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  • ==Problem == Simon and Theodore start <math>\frac{3-2}{3}(2526)=842</math> metres apart, as they are travelling in the same directio
    929 bytes (140 words) - 09:20, 23 November 2023

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  • ...ed States at the [[International Mathematics Olympiad]] (IMO). While most AIME participants are high school students, some bright middle school students a High scoring AIME students are invited to take the prestigious [[United States of America Mat
    8 KB (1,062 words) - 18:04, 17 January 2025
  • label("$a/2$",(B+D)/2,(0,-1));label("$R$",(3*O+2*B)/5,(-1,1));label("$\theta$",O,(-0.8,-1.2));label("$\theta$",A,(0,-1.5 ==Problems==
    4 KB (658 words) - 15:19, 28 April 2024
  • ...ake. Sometimes, the administrator may ask other people to sign up to write problems for the contest. ...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 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
  • ...= 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
  • ==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>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== .../math>, <math>10^6 \equiv 1 \pmod{13}</math> and the order of 10 mod 13 is 6. Thus, we get one value of <math>m</math> each time <math>n = 6j + 1</math
    2 KB (249 words) - 17:14, 3 April 2012
  • ==Problem== ...d_{3}=3</math>, <math>d_{4}=-7</math>, <math>d_{5}=13</math>, and <math>d_{6}=-16</math>, find <math>d_{7}</math>.
    3 KB (568 words) - 14:50, 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 == ...>, which can be split into 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
    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 == ...ius <math>51</math>, we must have that <math>\angle AOB=\frac{360^{\circ}}{3}=120^{\circ}</math>. We know that <math>AO=OB=51</math>, so we can use the
    1 KB (231 words) - 17:10, 10 July 2014
  • ==Problem== Find the remainder when <math>3^{3^{3^3}}</math> is divided by 1000.
    1 KB (127 words) - 23:15, 4 January 2010
  • ==Problem== ...h>\triangle ABC</math> is 13 and the area of <math>\triangle ACF</math> is 3. If <math>\frac{CE}{EA}=\frac{p+\sqrt{q}}{r}</math>, where <math>p</math>,
    2 KB (325 words) - 18:33, 9 February 2017
  • ==Problem== ...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== ...3sqrt3.Then the coordinate of point D are 24, -18sqrt3. The answer is then 6 + sqrt43, which yields 12.
    5 KB (734 words) - 13:46, 27 December 2024
  • In the context of problem-solving, the characteristic polynomial is often used to find closed forms f ...can be solved for each constant. Refer to the [[#Introductory|introductory problems]] below to see an example of how to do this. In particular, for the Fibonac
    19 KB (3,412 words) - 13:57, 21 September 2022

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