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• 2004 AIME II Problems/Problem 11
  == Problem == {{AIME box|year=2004|n=II|num-b=10|num-a=12}}
  2 KB (268 words) - 22:20, 23 March 2023

Page text matches

• Casework
  ..., then adding together the totals of each part. Casework is a very general problem-solving approach, and as such has wide applicability. '''Solution''': We divide the problem into cases, based on how long the word is.
  5 KB (709 words) - 17:40, 24 September 2024
• 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 14
  == Problem == ...nting <math>b</math>). Then <math>11b + 111c \le 855 + 2b + 3c \le 855 + 2(11) + 3(7) = 898 < 1000</math>, so these values work.
  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 10
  == Problem == {{AIME box|year=2004|n=II|num-b=9|num-a=11}}
  8 KB (1,283 words) - 19:19, 8 May 2024
• 2004 AIME II Problems/Problem 9
  == Problem == #<math>a_{10} < a_{11}</math> (note that the sequence must be increasing on all terms, not monoto
  3 KB (535 words) - 22:25, 5 March 2025
• 2004 AIME II Problems/Problem 6
  == Problem == ...math>, the second monkey got <math>\frac{1}{8}b_1 + \frac{1}{4}b_2 + \frac{11}{24}b_3</math>, and the third monkey got <math>\frac{1}{8}b_1 + \frac{3}{8}
  6 KB (950 words) - 14:18, 15 January 2024
• 2004 AIME II Problems/Problem 4
  == Problem == ...0, AA00, A0A0, A00A, AAA0, AA0A, A0AA</math>). This gives us <math>9\cdot 11 = 99</math> numbers of this form.
  3 KB (508 words) - 01:16, 19 January 2024
• 2004 AIME II Problems/Problem 3
  == Problem == ...s possible, which occurs when the smaller block is <math>3 \times 7 \times 11</math>. Then the extra layer makes the entire block <math>4\times8\times12
  2 KB (377 words) - 11:53, 10 March 2014
• 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
• 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