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  • == Problem == Let <cmath>(1+x^3)\left(1+2x^{3^2}\right)\cdots \left(1+kx^{3^k}\right) \cdots \left(1+1997x^{3^{1997}}\right
    2 KB (232 words) - 00:22, 1 January 2021

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  • 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== ...math> balls into these urns. Using the ball-and-urn argument, having <math>9</math> urns is equivalent to <math>8</math> dividers, and there are <math>{
    950 bytes (137 words) - 10:16, 29 November 2019
  • ==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== <math>\left|a_1^{2} - a_2^{2}\right| = 10</math>
    3 KB (520 words) - 12:55, 11 January 2019
  • ==Problem== <math>a_{n} = 2a_{n-1} + n^2</math>
    2 KB (306 words) - 10:36, 4 April 2012
  • ==Problem== ...rest being the previous: <math>+2, +5, +1, +15, +3, +19, +3, +15, +1, +5, +2</math>. This sequence then repeats itself. We hence find that there are a t
    714 bytes (105 words) - 23:59, 24 April 2013
  • ==Problem== <math>\left(\frac{2}{3}\right)^{2005} \cdot \sum_{k=1}^{2005} \frac{k^2}{2^k} \cdot {2005 \choose k}</math>
    3 KB (502 words) - 14:53, 19 July 2020
  • ==Problem== ...ath>QP</math>. Thus, <math>O_3R \bot PQ</math> and <math>|PR|=\frac{|PQ|}{2}=16</math>
    3 KB (563 words) - 02:05, 25 November 2023
  • == Problem == ...th> and <math>f(101) = 0</math>). Evaluate the remainder when <math>f(1)+f(2)+\cdots+f(99)</math> is divided by <math>1000</math>.
    2 KB (209 words) - 12:43, 10 August 2019
  • == Problem == ...d units digits. Thus the sum of the hundreds places is <math>(1+2+3+\cdots+9)(72) \times 100 = 45 \cdot 72 \cdot 100 = 324000</math>.
    1 KB (194 words) - 13:44, 5 September 2012
  • == Problem == ...hat is the smallest possible value of <math>AP^2 + BP^2 + CP^2 + DP^2 + EP^2</math>?
    1 KB (217 words) - 06:18, 2 July 2015
  • == Problem == Thus, the height of <math>P</math> is <math>\sqrt [3]{8} = 2</math> times the height of <math>P'</math>, and thus the height of each is
    3 KB (446 words) - 00:18, 10 February 2020
  • == Problem 1 == If <math>g(9) + g(26) + g(126) + g(401) = \frac {m}{n}</math> where <math>m</math> and <
    6 KB (909 words) - 07:27, 12 October 2022
  • == Problem == ...\cdot DP \cdot EP \cdot FP \cdot GP</math>. Determine the value of <math>p^2</math>.
    540 bytes (96 words) - 00:28, 23 December 2023
  • ==Problem 1== [[Mock AIME 4 Pre 2005/Problems/Problem 1 | Solution]]
    7 KB (1,094 words) - 15:39, 24 March 2019
  • == Problem 1 == [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]]
    6 KB (1,052 words) - 13:52, 9 June 2020

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