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- {{AIME box|year=2000|n=I|num-b=6|num-a=8}} [[Category:Intermediate Algebra Problems]]5 KB (781 words) - 15:02, 20 April 2024
- {{AIME box|year=2000|n=I|num-b=5|num-a=7}} [[Category:Intermediate Algebra Problems]]6 KB (966 words) - 11:31, 24 June 2024
- {{AIME box|year=2000|n=I|num-b=4|num-a=6}} [[Category:Intermediate Number Theory Problems]]7 KB (1,011 words) - 20:09, 4 January 2024
- {{AIME box|year=2000|n=I|num-b=3|num-a=5}} [[Category:Intermediate Geometry Problems]]3 KB (485 words) - 00:31, 19 January 2024
- In the expansion of <math>(ax + b)^{2000},</math> where <math>a</math> and <math>b</math> are [[relatively prime]] p Using the [[binomial theorem]], <math>\binom{2000}{1998} b^{1998}a^2 = \binom{2000}{1997}b^{1997}a^3 \Longrightarrow b=666a</math>.679 bytes (98 words) - 00:51, 2 November 2023
- {{AIME box|year=2000|n=I|num-b=1|num-a=3}} [[Category:Intermediate Geometry Problems]]3 KB (434 words) - 22:43, 16 May 2021
- {{AIME box|year=2000|n=I|before=First Question|num-a=2}} [[Category:Introductory Number Theory Problems]]1 KB (163 words) - 17:44, 16 December 2020
- {{AIME box|year=2000|n=II|num-b=14|after=Last Question}} [[Category:Intermediate Trigonometry Problems]]3 KB (469 words) - 21:14, 7 July 2022
- ...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!)\\7 KB (1,131 words) - 14:49, 6 April 2023
- {{AIME box|year=2000|n=II|num-b=12|num-a=14}} [[Category:Intermediate Algebra Problems]]6 KB (1,060 words) - 17:36, 26 April 2024
- {{AIME box|year=2000|n=II|num-b=11|num-a=13}} [[Category:Intermediate Geometry Problems]]3 KB (532 words) - 13:14, 22 August 2020
- {{AIME box|year=2000|n=II|num-b=10|num-a=12}} [[Category:Intermediate Geometry Problems]]4 KB (750 words) - 22:55, 5 February 2024
- {{AIME box|year=2000|n=II|num-b=9|num-a=11}} [[Category:Intermediate Geometry Problems]]2 KB (399 words) - 17:37, 2 January 2024
- ...th>, find the least integer that is greater than <math>z^{2000}+\frac 1{z^{2000}}</math>. Using [[De Moivre's Theorem]] we have <math>z^{2000} = \cos 6000^\circ + i\sin 6000^\circ</math>, <math>6000 = 16(360) + 240</m4 KB (675 words) - 13:42, 4 April 2024
- {{AIME box|year=2000|n=II|num-b=7|num-a=9}} [[Category:Intermediate Geometry Problems]]4 KB (584 words) - 19:35, 7 December 2019
- {{AIME box|year=2000|n=II|num-b=6|num-a=8}} [[Category:Intermediate Combinatorics Problems]]2 KB (281 words) - 12:09, 5 April 2024
- {{AIME box|year=2000|n=II|num-b=5|num-a=7}} [[Category:Intermediate Geometry Problems]]3 KB (433 words) - 19:42, 20 December 2021
- {{AIME box|year=2000|n=II|num-b=4|num-a=6}} [[Category:Intermediate Combinatorics Problems]]1 KB (184 words) - 21:13, 12 September 2020
- {{AIME box|year=2000|n=II|num-b=3|num-a=5}} [[Category:Intermediate Number Theory Problems]]2 KB (397 words) - 15:55, 11 May 2022
- {{AIME box|year=2000|n=II|num-b=2|num-a=4}} [[Category:Intermediate Combinatorics Problems]]1 KB (191 words) - 04:27, 4 November 2022
Page text matches
- ...ke the more challenging [[American Invitational Mathematics Examination]] (AIME). The AHSME was replaced with the [[AMC 10]] and [[AMC 12]] in 2000.3 KB (392 words) - 01:53, 4 July 2024
- ...olving.com/Forum/resources.php?c=182 AoPS Resources] section. If you find problems that are in the Resources section which are not in the AoPSWiki, please con | 2024 || [[2024 AIME I | AIME I]] || [[2024 AIME II | AIME II]]3 KB (391 words) - 16:00, 21 February 2024
- <cmath>2000 + 180 m^2 = 10(10+2\sqrt{2}m)^{2} + 1000</cmath> {{AIME box|year=2005|n=I|num-b=14|after=Last Question}}5 KB (906 words) - 23:15, 6 January 2024
- ...I''' problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution. * [[2000 AIME II Problems]]1 KB (139 words) - 08:41, 7 September 2011
- ...I''' problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution. * [[2001 AIME I Problems]]1 KB (139 words) - 08:41, 7 September 2011
- ...I''' problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution. * [[2000 AIME I Problems]]1 KB (135 words) - 18:05, 30 May 2015
- ...E''' problems and solutions. The first link contains the full set of test problems. The rest contain each individual problem and its solution. * [[1999 AIME Problems]]1 KB (118 words) - 08:41, 7 September 2011
- {{AIME Problems|year=1987}} [[1987 AIME Problems/Problem 1|Solution]]6 KB (869 words) - 15:34, 22 August 2023
- {{AIME Problems|year=1992}} [[1992 AIME Problems/Problem 1|Solution]]8 KB (1,117 words) - 05:32, 11 November 2023
- {{AIME Problems|year=1999}} [[1999 AIME Problems/Problem 1|Solution]]7 KB (1,094 words) - 13:39, 16 August 2020
- {{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 (1,212 words) - 22:16, 17 December 2023
- {{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
- ...ch <math>[a,b] = 1000</math>, <math>[b,c] = 2000</math>, and <math>[c,a] = 2000</math>. ...CM#Using prime factorization|looking at the prime factorization]] of <math>2000</math>, <math>c</math> must have a [[factor]] of <math>2^4</math>. If <math3 KB (547 words) - 10:00, 18 June 2024
- <math>n = 1: 2000+60*119 = 9140</math> {{AIME box|year=1989|num-b=10|num-a=12}}5 KB (851 words) - 18:01, 28 December 2022
- 0 &= 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000\\ t &= \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{100}{3}.\end{align*}</cmath>6 KB (980 words) - 15:08, 14 May 2024
- For how many pairs of consecutive integers in <math>\{1000,1001,1002,\ldots,2000\}</math> is no carrying required when the two integers are added? ...ath>. We can see that <math>1100, 1200, 1300, 1400, 1500</math>, and <math>2000</math> all work, giving a grand total of <math>150 + 6 = \boxed{156}</math>3 KB (455 words) - 02:03, 10 July 2021
- ...t on the circle if <math>\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}</math>. ...e that <math>(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{2000}</math>. Therefore, one of <math>1993 - n</math> or <math>1994 + n</math> i3 KB (488 words) - 02:06, 22 September 2023
- ...2} \cdot 1995^{1 - \sqrt{2}/2} = 1995^2</math>, making the solution <math>(2000-5)^2 \equiv \boxed{025} \pmod{1000}</math>. ...+\sqrt{8}}{2}}\right)\left(1995^{\frac{2-\sqrt{8}}{2}}\right)=1995^2=\left(2000-5\right)^2</math>, so the last three digits are <math>\boxed{025}</math>.2 KB (362 words) - 00:40, 29 January 2021
- ...- x</math> || <math>2x - 1000</math><font color="white">a</font> || <math>2000 - 3x</math> || <math>5x - 3000</math> || <math>5000 - 8x</math> #<math>2000 - 3x > 0 \Longrightarrow x < 666.\overline{6}</math>2 KB (354 words) - 19:37, 24 September 2023
