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- ==Modified Problem== <math> ny-p\left\lfloor \frac{ny}{p}\right\rfloor=1 </math>2 KB (340 words) - 15:52, 3 April 2012
- == Problem == ...5</math> and <math>f(101) = 0</math>). Evaluate the remainder when <math>f(1)+f(2)+\cdots+f(99)</math> is divided by <math>1000</math>.2 KB (209 words) - 12:43, 10 August 2019
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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 Mat8 KB (1,057 words) - 12:02, 25 February 2024
- === Method 1 === real r = 1;4 KB (658 words) - 16:19, 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) - 20:58, 6 December 2023
- 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) - 17:41, 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) - 16:06, 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) - 10:55, 4 April 2012
- ...us <math>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) - 15:53, 3 April 2012
- ...vely. If the <math>x</math>-coordinate of the triangle's centroid is <math>1</math>, find the area of <math>\triangle ABC</math>. *[[Mock AIME 1 2006-2007 Problems/Problem 3 | Previous Problem]]1 KB (244 words) - 15:21, 5 November 2012
- ==Problem== ...math> for [[positive integer]]s <math>a,b,c</math> where <math>\gcd(a,b,c)=1</math>, find <math>a+b+c</math>.3 KB (460 words) - 15:52, 3 April 2012
- ==Problem== ...3</math>. Point <math>E</math> is such that <math>CE=1</math> and <math>AE=5</math>. Construct point <math>F</math> on segment <math>BC</math> such that3 KB (518 words) - 16:54, 25 November 2015
- ==Problem== .../2), B=expi(pi*5/12), C=(0,0), D=expi(0), E=expi(0)+expi(pi/12), P=expi(pi*5/12)+expi(0);1 KB (244 words) - 14:54, 21 August 2020
- ==Problem== ...nce]] of [[complex number]]s with <math>a_{0}=1024</math> and <math>a_{10}=1</math>, and let <math>S</math> denote the [[infinite]] sum <math>S = a_{10}5 KB (744 words) - 19:46, 20 October 2020
- ==Problem== ...and <math>CA</math> have lengths <math>3</math>, <math>4</math>, and <math>5</math>, respectively. Let the incircle, circle <math>I</math>, of <math>\tr1 KB (236 words) - 23:58, 24 April 2013
- ==Problem== ...math>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) - 15:50, 3 April 2012
- ==Problem 1== [[Mock AIME 1 2006-2007 Problems/Problem 1|Solution]]8 KB (1,355 words) - 14:54, 21 August 2020
- == Problem == ...</math>, where <math>n</math> is as small as possible and <math>i = \sqrt{-1}</math>. Compute <math>\frac{b^2}{a^2}</math>.1 KB (240 words) - 10:50, 4 April 2012
- == Problem == [[Image:Mock AIME 2 2007 Problem14.jpg]]2 KB (284 words) - 10:53, 4 April 2012
- ...common ratio of <math>g_n</math>. Or, <math>\frac{a_ng_{n+1}-x_1-drS_g}{r-1}</math>, where <math>S_g</math> is the sum of the first <math>n</math> term <math>x_n=(a_1+d(n-1))(g_1\cdot r^{n-1})</math>2 KB (477 words) - 19:39, 17 August 2020
- ==Problem== ...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) - 20:34, 27 September 2019
- == Problem == ...5</math> and <math>f(101) = 0</math>). Evaluate the remainder when <math>f(1)+f(2)+\cdots+f(99)</math> is divided by <math>1000</math>.2 KB (209 words) - 12:43, 10 August 2019
- == Problems == ...emainder when <math>2^{202} +202</math> is divided by <math>2^{101}+2^{51}+1</math>? (2020 AMC10 B)2 KB (222 words) - 15:04, 30 December 2023
- In the context of problem-solving, the characteristic polynomial is often used to find closed forms f ...Indeed, if we define <math>T = \lambda I - A</math> and let <math>\bold{T}^1, \bold{T}^2, \ldots, \bold{T}^n</math> denote the column vectors of <math>T19 KB (3,412 words) - 14:57, 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) - 14:37, 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) - 14:39, 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) - 14:44, 3 July 2012
- ==Problem 1== [[Mock AIME 6 2006-2007 Problems/Problem 1|Solution]]7 KB (1,173 words) - 21:04, 7 December 2018
- == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=1|num-a=3}}752 bytes (117 words) - 21:16, 8 October 2014
- == Problem == ...even and <math>d_m > d_{m-1}</math> if <math>m</math> is odd for <math>m = 1,2,\ldots,k</math> (and <math>d_0 = 0</math>). Let <math>a</math> be the num795 bytes (133 words) - 08:14, 19 July 2016
- == Problem == Let <math>m</math> and <math>n</math> be integers such that <math>1 < m \le 10</math> and <math>m < n \le 100</math>. Given that <math>x = \log645 bytes (109 words) - 20:41, 22 March 2016
- == Problem == : <math>P_1(x) = 1+x+x^3+x^4+\cdots+x^{96}+x^{97}+x^{99}+x^{100}</math>522 bytes (77 words) - 21:17, 8 October 2014
- == Problem == A coin of radius <math>1</math> is flipped onto an <math>500 \times 500</math> square grid divided i853 bytes (134 words) - 21:18, 8 October 2014
- == Problem == ...Define <math>\mu(A) = \sum_{k \in A} \tau(k)</math>, where <math>\tau(k) = 1</math> if <math>k</math> has an odd number of divisors and <math>\tau(k) =786 bytes (131 words) - 21:19, 8 October 2014
- == Problem == So <math>Sin^2B=1-Cos^2B=\frac{85^2-13^2}{85^2}=\frac{84^2}{85^2}</math>2 KB (282 words) - 10:06, 9 August 2022
