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  • ==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

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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 ...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== <cmath>2f\left(\frac 1x\right) + f\left(x\right) = \frac{5}{x} + 4</cmath>
    1 KB (191 words) - 09:22, 4 April 2012
  • ==Problem== <cmath>\zeta_1^2+\zeta_2^2+\zeta_3^2=3</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 == ...e 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 == ...uivalent 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
  • == Problem == ...th> <tt>C</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
  • == 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
    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 == ...060657, \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
  • ...<math>3</math> bins. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. ...ach urn, then there would be <math>{n \choose k}</math> possibilities; the problem is that you can repeat urns, so this does not work.<math>n</math> and then
    5 KB (795 words) - 16:39, 31 December 2024
  • == Problem == Triangle <math>ABC</math> has an inradius of <math>5</math> and a circumradius of <math>16</math>. If <math>2\cos{B} = \cos{A} +
    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 == ...S = \{5^k | k \in \mathbb{Z}, 0 \le k \le 2004 \}</math>. Given that <math>5^{2004} = 5443 \cdots 0625</math> has <math>1401</math> digits, how many ele
    817 bytes (114 words) - 16:16, 4 August 2019
  • ==Problems== ...(n+1)}\qquad\textbf{(E)}\ \frac{3}{n(n+1)} </math> ([[1959 AHSME Problems/Problem 37|Source]])
    3 KB (558 words) - 15:37, 21 July 2024

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