Binary and Hexadecimal Number Systems: An IGCSE Computer Science Guide — IGCSE计算机科学：二进制与十六进制数制完全指南

一、什么是数制？计数系统的核心概念 | What is a Number System? Core Concepts of Counting Systems

在计算机科学的入门阶段，理解数制（Number System）是最基础也最关键的一步。简单来说，数制就是一套用符号表示数字的规则系统。我们日常生活中最常用的是十进制（Denary/Decimal），它使用0到9这十个数字符号。但计算机并不理解十进制 – 计算机内部的电路只能识别两种状态：开（1）和关（0），这决定了计算机必须使用二进制（Binary）来存储和处理所有信息。对于IGCSE计算机科学的同学来说，掌握二进制和十六进制不仅是考试必考的内容，更是理解计算机底层工作原理的钥匙。

Understanding number systems is the first and most fundamental step in studying computer science. Simply put, a number system is a set of rules for representing numbers using symbols. The system we use every day is the denary (decimal) system, which uses ten symbols: 0 through 9. However, computers do not understand denary – the circuits inside a computer can only detect two states: on (1) and off (0). This fundamental constraint means computers must use the binary system to store and process all information. For IGCSE Computer Science students, mastering binary and hexadecimal is not only essential for the exam but is also the key to understanding how computers work at the lowest level.

在IGCSE计算机科学课程中，你需要掌握的三种核心数制是：十进制（Denary，基数为10）、二进制（Binary，基数为2）和十六进制（Hexadecimal，基数为16）。这三种数制之间的转换是考试中反复出现的题型。本篇文章将系统地讲解每一种数制的原理、转换方法和实际应用，同时结合CIE IGCSE Computer Science考试大纲的要求，帮助你彻底掌握这一核心知识模块。

In the IGCSE Computer Science syllabus, the three core number systems you need to master are: Denary (base-10), Binary (base-2), and Hexadecimal (base-16). Conversions between these three systems are recurring question types in the exam. This article will systematically explain the principles, conversion methods, and practical applications of each number system, aligned with the CIE IGCSE Computer Science specification, to help you thoroughly master this essential knowledge module.

二、二进制基础：只有0和1的世界 | Binary Basics: A World of Only 0s and 1s

二进制（Binary）是一种基数为2的数制，只使用两个数字：0和1。在计算机中，每一个0或1代表一个比特（bit，binary digit的缩写），这是计算机存储信息的最小单位。为什么计算机只用0和1？原因在于计算机的硬件基础 – 晶体管（transistor）只有两种稳定状态：导通（电流通过，代表1）和截止（电流阻断，代表0）。数百万甚至数十亿个晶体管组合在一起，通过0和1的序列就能表达极其复杂的信息。

Binary is a base-2 number system that uses only two digits: 0 and 1. In a computer, each 0 or 1 represents one bit (short for binary digit), which is the smallest unit of information storage. Why do computers only use 0 and 1? The answer lies in the hardware foundation of computers – transistors have only two stable states: conducting (current flows, representing 1) and non-conducting (current blocked, representing 0). Millions or even billions of transistors combined together can express extraordinarily complex information through sequences of 0s and 1s.

在二进制中，每一位的位置代表一个2的幂。从右向左，第一位是2⁰（=1），第二位是2¹（=2），第三位是2²（=4），第四位是2³（=8），以此类推。例如，二进制数1101的计算方式为：(1 × 2³) + (1 × 2²) + (0 × 2¹) + (1 × 2⁰) = 8 + 4 + 0 + 1 = 13。这就是二进制到十进制转换的核心原理。

In binary, each position represents a power of 2. From right to left, the first position is 2⁰ (=1), the second is 2¹ (=2), the third is 2² (=4), the fourth is 2³ (=8), and so on. For example, the binary number 1101 is calculated as: (1 × 2³) + (1 × 2²) + (0 × 2¹) + (1 × 2⁰) = 8 + 4 + 0 + 1 = 13. This is the core principle of binary to decimal conversion.

在IGCSE考试中，你可能会被要求给出某个十进制数在8位二进制表示下的结果。例如，十进制数29的8位二进制表示为00011101。注意：8位二进制意味着始终使用8个位（bits），当数字较小时，左边用0填充（称为leading zeros）。一个字节（byte）正好由8个比特组成，这是计算机存储的基本单位。

In IGCSE exams, you may be asked to give the 8-bit binary representation of a decimal number. For example, the decimal number 29 in 8-bit binary is 00011101. Note: 8-bit binary means always using 8 bits; when the number is small, the leftmost positions are filled with zeros (called leading zeros). One byte is exactly 8 bits, and this is the fundamental unit of computer storage.

三、二进制与十进制转换：位置权重法 | Binary-Decimal Conversion: The Place Value Method

将二进制转换为十进制的方法非常简单：将每一位上的数字乘以它所在位置的权重（2的幂），然后求总和。这个方法称为”位置权重法”（Place Value Method）。以8位二进制数10110101为例：从右往左，各位的权重分别是2⁰=1、2¹=2、2²=4、2³=8、2⁴=16、2⁵=32、2⁶=64、2⁷=128。然后计算总和：(1×128) + (0×64) + (1×32) + (1×16) + (0×8) + (1×4) + (0×2) + (1×1) = 128 + 0 + 32 + 16 + 0 + 4 + 0 + 1 = 181。

The method for converting binary to decimal is straightforward: multiply each digit by its place value (power of 2) and sum the results. This is called the “Place Value Method”. Take the 8-bit binary number 10110101 as an example: from right to left, the place values are 2⁰=1, 2¹=2, 2²=4, 2³=8, 2⁴=16, 2⁵=32, 2⁶=64, 2⁷=128. Then calculate the sum: (1×128) + (0×64) + (1×32) + (1×16) + (0×8) + (1×4) + (0×2) + (1×1) = 128 + 0 + 32 + 16 + 0 + 4 + 0 + 1 = 181.

反过来，将十进制转换为二进制则需要使用”除以2取余法”（Successive Division by 2）。具体步骤是：反复将十进制数除以2，记录每次的余数（0或1），直到商为0。然后将所有余数从下往上（从最后一个到第一个）排列，就得到了对应的二进制数。例如，将十进制数45转换为二进制：45÷2=22余1、22÷2=11余0、11÷2=5余1、5÷2=2余1、2÷2=1余0、1÷2=0余1。从下往上读余数：101101。因此45的二进制表示为101101（6位），或00101101（8位表示）。

Conversely, converting decimal to binary uses the “Successive Division by 2” method. The specific steps: repeatedly divide the decimal number by 2, recording the remainder each time (0 or 1), until the quotient is 0. Then arrange all remainders from bottom to top (last to first) to obtain the binary number. For example, converting decimal 45 to binary: 45÷2=22 remainder 1, 22÷2=11 remainder 0, 11÷2=5 remainder 1, 5÷2=2 remainder 1, 2÷2=1 remainder 0, 1÷2=0 remainder 1. Reading remainders from bottom to top: 101101. Thus 45 in binary is 101101 (6 bits), or 00101101 in 8-bit representation.

在IGCSE考试中，经常会出现”给出十进制数X的8位二进制表示”的题目。考生需要记得：如果转换结果不足8位，需要在左侧补充前导零。另外，8位二进制能够表示的范围是0到255（即00000000到11111111）。超出这个范围的数字需要用更多的位（如16位）来表示。

In IGCSE exams, questions like “Give the 8-bit binary representation of the decimal number X” frequently appear. Students must remember: if the conversion result has fewer than 8 bits, leading zeros must be added on the left. Additionally, the range that can be represented with 8 bits is 0 to 255 (i.e., 00000000 to 11111111). Numbers beyond this range require more bits (e.g., 16 bits) for representation.

四、十六进制：为什么计算机需要它？ | Hexadecimal: Why Do Computers Need It?

十六进制（Hexadecimal，简称Hex）是一种基数为16的数制。它使用16个符号：0到9代表数值0到9，A、B、C、D、E、F分别代表数值10、11、12、13、14、15。那么问题来了：既然计算机只理解二进制，为什么我们还需要学习十六进制？答案在于可读性和简洁性。

Hexadecimal (Hex for short) is a base-16 number system. It uses 16 symbols: 0 through 9 for values 0 through 9, and A, B, C, D, E, F for values 10, 11, 12, 13, 14, and 15 respectively. So the natural question is: if computers only understand binary, why do we need to learn hexadecimal? The answer lies in readability and conciseness.

想象一下：一个32位的二进制数，写出来是32个0和1的序列。人眼几乎不可能快速地阅读、比较或记忆这样的数字。但如果我们把它转换成十六进制，一个32位的二进制数只需要8个十六进制位就能表示 – 因为1个十六进制位恰好等于4个二进制位。例如，二进制数1101 1010 1111 0010用十六进制表示为DAF2，明显简洁得多。这就是为什么程序员在查看内存地址、颜色代码（如#FF5733）和机器码时，几乎总是使用十六进制。

Imagine this: a 32-bit binary number written out is a sequence of 32 zeros and ones. The human eye can hardly read, compare, or remember such a number quickly. But if we convert it to hexadecimal, a 32-bit binary number can be represented with just 8 hexadecimal digits – because one hexadecimal digit corresponds exactly to 4 binary digits. For example, the binary number 1101 1010 1111 0010 is represented as DAF2 in hexadecimal, which is clearly much more concise. This is why programmers almost always use hexadecimal when examining memory addresses, colour codes (like #FF5733), and machine code.

十六进制与二进制之间有一种天然的一一对应关系：每4个二进制位（称为一个nibble）恰好对应一个十六进制位。这个转换关系非常重要，也是IGCSE考试的热门考点。以下是对照表：0000=0, 0001=1, 0010=2, 0011=3, 0100=4, 0101=5, 0110=6, 0111=7, 1000=8, 1001=9, 1010=A, 1011=B, 1100=C, 1101=D, 1110=E, 1111=F。IGCSE考生不需要背诵整张表，但必须理解这个4位对1位的对应逻辑。

There is a natural one-to-one correspondence between hexadecimal and binary: every 4 binary bits (called a nibble) corresponds exactly to one hexadecimal digit. This conversion relationship is very important and is a hot topic in IGCSE exams. Here is the reference table: 0000=0, 0001=1, 0010=2, 0011=3, 0100=4, 0101=5, 0110=6, 0111=7, 1000=8, 1001=9, 1010=A, 1011=B, 1100=C, 1101=D, 1110=E, 1111=F. IGCSE students do not need to memorise the entire table, but must understand the logic of this 4-to-1 correspondence.

五、十六进制与二进制/十进制互相转换 | Hexadecimal-Binary-Decimal Conversion Methods

将十六进制转换为二进制是最简单的转换操作：每个十六进制位替换为对应的4位二进制即可。例如，将十六进制数5F3转换为二进制：5=0101, F=1111, 3=0011，因此5F3的二进制为0101 1111 0011。注意：不要省略前导零，比如5必须是0101而不是101。反过来，将二进制转换为十六进制：从右向左每4位一组（不足4位在左侧补0），然后将每组转换为对应的十六进制符号。例如，二进制数110 1001转换为十六进制：从右向左分组→0110 1001（左侧补两个0），然后0110=6, 1001=9，结果为69。

Converting hexadecimal to binary is the simplest conversion operation: replace each hexadecimal digit with its corresponding 4-bit binary. For example, converting hexadecimal 5F3 to binary: 5=0101, F=1111, 3=0011, so 5F3 in binary is 0101 1111 0011. Note: do not omit leading zeros – 5 must be 0101, not 101. Conversely, converting binary to hexadecimal: group every 4 bits from right to left (pad with zeros on the left if the last group has fewer than 4 bits), then convert each group to its hexadecimal symbol. For example, binary 110 1001 → group right to left → 0110 1001 (pad two zeros on the left), then 0110=6, 1001=9, result: 69.

将十六进制转换为十进制，同样使用位置权重法 – 只是这次的基数是16。从右向左，各位权重分别是16⁰=1、16¹=16、16²=256、16³=4096等。例如，十六进制数2AF转换为十进制：(2×16²) + (10×16¹) + (15×16⁰) = (2×256) + (10×16) + (15×1) = 512 + 160 + 15 = 687。

To convert hexadecimal to decimal, we again use the Place Value Method – but this time the base is 16. From right to left, the place values are 16⁰=1, 16¹=16, 16²=256, 16³=4096, and so on. For example, converting hexadecimal 2AF to decimal: (2×16²) + (10×16¹) + (15×16⁰) = (2×256) + (10×16) + (15×1) = 512 + 160 + 15 = 687.

将十进制转换为十六进制，与十进制转二进制的逻辑类似，使用”除以16取余法”：反复将十进制数除以16，记录余数（0到15，其中10到15用A到F表示），直到商为0。例如，将十进制数498转换为十六进制：498÷16=31余2, 31÷16=1余15(F), 1÷16=0余1。从下往上读余数：1F2。因此498的十六进制表示为1F2。IGCSE考试中，这类题目通常会要求先转二进制再转十六进制，利用4位一组的快速转换法。

Converting decimal to hexadecimal follows similar logic to decimal-to-binary conversion, using the “Successive Division by 16” method: repeatedly divide the decimal number by 16, recording remainders (0 to 15, where 10 to 15 are represented as A to F), until the quotient is 0. For example, converting 498 to hexadecimal: 498÷16=31 remainder 2, 31÷16=1 remainder 15 (F), 1÷16=0 remainder 1. Reading remainders from bottom to top: 1F2. Thus 498 in hexadecimal is 1F2. In IGCSE exams, questions of this type often ask students to convert to binary first and then to hexadecimal, using the quick 4-bit grouping method.

六、二进制运算：加法与逻辑运算基础 | Binary Arithmetic: Addition and Logic Operations

二进制加法遵循与十进制加法相同的原则，只是”逢二进一”而非”逢十进一”。基本规则有四个：0+0=0, 0+1=1, 1+0=1, 1+1=0（进位1，即carry 1）。让我们看一个例子：计算0110（十进制6）+ 0101（十进制5）。从右向左逐位相加：第0位：0+1=1；第1位：1+0=1；第2位：1+1=0，进位1；第3位：0+0+进位1=1。结果：1011（十进制11），6+5=11，正确！

Binary addition follows the same principles as decimal addition, except it uses “carry when reaching 2” rather than “carry when reaching 10”. There are four basic rules: 0+0=0, 0+1=1, 1+0=1, 1+1=0 (carry 1). Let us look at an example: calculate 0110 (decimal 6) + 0101 (decimal 5). Adding bit by bit from right to left: bit 0: 0+1=1; bit 1: 1+0=1; bit 2: 1+1=0, carry 1; bit 3: 0+0+carry 1=1. Result: 1011 (decimal 11). 6+5=11, correct!

在计算机中，当两个8位二进制数相加的结果超过8位时（即超过255），会发生溢出（Overflow）。溢出是计算机算术中一个重要的概念 – 如果结果需要多于可用位数的位来表示，最左边的进位会被丢弃，导致结果不正确。IGCSE考试可能会让你识别一个给定的加法是否产生了溢出。判断方法很简单：如果两个正数相加得到一个负数（在使用补码表示的情况下），或者进位超出了可用位数，就发生了溢出。

In computers, when the result of adding two 8-bit binary numbers exceeds 8 bits (i.e., exceeds 255), an overflow occurs. Overflow is an important concept in computer arithmetic – if the result requires more bits than are available, the leftmost carry is discarded, leading to an incorrect result. IGCSE exams may ask you to identify whether a given addition has produced an overflow. The simple way to judge: if adding two positive numbers yields a negative number (when using two’s complement notation), or if the carry exceeds the available number of bits, an overflow has occurred.

除了算术运算，IGCSE考试还涉及基本的逻辑运算（Logical Operations），包括AND、OR、NOT和XOR。这些运算在二进制位上逐位进行。例如，AND运算的真值表为：1 AND 1=1，其他所有组合（1 AND 0, 0 AND 1, 0 AND 0）都等于0。逻辑运算在计算机的电路中由逻辑门（Logic Gates）实现，是构建所有数字电路的基础。

Beyond arithmetic operations, the IGCSE syllabus also covers basic logical operations, including AND, OR, NOT, and XOR. These operations are performed bit by bit on binary values. For example, the truth table for AND: 1 AND 1=1, all other combinations (1 AND 0, 0 AND 1, 0 AND 0) equal 0. Logic operations are implemented in computer circuits by logic gates, which form the foundation of all digital circuitry.

七、数据存储单位：bit, byte, kilobyte到gigabyte | Data Storage Units: From Bits to Gigabytes

理解数据存储单位是IGCSE计算机科学的基础知识。数据存储单位的层次结构如下：1 bit（比特）= 一个二进制位（0或1）；1 nibble = 4 bits（半字节）；1 byte（字节）= 8 bits；1 kilobyte（KB）= 1024 bytes（注意：是2¹⁰=1024，不是1000！）；1 megabyte（MB）= 1024 KB；1 gigabyte（GB）= 1024 MB；1 terabyte（TB）= 1024 GB。

Understanding data storage units is foundational knowledge for IGCSE Computer Science. The hierarchy of data storage units is as follows: 1 bit = one binary digit (0 or 1); 1 nibble = 4 bits; 1 byte = 8 bits; 1 kilobyte (KB) = 1024 bytes (note: 2¹⁰=1024, not 1000!); 1 megabyte (MB) = 1024 KB; 1 gigabyte (GB) = 1024 MB; 1 terabyte (TB) = 1024 GB.

这里有一个IGCSE考试中常见的陷阱：在计算机领域，kilobyte等单位的换算因子是1024（2¹⁰）而不是1000。这与国际单位制（SI）中kilo-代表1000不同。因此，1KB的存储空间实际上可以存储1024个字符（假设每个字符占1 byte），而不是1000个。考试中的计算题务必使用1024进行换算。另外，一个字节（1 byte）可以表示256种不同的值（从00000000到11111111），这刚好足够表示英文字母表中的所有大写和小写字母、数字和常用标点符号 – 这也是ASCII编码系统的基础。

Here is a common trap in IGCSE exams: in the computing field, the conversion factor for kilobytes and similar units is 1024 (2¹⁰), not 1000. This differs from the International System of Units (SI) where kilo- means 1000. Therefore, 1KB of storage can actually hold 1024 characters (assuming 1 byte per character), not 1000. Always use 1024 for calculations in exam questions. Additionally, one byte can represent 256 different values (from 00000000 to 11111111), which is just enough to cover all uppercase and lowercase letters of the English alphabet, digits, and common punctuation marks – this is also the basis of the ASCII encoding system.

IGCSE考试中常见的计算题包括：给定文件大小（以KB或MB为单位），计算可以存储多少个字符，或者计算传输该文件所需的时间（结合数据传输速率）。例如：一张图片大小为2.5MB，如果网络下载速度为512Kbps（kilobits per second，注意是bits不是bytes），计算下载所需时间。解答：2.5MB = 2.5 × 1024 × 1024 × 8 = 20,971,520 bits。时间 = 20,971,520 / 512,000 ≈ 41秒。这类题目考察的是单位换算的熟练度。

Common calculation questions in IGCSE exams include: given a file size (in KB or MB), calculate how many characters can be stored, or calculate the time required to transmit the file (combined with data transfer rates). For example: an image is 2.5MB in size, and the network download speed is 512Kbps (kilobits per second – note, bits not bytes). Calculate the download time. Solution: 2.5MB = 2.5 × 1024 × 1024 × 8 = 20,971,520 bits. Time = 20,971,520 / 512,000 ≈ 41 seconds. Questions of this type test your proficiency with unit conversions.

八、二进制在计算机中的应用：文本、图像与指令 | Binary Applications: Text, Images and Instructions

二进制不仅用于数字表示，它在计算机中有着广泛的实际应用。首先是文本表示：计算机使用字符编码标准将文字转换为二进制。最常见的编码是ASCII（American Standard Code for Information Interchange），使用7位（扩展ASCII使用8位）来表示英文字符。例如，大写字母’A’的ASCII码是65（二进制01000001），小写字母’a’是97（二进制01100001）。在现代计算机中，Unicode编码被广泛使用以支持全球各种语言的字符，它使用可变长度编码，可以表示超过14万个字符。

Binary is not only used for representing numbers; it has extensive practical applications in computers. First, text representation: computers use character encoding standards to convert text into binary. The most common encoding is ASCII (American Standard Code for Information Interchange), which uses 7 bits (Extended ASCII uses 8 bits) to represent English characters. For example, the uppercase letter ‘A’ has ASCII code 65 (binary 01000001), and lowercase ‘a’ is 97 (binary 01100001). In modern computers, Unicode encoding is widely used to support characters from all languages worldwide; it uses variable-length encoding and can represent over 140,000 characters.

其次是图像表示：计算机将图像分解为微小的像素（pixels），每个像素的颜色用二进制数值来表示。在黑白图像中，每个像素只需要1位（0=黑，1=白）。在灰度图像中，每个像素可能需要8位，提供256个灰度级别。在彩色图像中，每个像素通常使用24位（RGB模型 – 红、绿、蓝各8位），可以表示约1670万种颜色。图像的分辨率（resolution，即像素数量）和色深（colour depth，即每像素的位数）共同决定了图像文件的大小。

Second, image representation: computers break down images into tiny pixels, with each pixel’s colour represented by a binary value. In a black and white image, each pixel needs only 1 bit (0=black, 1=white). In a greyscale image, each pixel may need 8 bits, providing 256 grey levels. In a colour image, each pixel typically uses 24 bits (the RGB model – 8 bits each for red, green, and blue), capable of representing approximately 16.7 million colours. Image resolution (the number of pixels) and colour depth (bits per pixel) together determine the file size of an image.

第三是指令表示：当你在编程时写的代码（例如Python程序），最终都会被翻译成机器码（Machine Code） – 一串二进制指令，由CPU直接执行。每一条机器指令由操作码（Opcode，告诉CPU做什么操作）和操作数（Operand，告诉CPU操作的数据在哪里）组成，都是二进制形式的。这意味着，从你打字输入的每一个字符，到屏幕上显示的每一帧画面，再到你点击鼠标触发的每一个操作，在计算机底层全部都是0和1的序列。

Third, instruction representation: when you write code while programming (for example, a Python program), it is ultimately translated into machine code – a sequence of binary instructions directly executed by the CPU. Each machine instruction consists of an opcode (telling the CPU what operation to perform) and an operand (telling the CPU where the data is), both in binary form. This means that every character you type, every frame displayed on the screen, and every action triggered by clicking your mouse is ultimately just a sequence of zeros and ones at the computer’s lowest level.

九、IGCSE考试常见题型与解题技巧 | IGCSE Exam Question Types and Problem-Solving Techniques

在CIE IGCSE Computer Science的考试中（Paper 1: Theory），数制转换是一个高频考点。常见的题型包括以下几种。第一种：直接转换题 – 例如”将十进制数154转换为8位二进制数”。解题步骤：使用除以2取余法得到10011010（8位刚好）。如果是8位二进制，记得验证结果是否正好8位。

In the CIE IGCSE Computer Science exam (Paper 1: Theory), number system conversion is a high-frequency topic. Common question types include the following. Type one: direct conversion – for example, “Convert the denary number 154 into an 8-bit binary number.” Solution steps: use successive division by 2 to get 10011010 (exactly 8 bits). For 8-bit binary, always verify the result has exactly 8 bits.

第二种：十六进制与二进制的快速转换 – 例如”将十六进制数3E8转换为二进制”。解题技巧：不要先转十进制再转二进制！直接使用4位一组的替换法：3=0011, E=1110, 8=1000，答案：0011 1110 1000。这种题考察的就是你是否掌握了十六进制和二进制之间4位一组的对应关系。

Type two: quick conversion between hexadecimal and binary – for example, “Convert the hexadecimal number 3E8 into binary.” Technique: do not convert to decimal first and then to binary! Use the direct 4-bit group substitution method: 3=0011, E=1110, 8=1000. Answer: 0011 1110 1000. This type of question tests whether you have mastered the 4-bit group correspondence between hexadecimal and binary.

第三种：溢出识别题 – 给出两个二进制数的加法运算，问结果是否正确以及是否发生了溢出。解题技巧：先正常做二进制加法，然后检查两个关键点：①当两个8位二进制数相加时，进位到了第9位（即超出8位范围），发生了溢出；②或者检查符号位的变化：如果两个正数（第7位=0）相加后最高位变成了1（在补码表示中代表负数），则发生了溢出。

Type three: overflow identification – given the addition of two binary numbers, determine whether the result is correct and whether overflow has occurred. Technique: first perform the binary addition normally, then check two key points: (1) when adding two 8-bit binary numbers, if the carry reaches a 9th bit (exceeding the 8-bit range), overflow has occurred; (2) alternatively, check the sign bit: if two positive numbers (bit 7=0) added together produce a result with the most significant bit as 1 (representing a negative number in two’s complement notation), overflow has occurred.

第四种：存储计算题 – 例如”一张分辨率为1024×768的彩色照片，色深为24位，计算文件大小（以KB为单位）”。解题步骤：总像素数 = 1024 × 768 = 786,432；总位数 = 786,432 × 24 = 18,874,368 bits；转为bytes = 18,874,368 / 8 = 2,359,296 bytes；转为KB = 2,359,296 / 1024 ≈ 2304 KB ≈ 2.25 MB。注意单位换算的每一步都要清晰标注，这在分步给分的IGCSE考试中非常重要。

Type four: storage calculation – for example, “A colour photograph has a resolution of 1024×768 and a colour depth of 24 bits. Calculate the file size in KB.” Solution steps: total pixels = 1024 × 768 = 786,432; total bits = 786,432 × 24 = 18,874,368 bits; convert to bytes = 18,874,368 / 8 = 2,359,296 bytes; convert to KB = 2,359,296 / 1024 ≈ 2304 KB ≈ 2.25 MB. Note: clearly label each step of the unit conversion – this is very important in the IGCSE exam, which awards marks step by step.

对所有题型的通用建议：①做题时始终展示你的步骤，IGCSE考试按步骤给分；②注意检查单位（bits vs bytes，MB vs KB）；③练习时给自己计时，Paper 1 Theory总时间有限，数制题不应花费超过2-3分钟；④考前务必复习16个十六进制符号与4位二进制的对照表，做到一眼识别。

General advice for all question types: (1) always show your working steps – the IGCSE exam awards marks step by step; (2) be careful to check units (bits vs bytes, MB vs KB); (3) time yourself when practising – Paper 1 Theory has limited total time, and number system questions should not take more than 2-3 minutes each; (4) before the exam, make sure to review the correspondence table of the 16 hexadecimal symbols and their 4-bit binary equivalents until you can recognise them instantly.

Summary | 总结

本文从数制的基本概念出发，系统地讲解了IGCSE计算机科学中最核心的知识模块：二进制和十六进制数制系统。我们深入探讨了二进制的原理与运算方法、十六进制与二进制的快速转换技巧、数据存储单位的层次结构，以及二进制在文本编码、图像表示和机器指令中的实际应用。通过对IGCSE考试常见题型的分析和解题技巧的总结，希望每一位读者都能建立起对数制转换的直觉理解 – 这不仅仅是死记硬背公式，而是真正理解计算机如何在最底层用0和1来表达一切信息。掌握了这些知识，你就拥有了理解整个计算机科学大厦的基石。

Starting from the fundamental concepts of number systems, this article has systematically explained the most essential knowledge module in IGCSE Computer Science: binary and hexadecimal number systems. We explored in depth the principles and arithmetic of binary, quick conversion techniques between hexadecimal and binary, the hierarchy of data storage units, and the practical applications of binary in text encoding, image representation, and machine instructions. Through analysis of common IGCSE exam question types and a summary of problem-solving techniques, I hope every reader can build an intuitive understanding of number system conversions – this is not just about memorising formulas, but about truly understanding how computers express all information with zeros and ones at the lowest level. With this knowledge, you possess the cornerstone for understanding the entire edifice of computer science.

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