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• 2005 AIME II Problems/Problem 11
  == Problem == ...</math>, <math>a_{m-1}a_{m-2} = 3</math>; from the recursion given in the problem <math>a_{m-p+1} = a_{m-p-1} - 3/a_{m-p}</math>, so <math>a_{m-p+1} = 3p/a_{
  3 KB (499 words) - 18:52, 21 November 2022

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• Ellipse
  ...latively prime integers, find <math> p+q. </math> ([[2005 AIME II Problems/Problem 15|Source]]) ...ly prime positive integers. Find <math>m+n</math>. ([[2001 AIME I Problems/Problem 5|Source]])
  5 KB (892 words) - 21:52, 1 May 2021
• 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 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
• 2006 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME I Problems]]
  1 KB (135 words) - 12:31, 22 March 2011
• 2005 AIME II
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME II Problems]]
  1 KB (135 words) - 12:30, 22 March 2011
• 2006 AIME I Problems
  {{AIME Problems|year=2006|n=I}} == Problem 1 ==
  7 KB (1,173 words) - 03:31, 4 January 2023
• 2005 AIME II Problems/Problem 4
  == Problem == ...ve integer]]s that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math>
  3 KB (377 words) - 18:36, 1 January 2024
• 2005 AIME II Problems/Problem 5
  == Problem == ...og_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math>
  3 KB (547 words) - 19:15, 4 April 2024
• 2005 AIME II Problems
  {{AIME Problems|year=2005|n=II}} == Problem 1 ==
  7 KB (1,119 words) - 21:12, 28 February 2020
• 2005 AIME II Problems/Problem 7
  == Problem == ...4}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)(y+1) = (y^{15}+y^{14}+y^{13}+y^{12}+y^{11}+y^{10}+y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=\frac{y^{16}-1}{y-1}</cmath>
  2 KB (279 words) - 12:33, 27 October 2019
• 2005 AIME II Problems/Problem 12
  == Problem == ...^2</math> equal and solving for <math>x</math> (it is helpful to scale the problem down by a factor of 50 first), we get <math>x = 250\pm 50\sqrt{7}</math>. S
  13 KB (2,080 words) - 13:14, 23 July 2024
• 2005 AIME II Problems/Problem 10
  == Problem == {{AIME box|year=2005|n=II|num-b=9|num-a=11}}
  3 KB (436 words) - 20:35, 13 August 2024
• 2005 AIME I Problems
  {{AIME Problems|year=2005|n=I}} == Problem 1 ==
  6 KB (983 words) - 05:06, 20 February 2019
• 2005 AIME II Problems/Problem 14
  == Problem == == Solution 11 (Ultimate Stewarts Bash) ==
  14 KB (2,340 words) - 16:38, 21 August 2024
• 2004 AIME II Problems
  {{AIME Problems|year=2004|n=II}} == Problem 1 ==
  9 KB (1,410 words) - 05:05, 20 February 2019
• Mock AIME 5 Pre 2005 Problems
  == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]
  6 KB (909 words) - 07:27, 12 October 2022
• 2009 AIME II Problems/Problem 7
  == Problem == ...</math> will yield an answer <math>>999</math>, which cannot happen on the AIME.) Once you calculate the value of <math>4010</math>, and divide by <math>10
  8 KB (1,312 words) - 16:23, 30 March 2024
• Mock AIME 2 Pre 2005 Problems
  == Problem 1 == [[Mock AIME 2 Pre 2005 Problems/Problem 1|Solution]]
  6 KB (1,052 words) - 13:52, 9 June 2020
• Know the Facts: Gmaas
  - 2000 - 2005: Reconstruction of Newton's Principia Mathematica Gmaasis. (In English, The 1. GMAAS's theorem states that for any math problem, GMAAS knows the answer to it. This theorem was proved by GMAAS. But then G
  89 KB (15,007 words) - 15:02, 19 February 2025

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