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- '''2000 AIME I''' problems and solutions. The first link contains the full set of test pr * [[2000 AIME I Problems]]1 KB (135 words) - 17:05, 30 May 2015
- {{AIME Problems|year=2000|n=I}} [[2000 AIME I Problems/Problem 1|Solution]]7 KB (1,204 words) - 02:40, 4 January 2023
- ...he labels on the cards are now in ascending order: <math>1,2,3,\ldots,1999,2000.</math> In the original stack of cards, how many cards were above the card ...e cards such that they go to their correct starting positions in the <math>2000</math> card case, where all of them are below <math>1999</math>. From this,15 KB (2,673 words) - 18:16, 6 January 2024
- ...2</math>. Since <math>4x<180</math>, this implies that <math>3x=30</math>, i.e. <math>x=10</math>. Thus <math>y=70</math> and <cmath>r=\frac{10+70}{2\cd {{AIME box|year=2000|n=I|num-b=13|num-a=15|t=721509}}9 KB (1,505 words) - 22:17, 31 December 2024
- ...(\frac{35}{31},\frac{35}{31}\right)</math>. The bounded region in Quadrant I is made up of a square and two triangles. <math>A=x^2+x(5-x)=5x</math>. By ...irie, normally I would go the straight line distance. Right? However, here I would choose to travel on the highway the greatest distance, then drive the6 KB (1,042 words) - 14:08, 1 January 2025
- {{AIME box|year=2000|n=I|num-b=11|num-a=13}}5 KB (921 words) - 15:36, 28 November 2024
- <math>\sum_{i | 10^6}^{} \frac{i}{1000}</math> ...the first list, list powers of 2 and 5 from 0 to 3. In this specific case I find it easier to augment every denominator to 1000 and then divide by 10006 KB (867 words) - 11:49, 17 November 2024
- {{AIME box|year=2000|n=I|num-b=9|num-a=11}}2 KB (319 words) - 21:26, 29 December 2022
- <cmath>\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &=& 3 - \log 2000\ ...note from different author: <math>-(3 - \log 2000) = \log 2000 - 3 = \log 2000 - \log 1000 = \log 2.</math>5 KB (812 words) - 21:19, 27 November 2024
- {{AIME box|year=2000|n=I|num-b=7|num-a=9}}4 KB (677 words) - 15:33, 30 December 2023
- ...</math> Notice that the LHS of this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus <math Erm, this is very similar to 2000 AMC 12 Q20 ackthually5 KB (766 words) - 23:46, 8 November 2024
- {{AIME box|year=2000|n=I|num-b=5|num-a=7}}6 KB (970 words) - 09:17, 11 January 2025
- {{AIME box|year=2000|n=I|num-b=4|num-a=6}}7 KB (1,011 words) - 19:09, 4 January 2024
- {{AIME box|year=2000|n=I|num-b=3|num-a=5}}3 KB (485 words) - 23:31, 18 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) - 23:51, 1 November 2023
- {{AIME box|year=2000|n=I|num-b=1|num-a=3}}4 KB (735 words) - 13:31, 22 October 2024
- {{AIME box|year=2000|n=I|before=First Question|num-a=2}}1 KB (174 words) - 11:31, 3 January 2025
- Return to [[2000 AIME I]] ([[2000 AIME I Problems]]) ...|before=[[1999 AIME Answer Key|1999 AIME]]|after=[[2000 AIME II Answer Key|2000 AIME II]]}}261 bytes (24 words) - 18:17, 27 May 2016
Page text matches
- ...eamnyc.weebly.com/mathcounts.html List of all MathCounts competitions from 2000 to 2017]. * [http://www.artofproblemsolving.com/community/c5h447454 What should I be doing?]17 KB (2,329 words) - 04:01, 3 February 2025
- ==== AIME I (February 6, 2025)==== ====AIME I (February 1, 2024)====19 KB (2,028 words) - 03:31, 22 February 2025
- In 2000, the AHSME was split into the AMC 10 and 12, reduced to only 25 questions, * The Art of Problem Solving Volume I by [[Sandor Lehoczky]] and [[Richard Rusczyk]]. [http://www.artofproblemso5 KB (696 words) - 02:47, 24 December 2019
- * [[2001 AIME I Problems/Problem 13]] * [[2000 AIME I Problems/Problem 14]]1 KB (179 words) - 18:41, 3 January 2025
- 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) - 21:22, 15 January 2025
- where <math>n</math> is any natural number, the <math>p_{i}</math> are prime numbers, and the <math>e_i</math> are their positive inte * [[2000 AIME I Problems/Problem 1]]3 KB (496 words) - 21:14, 5 January 2024
- ...} </math><div style="text-align:right">([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])</div> ...e [[probability]] that heads never occur on consecutive tosses. Find <math>i+j_{}^{}</math>. <div style="text-align:right">([[1990 AIME Problems/Problem7 KB (1,111 words) - 13:57, 24 June 2024
- ...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) - 14:36, 10 December 2023
- The AHSME was replaced with the [[AMC 10]] and [[AMC 12]] in 2000. * The Art of Problem Solving Volume I by [[Sandor Lehoczky]] and [[Richard Rusczyk]]. [http://www.artofproblemso3 KB (390 words) - 11:01, 15 October 2024
- ! Year || Test I || Test II | 2025 || [[2025 AIME I | AIME I]] || [[2025 AIME II | AIME II]]3 KB (400 words) - 19:21, 13 February 2025
- pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C); <cmath>2000 + 180 m^2 = 10(10+2\sqrt{2}m)^{2} + 1000</cmath>5 KB (906 words) - 22:15, 6 January 2024
- '''2000 AIME II''' problems and solutions. The first link contains the full set of * [[2000 AIME II Problems]]1 KB (139 words) - 07:41, 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) - 07:41, 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) - 17:05, 30 May 2015
- {{AIME box|year=1999|before=[[1998 AIME]]|after=[[2000 AIME I]], [[2000 AIME II|II]]}}1 KB (118 words) - 07:41, 7 September 2011
- ...d so are all the other switches whose labels divide the label on the <math>i</math>-th switch. After step 1000 has been completed, how many switches wi {{AIME box|year = 1999|before=[[1998 AIME Problems]]|after=[[2000 AIME I Problems]]}}7 KB (1,094 words) - 12:39, 16 August 2020
- {{AIME Problems|year=2000|n=I}} [[2000 AIME I Problems/Problem 1|Solution]]7 KB (1,204 words) - 02:40, 4 January 2023
- {{AIME Problems|year=2001|n=I}} [[2001 AIME I Problems/Problem 1|Solution]]7 KB (1,220 words) - 13:05, 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) - 20:11, 19 February 2019
- ...he labels on the cards are now in ascending order: <math>1,2,3,\ldots,1999,2000.</math> In the original stack of cards, how many cards were above the card ...e cards such that they go to their correct starting positions in the <math>2000</math> card case, where all of them are below <math>1999</math>. From this,15 KB (2,673 words) - 18:16, 6 January 2024