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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== ...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 == ...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 \cdot
    1 KB (194 words) - 12:44, 5 September 2012
  • == Problem == When <math>1 + 7 + 7^2 + \cdots + 7^{2004}</math> is divided by <math>1000</math>, a remainder of <math>N</math
    685 bytes (81 words) - 09:51, 11 June 2013
  • == 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 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]
    6 KB (909 words) - 06:27, 12 October 2022
  • == Problem == ...0657, \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) - 19:01, 23 March 2017
  • == Problem == ...}</math>. Since <math>\sin B = \frac{b}{2R}</math>, we have that <math>b = 6\sqrt{23}</math>. By the Law of Cosines, we have that:
    2 KB (340 words) - 00:44, 3 March 2020
  • ==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