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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== ...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) - 01:05, 25 November 2023
  • == Problem == ...l of the prime numbers less than <math>50</math> are <math>2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,</math> and <math>47</math>. So we just nee
    2 KB (209 words) - 11:43, 10 August 2019
  • <div style="text-align:center;"><math>a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)</math></div> 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 (225 words) - 04:42, 30 January 2025
  • == Problem == ...<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) - 05:18, 2 July 2015
  • == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]
    6 KB (909 words) - 06:27, 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) - 19:01, 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) - 16:39, 31 December 2024
  • == Problem == ...{6}</math>. Thus, the area of <math>ABCD</math> is <math>(10\sqrt{6} + 23)(4\sqrt{6}) = 92\sqrt{6} + 240</math>, and our final answer is <math>92 + 6 +
    2 KB (376 words) - 21:41, 26 December 2016
  • == Problem == ...assigned to these houses such that there is at least one house with <math>4</math> residents?
    461 bytes (62 words) - 20:18, 8 October 2014
  • == 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>
    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
  • == Problem == In a box, there are <math>4</math> green balls, <math>4</math> blue balls, <math>2</math> red balls, a brown ball, a white ball, an
    1 KB (170 words) - 16:15, 4 August 2019
  • == Problem == ...7 \cdot 3^3 \cdot 37,</cmath> the number <math>10^{12} -1</math> has <math>4 \cdot 2^6 = 256</math> divisors and our answer is <math>256 - 1 = \boxed{25
    1 KB (171 words) - 16:38, 4 August 2019