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• 2004 AIME I
  * [[2004 AIME I Problems/Problem 4]] {{AIME box|year=2004|n=I|before=[[2003 AIME I]], [[2003 AIME II|II]]|after=[[2004 AIME II]]}}
  1 KB (135 words) - 18:15, 19 April 2021
• Collatz Problem
  ...sitive integer]] <math>n</math>, the sequence <math>\{n,f(n),f(f(n)),f(f(f(n))),\ldots\}</math> contains 1. This conjecture is still open. Some people h ==Properties of <math>f(n)</math> ==
  1 KB (231 words) - 19:45, 24 February 2020
• 2004 AIME II
  * [[2004 AIME II Problems/Problem 4]] {{AIME box|year=2004|n=II|before=[[2004 AIME I]]|after=[[2005 AIME I]], [[2005 AIME II|II]]}}
  1 KB (135 words) - 12:24, 22 March 2011
• 2005 AIME I
  * [[2005 AIME I Problems/Problem 4 | Problem 4]] {{AIME box|year=2005|n=I|before=[[2004 AIME I|2004 AIME I]], [[2004 AIME II|II]]|after=[[2005 AIME
  1 KB (154 words) - 12:30, 22 March 2011
• 2006 AIME I
  * [[2006 AIME I Problems/Problem 4]] {{AIME box|year=2006|n=I|before=[[2005 AIME I]], [[2005 AIME II|II]]|after=[[2006 AIME II]]}}
  1 KB (135 words) - 12:31, 22 March 2011
• 2005 AIME II
  * [[2005 AIME II Problems/Problem 4]] {{AIME box|year=2005|n=II|before=[[2005 AIME I]]|after=[[2006 AIME I]], [[2006 AIME II|II]]}}
  1 KB (135 words) - 12:30, 22 March 2011
• Mock AMC
  | n/a | n/a
  51 KB (6,175 words) - 20:58, 6 December 2023
• Bertrand's Postulate
  ...em. A more widely known version states that there is a prime between <math>n</math> and <math>2n</math>. ...closer look at the [[combinations|binomial coefficient]] <math>\binom{2n}{n}</math>. Assuming that the reader is familiar with that proof, the Bertrand
  2 KB (309 words) - 21:43, 11 January 2010
• Riemann zeta function
  <cmath>\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}= 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots</cmath>
  9 KB (1,547 words) - 03:04, 13 January 2021
• Modular arithmetic/Introduction
  draw((0,0), linewidth(4)); <math>1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, \ldots </math>
  15 KB (2,396 words) - 20:24, 21 February 2024
• Perfect square
  ...so that <math>m^2=n</math>. The first few perfect squares are <math>0, 1, 4, 9, 16, 25, 36</math>. ...th>n</math> square numbers (starting with <math>1</math>) is <math>\frac{n(n+1)(2n+1)}{6}</math>
  954 bytes (155 words) - 01:14, 29 November 2023
• Modular arithmetic/Intermediate
  ...th>n</math>), if the difference <math>{a - b}</math> is divisible by <math>n</math>. ...ntegers modulo <math>n</math>''' (usually known as "the integers mod <math>n</math>," or <math>\mathbb{Z}_n</math> for short). This structure gives us
  14 KB (2,317 words) - 19:01, 29 October 2021
• 2006 AIME I Problems
  {{AIME Problems|year=2006|n=I}} == Problem 4 ==
  7 KB (1,173 words) - 03:31, 4 January 2023
• 2006 AIME I Problems/Problem 15
  === Solution 4 === {{AIME box|year=2006|n=I|num-b=14|after=Last Question}}
  6 KB (910 words) - 19:31, 24 October 2023
• 2006 AIME I Problems/Problem 14
  ...> is not divisible by the square of any prime. Find <math> \lfloor m+\sqrt{n}\rfloor. </math> (The notation <math> \lfloor x\rfloor </math> denotes the triple M=(B+C)/2,S=(4*A+T)/5;
  6 KB (980 words) - 21:45, 31 March 2020
• 2006 AIME I Problems/Problem 13
  ...h> S_n=\sum_{k=1}^{2^{n-1}}g(2k). </math> Find the greatest integer <math> n </math> less than 1000 such that <math> S_n </math> is a [[perfect square]] .../math> but not by <math>2^n, \ldots,</math> and <math>2^{n-1}-2^{n-2} = 2^{n-2}</math> elements of <math>S</math> that are divisible by <math>2^1</math>
  10 KB (1,702 words) - 00:45, 16 November 2023
• 2006 AIME I Problems/Problem 10
  ...ough the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra {{AIME box|year=2006|n=I|num-b=9|num-a=11}}
  4 KB (731 words) - 17:59, 4 January 2022
• 2006 AIME I Problems/Problem 8
  pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1, label("$\mathcal{Q}$",(4.2,-1),NW);
  5 KB (730 words) - 15:05, 15 January 2024
• 2006 AIME I Problems/Problem 7
  for(int i=0; i<4; i=i+1) { pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10);
  4 KB (709 words) - 01:50, 10 January 2022
• 2006 AIME I Problems/Problem 5
  <math> b = 4 </math> <math>b=4</math>
  3 KB (439 words) - 18:24, 10 March 2015

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