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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 ...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
- 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
- == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,100 words) - 21:35, 9 January 2016
- ==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) - 15:03, 17 October 2021
- == Problem == ...igits, then tens digits, then units digits. Every one of <math>\{1,2,3,4,5,6,7,8,9\}</math> may appear as the hundreds digit, and there are <math>9 \cdo1 KB (194 words) - 12:44, 5 September 2012
- == Problem == ...top of the frustum is a rectangle <math>A'B'C'D'</math> with <math>A'B' = 6</math> and <math>B'C' = 8</math>.3 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
- == Problem == ...1001}\equiv 3 \pmod{500}</math>. Then <math>2\cdot3^{1001}=2(500k+3)=1000k+6</math>. Thus, our answer is <math>\boxed{006}</math>.2 KB (272 words) - 09:51, 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) - 16:39, 31 December 2024
- == Problem == ...rac15, GE = \frac{115}{\sqrt{24}}</math>, and <math>AG = AE - GE = 10\sqrt{6} - \frac{115}{\sqrt{24}} = \frac{5}{\sqrt{24}}</math>. Note that <math>\tri2 KB (376 words) - 21:41, 26 December 2016
- ==Problem 1== For how many integers <math>n>1</math> is it possible to express <math>2005</math> as the sum of <math>n</math> distinct positive integers?7 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 == ...3 \cdot 37,</cmath> the number <math>10^{12} -1</math> has <math>4 \cdot 2^6 = 256</math> divisors and our answer is <math>256 - 1 = \boxed{255}.</math>1 KB (171 words) - 16:38, 4 August 2019