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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 ...ficulty=7-9|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
  • == Problems == ...rs. Determine <math>p + q</math>. ([[Mock AIME 3 Pre 2005 Problems/Problem 7|Source]])
    3 KB (543 words) - 18:35, 29 October 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 i
    950 bytes (137 words) - 09:16, 29 November 2019
  • ==Problem== <cmath>\zeta_1^3+\zeta_2^3+\zeta_3^3=7</cmath>
    2 KB (221 words) - 01:49, 19 March 2015
  • ==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== {{Mock AIME box|year=Pre 2005|n=3|num-b=5|num-a=7|source=19349}}
    3 KB (501 words) - 13:48, 29 November 2019
  • == 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 == {{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 == [[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>2\cos B = \cos A + \cos C</math>, we have that <math>\cos B = \frac{7}{16}</math> and <math>\sin B = \frac{3\sqrt{23}}{16}</math>. Since <math>\s
    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 == ...^2}\right)^2 - 2 = 7^2 - 2 = 47.</cmath> Finally, <cmath>x^7 + \dfrac{1}{x^7} = \left(x^3 + \dfrac{1}{x^3}\right) \left(x^4 + \dfrac{1}{x^4}\right) - \l
    883 bytes (128 words) - 15:14, 4 August 2019
  • == Problem == ...+ 1)(10^3 + 1)(10^3 - 1) = 101 \cdot 9901 \cdot 37 \cdot 11 \cdot 13 \cdot 7 \cdot 3^3 \cdot 37,</cmath> the number <math>10^{12} -1</math> has <math>4
    1 KB (171 words) - 16:38, 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) - 23:22, 31 December 2020

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