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  • == Problem == ...</tt>s amongst the middle five numbers, and so there are <math>6-(5-k) = k+1</math> <tt>C</tt>s amongst the first four numbers.
    1 KB (221 words) - 16:27, 23 February 2013

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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
  • Substituting <math>\sin^2B=1-\cos^2B</math> results in <cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath>
    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+\zeta_2+\zeta_3=1</cmath>
    2 KB (221 words) - 01:49, 19 March 2015
  • ==Problem== === Solution 1 (recursive) ===
    5 KB (795 words) - 15:03, 17 October 2021
  • ==Problem== <cmath>\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}</cmath>
    3 KB (501 words) - 13:48, 29 November 2019
  • == Problem == ...r the tens and 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 == 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 == <cmath>6g(1 + (1/y)) + 12g(y + 1) = \log_{10} y</cmath>
    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 == ...he property that <math>x+\tfrac1x = 3</math>. Let <math>S_m = x^m + \tfrac{1}{x^m}</math>. Determine the value of <math>S_7</math>.
    883 bytes (128 words) - 15:14, 4 August 2019
  • == Problem == Let <math>S</math> be the set of integers <math>n > 1</math> for which <math>\tfrac1n = 0.d_1d_2d_3d_4\ldots</math>, an infinite
    1 KB (171 words) - 16:38, 4 August 2019
  • == Problem == ...ht)\cdots \left(1+kx^{3^k}\right) \cdots \left(1+1997x^{3^{1997}}\right) = 1+a_1 x^{k_1} + a_2 x^{k_2} + \cdots + a_m x^{k_m}</cmath> where <math>a_i \n
    2 KB (232 words) - 23:22, 31 December 2020

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