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• 2000 AIME II Problems/Problem 2
  ...tice point. How many lattice points lie on the hyperbola <math>x^2 - y^2 = 2000^2</math>? <cmath>(x-y)(x+y)=2000^2=2^8 \cdot 5^6</cmath>
  804 bytes (126 words) - 20:30, 4 July 2013
• Mock AIME 1 2006-2007 Problems/Problem 3
  ...= 400</math> so <math>3^{400} \equiv 1 \pmod{1000}</math> and so <math>3^{2000}\equiv 1 \pmod{1000}</math> and so <math>3^{2007} \equiv 3^7 \equiv 2187 \e *[[Mock AIME 1 2006-2007 Problems/Problem 2 | Previous Problem]]
  963 bytes (135 words) - 15:53, 3 April 2012
• Parity
  ...asic, checking the parity of numbers is often an useful tactic for solving problems, especially with [[proof by contradiction]]s and [[casework]]. == Problems ==
  4 KB (694 words) - 22:00, 12 January 2024
• 1999 AIME Answer Key
  Return to [[1999 AIME]] ([[1999 AIME Problems]]) ...ore=[[1998 AIME Answer Key|1998 AIME]]|after=[[2000 AIME I Answer Key|2000 AIME I]]}}
  251 bytes (20 words) - 18:29, 13 March 2009
• 2008 AIME II Problems/Problem 15
  ...n AIME problem so it is implicit that <math>n < 1000</math>, so <math>2n < 2000</math>. It is easy to see that <math>a_n</math> is strictly increasing by i ...21, 801</math>. <math>N</math> cannot exceed <math>1000</math> since it is AIME problem. Now take the first criterion, let <math>a</math> be the smaller co
  4 KB (628 words) - 16:23, 2 January 2024
• Cone
  == Problems == ...hat is the ratio of the cone’s height to its [[radius]]? ([[2003 AMC 12B Problems/Problem 13]])
  7 KB (1,128 words) - 20:12, 27 September 2022
• 2001 AIME I Answer Key
  Return to [[2001 AIME I]] ([[2001 AIME I Problems]]) ...000 AIME II Answer Key|2000 AIME II]]|after=[[2001 AIME II Answer Key|2001 AIME II]]}}
  267 bytes (26 words) - 18:16, 19 June 2008
• 2000 AIME I Answer Key
  Return to [[2000 AIME I]] ([[2000 AIME I Problems]]) ...re=[[1999 AIME Answer Key|1999 AIME]]|after=[[2000 AIME II Answer Key|2000 AIME II]]}}
  261 bytes (24 words) - 19:17, 27 May 2016
• 2000 AIME II Answer Key
  Return to [[2000 AIME II]] ([[2000 AIME II Problems]]) ...[[2000 AIME I Answer Key|2000 AIME I]]|after=[[2001 AIME I Answer Key|2001 AIME I]]}}
  266 bytes (26 words) - 13:09, 24 December 2020
• 2009 AIME II Problems
  {{AIME Problems|year=2009|n=II}} [[2009 AIME II Problems/Problem 1|Solution]]
  8 KB (1,366 words) - 13:59, 24 June 2024
• 2010 AIME I Problems/Problem 10
  ...math>. There are only 2 possible cases where <math>2010 - 10a_1 - a_0 \geq 2000</math>, namely <math>(a_1, a_0) = (1,0), (10,0)</math>. Thus, there are <ma *If <math>1000 \leq 2010 - 10a_1 - a_0 < 2000</math>, then there are 2 valid choices for <math>a_3</math>. Since there ar
  7 KB (1,200 words) - 12:38, 19 June 2024
• 2012 AIME II Problems/Problem 1
  ...<math>1, 6, ...</math> down into <math>0, 5, 10, ...</math>. So <math>20m=2000-12k</math>, where <math>k</math> is a member of the set <math>{0, 5, 10, 15 ...6</math>. Each of this is even (notice that when <math>k=25</math>, <math>2000-12k=1700</math>, it cycles again).
  3 KB (519 words) - 14:56, 2 June 2023
• Mock AIME 6 2006-2007 Problems
  [[Mock AIME 6 2006-2007 Problems/Problem 1|Solution]] [[Mock AIME 6 2006-2007 Problems/Problem 2|Solution]]
  7 KB (1,173 words) - 21:04, 7 December 2018
• 2014 AIME I Problems/Problem 8
  <math>2000 \cdot 9+1000 \equiv 9000\equiv9 \cdot 1000 \pmod {1000}</math> {{AIME box|year=2014|n=I|num-b=7|num-a=9}}
  12 KB (1,962 words) - 08:40, 4 November 2022
• Olympiad Archive
  ...ww.oma.org.ar/enunciados/index.htm ''Full Archive of Mathematical Olympiad Problems''] ** Problems w/o Solutions
  16 KB (1,987 words) - 11:39, 17 February 2024
• 2019 AMC 10C Problems
  Here are the problems from the 2019 AMC 10C, a mock contest created by the AoPS user fidgetboss_4 [[2019 AMC 10C Problems/Problem 1|Solution]]
  12 KB (1,917 words) - 12:14, 29 November 2021
• 2019 AIME II Problems/Problem 9
  ...<math>d(k) = 1</math>, and there is one solution <math>n = 2^4 \cdot 5^3 = 2000</math>. Then we have <math>\frac{S}{20} = \frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \boxed{472}</math>.
  4 KB (609 words) - 14:46, 4 January 2024
• 2020 AIME II Problems
  {{AIME Problems|year=2020|n=II}} [[2020 AIME II Problems/Problem 1 | Solution]]
  7 KB (1,199 words) - 16:17, 12 March 2021
• 2020 AIME II Problems/Problem 3
  ...equation in, we get <math>8^n\cdot3^{20}=3^{2020}</math>, so <math>8^n=3^{2000}</math>. Taking the 100th root, we get <math>8^{\frac{n}{100}}=3^{20}</math {{AIME box|year=2020|n=II|num-b=2|num-a=4}}
  4 KB (502 words) - 04:23, 23 January 2023
• Know the Facts: Gmaas
  - 2000 - 2005: Reconstruction of Newton's Principia Mathematica Gmaasis. (In Engli ...roblem. Therefore, I proved that you cannot use the Gmaas theorem to solve problems. But using theorem 3.141592653589793238462643383 "2. Gmaas can turn things
  88 KB (14,928 words) - 13:54, 29 April 2024

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