AMC8 Mathematics Competition: Syllabus Scope and Past Problem Analysis | AMC8数学竞赛：考纲范围与真题解析

📚 AMC8 Mathematics Competition: Syllabus Scope and Past Problem Analysis | AMC8数学竞赛：考纲范围与真题解析

The AMC8 (American Mathematics Competitions 8) is a 25-question, 40-minute multiple-choice contest designed for middle school students, typically in grades 8 and below. It aims to foster an interest in mathematics and to develop problem-solving skills through a broad range of topics. The exam covers arithmetic, algebra, geometry, counting and probability, and number theory. Understanding the syllabus scope and practising with past problems is essential for success.

AMC8（美国数学竞赛8年级组）是一场包含25道选择题、时长40分钟的竞赛，面向8年级及以下的中学生。其目的是通过广泛的主题激发数学兴趣并培养解题能力。考试内容涵盖算术、代数、几何、计数与概率以及数论。熟悉考纲范围并结合真题进行练习是取得成功的关键。

1. What is AMC8? | 什么是AMC8？

The AMC8, administered by the Mathematical Association of America (MAA), serves as an entry point into the world of competitive mathematics. It is held annually in November, with students participating from schools and learning centres worldwide. No calculators are allowed, so the problems emphasise reasoning, mental arithmetic, and logical thinking rather than heavy computation.

AMC8 由美国数学协会（MAA）举办，是进入数学竞赛世界的门户。它每年11月举行，来自世界各地的学校和教学中心的学生均可参加。考试不允许使用计算器，因此题目侧重于推理、心算和逻辑思维，而非繁重的计算。

The test consists of 25 questions arranged in increasing order of difficulty. The first 10 questions are generally straightforward, the next 10 require more thought, and the final 5 are designed to challenge the top scorers. A perfect score is 25 points. Correct answers earn 1 point each, and there is no penalty for incorrect answers, so guessing is encouraged.

试卷共25道题，难度逐渐递增。前10题通常较为简单，中间10题需要更多思考，最后5题旨在挑战高分段选手。满分25分，答对一题得1分，答错不扣分，因此鼓励猜测。

2. Syllabus: Arithmetic and Pre-Algebra | 考纲：算术与预备代数

Arithmetic forms the backbone of the AMC8, covering integer operations, fractions, decimals, percentages, ratios, and proportions. Students must be comfortable with the order of operations, estimating results, and converting between different representations. Pre-algebra topics such as simple equations, factors, multiples, and basic number properties also appear frequently.

算术是AMC8的基础，涵盖整数运算、分数、小数、百分数、比和比例。学生必须熟练掌握运算顺序、估算结果以及在不同表示形式之间的转换。简单方程、因数、倍数和基本数论性质等预备代数主题也经常出现。

Typical problems ask for the sale price after a discount, the average of a set, or the number of ways to combine ingredients from a recipe. Skills like finding the least common multiple (LCM) or the greatest common divisor (GCD) are tested in word problem contexts. A strong foundation in mental arithmetic saves valuable time.

典型题目会问折扣后的售价、一组数字的平均值，或从食谱组合配料的方法。诸如求最小公倍数（LCM）或最大公约数（GCD）的技巧会在文字题中被考查。扎实的心算基础可以节省宝贵时间。

3. Syllabus: Algebra | 考纲：代数

The algebraic content of AMC8 stays within elementary algebra. Students should be able to solve linear equations in one variable, interpret simple inequalities, and work with algebraic expressions. Patterns and sequences, such as arithmetic progressions, are also common. Some problems require setting up and solving an equation from a verbal description.

AMC8 的代数内容保持在初等代数水平。学生应能解一元一次方程、理解简单不等式以及处理代数式。模式与数列（如等差数列）也很常见。有些题目需要根据文字描述列出并求解方程。

For example, a problem may state: ‘Three times a number decreased by 7 equals 20. Find the number.’ Translating words into symbols is a crucial skill. In addition, students may be asked to evaluate expressions such as (a + b)² given specific values, using substitution and the distributive property.

例如，题目可能描述：“一个数的3倍减去7等于20，求该数。” 将文字转化为数学符号是一项关键技能。此外，学生也可能需要根据给定数值计算如 (a + b)² 的表达式，运用代入和分配律。

4. Syllabus: Geometry | 考纲：几何

Geometry questions on the AMC8 involve perimeter, area, and volume of basic shapes including triangles, rectangles, circles, and rectangular prisms. The Pythagorean theorem is a must-know tool, as are angle relationships in parallel lines and triangles. Symmetry, coordinate geometry, and the visualisation of transformations also appear.

AMC8 中的几何题涉及三角形、矩形、圆和长方体等基本图形的周长、面积与体积。勾股定理是必会工具，平行线与三角形的角度关系亦然。对称、坐标几何以及变换的可视化也会出现。

Often, figures are provided, but occasionally students must interpret a verbal description. Familiarity with formulas for the area of a triangle (½ × base × height), the circumference of a circle (2πr), and the volume of a prism is expected. Composite figures that require subtracting areas or breaking shapes into simpler parts are frequent.

通常题目会给出图示，但偶尔学生需要根据文字描述进行解读。需熟悉三角形面积公式（½ × 底 × 高）、圆的周长（2πr）和棱柱体积公式。需要扣除面积或将图形分解为简单部分的组合图形题经常出现。

5. Syllabus: Counting and Probability | 考纲：计数与概率

Counting problems ask ‘in how many ways’ something can occur. The fundamental counting principle, permutations, and combinations are core tools. Students must distinguish between cases where order matters and where it does not. Tree diagrams and organised lists help solve problems without formal formulas.

计数题会问某件事可能发生的方式数。基本计数原理、排列与组合是核心工具。学生必须区分顺序重要与不重要的情形。树状图和有序列表有助于在不使用正式公式的情况下解题。

Probability questions typically involve simple fractions: the number of favourable outcomes divided by the total number of possible outcomes. Dice, coins, and spinners are common props. Even without a formal probability course, a clear understanding of chance and equally likely outcomes is enough to tackle most AMC8 probability items.

概率题通常涉及简单分数：有利结果数除以所有可能结果数。骰子、硬币和转盘是常见道具。即便未正式学习过概率，只要对机会和等可能结果有清晰理解，就足以应对大多数AMC8概率题。

Problems sometimes combine counting and probability, for instance: ‘What is the probability of drawing a red marble from a bag containing 3 red, 5 blue, and 2 green marbles?’ The answer would be 3/(3+5+2) = 3/10.

有时问题会将计数与概率结合，例如：“从装有3个红色、5个蓝色和2个绿色弹珠的袋子里摸出一个红色弹珠的概率是多少？” 答案就是 3/(3+5+2) = 3/10。

6. Syllabus: Number Theory | 考纲：数论

Number theory topics include divisibility rules, prime and composite numbers, factorisation, remainders, and modular arithmetic at a basic level. Students should be able to identify the number of positive divisors of an integer by using its prime factorisation. The concept of least common multiple and greatest common divisor reappears here in more abstract settings.

数论主题包括整除规则、质数与合数、因式分解、余数以及初级模运算。学生应能利用质因数分解来确定一个整数的正因数个数。最小公倍数和最大公约数的概念会以更抽象的形式出现在此处。

For example, knowing that a number ending in 0 or 5 is divisible by 5, or that the sum of digits determines divisibility by 3, can quickly eliminate options in multiple-choice questions. Problems may ask: ‘How many of the numbers from 1 to 50 are multiples of 3 or 5?’ Inclusion-exclusion is often needed.

例如，知道以0或5结尾的数能被5整除，或者各位数字之和决定能否被3整除，可以快速排除选择题中的选项。问题可能问：“从1到50的数中，有多少个是3或5的倍数？” 通常需要用到容斥原理。

7. Sample Problem 1 – Percentages | 真题解析1 – 百分数

Problem: A shirt originally priced at $40 is on sale for 25% off. An additional 10% discount is applied on the already reduced price. What is the final sale price, in dollars?

题目：一件原价40美元的衬衫打25%的折扣。再在已减价的基础上打额外10%的折扣。求最终售价（美元）。

Solution: First, a 25% discount on $40 means the price becomes 75% of $40. So the first reduced price is 0.75 × 40 = $30. The extra 10% discount is applied on $30, meaning we pay 90% of $30. Thus, final price = 0.9 × 30 = $27. Alternatively, the overall multiplier is 0.75 × 0.9 = 0.675, and 0.675 × 40 = 27. The final sale price is $27.

解答：首先，对40美元打25%的折扣意味着价格变为40美元的75%。因此第一次减价后为 0.75 × 40 = 30美元。额外的10%折扣用在30美元上，意味着我们支付30美元的90%。因此，最终价格 = 0.9 × 30 = 27美元。另一种方法，整体乘数为 0.75 × 0.9 = 0.675，0.675 × 40 = 27。最终售价为27美元。

8. Sample Problem 2 – Linear Equation | 真题解析2 – 线性方程

Problem: Sam is three times as old as Tom will be in 5 years. If Sam is currently 36 years old, how old is Tom now?

题目：山姆的年龄是5年后汤姆年龄的3倍。如果山姆现在36岁，那么汤姆现在几岁？

Solution: Let Tom’s current age be x. In 5 years, Tom will be x + 5. According to the statement, Sam’s current age (36) equals three times Tom’s age in 5 years. Set up the equation: 36 = 3(x + 5). Divide both sides by 3: 12 = x + 5. Subtract 5 from both sides: x = 7. Therefore, Tom is 7 years old now.

解答：设汤姆现在的年龄为 x。5年后汤姆的年龄为 x + 5。根据题意，山姆现在的年龄（36岁）等于5年后汤姆年龄的3倍。列出方程：36 = 3(x + 5)。两边除以3得 12 = x + 5。两边减去5得 x = 7。因此，汤姆现在7岁。

9. Sample Problem 3 – Area of a Triangle | 真题解析3 – 三角形面积

Problem: In triangle ABC, base BC measures 14 cm and the altitude from vertex A to base BC is 9 cm. Point D lies on BC such that BD = 6 cm. Find the area of triangle ABD.

题目：在三角形 ABC 中，底边 BC 长14 cm，顶点 A 到底边 BC 的高为9 cm。点 D 在 BC 上，且 BD = 6 cm。求三角形 ABD 的面积。

Solution: The area of triangle ABD can be computed using the formula ½ × base × height. Here, the base of triangle ABD is BD = 6 cm, and the height is the same as that of the large triangle ABC because the perpendicular distance from A to the line containing BC remains unchanged. This height is given as 9 cm. Hence, area = ½ × 6 × 9 = 27 cm².

解答：三角形 ABD 的面积可以用公式 ½ × 底 × 高计算。此处三角形 ABD 的底为 BD = 6 cm，高与整个大三角形 ABC 的高相同，因为 A 到 BC 所在直线的垂直距离不变。该高为 9 cm。因此，面积 = ½ × 6 × 9 = 27 cm²。

Note that we did not need to find the area of the whole triangle or the position of C. The problem tests the concept that triangles sharing the same altitude have areas proportional to their bases. As BD = 6 and BC = 14, the ratio of areas is 6/14, and indeed (6/14)×(½×14×9 = 63) = 27.

注意我们不需要求整个三角形的面积或点 C 的位置。本题考察的是等高的三角形面积与底边成比例的概念。由于 BD = 6, BC = 14，面积比为 6/14，实际上 (6/14)×(½×14×9=63) = 27。

10. Sample Problem 4 – Factorials and Permutations | 真题解析4 – 阶乘与排列

Problem: In how many different ways can the letters of the word ‘CODE’ be arranged? (Permutations of four distinct letters.)

题目：单词 ‘CODE’ 中的字母可以排列成多少种不同的方式？（四个不同字母的排列。）

Solution: There are 4 distinct letters. The number of permutations of n distinct objects is n!. Here n = 4, so the number of arrangements is 4! = 4 × 3 × 2 × 1 = 24. If the letters had repetitions, we would divide by the factorial of the number of repetitions, but since all letters are distinct, it is simply 24.

解答：有4个不同的字母。n 个不同对象的排列数是 n!。此处 n = 4，因此排列方式数为 4! = 4 × 3 × 2 × 1 = 24。如果字母有重复，我们需要除以重复次数的阶乘，但既然所有字母都不相同，答案就是24。

A related AMC8 problem might ask for the number of arrangements of the letters in ‘MATH’ where vowels are together. Such questions require treating a group as a single entity and then multiplying by the internal arrangements. This level of thinking is typical for the competition.

一道相关的 AMC8 题目可能会问单词 ‘MATH’ 中元音字母相邻的排列方式数。这类问题需要将一组字母视为一个整体，然后乘以内部的排列数。这种层次的思维在竞赛中十分典型。

11. Sample Problem 5 – Factors and Multiples | 真题解析5 – 因数与倍数

Problem: How many positive integer factors does the number 72 have?

题目：正整数72有多少个正因数？

Solution: First, write the prime factorisation of 72: 72 = 2³ × 3². To find the total number of positive factors, add 1 to each exponent in the prime factorisation and multiply the results: (3 + 1) × (2 + 1) = 4 × 3 = 12. Therefore, 72 has 12 positive factors. Listing them: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.

解答：首先，写出72的质因数分解：72 = 2³ × 3²。要计算正因数的总数，将每个指数加1再相乘：(3 + 1) × (2 + 1) = 4 × 3 = 12。因此72有12个正因数。列出为：1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72。

This method works because any factor is of the form 2ᵃ × 3ᵇ, where a ∈ {0,1,2,3} and b ∈ {0,1,2}, giving 4 × 3 choices. Problems often extend this to find the sum of factors or to identify perfect squares among the factors.

该方法之所以有效，是因为任何因数都可表示为 2ᵃ × 3ᵇ，其中 a ∈ {0,1,2,3}，b ∈ {0,1,2}，共 4 × 3 种选择。题目经常进一步要求求因数之和，或找出其中的完全平方数。

12. Preparation Strategies | 备考策略

To excel in the AMC8, consistent practice with past papers is crucial. Work through problems from the last 10 years, timing yourself to build speed. Review mistakes thoroughly – each error highlights a concept that needs reinforcement. Focus on understanding the solution process rather than just getting the answer right.

要在 AMC8 中取得优异成绩，持续练习历年真题至关重要。做近10年的题目并计时以提升速度。仔细回顾错题——每一个错误都指明了需要强化的概念。重点在于理解解题过程，而不仅仅是得到正确答案。

Strengthen mental arithmetic and estimation skills, as calculators are not allowed. Memorise key squares, cubes, and fraction-decimal-percent equivalents. Create a formula sheet with area, volume, and number theory formulas. Engage in mathematical discussions or join a math circle to develop logical reasoning and the ability to articulate ideas.

加强心算和估算能力，因为考试不允许使用计算器。熟记重要的平方数、立方数以及分数、小数与百分数的转换。制作一张公式表，涵盖面积、体积和数论公式。参与数学讨论或加入数学社团，以培养逻辑推理能力和表达思路的能力。

On the day of the contest, read each question carefully, eliminate clearly wrong choices, and avoid spending too long on a single problem. The AMC8 rewards strategic guessing and efficient time management. Approach the test with confidence and a curious mind.

竞赛当天，仔细审题，排除明显错误的选项，避免在单个问题上花费过长时间。AMC8 奖励策略性猜测和高效的时间管理。带着自信和好奇心迎接考试。

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