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- == 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 eac3 KB (446 words) - 23:18, 9 February 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 ...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
- Substituting <math>\sin^2B=1-\cos^2B</math> results in <cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath>3 KB (543 words) - 18:35, 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) - 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== .../math> urns is equivalent to <math>8</math> dividers, and there are <math>{8 + 7 \choose 7} = {15 \choose 7} = 6435 \equiv \boxed{435} \pmod{1000}</math950 bytes (137 words) - 09:16, 29 November 2019
- ==Problem== === Solution 1 (recursive) ===5 KB (795 words) - 15:03, 17 October 2021
- ==Problem== ...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> can be e2 KB (330 words) - 09:23, 4 April 2012
- ==Problem== Let <math>N</math> denote the number of <math>8</math>-tuples <math>(a_1, a_2, \dots, a_8)</math> of real numbers such that3 KB (520 words) - 11:55, 11 January 2019
- ==Problem== ==Solution 1==2 KB (278 words) - 15:32, 27 December 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 == ...r 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
- == Problem == ...</tt>s amongst the middle five numbers, and so there are <math>6-(5-k) = k+1</math> <tt>C</tt>s amongst the first four numbers.1 KB (221 words) - 16:27, 23 February 2013
- == Problem == <cmath> \frac{1}{p} + \frac{1}{q} + \frac{1}{r} + \frac{360}{pqr} = 1</cmath>864 bytes (127 words) - 20:47, 21 February 2010
- == 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 eac3 KB (446 words) - 23:18, 9 February 2020
- == Problem 1 == <cmath>6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y</cmath>6 KB (909 words) - 06:27, 12 October 2022