📚 Highly Recognised Math Competition Problems Collection | 高认同度数学竞赛真题集锦
This article brings together a curated selection of problems from globally respected mathematics contests, including the AMC series, UKMT challenges, Waterloo competitions, AIME, and the Math Kangaroo. Each problem is chosen to illustrate a key concept or an elegant trick, with bilingual step‑by‑step solutions that sharpen both reasoning and competition technique.
本文精选来自全球高认同度数学竞赛的真题，涵盖 AMC 系列、UKMT 挑战赛、滑铁卢竞赛、AIME 和袋鼠数学等。每道题都展示一个核心概念或巧妙技巧，并配以中英双语的分步解析，帮助你打磨推理能力和竞赛技术。
1. Introduction | 引言
Mathematics competitions such as the AMC, UKMT, Waterloo, and Kangaroo are taken by millions of students every year. Their problems are carefully designed to test creativity, logical insight, and depth of understanding rather than routine calculation. Working through a collection of authentic contest problems is one of the most effective ways to build competitive thinking. In this article we present a set of genuine‑style problems, each accompanied by a clear solution and commentary.
像 AMC、UKMT、滑铁卢和袋鼠这样的数学竞赛每年吸引数百万学生参加。它们的题目精心设计，旨在考察创造力、逻辑洞察力和理解的深度，而不是机械计算。通过研读一套真实的竞赛真题，是培养竞赛思维最有效的方式之一。本文精选一组真题风格的题目，每道题都配有清晰的解答与点评。
2. AMC 8 and UKMT Junior Challenges | AMC 8 与 UKMT 初级竞赛
Problem (AMC 8 style): How many positive integers n less than 50 have the property that n² + n is divisible by 6?
真题 (AMC 8 风格)：有多少个小于 50 的正整数 n 满足 n² + n 能被 6 整除？
Solution: Notice that n² + n = n(n + 1) is always even, so divisibility by 2 is automatic. Divisibility by 3 requires that n(n + 1) be a multiple of 3, which happens exactly when n ≡ 0 (mod 3) or n ≡ 2 (mod 3). Thus among every three consecutive integers, two satisfy the condition. From 1 to 49 there are 49 numbers. 49 ÷ 3 = 16 remainder 1, so there are 16 full cycles of three, giving 16 × 2 = 32 numbers. The remaining number 49 is 1 mod 3, so it does not qualify. Hence the answer is 32.
解答：注意到 n² + n = n(n + 1) 总是偶数，因此能被 2 整除是自动成立的。能被 3 整除要求 n(n + 1) 是 3 的倍数，而这恰好发生在 n ≡ 0 (mod 3) 或 n ≡ 2 (mod 3) 时。因此在每三个连续整数中，有两个满足条件。从 1 到 49 共有 49 个数。49 ÷ 3 = 16 余 1，所以有 16 个完整的三个数的循环，给出 16 × 2 = 32 个数。余下的 49 模 3 余 1，不符合要求。故答案为 32。
3. AMC 10 and AMC 12 | AMC 10 与 AMC 12
Problem (AMC 10/12 style): Simplify the expression ∛(20 + 14√2) + ∛(20 – 14√2).
真题 (AMC 10/12 风格)：化简表达式 ∛(20 + 14√2) + ∛(20 – 14√2)。
Solution: Assume the cube roots can be written in the form a + b√2 and a – b√2. Then (a + b√2)³ = a³ + 6ab² + (3a²b + 2b³)√2. We want this to equal 20 + 14√2. By matching rational and irrational parts we try a = 2, b = 1: (2 + √2)³ = 8 + 12 + (12 + 2)√2 = 20 + 14√2. So ∛(20 + 14√2) = 2 + √2 and ∛(20 – 14√2) = 2 – √2. Their sum is (2 + √2) + (2 – √2) = 4.
解答：假设这两个立方根可以写成 a + b√2 和 a – b√2 的形式。那么 (a + b√2)³ = a³ + 6ab² + (3a²b + 2b³)√2。我们希望它等于 20 + 14√2。对比有理和无理部分，尝试 a = 2, b = 1： (2 + √2)³ = 8 + 12 + (12 + 2)√2 = 20 + 14√2。所以 ∛(20 + 14√2) = 2 + √2，∛(20 – 14√2) = 2 – √2。它们的和为 (2 + √2) + (2 – √2) = 4。
4. AIME – American Invitational Mathematics Examination | AIME 美国数学邀请赛
Problem (AIME style): How many positive integers n ≤ 100 have the property that n² – 1 is divisible by 24?
真题 (AIME 风格)：有多少个不大于 100 的正整数 n 具有性质： n² – 1 能被 24 整除？
Solution: Write n² – 1 = (n – 1)(n + 1). For the product to be a multiple of 24 = 2³ × 3, we need enough factors of 2 and a factor of 3. If n is odd, n – 1 and n + 1 are consecutive even numbers; one of them is a multiple of 4, the other is a multiple of 2 but not 4, so their product contains at least 2³. If n is even, one factor is odd and the product lacks the required power of 2. Thus n must be odd. Now we need divisibility by 3. If n is a multiple of 3, n² – 1 is not divisible by 3. If n ≡ 1 or 2 (mod 3), one of n – 1, n + 1 is a multiple of 3. So n must be odd and not a multiple of 3. Among the odd numbers from 1 to 100 there are 50. Every third odd number is a multiple of 3: 3, 9, 15, …, 99. The number of such multiples is (99 – 3)/6 + 1 = 17. Therefore valid n count = 50 – 17 = 33.
解答：把 n² – 1 写成 (n – 1)(n + 1)。要使乘积为 24 = 2³ × 3 的倍数，需要足够多的因子 2 和一个因子 3。若 n 为奇数，n – 1 和 n + 1 是两个连续的偶数；其中一个是 4 的倍数，另一个是 2 的倍数但不是 4 的倍数，因此乘积至少含有 2³。若 n 为偶数，其中一个因子为奇数，乘积缺少足够的 2 的幂。因此 n 必须是奇数。接下来看被 3 整除的条件。如果 n 是 3 的倍数，n² – 1 不能被 3 整除。如果 n ≡ 1 或 2 (mod 3)，n – 1 与 n + 1 中必有一个是 3 的倍数。所以 n 必须是奇数且不是 3 的倍数。从 1 到 100 的奇数共有 50 个。每隔两个奇数就会出现一个 3 的倍数：3, 9, 15, …, 99。这样的倍数个数为 (99 – 3)/6 + 1 = 17。因此满足条件的 n 有 50 – 17 = 33 个。
5. UKMT Intermediate and Senior Challenges | UKMT中级与高级挑战赛
Problem (UKMT Intermediate style): In an isosceles triangle ABC with AB = AC, point D lies on BC such that ∠BAD = 30° and ∠CAD = 20°. Determine the size of ∠AEB, where E is the point where the angle bisector of ∠B meets AC.
真题 (UKMT 中级风格)：在等腰三角形 ABC 中，AB = AC，点 D 在 BC 上，且有 ∠BAD = 30°，∠CAD = 20°。设 ∠B 的平分线交 AC 于点 E，求 ∠AEB 的大小。
Solution: Since AB = AC, ∠B = ∠C. The whole angle at A is ∠BAD + ∠CAD = 50°. Thus the sum of the base angles is 180° – 50° = 130°, so ∠B = ∠C = 65°. In triangle ABE, BE bisects ∠B, so ∠ABE = ∠B / 2 = 32.5°, and ∠BAE = ∠BAD = 30° (E lies on AC, so ∠BAE is the same as ∠BAD as D and E are on BC and AC respectively but we need careful; point E is on AC, so ray AE is inside angle A. However the angle ∠BAE is the whole angle at A, which is 50°, not necessarily 30°. Wait: The problem defined D on BC, but the question asks for ∠AEB with E being the intersection of the angle bisector of B with AC. This is independent of D. So we have triangle ABC with A = 50°, B = 65°, C = 65°. The bisector of B meets AC at E. We want ∠AEB. In triangle ABE, ∠ABE = 32.5°, ∠BAE = 50° (since E is on AC), so ∠AEB = 180° – 32.5° – 50° = 97.5°. But the mention of D is just to give the angles at A? Let’s re‑read: D on BC such that ∠BAD=30°, ∠CAD=20°. Indeed this only tells us ∠A = 50°. D is not used afterwards. So E is defined only via the angle bisector. Thus ∠AEB = 97.5°. Alternatively if they wanted the angle formed with D, but the problem statement as given says: determine ∠AEB where E is the point where the angle bisector of ∠B meets AC. So yes. The answer is 97.5°.
解答：由于 AB = AC，有 ∠B = ∠C。点 D 给出 ∠BAD = 30°，∠CAD = 20°，因此 ∠A = 50°。底角之和为 180° – 50° = 130°，故 ∠B = ∠C = 65°。∠B 的平分线交 AC 于点 E。在三角形 ABE 中，∠ABE = 32.5°，∠BAE = 50°（E 在 AC 上），所以 ∠AEB = 180° – 32.5° – 50° = 97.5°。注意 D 的作用仅仅是给出 ∠A 的大小，后续与 D 无关。答案为 97.5°。
6. Waterloo Contests (Gauss, Pascal, Cayley, Fermat) | 滑铁卢系列竞赛
Problem (Gauss/Pascal style): An operation ★ is defined by a ★ b = a² – b². What is the value of (3 ★ 2) ★ 1?
真题 (Gauss/Pascal 风格)：定义运算 ★ 为 a ★ b = a² – b²。求 (3 ★ 2) ★ 1 的值。
Solution: First compute 3 ★ 2 = 3² – 2² = 9 – 4 = 5. Then replace the expression with 5 ★ 1 = 5² – 1² = 25 – 1 = 24. Therefore the answer is 24.
解答：先计算 3 ★ 2 = 3² – 2² = 9 – 4 = 5。然后将表达式替换为 5 ★ 1 = 5² – 1² = 25 – 1 = 24。所以答案为 24。
7. Euclid Contest | 欧几里得竞赛
Problem (Euclid style): If 2x + 3y = 6, find the minimum value of x² + y² for real numbers x and y.
真题 (欧几里得风格)：若实数 x, y 满足 2x + 3y = 6，求 x² + y² 的最小值。
Solution: Consider the geometric interpretation: x² + y² is the squared distance from the origin to the point (x, y) on the line 2x + 3y = 6. The minimum distance occurs at the foot of the perpendicular from the origin to the line. The distance d from (0,0) to the line 2x + 3y – 6 = 0 is |–6|/√(2² + 3²) = 6/√13. Then the minimum of x² + y² is d² = 36/13. Alternatively, use y = (6 – 2x)/3 and minimise f(x) = x² + (6 – 2x)²/9 = (13x² – 24x + 36)/9. The vertex x = 24/(2×13) = 12/13, giving minimum f = 36/13. Thus the minimum value is 36/13.
解答：几何解释：x² + y² 表示原点到直线 2x + 3y = 6 上点 (x, y) 的距离的平方。最小距离出现在原点到直线的垂足处。点 (0,0) 到直线 2x + 3y – 6 = 0 的距离 d = |–6|/√(2² + 3²) = 6/√13。因此 x² + y² 的最小值为 d² = 36/13。也可用代数方法：y = (6 – 2x)/3，最小化 f(x) = x² + (6 – 2x)²/9 = (13x² – 24x + 36)/9，顶点处 x = 24/(2×13) = 12/13，得最小值 36/13。
8. Math Kangaroo | 袋鼠数学竞赛
Problem (Kangaroo style): How many three‑digit numbers have the property that the product of their three digits equals 0?
真题 (袋鼠风格)：有多少个三位数，其三个数字的乘积等于 0？
Solution: A three‑digit number ranges from 100 to 999, total 900 numbers. The product of digits is zero if and only if at least one digit is 0. The hundreds digit cannot be 0, so the zero must appear in the tens or ones place. Count the complement: numbers with no zero digits. The hundreds digit can be 1–9 (9 choices), the tens digit 1–9 (9 choices), and the ones digit 1–9 (9 choices), giving 9 × 9 × 9 = 729 numbers. Hence the numbers containing at least one zero are 900 – 729 = 171.
解答：三位数从 100 到 999 共 900 个。数字乘积为 0 当且仅当至少有一个数字是 0。百位不能为 0，因此 0 只能出现在十位或个位。计算补集：没有数字 0 的三位数。百位可取 1–9 (9 种)，十位 1–9 (9 种)，个位 1–9 (9 种)，共有 9 × 9 × 9 = 729 个。因此至少含有一个 0 的三位数有 900 – 729 = 171 个。
9. Common Strategies | 常用策略
Across all these contests, successful participants rely on a handful of powerful strategies. Factorisation and modular arithmetic simplify divisibility problems, as seen in the AMC 8 and AIME tasks. Algebraic manipulation, such as expressing cube roots as binomials, turns surds into neat integers. Geometric reasoning, as in the Euclid question, often yields an elegant shortcut. Counting complements, used in the Kangaroo problem, avoids overcounting. Always sketch a diagram, test small cases, and look for symmetry.
在所有这些竞赛中，成功的选手都依赖几项有力的策略。因式分解和模运算能简化整除问题，正如 AMC 8 和 AIME 题目中所示。代数变形，比如将立方根写成二项式，把无理式化为整齐的整数。几何推理，如欧几里得风格之题，常给出优雅的捷径。计算补集的方法，在袋鼠题中避免了重复计数。始终要画草图、检验小情形并寻找对称性。
10. Conclusion | 结语
Working through a diverse set of competition problems trains the mind to spot hidden structure and to approach unfamiliar challenges with confidence. This collection is just a starting point – each contest has its own flavour, and regular practice with past papers is the surest path to improvement. We hope these bilingual solutions inspire you to explore further and to enjoy the elegance of mathematical thinking.
练习多样化的竞赛题目能训练你发现隐藏结构的眼光，并充满信心地应对陌生挑战。本集锦只是一个起点——每种竞赛都有其独特风味，定期刷历年真题是提高的最可靠途径。我们希望这些双语解答能激励你进一步探索，并享受数学思维的优雅。
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