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  • 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
  • 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
  • ==Problem== ...d{13}</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<
    2 KB (249 words) - 17:14, 3 April 2012
  • ==Problem== ...h>d_{2}=2</math>, <math>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== ...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== ...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== ...ath>\overline{BC}</math>, and <math>\overline{CA}</math> of [[length]] 43, 13, and 48, respectively. Let <math>\omega</math> be the [[circle]] [[circumsc
    5 KB (734 words) - 13:46, 27 December 2024
  • == 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
  • 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
  • ...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
  • == 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