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- == Problem == ...an make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the poss4 KB (731 words) - 17:59, 4 January 2022
- == Problem == {{AIME box|year=2005|n=II|num-b=9|num-a=11}}3 KB (436 words) - 20:35, 13 August 2024
- == Problem == {{AIME box|year=2005|n=I|num-b=9|num-a=11}}5 KB (852 words) - 21:23, 4 October 2023
- == Problem == [[Image:2004_I_AIME-10.png]]5 KB (836 words) - 07:53, 15 October 2023
- == Problem == {{AIME box|year=2004|n=II|num-b=9|num-a=11}}8 KB (1,283 words) - 19:19, 8 May 2024
- == Problem == ...possible sequences with one digit repeated twice, and then divide by <math>10</math>.5 KB (855 words) - 20:26, 14 January 2023
- == Problem == ...h>w</math> is the number of wrong answers. (Students are not penalized for problems left unanswered.)7 KB (1,181 words) - 01:53, 21 November 2024
- == Problem == ...answer will be 100 times the number of integers we can reach between 1 and 10.12 KB (1,859 words) - 18:16, 28 March 2022
- == Problem == ...</math>, <math>b</math>, and <math>c</math> represent digits in base <math>10</math> in the order indicated. The magician then asks this person to form t3 KB (565 words) - 16:51, 1 October 2023
- == Problem == Then this can be interpreted as a classic chasing problem: Bob is "behind" by <math>x</math> steps, and since he moves at a pace of <7 KB (1,187 words) - 16:21, 27 January 2024
- == Problem == We know that all vertices look the same (from the problem statement), so we should find the number of line segments originating from6 KB (906 words) - 13:25, 19 November 2024
- == Problem == since <math>a^2 + b^2 = 1989c^2</math> from the problem and that there is another <math>-\frac{2xy}{2}</math> after the <math>h^2</8 KB (1,401 words) - 21:41, 20 January 2024
- == Problem == By the property of [[Diophantine equation]], given a problem to find integers x and y so that ax + by = c for some integer constants a,3 KB (564 words) - 04:47, 4 August 2023
- == Problem == {{AIME box|year=1991|num-b=9|num-a=11}}5 KB (813 words) - 06:10, 25 February 2024
- == Problem == {{AIME box|year=1992|num-b=9|num-a=11}}2 KB (323 words) - 12:05, 16 July 2019
- == Problem == The convex polyhedron of the problem can be easily visualized; it corresponds to a [[dodecahedron]] (a regular s4 KB (726 words) - 01:11, 12 January 2025
- == Problem == {{AIME box|year=1994|num-b=9|num-a=11}}3 KB (536 words) - 09:54, 5 November 2024
- == Problem == {{AIME box|year=1995|num-b=9|num-a=11}}3 KB (436 words) - 12:34, 14 June 2024
- == Problem == We know that <math>38 \times 9=342</math> and <math>38 \times 10 \equiv 20 \pmod{360}</math> (by simple arithmetic).5 KB (757 words) - 21:59, 23 December 2024
- == Problem == {{AIME box|year=1997|num-b=9|num-a=11}}3 KB (585 words) - 19:37, 25 April 2022
Page text matches
- ...e many tools in a good geometer's arsenal. A very large number of geometry problems can be solved by building right triangles and applying the Pythagorean Theo A = (-10,10);6 KB (978 words) - 12:02, 6 March 2025
- **[[AMC 10]] **[[AIME]]6 KB (625 words) - 23:27, 13 January 2025
- ...idual articles often have sample problems and solutions for many levels of problem solvers. Many also have links to books, websites, and other resources rele * CompetifyHub's Problem Sets [//competifyhub.com/resources/ Free Competition Resources for Grades 114 KB (1,913 words) - 23:52, 6 March 2025
- The '''American Mathematics Contest 10''' ('''AMC 10'''), along with the [[AMC 12]], are the first exams in the series of exams ...are invited to take the [[American Invitational Mathematics Examination]] (AIME).4 KB (596 words) - 04:53, 3 February 2025
- ...'''American Mathematics Contest 12''' ('''AMC 12'''), along with the [[AMC 10]], are the first exams in the series of exams used to challenge bright stud ...ke the more challenging [[American Invitational Mathematics Examination]] (AIME).5 KB (646 words) - 04:52, 3 February 2025
- ...ed States at the [[International Mathematics Olympiad]] (IMO). While most AIME participants are high school students, some bright middle school students a ...r Mathematical Olympiad]] (USAJMO) for qualification from taking the [[AMC 10]].5 KB (669 words) - 17:19, 11 March 2025
- ...th Jam''' is a free online class or information session hosted by [[Art of Problem Solving]] in the [[AoPS Schoolhouse|classroom]]. ...discussion of the problems from each year's [[AMC 10]], [[AMC 12]], and [[AIME]] exams.891 bytes (116 words) - 18:36, 31 January 2025
- ...an abbreviation for American Math Contest, used to refer to the AMC 8, AMC 10, and AMC 12. * [[AMC 10]] — for students grades 10 and under.5 KB (696 words) - 03:47, 24 December 2019
- ...ree top teams usually all place in the top 20, often even in the top 15 or 10. * Performance on problems at practice sessions.22 KB (3,532 words) - 11:25, 27 September 2024
- ...nd make the equation factorable. It can be used to solve more than algebra problems, sometimes going into other topics such as number theory. == Fun Practice Problems ==4 KB (682 words) - 10:25, 18 February 2025
- ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ==== Problem ====9 KB (1,703 words) - 01:20, 7 December 2024
- * <math>10! = 3628800</math> ==Problems==10 KB (809 words) - 16:40, 17 March 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME II Problems]]1 KB (133 words) - 12:32, 22 March 2011
- ==Problems== ...style="text-align:right;">([[2006 AMC 10A Problems/Problem 8|2006 AMC 10A, Problem 8]])</div>3 KB (551 words) - 16:22, 13 September 2023
- ...ve results. Examples include the [[Monty Hall paradox]] and the [[birthday problem]]. Probability can be loosely defined as the chance that an event will hap ...an understanding of the differences between different kinds of probability problems.4 KB (590 words) - 11:52, 28 September 2024
- An example of a classic problem is as follows: ...a coincidence that 67 is close to two-thirds of 100! We can approach this problem in a constructive way, building the set based on the remainders when divide4 KB (635 words) - 12:19, 2 January 2022
- ...es for each digit to arrive at our answer: <math>9 \cdot 10 \cdot 10 \cdot 10 = 9000</math>. <math>\square</math> This is a problem where constructive counting is not the simplest way to proceed. This next e13 KB (2,018 words) - 15:31, 10 January 2025
- ..., then adding together the totals of each part. Casework is a very general problem-solving approach, and as such has wide applicability. ...emonstrate casework in action. Unlike the selections in this article, most problems cannot be completely solved through casework. However, it is crucial as an5 KB (709 words) - 17:40, 24 September 2024
- * [[2020 AMC 10A Problems/Problem 24]] * [[1985 AIME Problems/Problem 13]]6 KB (923 words) - 17:39, 30 September 2024
- ...ng may lead to a quick solution is the phrase "not" or "at least" within a problem statement. ...integers. What we don't want are the multiples of five. These are <math>5, 10,..., 95</math> or <math>1 \cdot 5, 2 \cdot 5,..., 19 \cdot 5</math>; it's e8 KB (1,192 words) - 17:20, 16 June 2023
