Search results
Create the page "Mock AIME 2 2005-2006/Problem 3" on this wiki! See also the search results found.
Page title matches
- ==Problem== ...ets of <math>7</math> digits, consider <math>9</math> urns labeled <math>1,2,\cdots,9</math> (note that <math>0</math> is not a permissible digit); then950 bytes (137 words) - 10:16, 29 November 2019
- == Problem == ...ox, there are <math>4</math> green balls, <math>4</math> blue balls, <math>2</math> red balls, a brown ball, a white ball, and a black ball. These balls1 KB (170 words) - 17:15, 4 August 2019
Page text matches
- ...th>, <math>d</math> are the four side lengths and <math>s = \frac{a+b+c+d}{2}</math>. .../math>. Hence, <math>[ABCD]=\frac{\sin B(ab+cd)}{2}</math>. Multiplying by 2 and squaring, we get:3 KB (543 words) - 19:35, 29 October 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) - 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== ...rcles are mutually externally tangent. Two of the circles have radii <math>3</math> and <math>7</math>. If the area of the triangle formed by connecting7 KB (1,135 words) - 23:53, 24 March 2019
- ==Problem== ...rcles are mutually externally tangent. Two of the circles have radii <math>3</math> and <math>7</math>. If the area of the triangle formed by connecting795 bytes (129 words) - 10:22, 4 April 2012
- ==Problem== ...ets of <math>7</math> digits, consider <math>9</math> urns labeled <math>1,2,\cdots,9</math> (note that <math>0</math> is not a permissible digit); then950 bytes (137 words) - 10:16, 29 November 2019
- ==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== <cmath>\zeta_1^2+\zeta_2^2+\zeta_3^2=3</cmath>2 KB (221 words) - 02:49, 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) - 16:03, 17 October 2021
- ==Problem== <cmath>\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}</cmath>3 KB (501 words) - 14:48, 29 November 2019
- ==Problem== ...ect at <math>P</math>. If <math>AB = 1, CD = 4,</math> and <math>BP : DP = 3 : 8,</math> then the area of the inscribed circle of <math>ABCD</math> can2 KB (330 words) - 10:23, 4 April 2012
- ==Problem== <math>\left|a_1^{2} - a_2^{2}\right| = 10</math>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 as2 KB (278 words) - 16:32, 27 December 2019
- ==Problem== <math>a_{n} = 2a_{n-1} + n^2</math>2 KB (306 words) - 10:36, 4 April 2012
- ==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 consequ2 KB (379 words) - 01:27, 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 t714 bytes (105 words) - 23:59, 24 April 2013
- ==Problem== <math>\left(\frac{2}{3}\right)^{2005} \cdot \sum_{k=1}^{2005} \frac{k^2}{2^k} \cdot {2005 \choose k}</math>3 KB (502 words) - 14:53, 19 July 2020