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- ==Problem== *[[Mock AIME 1 2006-2007 Problems/Problem 6 | Previous Problem]]3 KB (518 words) - 15:54, 25 November 2015
- == Problem == ...ot 2\pi=34\pi</math>. Let the vertex of this sector be <math>O</math>. The problem is then reduced to finding the shortest distance between the two points <ma1 KB (231 words) - 17:10, 10 July 2014
- ==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 == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=6|num-a=8}}853 bytes (134 words) - 20:18, 8 October 2014
- ==Problem ==3 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}</ ...box and 1 into each of two others. Thus, this gives us <math>m = 18564 - 7 - 42 - 42 - 105 = 18368</math> so <math>\star(m) = 1 + 8 + 3 + 6 + 8 = 026<1 KB (188 words) - 14:53, 3 April 2012
- Let <math>\triangle ABC</math> have <math>BC=\sqrt{7}</math>, <math>CA=1</math>, and <math>AB=3</math>. If <math>\angle A=\frac{ ...o <math>3^{2000}\equiv 1 \pmod{1000}</math> and so <math>3^{2007} \equiv 3^7 \equiv 2187 \equiv 187 \pmod{1000}</math>963 bytes (135 words) - 14:53, 3 April 2012
- ...> and <math>a = (a + b) - b = 12</math>, so our three vertices are <math>(-7, 49), (-2, 4)</math> and <math>(12, 144)</math>. *[[Mock AIME 1 2006-2007 Problems/Problem 3 | Previous Problem]]1 KB (244 words) - 14:21, 5 November 2012
- ==Problem== *[[Mock AIME 1 2006-2007 Problems/Problem 5 | Previous Problem]]3 KB (460 words) - 14:52, 3 April 2012
- ==Problem== *[[Mock AIME 1 2006-2007 Problems/Problem 7 | Previous Problem]]1 KB (244 words) - 13:54, 21 August 2020
- ==Problem== Radius <math>a=\frac{3}{7}</math>, radius <math>b=\frac{6}{11}</math>, radius <math>c=\frac{2}{5}</ma1 KB (236 words) - 22:58, 24 April 2013
- ==Problem== We will solve this problem by constructing a [[recursion]] satisfied by <math>\mathcal{S}_n</math>.2 KB (424 words) - 14:51, 3 April 2012
- ==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 == ...tegers. Then <math>x_1 = 8 - x_2 = 7</math> and our [[sequence]] is <math>7, 1, 1, 8, 16, 144, 144(16 + 8) = 3456</math>.3 KB (470 words) - 23:33, 9 August 2019
- ...istered to the AoPS community, while others may be sourced from a group of problem writers. Different users may have a different way of participating; some ma ...COUNTS competition. There is no guarantee that community members will make Mock MATHCOUNTS in any given year, but it's usually a good bet that someone will26 KB (3,260 words) - 18:28, 15 August 2024
- ==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>, by [[Euler's Totient Theorem]] <math>2^{20 \cdot 100 + 7} \equiv 2^7 \equiv 3 \pmod{125}</math>. Combining, we have <math>2^{2007} \equiv 128 \p4 KB (595 words) - 11:14, 25 November 2023
- ==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
- == Problem == So all of the prime numbers less than <math>50</math> are <math>2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,</math> and <math>47</math>. So we2 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 Fibonac19 KB (3,412 words) - 13:57, 21 September 2022