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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== ...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 == ...its, 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 \cdot1 KB (194 words) - 13:44, 5 September 2012
- == Problem == ...m over triangle <math>ABC</math> to be <math>20</math>. Thus, <math>A'C' = 10</math>. Since the height of this trapezoid is <math>12</math>, and <math>AC3 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 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 == ...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
- == Problem == <cmath>k_{1997}={3^{11}}+{3^{10}}+{3^9}+{3^8}+{3^7}+{3^4}+{3^3}+{3^1}.</cmath>2 KB (232 words) - 00:22, 1 January 2021
