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• 2004 AIME II Problems/Problem 13
  == Problem == {{AIME box|year=2004|n=II|num-b=12|num-a=14}}
  3 KB (486 words) - 22:15, 7 April 2023

Page text matches

• Complex number
  *[[2007 AMC 12A Problems/Problem 18]] *[[1984 AIME Problems/Problem 8|1984 AIME Problem 8]]
  5 KB (860 words) - 15:36, 10 December 2023
• 2004 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME I Problems]]
  1 KB (135 words) - 18:15, 19 April 2021
• 2004 AIME II
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME II Problems]]
  1 KB (135 words) - 12:24, 22 March 2011
• 2005 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME I Problems]]
  1 KB (154 words) - 12:30, 22 March 2011
• 2005 AIME I Problems
  {{AIME Problems|year=2005|n=I}} == Problem 1 ==
  6 KB (983 words) - 05:06, 20 February 2019
• 2004 AIME I Problems
  {{AIME Problems|year=2004|n=I}} == Problem 1 ==
  9 KB (1,434 words) - 13:34, 29 December 2021
• 2004 AIME II Problems/Problem 15
  == Problem == Now, consider the strip of length <math>1024</math>. The problem asks for <math>s_{941, 10}</math>. We can derive some useful recurrences f
  6 KB (899 words) - 20:58, 12 May 2022
• 2004 AIME II Problems/Problem 14
  == Problem == ...n't need to be nearly as rigorous). A more natural manner of attacking the problem is to think of the process in reverse, namely seeing that <math>n \equiv 1
  11 KB (1,857 words) - 12:57, 18 July 2024
• 2004 AIME II Problems/Problem 12
  == Problem == {{AIME box|year=2004|n=II|num-b=11|num-a=13}}
  3 KB (431 words) - 23:21, 4 July 2013
• 2004 AIME II Problems/Problem 6
  == Problem == ...each fraction is equal to <math>8</math>, and the solution is <math>8(11 + 13 + 27) = \boxed{408}</math>.
  6 KB (950 words) - 14:18, 15 January 2024
• 2004 AIME II Problems
  {{AIME Problems|year=2004|n=II}} == Problem 1 ==
  9 KB (1,410 words) - 05:05, 20 February 2019
• 2003 AIME II Problems
  {{AIME Problems|year=2003|n=II}} == Problem 1 ==
  7 KB (1,127 words) - 09:02, 11 July 2023
• Mock AIME 2 Pre 2005 Problems
  == Problem 1 == ...largest integer <math>k</math> such that <math>2004^k</math> divides <math>2004!</math>.
  6 KB (1,052 words) - 13:52, 9 June 2020
• 2020 AIME II Problems/Problem 14
  ==Problem== Hence, <cmath>N = \frac{2005 \cdot 2004 \cdot 2003}{3 \cdot 2\cdot 1} \equiv \boxed{010} (\mathrm{mod} \hskip .2cm
  7 KB (1,187 words) - 05:58, 3 February 2025
• Mathboy282
  [[2019 AIME II Problems/Problem 15]] Solution 5 [[2023 USAJMO Problems/Problem 6]] Solution 1
  10 KB (1,116 words) - 12:37, 11 June 2024