Search results
Create the page "Mock AIME 3 2005-2006/Problem 6" on this wiki! See also the search results found.
Page title matches
- ==Problem== {{Mock AIME box|year=Pre 2005|n=3|num-b=5|num-a=7|source=19349}}3 KB (501 words) - 14:48, 29 November 2019
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) - 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== ...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== ...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== Here are some thoughts on the problem:3 KB (520 words) - 12:55, 11 January 2019
- ==Problem== <cmath> C = \sum_{k=0}^{n-1} k^2 2^k = E+F = (n^2-4n+6)2^n - 6 </cmath>2 KB (306 words) - 10:36, 4 April 2012
- ==Problem== ...> at <math>C</math> and <math>D</math> respectively. If <math>AD = 3, AP = 6, DP = 4,</math> and <math>PQ = 32</math>, then the area of triangle <math>P3 KB (563 words) - 02:05, 25 November 2023
- == Problem == ...e 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 == ...t to finding <math>\frac{7^{400 \cdot 5 + 5} - 1}{6} \equiv \frac{7^5 - 1}{6} \equiv \boxed{801} \pmod{1000}</math>.685 bytes (81 words) - 10:51, 11 June 2013
- == Problem == .../math> <tt>C</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) - 17:27, 23 February 2013
- == 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 of3 KB (446 words) - 00:18, 10 February 2020
- == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]6 KB (909 words) - 07:27, 12 October 2022
- == Problem == ...04060657, \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) - 20:01, 23 March 2017
- == Problem == Let <math>\omega^3=1</math> with <math>\omega\neq 1</math>. We have2 KB (272 words) - 10:51, 2 July 2015
- .../math> bins. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. ...ach urn, then there would be <math>{n \choose k}</math> possibilities; the problem is that you can repeat urns, so this does not work.<math>n</math> and then5 KB (795 words) - 17:39, 31 December 2024
- == Problem == ...}</math>. Since <math>\sin B = \frac{b}{2R}</math>, we have that <math>b = 6\sqrt{23}</math>. By the Law of Cosines, we have that:2 KB (340 words) - 01:44, 3 March 2020
