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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 ...down=<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== Let <math>N</math> denote the number of <math>7</math> digit positive integers have the property that their digits are in i950 bytes (137 words) - 09:16, 29 November 2019
- ==Problem== ...VC</tt> - the only other combination, two vowels, is impossible due to the problem statement). Then, note that:5 KB (795 words) - 15:03, 17 October 2021
- == Problem == ...d 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) - 12:44, 5 September 2012
- == Problem == {{Mock AIME box|year=Pre 2005|n=1|num-b=7|num-a=9|source=14769}}3 KB (446 words) - 23:18, 9 February 2020
- == Problem 1 == If <math>g(9) + g(26) + g(126) + g(401) = \frac {m}{n}</math> where <math>m</math> and <6 KB (909 words) - 06:27, 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) - 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 == <cmath>k_{1997}={3^{11}}+{3^{10}}+{3^9}+{3^8}+{3^7}+{3^4}+{3^3}+{3^1}.</cmath>2 KB (232 words) - 23:22, 31 December 2020