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  • ...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
  • 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
  • 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
  • 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
  • == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]
    6 KB (1,100 words) - 22:35, 9 January 2016
  • ==Problem 1== [[Mock AIME 3 Pre 2005/Problem 1|Solution]]
    7 KB (1,135 words) - 23:53, 24 March 2019
  • ==Problem== <cmath>2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4</cmath>
    1 KB (191 words) - 10:22, 4 April 2012
  • ==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
  • ==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
  • ==Problem== Here are some thoughts on the problem:
    3 KB (520 words) - 12:55, 11 January 2019
  • ==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
  • == Problem == ...sors of <math>n</math> less than <math>50</math> (e.g. <math>f(12) = 2+3 = 5</math> and <math>f(101) = 0</math>). Evaluate the remainder when <math>f(1)
    2 KB (209 words) - 12:43, 10 August 2019
  • <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
  • == Problem == ...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 \c
    1 KB (194 words) - 13:44, 5 September 2012
  • == Problem == ...>=r, <math>BP = |r - 1|</math>, <math>CP = |r - 2|</math>, <math>DP = |r - 4|</math>, and <math>EP = |r - 13|.</math> Squaring each of these gives:
    1 KB (217 words) - 06:18, 2 July 2015
  • == Problem == ...th> <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
  • == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]
    6 KB (909 words) - 07:27, 12 October 2022
  • == 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
  • ...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
  • == Problem == ...h> such that <math>\angle{AB'C'} \cong \angle{B'EA}</math>. If <math>AB' = 5</math> and <math>BE = 23</math>, then the area of <math>ABCD</math> can be
    2 KB (376 words) - 22:41, 26 December 2016

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