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• 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

Page text matches

• Ellipse
  pair P=(3,12/5), F1=(-4,0), F2=(4,0); ...latively prime integers, find <math> p+q. </math> ([[2005 AIME II Problems/Problem 15|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
• Geometric sequence
  * [[1965 AHSME Problems/Problem 36 | 1965 AHSME Problem 36]] * [[2005_AIME_II_Problems/Problem_3 | 2005 AIME II Problem 3]]
  4 KB (649 words) - 21:09, 19 July 2024
• 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 == <math>18^{11} = 2^{11}\cdot3^{22}</math> so <math>18^{11}</math> has <math>12\cdot23 = 276</math> divisors.
  3 KB (377 words) - 18:36, 1 January 2024
• 2005 AIME II Problems/Problem 13
  == Problem == ...olutions so <math>P(x)</math> is at least quadratic. Let us first try this problem out as if <math>P(x)</math> is a quadratic polynomial. Thus <math>P(n)-(n+3
  4 KB (642 words) - 02:14, 1 June 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 == <cmath>(y^8+1)(y^4+1) =(y^{12}+y^8+y^4+1)</cmath>
  2 KB (279 words) - 12:33, 27 October 2019
• 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
• 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 15
  == Problem == ...equations as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>.
  12 KB (2,001 words) - 20:26, 23 July 2024
• 2005 AIME II Problems/Problem 14
  == Problem == ...H = \frac{56}{5}</math> and <math>HC = \frac{42}{5} \Rightarrow HD = \frac{12}{5}</math>.
  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
• Vieta's formulas
  * [[2005 AMC 12B Problems/Problem 12 | 2005 AMC 12B Problem 12]] * [[2007 AMC 12A Problems/Problem 21 | 2007 AMC 12A Problem 21]]
  3 KB (528 words) - 21:24, 21 October 2024
• 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

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