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- == Problem == Let <cmath>(1+x^3)\left(1+2x^{3^2}\right)\cdots \left(1+kx^{3^k}\right) \cdots \left(1+1997x^{3^{1997}}\right2 KB (232 words) - 23:22, 31 December 2020
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- ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC and of the recent expansion ...down=<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
- 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== ...math> balls into these urns. Using the ball-and-urn argument, having <math>9</math> urns is equivalent to <math>8</math> dividers, and there are <math>{950 bytes (137 words) - 09:16, 29 November 2019
- ==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== <math>\left|a_1^{2} - a_2^{2}\right| = 10</math>3 KB (520 words) - 11:55, 11 January 2019
- ==Problem== <math>a_{n} = 2a_{n-1} + n^2</math>2 KB (306 words) - 09:36, 4 April 2012
- ==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) - 22:59, 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) - 13:53, 19 July 2020
- ==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
- == Problem == ...d 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
- == Problem == ...hat is the smallest possible value of <math>AP^2 + BP^2 + CP^2 + DP^2 + EP^2</math>?1 KB (217 words) - 05:18, 2 July 2015
- == 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 is3 KB (446 words) - 23:18, 9 February 2020
- == Problem 1 == If <math>g(9) + g(26) + g(126) + g(401) = \frac {m}{n}</math> where <math>m</math> and <6 KB (909 words) - 06:27, 12 October 2022
- == Problem == ...\cdot DP \cdot EP \cdot FP \cdot GP</math>. Determine the value of <math>p^2</math>.540 bytes (96 words) - 23:28, 22 December 2023
- ==Problem 1== [[Mock AIME 4 Pre 2005/Problems/Problem 1 | Solution]]7 KB (1,094 words) - 14:39, 24 March 2019