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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== ...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 t714 bytes (105 words) - 22:59, 24 April 2013
- ==Problem== <math>13.</math> Let <math>S</math> denote the value of the sum3 KB (502 words) - 13:53, 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) - 01:05, 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) - 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 == ...hat is the smallest possible value of <math>AP^2 + BP^2 + CP^2 + DP^2 + EP^2</math>?1 KB (217 words) - 05:18, 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 is3 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 == ...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, \end{aligned}<517 bytes (55 words) - 19:01, 23 March 2017
- ...s. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. == Reasoning 2 ==5 KB (795 words) - 16:39, 31 December 2024
- == Problem == ...in \angle{EAB'} = \sin(90^{\circ} - 2 \theta) = \cos 2 \theta = 1 - 2 \sin^2 \theta</math>. Now, we use law of sines, which gives us the following:2 KB (376 words) - 21:41, 26 December 2016
- == 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 == <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