LCM is used to find a common meeting time or the next time two events occur together. For example, if a bus arrives every 12 minutes and a train every 18 minutes, the LCM gives the number of minutes until they arrive together again.
- LCM用于寻找共同的相遇时间或两个事件同时发生的下一次时间。例如，公共汽车每12分钟一班，火车每18分钟一班，LCM给出它们再次同时到站的分钟数。
- Large prime numbers are used in computer encryption, such as the RSA algorithm, where the difficulty of factorising a large product of two primes keeps data secure.
- 大质数用于计算机加密，例如RSA算法，其安全性依赖于将两个大质数的乘积进行因数分解的难度。
9. Exam Tips for Prime Numbers | 质数考点考试技巧
In the Edexcel IGCSE Mathematics exam, questions on primes, HCF and LCM often appear in both paper 1 and paper 2. Here are some useful tips.
在Edexcel IGCSE数学考试中，关于质数、HCF和LCM的题目通常会在试卷1和试卷2中出现。以下是一些实用技巧。
- Lear the first 15 primes by heart: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
- 熟记前15个质数：2、3、5、7、11、13、17、19、23、29、31、37、41、43、47。
- Use divisibility rules to test small primes quickly.
- 利用整除规则快速检验较小的质数。
- In prime factorisation, always check that your final product expands to the original number.
- 在进行质因数分解时，务必检查最终乘积是否展开后等于原数。
- When asked for HCF or LCM, clearly state which one you are calculating; do not mix up the two methods.
- 当题目要求HCF或LCM时，明确说明你计算的是哪一个；不要混淆两者的方法。
10. Common Mistakes to Avoid | 避免常见错误
Many students lose marks due to small but repeated errors. Let us highlight the most common pitfalls.
许多学生因为一些细小但反复出现的错误而丢分。下面我们指出最常见的几个陷阱。
- Thinking that 1 is a prime number. It is not.
- 误认为1是质数。1不是质数。
- Thinking that all odd numbers are prime. For example, 9, 15 and 21 are odd but composite.
- 误认为所有奇数都是质数。例如，9、15、21都是奇数但为合数。
- Forgetting that 2 is prime. Since 2 is also even, it is easily overlooked.
- 忘记2是质数。因为2也是偶数，所以容易被忽略。
- When finding LCM, forgetting to include a prime that appears in only one of the factorisations.
- 求LCM时，忘记包含只出现在某一个分解中的质数。
- Writing factors incorrectly for negative numbers. Factors are usually positive unless otherwise stated.
- 对负数因数的书写错误。除非题中另有说明，因数通常为正。
11. Practice Questions | 练习题
Try these questions to check your understanding.
尝试以下题目来检验你的理解。
Q1. Write 360 as a product of prime factors.
第1题：将360写成质因数的乘积。
Answer: 360 = 2³ × 3² × 5
答案：360 = 2³ × 3² × 5
Q2. Find the HCF of 72 and 108.
第2题：求72和108的HCF。
Answer: HCF = 36
答案：HCF = 36
², 108 = 2² × 3³. Common primes: 2 and 3. Lowest powers: 2² and 3², so HCF = 2² × 3² = 36.
解析：72 = 2³ × 3²，108 = 2² × 3³。共同质数为2和3，最低次幂为2²和3²，因此HCF = 2² × 3² = 36。
Q3. Find the LCM of 15, 20 and 25.
第3题：求15、20和25的LCM。
Answer: LCM = 300
答案：LCM = 300
ion: 15 = 3 × 5, 20 = 2² × 5, 25 = 5². Highest powers: 2², 3, 5². LCM = 2² × 3 × 5² = 300.
解析：15 = 3 × 5，20 = 2² × 5，25 = 5²。最高次幂为2²、3、5²，因此LCM = 2² × 3 × 5² = 300。
12. Conclusion | 总结
In this article, we used the prime number 179 to explore the fundamental ideas of primes, factors, multiples and prime factorisation. We then applied these ideas to HCF and LCM, looked at real-life examples, and discussed common exam traps.
在本文中，我们以质数179为线索，探索了质数、因数、倍数和质因数分解等基本概念。随后将这些概念应用于HCF和LCM，讨论了实际应用，并分析了常见的考试陷阱。
Remember that prime factorisation is the key tool for both HCF and LCM. Practise it until it becomes automatic, and always read each exam question carefully to avoid losing easy marks.
请牢记：质因数分解是求解HCF和LCM的关键工具。反复练习直到熟练，并细心阅读每一道考试题目，以避免丢失容易得到的分数。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图