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- 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
- == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,100 words) - 22:35, 9 January 2016
- == 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
- .../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 (775 words) - 23:53, 13 April 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) - 22:41, 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) - 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 == ...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) - 17:38, 4 August 2019
