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  • *[[Mock AIME 1 2006-2007 Problems/Problem 4 | Previous Problem]] *[[Mock AIME 1 2006-2007 Problems/Problem 6 | Next Problem]]
    2 KB (340 words) - 14:52, 3 April 2012
  • Given that <math> iz^2=1+\frac 2z + \frac{3}{z^2}+\frac{4}{z ^3}+\frac{5}{z^4}+\cdots</math> and <math>z=n\pm \sqrt{-i},</math> find <math> \lfloor {{Mock AIME box|year=2006-2007|n=2|num-b=4|num-a=6}}
    912 bytes (145 words) - 09:51, 4 April 2012
  • ...sors 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>f(1) 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
    2 KB (209 words) - 11:43, 10 August 2019
  • {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=1|num-a=3}} [[Category:Introductory Geometry Problems]]
    752 bytes (117 words) - 20:16, 8 October 2014
  • {{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
  • {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=3|num-a=5}} [[Category:Introductory Algebra Problems]]
    645 bytes (109 words) - 19:41, 22 March 2016
  • {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=5|num-a=7}} [[Category:Intermediate Algebra Problems]]
    522 bytes (77 words) - 20:17, 8 October 2014
  • {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=6|num-a=8}} [[Category:Intermediate Number Theory Problems]]
    853 bytes (134 words) - 20:18, 8 October 2014
  • {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=7|num-a=9}} [[Category:Intermediate Geometry Problems]]
    511 bytes (79 words) - 20:18, 8 October 2014
  • {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=8|num-a=10}} [[Category:Intermediate Combinatorics Problems]]
    461 bytes (62 words) - 20:18, 8 October 2014
  • {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=9|num-a=11}} [[Category:Intermediate Number Theory Problems]]
    328 bytes (45 words) - 13:15, 11 April 2018
  • {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=10|num-a=12}} [[Category:Intermediate Number Theory Problems]]
    786 bytes (131 words) - 20:19, 8 October 2014
  • By pythagoras theorem on <math>\triangle ABD</math> , We get <math>BD=5=CE</math> So, <math>\frac{[AEC]}{[ABC]}=\frac{EC}{BC}=\frac{5}{14}</math>
    2 KB (294 words) - 15:24, 24 August 2022
  • ...<math>x \in S</math> with <math>n</math> digits must be divisible by <math>5^n</math>. Let <math>A</math> be the sum of the <math>20</math> smallest ele {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=12|num-a=14}}
    539 bytes (83 words) - 20:20, 8 October 2014
  • {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=13|num-a=15}} [[Category:Intermediate Geometry Problems]]
    2 KB (282 words) - 09:06, 9 August 2022
  • {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=14|num-a=15}} [[Category:Intermediate Combinatorics Problems]]
    955 bytes (157 words) - 20:20, 8 October 2014
  • '''Case 1:''' <math>n</math> has 5 digits or more. when <math>d \ge 5</math>,
    6 KB (974 words) - 22:02, 24 November 2023

Page text matches

  • ...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
  • pair A=(-1,5), B=(-4,-1), C=(4,-1), D, O; ...*O+2*B)/5,(-1,1));label("$\theta$",O,(-0.8,-1.2));label("$\theta$",A,(0,-1.5));
    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
  • ...th> place 10 into one box and 2 into a second. <math>7 \cdot \frac{6\cdot 5}{2} = 105</math> place 10 into one box and 1 into each of two others. Thus *[[Mock AIME 1 2006-2007 Problems/Problem 1 | Previous Problem]]
    1 KB (188 words) - 14:53, 3 April 2012
  • *[[Mock AIME 1 2006-2007 Problems/Problem 3 | Previous Problem]] *[[Mock AIME 1 2006-2007 Problems/Problem 5 | Next Problem]]
    1 KB (244 words) - 14:21, 5 November 2012
  • ...eatest common divisor]] of the three numbers is 1, <math>a = 1, b = 5, c = 5</math> and <math>a + b + c = 011</math>. *[[Mock AIME 1 2006-2007 Problems/Problem 5 | Previous Problem]]
    3 KB (460 words) - 14:52, 3 April 2012
  • ...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 that <cmath>\frac{5}{1}\cdot \frac{2}{1}\cdot \frac{BD}{DA}=1</cmath>
    3 KB (518 words) - 15:54, 25 November 2015
  • .../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); *[[Mock AIME 1 2006-2007 Problems/Problem 7 | Previous Problem]]
    1 KB (244 words) - 13:54, 21 August 2020
  • ...h>1023</math> can be rewritten as <math>1023=(1024-1)=(2^{10}-1)=(2^5-1)(2^5+1)=31\cdot33=3\cdot11\cdot31</math>, and the final answer is <math>3+11+31= *[[Mock AIME 1 2006-2007 Problems/Problem 8 | Previous Problem]]
    5 KB (744 words) - 18:46, 20 October 2020
  • ...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>\tr ...{3}{7}</math>, radius <math>b=\frac{6}{11}</math>, radius <math>c=\frac{2}{5}</math> and <math>r=1</math>, see picture.
    1 KB (236 words) - 22:58, 24 April 2013
  • ...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>. *[[Mock AIME 1 2006-2007 Problems/Problem 12 | Previous Problem]]
    3 KB (568 words) - 14:50, 3 April 2012
  • [[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
  • {{Mock AIME box|year=2006-2007|n=2|num-b=3|num-a=5}} [[Category:Intermediate Algebra Problems]]
    1 KB (240 words) - 09:50, 4 April 2012
  • [[Image:Mock AIME 2 2007 Problem14.jpg]] ...or Theorem]], <math>\frac{CD}{BD} = \frac{AC}{AB} = \frac{35}{308} = \frac{5}{44}</math>.
    2 KB (284 words) - 09:53, 4 April 2012
  • == Example Problems == * [[Mock_AIME_2_2006-2007/Problem_5 | Mock AIME 2 2006-2007 Problem 5]]
    2 KB (477 words) - 18:39, 17 August 2020
  • ...es 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, *[[Mock AIME 4 2006-2007 Problems/Problem 13| Next Problem]]
    992 bytes (156 words) - 19:34, 27 September 2019

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