Search results
Create the page "Mock AIME 1 2005-2006/Problem 11" on this wiki! See also the search results found.
Page title matches
- == Problem == <cmath>\sum_{n=0}^{668} (-1)^{n} {2004 \choose 3n}</cmath>2 KB (272 words) - 09:51, 2 July 2015
Page text matches
- ...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>\zeta_1+\zeta_2+\zeta_3=1</cmath>2 KB (221 words) - 01:49, 19 March 2015
- ==Problem== Here are some thoughts on the problem:3 KB (520 words) - 11:55, 11 January 2019
- ==Problem== <math>\{A_n\}_{n \ge 1}</math> is a sequence of positive integers such that2 KB (306 words) - 09:36, 4 April 2012
- ==Problem== ...<math>n</math> such that <math>1 \le n \le 1000</math> and <math>n^{12} - 1</math> is divisible by <math>73</math>.714 bytes (105 words) - 22:59, 24 April 2013
- == Problem == ...5</math> 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 1 == <cmath>6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y</cmath>6 KB (909 words) - 06:27, 12 October 2022
- == Problem == ...56 = 104060657, \end{aligned}</math> and so the sum of the digits is <math>1+4+6+6+5+7 = \boxed{29}.</math>517 bytes (55 words) - 19:01, 23 March 2017
- == Problem == {{Mock AIME box|year=Pre 2005|n=1|num-b=9|num-a=11|source=14769}}540 bytes (96 words) - 23:28, 22 December 2023
- ...ath>3</math> bins. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. ...ath>n</math> bins is <math>{n+k-1 \choose n-1}</math> or <math>\dbinom{n+k-1}k</math>.5 KB (795 words) - 16:39, 31 December 2024
- == Problem == ...at <math>\sin \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 foll2 KB (376 words) - 21:41, 26 December 2016
- == Problem == ...xist an <math>A</math> such that <math>|A| = 620</math> and <math>\mu(A) = 11</math>? (<math>|X|</math> denotes the cardinality of the set <math>X</math>786 bytes (131 words) - 20:19, 8 October 2014
- ==Problem 1== For how many integers <math>n>1</math> is it possible to express <math>2005</math> as the sum of <math>n</m7 KB (1,094 words) - 14:39, 24 March 2019
- == Problem 1 == [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,052 words) - 12:52, 9 June 2020
- == Problem == Let <math>S</math> be the set of integers <math>n > 1</math> for which <math>\tfrac1n = 0.d_1d_2d_3d_4\ldots</math>, an infinite1 KB (171 words) - 16:38, 4 August 2019