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- ==Problem== {{Mock AIME box|year=Pre 2005|n=3|num-b=9|num-a=11}}2 KB (306 words) - 10:36, 4 April 2012
- #REDIRECT [[Mock AIME 3 Pre 2005 Problems/Problem 10]]54 bytes (6 words) - 10:25, 4 April 2012
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- ...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== 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== ...ear at least two consonants. Let <math>N</math> denote the number of <math>10</math>-letter Zuminglish words. Determine the remainder obtained when <math5 KB (795 words) - 16:03, 17 October 2021
- ==Problem== ...<math>(a_1, a_2, \dots, a_8)</math> of real numbers such that <math>a_1 = 10</math> and3 KB (520 words) - 12:55, 11 January 2019
- ==Problem== {{Mock AIME box|year=Pre 2005|n=3|num-b=8|num-a=10}}2 KB (278 words) - 16:32, 27 December 2019
- ==Problem== {{Mock AIME box|year=Pre 2005|n=3|num-b=10|num-a=12}}2 KB (379 words) - 01:27, 6 December 2024
- == 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 == 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 == <cmath>6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y</cmath>6 KB (909 words) - 07:27, 12 October 2022
- == Problem == Let <math>\omega^3=1</math> with <math>\omega\neq 1</math>. We have2 KB (272 words) - 10:51, 2 July 2015
- ...<math>3</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 1== [[Mock AIME 4 Pre 2005/Problems/Problem 1 | Solution]]7 KB (1,094 words) - 15:39, 24 March 2019
- == Problem 1 == [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,052 words) - 13:52, 9 June 2020
- == Problem == ...37 \cdot 11 \cdot 13 \cdot 7 \cdot 3^3 \cdot 37,</cmath> the number <math>10^{12} -1</math> has <math>4 \cdot 2^6 = 256</math> divisors and our answer i1 KB (171 words) - 17:38, 4 August 2019
