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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
  • 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== 0 &= x^2 - \frac{3 \times 2004 - 4}{10}x + \frac 52\end{align*}</cmath>
    1 KB (191 words) - 09:22, 4 April 2012
  • ==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 <math
    5 KB (795 words) - 15:03, 17 October 2021
  • == Problem == ...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 == Thus, the height of <math>P</math> is <math>\sqrt [3]{8} = 2</math> times the height of <math>P'</math>, and thus the height of each is
    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 == ...+f(\omega)+f(\omega^2)}{3} &= \frac{(1-1)^{2004}+(\omega-1)^{2004}+(\omega^2-1)^{2004}}{3} \
    2 KB (272 words) - 09:51, 2 July 2015
  • ...s. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. == Reasoning 2 ==
    5 KB (792 words) - 00:11, 5 October 2024
  • == Problem == ...in \angle{EAB'} = \sin(90^{\circ} - 2 \theta) = \cos 2 \theta = 1 - 2 \sin^2 \theta</math>. Now, we use law of sines, which gives us the following:
    2 KB (376 words) - 21:41, 26 December 2016
  • ==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 == ...3^3 \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}.</mat
    1 KB (171 words) - 16:38, 4 August 2019
  • == Problem == Let <cmath>(1+x^3)\left(1+2x^{3^2}\right)\cdots \left(1+kx^{3^k}\right) \cdots \left(1+1997x^{3^{1997}}\right
    2 KB (232 words) - 23:22, 31 December 2020