Mastering IGCSE Mathematics with 201 | 用201掌握IGCSE数学

📚 Mastering IGCSE Mathematics with 201 | 用201掌握IGCSE数学

Welcome to this IGCSE Mathematics revision session for Edexcel students. In this article, we use the number 201 as a single unifying example to explore key topics from the syllabus. By following the same number through arithmetic, algebra, geometry, trigonometry and statistics, you will see how concepts connect and how to apply them in exam questions.

欢迎阅读本篇文章，这是为爱德思（Edexcel）IGCSE数学考生准备的复习指南。我们将以数字201作为贯穿全文的例子，系统梳理考纲中的重要知识点。通过同一个数字在算术、代数、几何、三角与统计中的反复运用，你会更清楚地理解各部分之间的联系，并知道如何在考试中灵活应用。

1. Prime Factorisation and Divisibility | 质因数分解与整除性

The number 201 is composite, not prime. To factorise it, first check small primes. The sum of the digits of 201 is 2 + 0 + 1 = 3, so 201 is divisible by 3. Dividing gives 201 ÷ 3 = 67. Since 67 has no positive divisors except 1 and 67, the prime factorisation is 201 = 3 × 67.

数字201是合数，不是质数。要分解它，先判断能否被较小的质数整除。201的各位数字之和为2 + 0 + 1 = 3，因此201能被3整除。计算得201 ÷ 3 = 67。由于67除了1和67以外没有其他正因数，所以201的质因数分解为201 = 3 × 67。

Factors of 201: 1, 3, 67, 201.
201的因数：1、3、67、201。
201 has exactly four factors because its prime factorisation is 3¹ × 67¹.
201之所以有四个因数，是因为它的质因数分解形式为3¹ × 67¹。

Use the factor tree method in the exam: write 201, split into 3 and 67, then circle the primes. This method is quick and reliable.

考试中可以使用因数分解树：先写201，分成3和67，然后圈出质数。这种方法既快速又可靠。

2. HCF and LCM Using Prime Factorisation | 用质因数分解求最大公因数与最小公倍数

Take two numbers: 201 and 99. Their prime factorisations are 201 = 3 × 67 and 99 = 3² × 11. To find the highest common factor (HCF), multiply the smallest power of each common prime. Here the only common prime is 3, so HCF = 3.

取两个数：201和99。它们的质因数分解分别是201 = 3 × 67和99 = 3² × 11。求最大公因数（HCF）时，把公有质数的最小指数相乘。这里公有的质数只有3，因此HCF = 3。

For the lowest common multiple (LCM), take the largest power of every prime that appears: 3² × 11 × 67 = 6633. Check: HCF × LCM = 3 × 6633 = 19899, and 201 × 99 = 19899. This verifies the rule a × b = HCF × LCM.

求最小公倍数（LCM）时，把所有出现的质数的最大指数相乘：3² × 11 × 67 = 6633。检验：HCF × LCM = 3 × 6633 = 19899，而201 × 99 = 19899。这验证了公式a × b = HCF × LCM。

Number	Prime factorisation
201	3 × 67
99	3² × 11

3. Algebraic Expressions and Substitution | 代数式与代入

Algebra often asks you to evaluate or simplify expressions. Use 201 as a coefficient. Evaluate 201a − 3b when a = 2 and b = 1: 201(2) − 3(1) = 402 − 3 = 399.

代数题经常要求计算或化简表达式。用201作为系数。当a = 2且b = 1时，计算201a − 3b：201(2) − 3(1) = 402 − 3 = 399。

Simplify 201x + x. Since both terms contain x, add the coefficients: 201 + 1 = 202, so the result is 202x. Similarly, expand 201(x + 4): multiply both terms inside the bracket by 201, giving 201x + 804.

化简201x + x。两项都含有x，所以系数相加：201 + 1 = 202，结果为202x。类似地，展开201(x + 4)：括号内每一项都乘以201，得到201x + 804。

Always collect like terms before simplifying.
化简前一定要先合并同类项。
When substituting, use brackets around negative numbers.
代入时，负数要用括号括起来。

4. Solving Linear Equations | 解线性方程

Equations containing 201 are excellent practice. Solve 3x + 201 = 417. Subtract 201 from both sides: 3x = 216. Then divide by 3: x = 72. Check: 3(72) + 201 = 216 + 201 = 417, correct.

含有201的方程很适合用来练习。解方程3x + 201 = 417。两边同时减去201：3x = 216。然后两边除以3：x = 72。检验：3(72) + 201 = 216 + 201 = 417，正确。

Another example: x ÷ 67 = 201. Multiply both sides by 67: x = 201 × 67 = 13467. Use the inverse operation each time: subtraction cancels addition, and division cancels multiplication.

另一个例子：x ÷ 67 = 201。两边同时乘以67：x = 201 × 67 = 13467。每一步都使用逆运算：减法抵消加法，除法抵消乘法。

3x + 201 = 417 → 3x = 216 → x = 72

5. Percentages and Ratio | 百分比与比

Percentages are a core IGCSE topic. Find 15% of 201: 15 ÷ 100 × 201 = 30.15. To increase 201 by 8%, multiply by 1.08: 201 × 1.08 = 217.08. To decrease 201 by 20%, multiply by 0.80: 201 × 0.80 = 160.8.

百分比是IGCSE的核心考点。求201的15%：15 ÷ 100 × 201 = 30.15。把201增加8%，乘以1.08：201 × 1.08 = 217.08。把201减少20%，乘以0.80：201 × 0.80 = 160.8。

Ratio questions also appear frequently. Divide 201 in the ratio 2 : 1. Total parts = 2 + 1 = 3. One part = 201 ÷ 3 = 67. So the parts are 2 × 67 = 134 and 1 × 67 = 67.

比的应用题也经常出现。把201按2 : 1分配。总份数 = 2 + 1 = 3。每份 = 201 ÷ 3 = 67。因此两部分分别是2 × 67 = 134和1 × 67 = 67。

6. Sequences and nth Term | 数列与第n项

Use sequences to see where 201 might appear. Consider the arithmetic sequence 3, 6, 9, 12, … . The nth term is 3n. Set 3n = 201, so n = 67. Therefore 201 is the 67th term of this sequence.

用数列来判断201是否会出现。观察等差数列3, 6, 9, 12, …。它的第n项为3n。令3n = 201，得n = 67。因此201是该数列的第67项。

Now examine the sequence 5, 11, 17, 23, … . Its nth term is 6n − 1. Set 6n − 1 = 201, giving 6n = 202, so n = 33.666…, which is not an integer. Thus 201 is not a term in this sequence.

再检验数列5, 11, 17, 23, …。它的第n项为6n − 1。令6n − 1 = 201，得6n = 202，所以n = 33.666…，不是整数。因此201不是这个数列中的项。

For arithmetic sequences, nth term = first term + (n − 1) × common difference.
等差数列的第n项 = 首项 + (n − 1) × 公差。
If n is not a positive integer, the value is not in the sequence.
如果n不是正整数，则该值不在数列中。

7. Coordinate Geometry and Straight Lines | 坐标几何与直线

Straight-line graphs link algebra and geometry. Find the gradient of the line through (0, 201) and (4, 189). Gradient m = (189 − 201) ÷ (4 − 0) = −12 ÷ 4 = −3. Using y = mx + c, with intercept 201, the equation is y = −3x + 201.

直线图连接了代数与几何。求经过(0, 201)和(4, 189)的直线斜率。斜率m = (189 − 201) ÷ (4 − 0) = −12 ÷ 4 = −3。由y = mx + c，截距为201，直线方程为y = −3x + 201。

To find the x-intercept, set y = 0: 0 = −3x + 201, so 3x = 201 and x = 67. This means the line crosses the x-axis at 67, linking back to prime factorisation.

求x轴截距时，令y = 0：0 = −3x + 201，所以3x = 201，x = 67。这意味着直线与x轴交于67，正好与质因数分解联系起来。

y = −3x + 201

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Mastering IGCSE Mathematics with 201 | 用201掌握IGCSE数学