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Create the page "2000 AIME I, II" on this wiki! See also the search results found.
- ...AMC historical results''' page. This page should include results for the [[AIME]] as well. For [[USAMO]] results, see [[USAMO historical results]]. == AIME Qualification ==21 KB (2,347 words) - 02:23, 12 March 2025
- * [[American Invitational Mathematics Examination]] (AIME) — high scorers from the AMC 10/12 exams. * [[United States of America Mathematics Olympiad]] (USAMO) — high AIME and AMC scorers.5 KB (696 words) - 03:47, 24 December 2019
- For <math>n,r\in\mathbb{N}, n>r,\sum^n_{i=r}{i\choose r}={n+1\choose r+1}</math>. for(int i=0;i<r;++i){12 KB (1,993 words) - 22:22, 15 January 2025
- ...ver, it is possible to define a number, <math> i </math>, such that <math> i = \sqrt{-1} </math>. If we add this new number to the reals, we will have ...rm <math> a + bi </math> where <math> a,b\in \mathbb{R} </math> and <math> i = \sqrt{-1} </math> is the [[imaginary unit]]. The set of complex numbers i5 KB (860 words) - 15:36, 10 December 2023
- ...ke the more challenging [[American Invitational Mathematics Examination]] (AIME). The AHSME was replaced with the [[AMC 10]] and [[AMC 12]] in 2000.3 KB (390 words) - 12:01, 15 October 2024
- This is a list of all [[AIME]] exams in the AoPSWiki. Many of these problems and solutions are also ava ! Year || Test I || Test II3 KB (400 words) - 20:21, 13 February 2025
- '''2000 AIME II''' problems and solutions. The first link contains the full set of test pr * [[2000 AIME II Problems]]1 KB (139 words) - 08:41, 7 September 2011
- '''2001 AIME I''' problems and solutions. The first link contains the full set of test pr * [[2001 AIME I Problems]]1 KB (139 words) - 08:41, 7 September 2011
- '''2000 AIME I''' problems and solutions. The first link contains the full set of test pr * [[2000 AIME I Problems]]1 KB (135 words) - 18:05, 30 May 2015
- '''1999 AIME''' problems and solutions. The first link contains the full set of test pr * [[1999 AIME Problems]]1 KB (118 words) - 08:41, 7 September 2011
- {{AIME Problems|year=2000|n=I}} [[2000 AIME I Problems/Problem 1|Solution]]7 KB (1,204 words) - 03:40, 4 January 2023
- {{AIME Problems|year=2001|n=I}} [[2001 AIME I Problems/Problem 1|Solution]]7 KB (1,220 words) - 14:05, 24 November 2024
- {{AIME Problems|year=2000|n=II}} <center><math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math></center>6 KB (947 words) - 21:11, 19 February 2019
- ...s the factorial base expansion of <math>16!-32!+48!-64!+\cdots+1968!-1984!+2000!</math>, find the value of <math>f_1-f_2+f_3-f_4+\cdots+(-1)^{j+1}f_j</math 16!-32!+48!-64!+\cdots+1968!-1984!+2000!&=16!+(48!-32!)+(80!-64!)\cdots+(2000!-1984!)\9 KB (1,398 words) - 00:34, 21 January 2025
- ...cients were symmetrical. What if we make the substitution <math>x = -\frac{i}{\sqrt{10}}y</math>? Then the equation becomes ...}{\sqrt{10}}\right)y^5 + \left(\frac{i}{\sqrt{10}}\right)y^3 - \left(\frac{i}{\sqrt{10}}\right)y - 2 = 0</math>.7 KB (1,098 words) - 00:33, 21 January 2025
- ...th>, find the least integer that is greater than <math>z^{2000}+\frac 1{z^{2000}}</math>. ...we have <math>z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}</math>.4 KB (671 words) - 15:36, 20 January 2025
- ...th integers. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without remainder; an odd number is an integer that is n <b>2000 AIME II Problem 2:</b>4 KB (694 words) - 22:00, 12 January 2024
- :(i) <math>n^2</math> can be expressed as the difference of two consecutive cub :(ii) <math>2n + 79</math> is a perfect square.4 KB (628 words) - 16:23, 2 January 2024
- Return to [[2000 AIME I]] ([[2000 AIME I Problems]]) ...1999 AIME Answer Key|1999 AIME]]|after=[[2000 AIME II Answer Key|2000 AIME II]]}}261 bytes (24 words) - 19:17, 27 May 2016
- Return to [[2000 AIME II]] ([[2000 AIME II Problems]]) ...0 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
