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  • ...e=USAMO|region=USA|type=Proof|difficulty=7-9|breakdown=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}} ...on the USAMO have considerable experience both solving highly challenging problems and [[writing proofs]].
    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
  • A '''Mock AIME''' is a contest that is intended to mimic the [[AIME]] competition. (In more recent years, recurring competitions will be listed * Pre 2006
    8 KB (901 words) - 19:45, 13 January 2025
  • 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
  • [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]] [[Mock AIME 1 Pre 2005 Problems/Problem 2|Solution]]
    6 KB (1,100 words) - 21:35, 9 January 2016
  • Let <math>N</math> denote the number of <math>7</math> digit positive integers have the property that their digits are in i ...ath>8</math> dividers, and there are <math>{8 + 7 \choose 7} = {15 \choose 7} = 6435 \equiv \boxed{435} \pmod{1000}</math>.
    950 bytes (137 words) - 09:16, 29 November 2019
  • <cmath>\zeta_1^3+\zeta_2^3+\zeta_3^3=7</cmath> Compute <math>\zeta_1^{7} + \zeta_2^{7} + \zeta_3^{7}</math>.
    2 KB (221 words) - 01:49, 19 March 2015
  • 7 & 544 & 208 & 160 \ In total, case <math>2</math> holds <math>(4+12+4)\cdot2^7=2560</math> possible words.
    5 KB (795 words) - 15:03, 17 October 2021
  • {{Mock AIME box|year=Pre 2005|n=3|num-b=5|num-a=7|source=19349}} [[Category:Intermediate Algebra Problems]]
    3 KB (501 words) - 13:48, 29 November 2019
  • ...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 Every one of <math>\{0,1,2,3,4,5,6,7,8,9\}</math> may appear as the tens digit; however, since <math>0</math> do
    1 KB (194 words) - 12:44, 5 September 2012
  • When <math>1 + 7 + 7^2 + \cdots + 7^{2004}</math> is divided by <math>1000</math>, a remainder of <math>N</math ...equivalent to finding <math>\frac{7^{400 \cdot 5 + 5} - 1}{6} \equiv \frac{7^5 - 1}{6} \equiv \boxed{801} \pmod{1000}</math>.
    685 bytes (81 words) - 09:51, 11 June 2013
  • {{Mock AIME box|year=Pre 2005|n=1|num-b=7|num-a=9|source=14769}} [[Category:Intermediate Geometry Problems]]
    3 KB (446 words) - 23:18, 9 February 2020
  • [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]] [[Mock AIME 5 Pre 2005 Problems/Problem 2|Solution]]
    6 KB (909 words) - 06:27, 12 October 2022
  • ...0657, \end{aligned}</math> and so the sum of the digits is <math>1+4+6+6+5+7 = \boxed{29}.</math> {{Mock AIME box|year=Pre 2005|n=5|num-b=11|num-a=13|source=28368}}
    517 bytes (55 words) - 19:01, 23 March 2017
  • ...<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 ...{8} = 36 \cdot 23</math>, we have that: <math>\frac{(a+c)(-2ac \cdot \frac{7}{16}+2ac)}{ac} = \frac{21\sqrt{23}}{2} \implies</math> <math>a+c = \frac{28
    2 KB (340 words) - 00:44, 3 March 2020
  • 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? [[Mock AIME 4 Pre 2005/Problems/Problem 1 | Solution]]
    7 KB (1,094 words) - 14:39, 24 March 2019
  • [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]] [[Mock AIME 2 Pre 2005 Problems/Problem 2|Solution]]
    6 KB (1,052 words) - 12:52, 9 June 2020
  • ...^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 {{Mock AIME box|year=Pre 2005|n=2|num-b=1|num-a=3|source=14769}}
    883 bytes (128 words) - 15:14, 4 August 2019
  • ...+ 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 {{Mock AIME box|year=Pre 2005|n=2|num-b=4|num-a=6|source=14769}}
    1 KB (171 words) - 16:38, 4 August 2019

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