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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
• User:Rowechen
  [[1987 AIME Problems/Problem 13|Solution]] [[1989 AIME Problems/Problem 14|Solution]]
  22 KB (3,778 words) - 19:29, 2 June 2020
• 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
• Template:AIME box
  ...('''[[{{{year}}} AIME {{{n|}}} Problems|Problems]]''' • '''[[{{{year}}} AIME {{{n|}}} Answer Key|Answer Key]]''' • [http://www.artofproblemsolving.com ...beforetext|Preceded&nbsp;by<br/>}}}'''{{{before|[[{{{year}}} AIME {{{n|}}} Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]]}}}'''
  3 KB (413 words) - 21:33, 15 August 2014
• User:Azjps/sandbox
  ...> - <s>[[2007 AIME II|2007]]</s> - <s>[[2008 AIME I|2008]]</s> - '''[[2008 AIME II|2008]]''' ...Problem 13|13]] - || [[1983 AIME Problems/Problem 14|14]] - || [[1983 AIME Problems/Problem 15|15]] - ||
  28 KB (3,170 words) - 17:54, 2 January 2009
• User:Azjps
  ...i>{{duplicate|[[200Y AMC 12A Problems|200Y AMC 12A #N]] and [[200Y AMC 10A Problems/Problem N|200Y AMC 10A #N]]}}</nowiki></code> | style="width:25%" | {{[[Template:AIME box|AIME box]]}} || <code><nowiki>{{AIME box|year=19YY|num-b=|num-a=}}</nowiki></code>
  3 KB (480 words) - 17:43, 19 May 2024
• 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
• User talk:1=2
  ...hem, mark them complete''') if they have been completed. Note that not all problems need solutions for the pages to be correctly formatted. In addition, there' ...toration Project] that is up and running, if you are interested in posting problems rather than formatting pages. Be sure to read the whole topic before you st
  13 KB (1,464 words) - 17:28, 6 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
• User:Gaussian
  [[1983 AIME Problems/Problem 1]] [[1983 AIME Problems/Problem 2]]
  24 KB (2,505 words) - 16:46, 19 March 2014

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