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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 1
  ==Problem== ..., or <math>84*84=r(10+r)*21</math>, or <math>84*4=r(10+r)</math>. <math>84*4=14*24</math>, so <math>r=14</math>. Thus the area of the circle is <math>\b
  795 bytes (129 words) - 10:22, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 3
  ==Problem== <cmath>2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4</cmath>
  1 KB (191 words) - 10:22, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 5
  ==Problem== ...ear at least two consonants. Let <math>N</math> denote the number of <math>10</math>-letter Zuminglish words. Determine the remainder obtained when <math
  5 KB (795 words) - 16:03, 17 October 2021
• Mock AIME 3 Pre 2005 Problems/Problem 8
  ==Problem== ...<math>(a_1, a_2, \dots, a_8)</math> of real numbers such that <math>a_1 = 10</math> and
  3 KB (520 words) - 12:55, 11 January 2019
• 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 == ...ds digits, 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
  1 KB (194 words) - 13:44, 5 September 2012
• Mock AIME 5 Pre 2005 Problems
  == Problem 1 == <cmath>6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y</cmath>
  6 KB (909 words) - 07:27, 12 October 2022
• Ball-and-urn
  ...h> objects in <math>3</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 == ...eta = \frac15, 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 <ma
  2 KB (376 words) - 22:41, 26 December 2016
• Mock AIME 5 2005-2006 Problems/Problem 3
  == Problem == ...here <math>d_i</math> denotes the <math>i</math>th digit in the base-<math>10</math> representation of <math>n</math> for <math>i = 1,2, \ldots,k</math>,
  795 bytes (133 words) - 08:14, 19 July 2016
• Mock AIME 5 2005-2006 Problems/Problem 9
  == Problem == ...assigned to these houses such that there is at least one house with <math>4</math> residents?
  461 bytes (62 words) - 21:18, 8 October 2014
• 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
• Mock AIME 2 Pre 2005 Problems
  == Problem 1 == [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]]
  6 KB (1,052 words) - 13:52, 9 June 2020
• Mock AIME 2 Pre 2005 Problems/Problem 5
  == Problem == ...7 \cdot 3^3 \cdot 37,</cmath> the number <math>10^{12} -1</math> has <math>4 \cdot 2^6 = 256</math> divisors and our answer is <math>256 - 1 = \boxed{25
  1 KB (171 words) - 17:38, 4 August 2019

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