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- ==Problem == <math>3, 6, 9, 12, 15,\cdots, 1002</math> gives '''334''' terms3 KB (442 words) - 19:51, 26 November 2023
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- ...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
- ...al{S}</math>, we have that <math>\star (n)=12</math> and <math>0\le n< 10^{7}</math>. If <math>m</math> is the number of elements in <math>\mathcal{S}</ ...= 18564 - 7 - 42 - 42 - 105 = 18368</math> so <math>\star(m) = 1 + 8 + 3 + 6 + 8 = 026</math>.1 KB (188 words) - 14:53, 3 April 2012
- ==Problem== Radius <math>a=\frac{3}{7}</math>, radius <math>b=\frac{6}{11}</math>, radius <math>c=\frac{2}{5}</math> and <math>r=1</math>, see pi1 KB (236 words) - 22:58, 24 April 2013
- ==Problem== ...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>x _2 + x_1 + 1</math>. The [[divisor | factors]] of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72 and 144 itself. As both <math>x_1</math>3 KB (470 words) - 23:33, 9 August 2019
- ==Problem== ...ion again, we have <math>\lambda(100)=20</math>, so <math>N=3^{27}\equiv 3^7\pmod{100}\equiv 87\pmod{100}</math>. Therefore <math>n=87</math>, and so we1 KB (127 words) - 23:15, 4 January 2010
- ==Problem== ...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
- 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 Fibonac19 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