📚 Prime Factorisation and the Number 226 | 质因数分解与数字226
In this article, we explore the fundamental ideas of factors, multiples, primes, and prime factorisation, using the number 226 as our worked example throughout. These concepts form the backbone of number theory and are heavily tested in the Edexcel IGCSE Mathematics syllabus.
在这篇文章中，我们以数字226为贯穿全篇的例子，深入探讨因数、倍数、质数以及质因数分解的基本概念。这些概念构成了数论的核心，同时也是Edexcel IGCSE数学考试中的重点考查内容。
1. What Are Prime Numbers? | 什么是质数？
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 the first six prime numbers.
质数是大于1的正整数，它只有两个不同的正因数：1和它本身。例如，2、3、5、7、11和13是前六个质数。
A composite number is a positive integer that has more than two factors. For instance, 4, 6, 8, 9, and 10 are composite because each has additional factors besides 1 and itself.
合数是具有两个以上因数的正整数。例如，4、6、8、9和10都是合数，因为除了1和它本身之外，它们还有其他因数。
The numbers 0 and 1 are special. They are neither prime nor composite. Also, 2 is the only even prime number; every other even number is divisible by 2.
数字0和1是特殊的，它们既不是质数也不是合数。此外，2是唯一的偶质数；所有其他偶数都能被2整除。
Prime: has exactly two factors.
质数：恰好有两个因数。
Composite: has more than two factors.
合数：有两个以上因数。
Neither prime nor composite: 0 and 1.
既不是质数也不是合数：0和1。
2. Divisibility Rules | 整除规则
Before factorising a number, we often use divisibility rules to quickly decide if a small prime divides it. For example, a number is divisible by 2 if its last digit is even, and divisible by 3 if the sum of its digits is divisible by 3.
在分解一个数之前，我们通常使用整除规则快速判断它能否被某个小质数整除。例如，一个数的末位数字是偶数则能被2整除；一个数的各位数字之和能被3整除，则这个数能被3整除。
Other useful rules: a number is divisible by 5 if it ends in 0 or 5; divisible by 9 if the sum of its digits is divisible by 9; divisible by 11 if the alternating sum of its digits is a multiple of 11.
其他有用的规则：如果一个数末位是0或5，则能被5整除；如果一个数各位数字之和能被9整除，则这个数能被9整除；如果一个数的奇偶位数字交替相加之差是11的倍数，则这个数能被11整除。
visible by 3, so 226 is not divisible by 3. Since the number does not end in 0 or 5, it is not divisible by 5.
让我们将这些规则应用于数字226。它的末位数字是6，是偶数，因此226能被2整除。它的各位数字之和是2+2+6=10，不能被3整除，所以226不能被3整除。由于这个数不以0或5结尾，因此它不能被5整除。
Divisible by 2: last digit is even (0, 2, 4, 6, 8).
能被2整除：末位数字为偶数（0、2、4、6、8）。
Divisible by 3: sum of digits divisible by 3.
能被3整除：各位数字之和能被3整除。
Divisible by 5: last digit is 0 or 5.
能被5整除：末位数字是0或5。
Divisible by 9: sum of digits divisible by 9.
能被9整除：各位数字之和能被9整除。
um of digits is a multiple of 11. For 226, (2 − 2 + 6) = 6, which is not a multiple of 11.
能被11整除：奇偶位数字交替相加之差是11的倍数。对于226，2−2+6=6，不是11的倍数。
3. Is 226 Prime or Composite? | 226是质数还是合数？
p to its square root. For 226, √226 ≈ 15.03, so we test the primes 2, 3, 5, 7, 11, and 13.
要判断一个数是否为质数，我们只需要检验不超过其平方根的质因数。对于226，√226≈15.03，因此我们只需检验质数2、3、5、7、11和13。
6 has at least one factor besides 1 and itself, it is not prime; it is a composite number.
我们已经知道226能被2整除。事实上，226÷2=113。由于226除了1和它本身之外至少还有一个因数，所以它不是质数，而是合数。
But wait – is 113 itself prime? ÷ 7 leaves a remainder, meaning 7 does not divide 113. Therefore, 113 has no divisors other than 1 and itself, so 113 is prime.
但是，113本身是质数吗？要检查这一点，我们需要检验不超过√113≈10.6的质数，即2、3、5和7。113的末位是3，所以不能被2或5整除。各位数字之和为1+1+3=5，不能被3整除。最后，7×16=112，所以113除以7有余数，说明7不能整除113。因此，113除了1和它本身之外没有其他因数，所以113是质数。
4. Prime Factorisation: Factor Tree Method | 质因数分解：因数树法
Prime factorisation is the process of writing a number as a product of prime numbers. A common and visual method is the factor tree.
质因数分解是将一个数写成质数乘积的过程。一种常见且直观的方法是因数树法。
For 226, we see that the smallest prime divisor is 2. Divide 226 by 2 to get 113. Since 2 and 113 are both prime, the tree stops here.
对于226，最小的质因数是2。用2除以226得到113。由于2和113都是质数，因数树到此结束。
The factor tree can be represented in words:
因数树可以用文字表示如下：
226 = 2 × 113
Here, 226 is the original number; it splits into two “branches”, 2 and 113. Because both branches end in prime numbers, the prime factorisation is complete.
这里，226是原始数字；它分成两个“分支”，2和113。因为两个分支都结束于质数，所以质因数分解完成。
Step 1: Find the smallest prime that divides the number.
步骤1：找出能整除该数的最小质数。
Step 2: Write the number as that prime times the quotient.
步骤2：将这个数写成该质数乘以商。
Step 3: Continue factorising the quotient until all factors are prime.
步骤3：继续分解商，直到所有因数都是质数。
5. Writing 226 as a Product of Prime Factors | 将226写成质因数的乘积
The prime factorisation of 226 can be written as a product without showing the multiplication sign:
226的质因数分解可以写成不带乘号的乘积形式：
226 = 2 × 113
or occurs more than once, we use exponents. For example, 72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3².
由于没有重复的质因数，我们不需要使用指数记法。然而，如果一个因数出现多次，我们就使用指数。例如，72=2×2×2×3×3=2³×3²。
In some exam questions, you may be asked to write your answer in “index form” or 226, the index form is simply 2¹ × 113¹, but we usually omit exponent 1 for simplicity.
在某些考试题目中，你可能被要求用“指数形式”写出答案。对于226，指数形式就是2¹×113¹，但通常为了简洁，我们省略指数1。
As a general rule, always write prime factors in ascending order – smaller primes first. This helps avoid missing any factors and makes your working easy to mark.
作为一般规则，始终按升序排列质因数——较小的质数在前。这有助于避免遗漏任何因数，并使你的解题过程易于评分。
6. Using Prime Factorisation to Find HCF and LCM | 用质因数分解求HCF与LCM
Prime factorisation is a powerful tool for finding the Highest Common Factor (HCF) and the Lowest Common Multiple (LCM) of two numbers.
质因数分解是求两个数的最大公因数（HCF）和最小公倍数（LCM）的强大工具。
For the HCF, take each common prime factor with the smaller (lowest) exponent. For the LCM, take every prime factor that appears in either number with the larger (highest) exponent.
求HCF时，对每个公共质因数取较小的指数。求LCM时，对任一数字中出现的每个质因数取较大的指数。
80. We already know 226 = 2 × 113. Next, 180 = 18 × 10 = (2 × 3²) × (2 × 5) = 2² × 3² × 5.
让我们使用226和180的质因数分解。我们已经知道226=2×113。接下来，180=18×10=(2×3²)×(2×5)=2²×3²×5。
Compare the prime factors:
比较质因数：
| Prime | In 226 | In 180 | HCF exponent | LCM exponent |
| 2 | 2¹ | 2² | 1 | 2 |
| 3 | — | 3² | — | 2 |
| 5 | — | 5¹ | — | 1 |
| 113 | 113¹ | — | — | 1 |
, HCF(226, 180) = 2¹ = 2, and LCM(226, 180) = 2² × 3² × 5 × 113 = 4 × 9 × 5 × 113 = 20340.
因此，HCF(226,180)=2¹=2，LCM(226,180)=2²×3²×5×113=4×9×5×113=20340。
, b) = a × b. Here, 2 × 20340 = 40680, and 226 × 180 = 40680, so the answer is consistent.
你还可以验证这样一个关系：对于任意两个正整数a和b，HCF(a,b)×LCM(a,b)=a×b。这里，2×20340=40680，而226×180=40680，所以答案是一致的。
7. Powers and Roots Involving 226 | 与226相关的幂与根
derstand its square root and cube root. Because 226 = 2 × 113, and since neither 2 nor 113
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