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• Mock AMC
  ...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) - 21:41, 27 November 2024
• Mock AIME 2 Pre 2005
  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) - 10:58, 18 March 2015
• Mock AIME 7 Pre 2005
  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) - 16:33, 14 October 2022
• Mock AIME 1 2005-2006
  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) - 17:41, 21 January 2017
• Mock AIME 1 Pre 2005 Problems
  == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]
  6 KB (1,100 words) - 22:35, 9 January 2016
• Mock AIME 3 Pre 2005 Problems
  ==Problem 1== [[Mock AIME 3 Pre 2005/Problem 1|Solution]]
  7 KB (1,135 words) - 23:53, 24 March 2019
• Mock AIME 3 Pre 2005 Problems/Problem 5
  ==Problem== ...VC</tt> - the only other combination, two vowels, is impossible due to the problem statement). Then, note that:
  5 KB (795 words) - 16:03, 17 October 2021
• Mock AIME 3 Pre 2005 Problems/Problem 7
  ==Problem== ...ls of <math>ABCD</math> intersect at <math>P</math>. If <math>AB = 1, CD = 4,</math> and <math>BP : DP = 3 : 8,</math> then the area of the inscribed ci
  2 KB (330 words) - 10:23, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 8
  ==Problem== Here are some thoughts on the problem:
  3 KB (520 words) - 12:55, 11 January 2019
• Mock AIME 3 Pre 2005 Problems/Problem 10
  ==Problem== <cmath> C = \sum_{k=0}^{n-1} k^2 2^k = E+F = (n^2-4n+6)2^n - 6 </cmath>
  2 KB (306 words) - 10:36, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 14
  ==Problem== ...th>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 triangle <math>PBC</math
  3 KB (563 words) - 02:05, 25 November 2023
• Sophie Germain Identity
  <cmath>a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)</cmath> a^4 + 4b^4 &= a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 \ &= (a^2 + 2b^2)^2 - (2ab)^2 \ &= (a^2 + 2b^2 - 2ab) (a^2 + 2
  2 KB (210 words) - 17:05, 30 January 2025
• Mock AIME 1 Pre 2005 Problems/Problem 1
  == Problem == ...igits, then tens digits, then units digits. Every one of <math>\{1,2,3,4,5,6,7,8,9\}</math> may appear as the hundreds digit, and there are <math>9 \cdo
  1 KB (194 words) - 13:44, 5 September 2012
• Mock AIME 1 Pre 2005 Problems/Problem 7
  == Problem == .../math> <tt>C</tt>s amongst the middle five numbers, and so there are <math>6-(5-k) = k+1</math> <tt>C</tt>s amongst the first four numbers.
  1 KB (221 words) - 17:27, 23 February 2013
• Mock AIME 5 Pre 2005 Problems
  == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]
  6 KB (909 words) - 07:27, 12 October 2022
• Mock AIME 5 Pre 2005 Problems/Problem 12
  == Problem == Let <math>m = 101^4 + 256</math>. Find the sum of the digits of <math>m</math>.
  517 bytes (55 words) - 20:01, 23 March 2017
• Ball-and-urn
  .../math> bins. The 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 then
  5 KB (795 words) - 17:39, 31 December 2024
• Mock AIME 1 Pre 2005 Problems/Problem 12
  == Problem == ...rac15, GE = \frac{115}{\sqrt{24}}</math>, and <math>AG = AE - GE = 10\sqrt{6} - \frac{115}{\sqrt{24}} = \frac{5}{\sqrt{24}}</math>. Note that <math>\tri
  2 KB (376 words) - 22:41, 26 December 2016
• Mock AIME 1 Pre 2005 Problems/Problem 15
  == Problem == ...}</math>. Since <math>\sin B = \frac{b}{2R}</math>, we have that <math>b = 6\sqrt{23}</math>. By the Law of Cosines, we have that:
  2 KB (340 words) - 01:44, 3 March 2020
• Mock AIME 4 Pre 2005/Problems
  ==Problem 1== [[Mock AIME 4 Pre 2005/Problems/Problem 1 | Solution]]
  7 KB (1,094 words) - 15:39, 24 March 2019

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