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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) - 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== ...rcles are mutually externally tangent. Two of the circles have radii <math>3</math> and <math>7</math>. If the area of the triangle formed by connecting
    7 KB (1,135 words) - 22:53, 24 March 2019
  • ==Problem== ...</math>, and the 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 t
    714 bytes (105 words) - 22:59, 24 April 2013
  • ==Problem== <math>13.</math> Let <math>S</math> denote the value of the sum
    3 KB (502 words) - 13:53, 19 July 2020
  • ==Problem== ..._1</math> at <math>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
    3 KB (563 words) - 01:05, 25 November 2023
  • == Problem == ...divisors 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>
    2 KB (209 words) - 11:43, 10 August 2019
  • == 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
    3 KB (446 words) - 23:18, 9 February 2020
  • == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]
    6 KB (909 words) - 06:27, 12 October 2022
  • == Problem == ...\begin{aligned} 101^4 + 256 &= (100 + 1)^4 + 256 \ &= (100^4 + 4\cdot 100^3 + 6 \cdot 100^2 + 4 \cdot 100 + 1) + 256 \ &= 104060401 + 256 = 104060657
    517 bytes (55 words) - 19:01, 23 March 2017
  • ...<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) - 16:39, 31 December 2024
  • == Problem == <math>CosB=\frac{68^2+100^2-112^2}{2.68.100}=\frac{13}{85}</math>
    2 KB (282 words) - 09:06, 9 August 2022
  • ==Problem 1== [[Mock AIME 4 Pre 2005/Problems/Problem 1 | Solution]]
    7 KB (1,094 words) - 14:39, 24 March 2019
  • == Problem 1 == [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]]
    6 KB (1,052 words) - 12:52, 9 June 2020
  • == Problem == ...+ 1)(10^3 - 1) = 101 \cdot 9901 \cdot 37 \cdot 11 \cdot 13 \cdot 7 \cdot 3^3 \cdot 37,</cmath> the number <math>10^{12} -1</math> has <math>4 \cdot 2^6
    1 KB (171 words) - 16:38, 4 August 2019