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  • ...a</math> are the roots of <math>x(x-200)(x+1/4)-1/4 = x^3 - \frac{799}{4}x^2 - 50x - \frac{1}{4}</math>. By Vieta's formulas, we have: {{Mock AIME box|year=Pre 2005|n=2|num-b=10|num-a=12}}
    1 KB (192 words) - 17:03, 13 September 2020

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  • ...A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member ...AMC]] competition. There is no guarantee that community members will make Mock AMCs in any given year, but there probably will be one.
    51 KB (6,175 words) - 20:41, 27 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== <cmath>\zeta_1^2+\zeta_2^2+\zeta_3^2=3</cmath>
    2 KB (221 words) - 01:49, 19 March 2015
  • ==Problem== <math>\left|a_1^{2} - a_2^{2}\right| = 10</math>
    3 KB (520 words) - 11:55, 11 January 2019
  • ==Problem== <math>a_{n} = 2a_{n-1} + n^2</math>
    2 KB (306 words) - 09:36, 4 April 2012
  • ==Problem== ...equence then repeats itself. We hence find that there are a total of <math>11*15 - 1</math> or <math>\boxed{164}</math> numbers that satisfy the inequali
    714 bytes (105 words) - 22:59, 24 April 2013
  • == 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
  • == 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
  • == Problem == ...\cdot DP \cdot EP \cdot FP \cdot GP</math>. Determine the value of <math>p^2</math>.
    540 bytes (96 words) - 23:28, 22 December 2023
  • ...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 == <math>Or,\frac{AD.BC}{2}=84</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 == ...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{255}.</mat
    1 KB (171 words) - 16:38, 4 August 2019

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