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  • == Problem == ...y tangent to a larger circle <math>\omega_2</math> of radius <math>12\sqrt{2}</math> such that the center of <math>\omega_2</math> lies on <math>\omega_
    752 bytes (117 words) - 20:16, 8 October 2014
  • == Problem == ...3^3 \cdot 37,</cmath> the number <math>10^{12} -1</math> has <math>4 \cdot 2^6 = 256</math> divisors and our answer is <math>256 - 1 = \boxed{255}.</mat
    1 KB (171 words) - 16:38, 4 August 2019
  • If <math>a=2</math> and <math>b=0</math>, then <math>g=0=b</math>, a contradiction. If <math>a=2</math> and <math>b=1</math>, then <math>f=1=b</math>, a contradiction.
    2 KB (335 words) - 05:41, 5 September 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
  • ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC and of the recent expansion ...ficulty=7-9|breakdown=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}}
    6 KB (874 words) - 22:02, 10 November 2024
  • ...A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member ...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 2 Pre 2005''' was written by [[Art of Problem Solving]] community member Mildorf. * [[Mock AIME 2 Pre 2005 Problems|Entire Exam]]
    2 KB (181 words) - 09:58, 18 March 2015
  • The '''Mock AIME 7 Pre 2005''' was written by [[Art of Problem Solving]] community member Mildorf. * [[Mock AIME 7 Pre 2005 Problems|Entire Exam]]
    1 KB (146 words) - 15:33, 14 October 2022
  • 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
  • == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]
    6 KB (1,100 words) - 21:35, 9 January 2016
  • ==Problem 1== [[Mock AIME 3 Pre 2005/Problem 1|Solution]]
    7 KB (1,135 words) - 22:53, 24 March 2019
  • ==Problem== <cmath>2f\left(\frac 1x\right) + f\left(x\right) = \frac{5}{x} + 4</cmath>
    1 KB (191 words) - 09:22, 4 April 2012
  • ==Problem== <cmath>\zeta_1^2+\zeta_2^2+\zeta_3^2=3</cmath>
    2 KB (221 words) - 01:49, 19 March 2015
  • ==Problem== ...VC</tt> - the only other combination, two vowels, is impossible due to the problem statement). Then, note that:
    5 KB (795 words) - 15:03, 17 October 2021
  • ==Problem== <cmath>\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}</cmath>
    3 KB (501 words) - 13:48, 29 November 2019
  • ==Problem== ...algebraically or by inspection) to get that <math>BC=3</math> and <math>AD=2</math>.
    2 KB (330 words) - 09:23, 4 April 2012
  • ==Problem== <math>\left|a_1^{2} - a_2^{2}\right| = 10</math>
    3 KB (520 words) - 11:55, 11 January 2019
  • ==Problem== ...tion and rearranging variables gives <math>45x=8\cdot 15+bx=15^2\implies x=5</math>. Back-substitution yields <math>AC=21</math> and consequently <math>
    2 KB (379 words) - 00:27, 6 December 2024
  • ==Problem== ...rest being the previous: <math>+2, +5, +1, +15, +3, +19, +3, +15, +1, +5, +2</math>. This sequence then repeats itself. We hence find that there are a t
    714 bytes (105 words) - 22:59, 24 April 2013
  • ==Problem== ...ath>QP</math>. Thus, <math>O_3R \bot PQ</math> and <math>|PR|=\frac{|PQ|}{2}=16</math>
    3 KB (563 words) - 01:05, 25 November 2023
  • == Problem == ...th> 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) - 11:43, 10 August 2019
  • ...style="text-align:center;"><math>a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)</math></div> ...4a^2b^2 \ &= (a^2 + 2b^2)^2 - (2ab)^2 \ &= (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)
    2 KB (225 words) - 04:42, 30 January 2025
  • == Problem == ...the tens and units digits. Thus the sum of the hundreds places is <math>(1+2+3+\cdots+9)(72) \times 100 = 45 \cdot 72 \cdot 100 = 324000</math>.
    1 KB (194 words) - 12:44, 5 September 2012

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