Year 7 CAIE Advanced Mathematics: Case Study Practical Exercises | Year 7 CAIE 进阶数学：案例分析实战演练

📚 Year 7 CAIE Advanced Mathematics: Case Study Practical Exercises | Year 7 CAIE 进阶数学：案例分析实战演练

Advanced Mathematics at Year 7 level goes beyond routine calculations and requires logical thinking, pattern recognition, and the ability to apply concepts to unfamiliar situations. In this article, we present ten carefully selected case studies that reflect the style of CAIE problem-solving papers. Each case study breaks down the problem step by step, allowing you to practise the essential skills needed for higher-level mathematics.

Year 7 阶段的进阶数学超越了常规计算，要求逻辑思维、模式识别以及将概念应用于陌生情境的能力。本文精选了十个符合 CAIE 解题风格的案例。每个案例都一步步分解问题，帮助你练习高阶数学所必需的核心技能。

1. Number Theory Puzzle: The Missing Digit | 数论谜题：缺失的数字

A five-digit number ‘7A2B8’ is given. It is known that the number is divisible by 9 and the sum of its digits equals 18. When the number is divided by 5, the remainder is 3. Your task is to find the digits A and B.

有一个五位数 ‘7A2B8’。已知该数能被 9 整除，且各位数字之和为 18。当这个数除以 5 时，余数为 3。你需要找出 A 和 B。

We first use the divisibility rule for 5. A number ending in 8 divided by 5 leaves a remainder of 3 (since 8 = 5×1 + 3), so the condition is already satisfied regardless of A and B. Next, the sum of digits: 7 + A + 2 + B + 8 = 17 + A + B = 18, thus A + B = 1. Also, the number must be divisible by 9, meaning the digit sum must be a multiple of 9. Here sum is 18, which is a multiple of 9, so the condition holds automatically. Therefore A and B are digits (0-9) such that A + B = 1. The possible pairs are (1,0) or (0,1). Since A is the second digit and B is the fourth, both are valid. So the numbers could be 71208 or 70218.

我们先利用 5 的整除规则。尾数为 8 的数字除以 5 余数为 3（因为 8 = 5×1 + 3

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