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  • ...ABC</math> have <math>BC=\sqrt{7}</math>, <math>CA=1</math>, and <math>AB=3</math>. If <math>\angle A=\frac{\pi}{n}</math> where <math>n</math> is an i ...so <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
  • ...>, 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 <math>n + m + 6 = 681</math> and <math>n - m + 6 = 3</math> give <math>2n + 12 = 684</math> and <math>n = 336</math>.
    1 KB (198 words) - 09:50, 4 April 2012
  • Substitute <math>y = \frac{9 - x^2}{3}</math> into the second to get the quartic polynomial: <cmath>x^4 - 18x^2 - ...of <math>y</math> by making the substitution for <math>x = \frac{9 - y^2}{3}</math> and the polynomial <cmath>y^4 - 6y^2 + 27y = 0</cmath> results.
    907 bytes (149 words) - 00:14, 30 December 2009
  • ...b - bc - ca + a + b + c - 1= 2007 \Longrightarrow (a-1)(b-1)(c-1) = 2007 = 3^2\cdot 223</math>. From factorization, let <math>a = 4, b = 4, c = 224</mat *[[Mock AIME 3 2006-2007/Problem 3 | Next Problem]]
    673 bytes (107 words) - 00:25, 30 December 2009
  • .... The particle reaches an edge (stops) if the coordinates satisfy 2 of the 3 following conditions: 3) <math>z</math> is an integer
    1 KB (213 words) - 19:48, 15 February 2015
  • <cmath>8x^4 - 8(a + 3)x^2 + (a + 6)^2 = 0</cmath>. ...math> have no real solutions). Since <math>a > 0</math>, both <math>-8(a + 3) > 0</math> and <math>(a + 6)^2 > 0</math>, so <math>r, s > 0</math>.
    4 KB (699 words) - 00:53, 30 April 2022
  • {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=2|num-a=4}} [[Category:Introductory Number Theory Problems]]
    795 bytes (133 words) - 07:14, 19 July 2016
  • Solving, we get <math>S_{59}(a) = 59\cdot 3 = \boxed{177}</math>.
    769 bytes (139 words) - 22:38, 15 February 2015
  • 603 bytes (105 words) - 22:35, 15 February 2015
  • 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
  • Let N be the number of integers n such that 0<=n<k where k=2^3*5^2*3^4 such that there exists an integer a such that when a^2 is divided by k it
    210 bytes (41 words) - 08:51, 18 June 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
  • A '''Mock AIME''' is a contest that is intended to mimic the [[AIME]] competition. (In more recent years, recurring competitions will be listed ...xOGY2Y2QwOTc3NWZiYjY0LnBkZg==&rn=TWlsZG9yZiBNb2NrIEFJTUUucGRm Mildorf Mock AIME 1]
    8 KB (901 words) - 19:45, 13 January 2025
  • The '''Mock AIME 1 2005-2006''' was written by [[Art of Problem Solving]] community member p * [[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
  • ...h>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) - 14:53, 3 April 2012
  • ...ABC</math> have <math>BC=\sqrt{7}</math>, <math>CA=1</math>, and <math>AB=3</math>. If <math>\angle A=\frac{\pi}{n}</math> where <math>n</math> is an i ...so <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
  • ...^2)</math>. Then we have by the centroid condition that <math>a + b + c = 3</math>. From the first [[slope]] condition we have <math>10 = \frac{b^2 - *[[Mock AIME 1 2006-2007 Problems/Problem 3 | Previous Problem]]
    1 KB (244 words) - 14:21, 5 November 2012
  • ...h by <math>-a^3</math> and let <math>\frac b a = q</math> to get <math>25q^3 +101q^2 - 1125q - 25 = 0</math>. Since we're searching for a rational root *[[Mock AIME 1 2006-2007 Problems/Problem 5 | Previous Problem]]
    3 KB (460 words) - 14:52, 3 April 2012
  • 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< ...ath>CE\cdot CA=CF\cdot CB</math>. In other words, <math>1\cdot 6 = CF\cdot 3</math>. <math>CF</math> is then 2, and <math>BF</math> is 1. We can now use
    3 KB (518 words) - 15:54, 25 November 2015
  • ...\cdot10/(2^{10}-1)=2^{11}\cdot5/(2^{10}-1)</math> and <math>2^{10}-1=33*31=3*11*31</math>. ...-1)(2^5+1)=31\cdot33=3\cdot11\cdot31</math>, and the final answer is <math>3+11+31=45</math>.
    5 KB (744 words) - 18:46, 20 October 2020
  • ..., <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, Radius <math>a=\frac{3}{7}</math>, radius <math>b=\frac{6}{11}</math>, radius <math>c=\frac{2}{5}<
    1 KB (236 words) - 22:58, 24 April 2013
  • ...] of strings with only 0's or 1's with length <math>n</math> such that any 3 adjacent place numbers sum to at least 1. For example, <math>00100</math> w ...+ A_1(n - 2) = \mathcal{S}_{n - 1} + \mathcal{S}_{n -2} + \mathcal{S}_{n - 3}</math>. Then using the initial values <math>\mathcal{S}_1 = 2, \mathcal{S
    2 KB (424 words) - 14:51, 3 April 2012
  • ...d {13}</math>. Now <math>10^2 = 100 \equiv 9 \pmod{13}</math> so <math>10^3 \equiv 90 \equiv -1 \pmod{13}</math>, <math>10^6 \equiv 1 \pmod{13}</math> *[[Mock AIME 1 2006-2007 Problems/Problem 11 | Previous Problem]]
    2 KB (249 words) - 17:14, 3 April 2012
  • ...<math>n</math>. If <math>d_{1}=1</math>, <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</ ...ath>, and <math>\alpha\beta\gamma=r</math>. Then we can write <math>P(x)=x^3 - px^2 - qx - r</math>.
    3 KB (568 words) - 14:50, 3 April 2012
  • ...nd <math>n</math> are relatively prime positive integers. Compute the last 3 digits of <math>m+n</math> *[[Mock AIME 1 2006-2007 Problems/Problem 14 | Previous Problem]]
    3 KB (414 words) - 12:45, 19 February 2016
  • [[Mock AIME 1 2006-2007 Problems/Problem 1|Solution]] [[Mock AIME 1 2006-2007 Problems/Problem 2|Solution]]
    8 KB (1,355 words) - 13:54, 21 August 2020
  • ...>, 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 <math>n + m + 6 = 681</math> and <math>n - m + 6 = 3</math> give <math>2n + 12 = 684</math> and <math>n = 336</math>.
    1 KB (198 words) - 09:50, 4 April 2012

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