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- ==Problem== <math>\sum_{k=1}^{40} \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}}\right)</math>2 KB (312 words) - 09:38, 4 April 2012
- #REDIRECT [[Mock AIME 3 Pre 2005 Problems/Problem 15]]54 bytes (6 words) - 18:42, 9 April 2012
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- ...ed States at the [[International Mathematics Olympiad]] (IMO). While most AIME participants are high school students, some bright middle school students a High scoring AIME students are invited to take the prestigious [[United States of America Mat8 KB (1,062 words) - 18:04, 17 January 2025
- ...e integers. Determine <math>p + q</math>. ([[Mock AIME 3 Pre 2005 Problems/Problem 7|Source]]) ...prime factor. What is <math>a+b+c?</math> ([[2022 AMC 10A Problems/Problem 15|Source]])3 KB (543 words) - 18:35, 29 October 2024
- ...A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member ...AMC]] competition. There is no guarantee that community members will make Mock AMCs in any given year, but there probably will be one.51 KB (6,175 words) - 20:41, 27 November 2024
- The '''Mock AIME 2 Pre 2005''' was written by [[Art of Problem Solving]] community member Mildorf. * [[Mock AIME 2 Pre 2005 Problems|Entire Exam]]2 KB (181 words) - 09:58, 18 March 2015
- The '''Mock AIME 7 Pre 2005''' was written by [[Art of Problem Solving]] community member Mildorf. * [[Mock AIME 7 Pre 2005 Problems|Entire Exam]]1 KB (146 words) - 15:33, 14 October 2022
- The '''Mock AIME 1 2005-2006''' was written by [[Art of Problem Solving]] community member paladin8. * [[Mock AIME 1 2005-2006/Answer Key|Answer Key]]1 KB (135 words) - 16:41, 21 January 2017
- == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,100 words) - 21:35, 9 January 2016
- ==Problem 1== ...rcles are mutually externally tangent. Two of the circles have radii <math>3</math> and <math>7</math>. If the area of the triangle formed by connecting7 KB (1,135 words) - 22:53, 24 March 2019
- ==Problem== ...alent to <math>8</math> dividers, and there are <math>{8 + 7 \choose 7} = {15 \choose 7} = 6435 \equiv \boxed{435} \pmod{1000}</math>.950 bytes (137 words) - 09:16, 29 November 2019
- ==Problem== Here are some thoughts on the problem:3 KB (520 words) - 11:55, 11 January 2019
- ==Problem== ...<math>D</math> such that <math>\angle{BAE}</math> is right. If <math>BD = 15, DE = 2,</math> and <math>BC = 16</math>, then <math>CD</math> can be expre2 KB (278 words) - 15:32, 27 December 2019
- ==Problem== ...the second equation and rearranging variables gives <math>45x=8\cdot 15+bx=15^2\implies x=5</math>. Back-substitution yields <math>AC=21</math> and conse2 KB (379 words) - 00:27, 6 December 2024
- ==Problem== ...ence then repeats itself. We hence find that there are a total of <math>11*15 - 1</math> or <math>\boxed{164}</math> numbers that satisfy the inequality.714 bytes (105 words) - 22:59, 24 April 2013
- ==Problem== ..._1</math> at <math>C</math> and <math>D</math> respectively. If <math>AD = 3, AP = 6, DP = 4,</math> and <math>PQ = 32</math>, then the area of triangle3 KB (563 words) - 01:05, 25 November 2023
- == Problem == Let <math>N</math> denote the number of permutations of the <math>15</math>-character string <math>AAAABBBBBCCCCCC</math> such that1 KB (221 words) - 16:27, 23 February 2013
- == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]6 KB (909 words) - 06:27, 12 October 2022
- ...e number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. ...ach urn, then there would be <math>{n \choose k}</math> possibilities; the problem is that you can repeat urns, so this does not work.<math>n</math> and then5 KB (795 words) - 16:39, 31 December 2024
- #REDIRECT [[Mock AIME 3 Pre 2005 Problems/Problem 15]]54 bytes (6 words) - 18:42, 9 April 2012
- == Problem == ..., we have that <math>\cos B = \frac{7}{16}</math> and <math>\sin B = \frac{3\sqrt{23}}{16}</math>. Since <math>\sin B = \frac{b}{2R}</math>, we have tha2 KB (340 words) - 00:44, 3 March 2020
- == Problem == <math>2R=\frac{AC}{SinB}=\frac{112}{\frac{84}{85}}=\frac{340}{3}</math>2 KB (282 words) - 09:06, 9 August 2022