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  • 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
  • The '''Mock AIME 1 2006-2007''' was written by [[Art of Problem Solving]] community member Altheman. * [[Mock AIME 1 2006-2007/Problems|Entire Exam]]
    1 KB (155 words) - 15:06, 3 April 2012
  • The '''Mock AIME 2 2006-2007''' was written by [[Art of Problem Solving]] community member 4everwise. * [[Mock AIME 2 2006-2007 Problems|Entire Exam]]
    1 KB (145 words) - 09:55, 4 April 2012
  • ...h>m = 18564 - 7 - 42 - 42 - 105 = 18368</math> so <math>\star(m) = 1 + 8 + 3 + 6 + 8 = 026</math>. *[[Mock AIME 1 2006-2007 Problems/Problem 1 | Previous Problem]]
    1 KB (188 words) - 14:53, 3 April 2012
  • ...opes of <math>AB</math> and <math>BC</math> are <math>10</math> and <math>-9</math>, respectively. If the <math>x</math>-coordinate of the triangle's ce ...a^2}{b - a} = b + a</math> and from the second slope condition that <math>-9 = \frac{c^2 - b^2}{c - b} = c + b</math>. Then <math>c = (a + b + c) - (a
    1 KB (244 words) - 14:21, 5 November 2012
  • ==Problem== Let <math>\triangle ABC</math> have <math>AC=6</math> and <math>BC=3</math>. Point <math>E</math> is such that <math>CE=1</math> and <math>AE=5<
    3 KB (518 words) - 15:54, 25 November 2015
  • ==Problem== ...</math>, so we don't need to worry about the distinctness condition in the problem. Then the value we want is <math>\sum_{\omega^{10} = 1} \sum_{i = 10}^\inf
    5 KB (744 words) - 18:46, 20 October 2020
  • ==Problem== ..., <math>AB</math>, <math>BC</math>, and <math>CA</math> have lengths <math>3</math>, <math>4</math>, and <math>5</math>, respectively. Let the incircle,
    1 KB (236 words) - 22:58, 24 April 2013
  • ==Problem== ...d {13}</math>. Now <math>10^2 = 100 \equiv 9 \pmod{13}</math> so <math>10^3 \equiv 90 \equiv -1 \pmod{13}</math>, <math>10^6 \equiv 1 \pmod{13}</math>
    2 KB (249 words) - 17:14, 3 April 2012
  • ==Problem 1== [[Mock AIME 1 2006-2007 Problems/Problem 1|Solution]]
    8 KB (1,355 words) - 13:54, 21 August 2020
  • == Problem == ...to two [[factor]]s in 3 ways, <math>2043 \cdot 1 = 3 \cdot 681 = 227 \cdot 9</math>. This gives us three pairs of [[equation]]s to solve for <math>n</m
    1 KB (198 words) - 09:50, 4 April 2012
  • == Problem == .../math> and <math>x_{n+3} = x_{n+2}(x_{n+1}+x_n)</math> for <math>n = 1, 2, 3, 4</math>. Find the last three [[digit]]s of <math>x_7</math>.
    3 KB (470 words) - 23:33, 9 August 2019
  • ==Problem== ...e the [[remainder]] when <center><p><math>{2007 \choose 0} + {2007 \choose 3} + \cdots + {2007 \choose 2007}</math></p></center> is divided by 1000.
    4 KB (595 words) - 11:14, 25 November 2023
  • ==Problem== ...a few small values for <math>x</math>, we see that <math>f(1)=0, f(2)=0, f(3)=0, f(4)=1, f(5)=1, f(6)=2, f(7)=2,
    992 bytes (156 words) - 19:34, 27 September 2019
  • == 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
  • ...contains the full set of test problems. The rest contain each individual problem and its solution. The Mock AIME 5 2006-2007 was written by Art of Problem Solving community member Altheman.
    1 KB (172 words) - 13:37, 3 July 2012
  • ...contains the full set of test problems. The rest contain each individual problem and its solution. The Mock AIME 6 2006-2007 was written by Art of Problem Solving community member paladin8.
    1 KB (172 words) - 13:39, 3 July 2012
  • ...contains the full set of test problems. The rest contain each individual problem and its solution. The Mock AIME 7 2006-2007 was written by Art of Problem Solving community member Altheman.
    1 KB (160 words) - 13:44, 3 July 2012
  • ==Problem 1== [[Mock AIME 6 2006-2007 Problems/Problem 1|Solution]]
    7 KB (1,173 words) - 20:04, 7 December 2018