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  • ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC and of the recent expansion ...down=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}}
    6 KB (874 words) - 22:02, 10 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== [[Mock AIME 3 Pre 2005/Problem 1|Solution]]
    7 KB (1,135 words) - 22:53, 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) - 09:16, 29 November 2019
  • ==Problem== === Solution 1 (recursive) ===
    5 KB (795 words) - 15:03, 17 October 2021
  • ==Problem== Here are some thoughts on the problem:
    3 KB (520 words) - 11:55, 11 January 2019
  • ==Problem== <math>\{A_n\}_{n \ge 1}</math> is a sequence of positive integers such that
    2 KB (306 words) - 09:36, 4 April 2012
  • ==Problem== ...<math>n</math> such that <math>1 \le n \le 1000</math> and <math>n^{12} - 1</math> is divisible by <math>73</math>.
    714 bytes (105 words) - 22:59, 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) - 13:53, 19 July 2020
  • ==Problem== <math>A_1=\frac{\sqrt{\left(13\right)\left(7\right)\left(1\right)\left(5\right)}}{4}=\frac{\sqrt{455}}{4}</math>
    3 KB (563 words) - 01:05, 25 November 2023
  • == Problem == ...5</math> 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) - 11:43, 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) - 12:44, 5 September 2012
  • == Problem == ...r in that order such that <math>AB = BC = 1, CD = 2,</math> and <math>DE = 9</math>. If <math>P</math> can be any point in space, what is the smallest p
    1 KB (217 words) - 05:18, 2 July 2015
  • == Problem == {{Mock AIME box|year=Pre 2005|n=1|num-b=7|num-a=9|source=14769}}
    3 KB (446 words) - 23:18, 9 February 2020
  • == Problem 1 == <cmath>6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y</cmath>
    6 KB (909 words) - 06:27, 12 October 2022
  • == Problem == {{Mock AIME box|year=Pre 2005|n=1|num-b=9|num-a=11|source=14769}}
    540 bytes (96 words) - 23:28, 22 December 2023
  • ==Problem 1== For how many integers <math>n>1</math> is it possible to express <math>2005</math> as the sum of <math>n</m
    7 KB (1,094 words) - 14:39, 24 March 2019

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