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- == Problem == {{AIME box|year=2006|n=I|num-b=11|num-a=13}}911 bytes (147 words) - 18:54, 13 December 2017
- == Problem == ...^2</math> equal and solving for <math>x</math> (it is helpful to scale the problem down by a factor of 50 first), we get <math>x = 250\pm 50\sqrt{7}</math>. S13 KB (2,080 words) - 13:14, 23 July 2024
- == Problem == {{AIME box|year=2005|n=I|num-b=11|num-a=13}}4 KB (647 words) - 02:29, 4 May 2021
- == Problem == {{AIME box|year=2004|n=I|num-b=11|num-a=13}}2 KB (303 words) - 18:43, 16 October 2024
- == Problem == {{AIME box|year=2004|n=II|num-b=11|num-a=13}}3 KB (431 words) - 23:21, 4 July 2013
- == Problem == {{AIME box|year=1983|num-b=11|num-a=13}}2 KB (412 words) - 18:23, 1 January 2024
- == Problem == Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent wit3 KB (588 words) - 14:37, 22 July 2020
- == Problem == ...math>(x+x^2+x^3)^n</math>, so the generating function of interest for this problem is <math>(x+x^2+x^3)^7</math>. Our goal is to find the coefficients of ever19 KB (3,128 words) - 21:38, 23 July 2024
- == Problem == ...ve more than 4 elements, otherwise its sum would be at most <math>15+14+13+12=54</math>.2 KB (382 words) - 19:52, 20 January 2025
- == Problem == {{AIME box|year=1987|num-b=11|num-a=13}}4 KB (673 words) - 19:48, 28 December 2023
- == Problem == [[Image:1988_AIME-12.png]]4 KB (743 words) - 21:02, 8 December 2024
- == Problem == pen p=fontsize(12)+linewidth(3);2 KB (376 words) - 13:49, 1 August 2022
- == Problem == ...us]] 12. The [[sum]] of the lengths of all sides and [[diagonal]]s of the 12-gon can be written in the form <math>a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6 KB (906 words) - 13:23, 5 September 2021
- == Problem == <center><asy>defaultpen(fontsize(12)+linewidth(1.3)); pair A=(0,28.8), B=(38.4,28.8), C=(38.4,0), D=(0,0), O, P8 KB (1,270 words) - 23:36, 27 August 2023
- == Problem == Therefore, the total number of such paths is <math>\binom{12}{5}=\boxed{792}</math>2 KB (443 words) - 22:41, 22 December 2021
- == Problem == {{AIME box|year=1993|num-b=11|num-a=13}}4 KB (611 words) - 13:59, 15 July 2023
- == Problem == {{AIME box|year=1994|num-b=11|num-a=13}}3 KB (473 words) - 17:06, 1 January 2024
- == Problem == {{AIME box|year=1995|num-b=11|num-a=13|t=394527}}8 KB (1,172 words) - 21:57, 22 September 2022
- == Problem == {{AIME box|year=1996|num-b=11|num-a=13|t=394253}}5 KB (879 words) - 11:23, 5 September 2021
- == Problem == In our problem <math>f^2(x) = x</math>. It follows that12 KB (2,229 words) - 19:58, 29 January 2025
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 <cmath>5-12-13</cmath>6 KB (978 words) - 12:02, 6 March 2025
- **[[AMC 12]] **[[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 ...ets [//competifyhub.com/resources/ Free Competition Resources for Grades 1-12]14 KB (1,913 words) - 23:52, 6 March 2025
- ...'''American Mathematics Contest 10''' ('''AMC 10'''), along with the [[AMC 12]], are the first exams in the series of exams used to challenge bright stud ...are invited to take the [[American Invitational Mathematics Examination]] (AIME).4 KB (596 words) - 04:53, 3 February 2025
- The '''American Mathematics Contest 12''' ('''AMC 12'''), along with the [[AMC 10]], are the first exams in the series of exams ...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 ...ica Mathematics Olympiad]] (USAMO) for qualification from taking the [[AMC 12]] or [[United States of America Junior Mathematical Olympiad]] (USAJMO) for5 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
- ...ion for American Math Contest, used to refer to the AMC 8, AMC 10, and AMC 12. * [[AMC 12]] — for students grades 12 and under.5 KB (696 words) - 03:47, 24 December 2019
- The top 12 students of the USAMO are invited to an awards ceremony in Washington, D.C. ...dministered by the [[Mathematical Association of America]] (MAA). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the MAA.6 KB (836 words) - 04:57, 3 February 2025
- * Performance on problems at practice sessions. * Performance on AMC and AIME, including current and previous years.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
- * <math>12! = 479001600</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
- In contest problems, Fermat's Little Theorem is often used in conjunction with the [[Chinese Re ...ts, x_n) = \left(a + \frac 12 - x_1, a + \frac 12 - x_2, \ldots, a + \frac 12 - x_p\right),</cmath>15 KB (2,618 words) - 12:03, 19 February 2025
- == Problems == * [[1991 AIME Problems/Problem 12]]1 KB (179 words) - 19:41, 3 January 2025
- An example of a classic problem is as follows: ...ivisible by 3. However, we note that we overcount several numbers, such as 12, which is divisible by both 2 and 3. To correct for this overcounting, we m4 KB (635 words) - 12:19, 2 January 2022
- This is a problem where constructive counting is not the simplest way to proceed. This next e ...proceed with the construction. If we were to go like before and break the problem down by each box, we'd get a fairly messy solution.13 KB (2,018 words) - 15:31, 10 January 2025
- ...ng may lead to a quick solution is the phrase "not" or "at least" within a problem statement. ''[[2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least one digit that is8 KB (1,192 words) - 17:20, 16 June 2023
- pair P=(3,12/5), F1=(-4,0), F2=(4,0); ==Problems==5 KB (892 words) - 21:52, 1 May 2021
- ...nds in a minute (they might have used it because it has so many multiples, 12 in fact, we wouldn't want any fractions). The [[Roman system]], which didn == Example Problems ==4 KB (547 words) - 17:23, 30 December 2020
