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  • ==Problem== ...utations of <math>(4, 5)</math>, so there are a total of <math>2 \cdot 1 = 2</math> ways for this case.
    4 KB (690 words) - 13:32, 6 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) - 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>, 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>\boxed{196}\pi</math>.
    795 bytes (129 words) - 09:22, 4 April 2012
  • ==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== ...k=1}^{40} \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}}\right)</math>
    2 KB (312 words) - 09:38, 4 April 2012
  • == 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) - 11:43, 10 August 2019
  • ...style="text-align:center;"><math>a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)</math></div> ...4a^2b^2 \ &= (a^2 + 2b^2)^2 - (2ab)^2 \ &= (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)
    2 KB (225 words) - 04:42, 30 January 2025
  • == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]
    6 KB (909 words) - 06:27, 12 October 2022
  • == Problem == ...inradius of <math>5</math> and a circumradius of <math>16</math>. If <math>2\cos{B} = \cos{A} + \cos{C}</math>, then the area of triangle <math>ABC</mat
    2 KB (340 words) - 00:44, 3 March 2020
  • == Problem == Let <math>ABC</math> be a triangle with <math>AB = 13</math>, <math>BC = 14</math>, and <math>AC = 15</math>. Let <math>D</math> be the foot of the alt
    2 KB (294 words) - 15:24, 24 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