Search results

  • ...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) - 21:41, 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) - 10:58, 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) - 16:33, 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) - 17:41, 21 January 2017
  • == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]
    6 KB (1,100 words) - 22:35, 9 January 2016
  • ==Problem 1== [[Mock AIME 3 Pre 2005/Problem 1|Solution]]
    7 KB (1,135 words) - 23:53, 24 March 2019
  • ==Problem== ...(10+r)(r)(7)(3)}</math>, or <math>84*84=r(10+r)*21</math>, or <math>84*4=r(10+r)</math>. <math>84*4=14*24</math>, so <math>r=14</math>. Thus the area of
    795 bytes (129 words) - 10:22, 4 April 2012
  • ==Problem== 0 &= x^2 - \frac{3 \times 2004 - 4}{10}x + \frac 52\end{align*}</cmath>
    1 KB (191 words) - 10:22, 4 April 2012
  • ==Problem== ...ear at least two consonants. Let <math>N</math> denote the number of <math>10</math>-letter Zuminglish words. Determine the remainder obtained when <math
    5 KB (795 words) - 16:03, 17 October 2021
  • ==Problem== ...<math>(a_1, a_2, \dots, a_8)</math> of real numbers such that <math>a_1 = 10</math> and
    3 KB (520 words) - 12:55, 11 January 2019
  • ==Problem== .../math> such that <math>\angle{BAE}</math> is right. If <math>BD = 15, DE = 2,</math> and <math>BC = 16</math>, then <math>CD</math> can be expressed as
    2 KB (278 words) - 16:32, 27 December 2019
  • ==Problem== ...second equation and rearranging variables gives <math>45x=8\cdot 15+bx=15^2\implies x=5</math>. Back-substitution yields <math>AC=21</math> and consequ
    2 KB (379 words) - 01:27, 6 December 2024
  • <cmath>a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)</cmath> ...4a^2b^2 \ &= (a^2 + 2b^2)^2 - (2ab)^2 \ &= (a^2 + 2b^2 - 2ab) (a^2 + 2b^2 + 2ab)
    2 KB (210 words) - 17:05, 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) - 13:44, 5 September 2012
  • == Problem == Thus, the height of <math>P</math> is <math>\sqrt [3]{8} = 2</math> times the height of <math>P'</math>, and thus the height of each is
    3 KB (446 words) - 00:18, 10 February 2020
  • == Problem 1 == <cmath>6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y</cmath>
    6 KB (909 words) - 07:27, 12 October 2022
  • == Problem == ...+f(\omega)+f(\omega^2)}{3} &= \frac{(1-1)^{2004}+(\omega-1)^{2004}+(\omega^2-1)^{2004}}{3} \
    2 KB (272 words) - 10:51, 2 July 2015
  • ...s. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. == Reasoning 2 ==
    5 KB (795 words) - 17:39, 31 December 2024
  • == Problem == ...in \angle{EAB'} = \sin(90^{\circ} - 2 \theta) = \cos 2 \theta = 1 - 2 \sin^2 \theta</math>. Now, we use law of sines, which gives us the following:
    2 KB (376 words) - 22:41, 26 December 2016
  • == Problem == ...en 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 numbe
    795 bytes (133 words) - 08:14, 19 July 2016

View (previous 20 | next 20) (20 | 50 | 100 | 250 | 500)