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Page title matches

• 2000 AIME I Problems
  {{AIME Problems|year=2000|n=I}} [[2000 AIME I Problems/Problem 1|Solution]]
  7 KB (1,204 words) - 03:40, 4 January 2023
• 2000 AIME II Problems
  {{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
• 2000 AIME I Problems/Problem 15
  ...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) - 19:16, 6 January 2024
• 2000 AIME I Problems/Problem 14
  {{AIME box|year=2000|n=I|num-b=13|num-a=15|t=721509}} [[Category:Intermediate Geometry Problems]]
  8 KB (1,275 words) - 03:04, 27 February 2022
• 2000 AIME I Problems/Problem 13
  {{AIME box|year=2000|n=I|num-b=12|num-a=14|t=721508}} [[Category:Intermediate Geometry Problems]]
  3 KB (571 words) - 00:38, 13 March 2014
• 2000 AIME I Problems/Problem 12
  {{AIME box|year=2000|n=I|num-b=11|num-a=13}} [[Category:Intermediate Algebra Problems]]
  1 KB (238 words) - 18:50, 10 March 2015
• 2000 AIME I Problems/Problem 11
  {{AIME box|year=2000|n=I|num-b=10|num-a=12}} [[Category:Intermediate Algebra Problems]]
  4 KB (667 words) - 13:58, 31 July 2020
• 2000 AIME I Problems/Problem 10
  {{AIME box|year=2000|n=I|num-b=9|num-a=11}} [[Category:Intermediate Algebra Problems]]
  2 KB (319 words) - 22:26, 29 December 2022
• 2000 AIME I Problems/Problem 9
  <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>
  4 KB (623 words) - 15:56, 8 May 2021
• 2000 AIME I Problems/Problem 8
  {{AIME box|year=2000|n=I|num-b=7|num-a=9}} [[Category:Intermediate Geometry Problems]]
  4 KB (677 words) - 16:33, 30 December 2023
• 2000 AIME I Problems/Problem 7
  {{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
• 2000 AIME I Problems/Problem 6
  {{AIME box|year=2000|n=I|num-b=5|num-a=7}} [[Category:Intermediate Algebra Problems]]
  6 KB (966 words) - 21:48, 29 January 2024
• 2000 AIME I Problems/Problem 5
  {{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
• 2000 AIME I Problems/Problem 4
  {{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
• 2000 AIME I Problems/Problem 3
  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
• 2000 AIME I Problems/Problem 2
  {{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
• 2000 AIME I Problems/Problem 1
  {{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
• 2000 AIME II Problems/Problem 15
  {{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
• 2000 AIME II Problems/Problem 14
  ...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
• 2000 AIME II Problems/Problem 13
  {{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
• 2000 AIME II Problems/Problem 12
  {{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
• 2000 AIME II Problems/Problem 11
  {{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
• 2000 AIME II Problems/Problem 10
  {{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
• 2000 AIME II Problems/Problem 9
  ...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</m
  4 KB (675 words) - 13:42, 4 April 2024
• 2000 AIME II Problems/Problem 8
  {{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
• 2000 AIME II Problems/Problem 7
  {{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
• 2000 AIME II Problems/Problem 6
  {{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
• 2000 AIME II Problems/Problem 5
  {{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
• 2000 AIME II Problems/Problem 4
  {{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
• 2000 AIME II Problems/Problem 3
  {{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
• 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
• 2000 AIME II Problems/Problem 1
  <center><math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math></center> <math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math>
  2 KB (292 words) - 13:33, 4 April 2024

Page text matches

• Wyoming mathematics competitions
  ...ex.php?f=298 middle school math forums] where students can discuss contest problems and mathematics. ...ndex.php?f=214 high school math forums] where students can discuss contest problems and mathematics
  3 KB (428 words) - 20:16, 2 April 2007
• Resources for mathematics competitions
  ...tions]]. Look around the AoPSWiki. Individual articles often have sample problems and solutions for many levels of problem solvers. Many also have links to == Math Competition Problems ==
  16 KB (2,152 words) - 21:46, 6 May 2024
• American Invitational Mathematics Examination
  ...ed States at the [[International Mathematics Olympiad]] (IMO). While most AIME participants are high school students, some bright middle school students a High scoring AIME students are invited to take the prestigious [[United States of America Mat
  8 KB (1,057 words) - 12:02, 25 February 2024
• American Mathematics Competitions
  * [[American Invitational Mathematics Examination]] (AIME) &mdash; high scorers from the AMC 10/12 exams. * [[United States of America Mathematics Olympiad]] (USAMO) &mdash; high AIME and AMC scorers.
  5 KB (696 words) - 03:47, 24 December 2019
• Simon's Favorite Factoring Trick
  This is important because this keeps showing up in number theory problems. Let's look at this problem below: This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually <math>x</math> a
  7 KB (1,107 words) - 07:35, 26 March 2024
• Casework
  ...emonstrate casework in action. Unlike the selections in this article, most problems cannot be completely solved through casework. However, it is crucial as an While there are problems where casework produces the most elegant solution, in those where a shorter
  5 KB (709 words) - 10:28, 19 February 2024
• Prime factorization
  ==Problems== * Practice Problems from [https://artofproblemsolving.com/alcumus/ Alcumus]
  3 KB (496 words) - 22:14, 5 January 2024
• Fibonacci sequence
  ==Problems== ...} </math><div style="text-align:right">([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])</div>
  6 KB (957 words) - 23:49, 7 March 2024
• Complex number
  == Problems == *[[2007 AMC 12A Problems/Problem 18]]
  5 KB (860 words) - 15:36, 10 December 2023
• Floor function
  ...the leftover stuff after the decimal point. This can greatly simplify many problems. == Problems ==
  3 KB (508 words) - 21:05, 26 February 2024
• American High School Mathematics Examination
  ...ke the more challenging [[American Invitational Mathematics Examination]] (AIME). The AHSME was replaced with the [[AMC 10]] and [[AMC 12]] in 2000.
  3 KB (388 words) - 23:07, 5 February 2024
• AIME Problems and Solutions
  ...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
• 2005 AIME I Problems/Problem 15
  <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
• 2000 AIME II
  ...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
• 2001 AIME I
  ...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
• 2000 AIME I
  ...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
• 1999 AIME
  ...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
• 1987 AIME Problems
  {{AIME Problems|year=1987}} [[1987 AIME Problems/Problem 1|Solution]]
  6 KB (869 words) - 15:34, 22 August 2023
• 1992 AIME Problems
  {{AIME Problems|year=1992}} [[1992 AIME Problems/Problem 1|Solution]]
  8 KB (1,117 words) - 05:32, 11 November 2023
• 1999 AIME Problems
  {{AIME Problems|year=1999}} [[1999 AIME Problems/Problem 1|Solution]]
  7 KB (1,094 words) - 13:39, 16 August 2020
• 2000 AIME I Problems
  {{AIME Problems|year=2000|n=I}} [[2000 AIME I Problems/Problem 1|Solution]]
  7 KB (1,204 words) - 03:40, 4 January 2023
• 2001 AIME I Problems
  {{AIME Problems|year=2001|n=I}} [[2001 AIME I Problems/Problem 1|Solution]]
  7 KB (1,212 words) - 22:16, 17 December 2023
• 2000 AIME II Problems
  {{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
• 1987 AIME Problems/Problem 7
  ...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 <math
  3 KB (547 words) - 22:54, 4 April 2016
• 1989 AIME Problems/Problem 11
  <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
• 1989 AIME Problems/Problem 6
  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>
  5 KB (864 words) - 19:55, 2 July 2023
• 1992 AIME Problems/Problem 6
  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
• 1993 AIME Problems/Problem 9
  ...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> i
  3 KB (488 words) - 02:06, 22 September 2023
• 1995 AIME Problems/Problem 2
  ...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
• 1998 AIME Problems/Problem 8
  ...- 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
• 2000 AIME I Problems/Problem 14
  {{AIME box|year=2000|n=I|num-b=13|num-a=15|t=721509}} [[Category:Intermediate Geometry Problems]]
  8 KB (1,275 words) - 03:04, 27 February 2022
• 2000 AIME I Problems/Problem 13
  {{AIME box|year=2000|n=I|num-b=12|num-a=14|t=721508}} [[Category:Intermediate Geometry Problems]]
  3 KB (571 words) - 00:38, 13 March 2014
• 2000 AIME I Problems/Problem 12
  {{AIME box|year=2000|n=I|num-b=11|num-a=13}} [[Category:Intermediate Algebra Problems]]
  1 KB (238 words) - 18:50, 10 March 2015
• 2000 AIME I Problems/Problem 11
  {{AIME box|year=2000|n=I|num-b=10|num-a=12}} [[Category:Intermediate Algebra Problems]]
  4 KB (667 words) - 13:58, 31 July 2020
• 2000 AIME I Problems/Problem 10
  {{AIME box|year=2000|n=I|num-b=9|num-a=11}} [[Category:Intermediate Algebra Problems]]
  2 KB (319 words) - 22:26, 29 December 2022
• 2000 AIME I Problems/Problem 9
  <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>
  4 KB (623 words) - 15:56, 8 May 2021
• 2000 AIME I Problems/Problem 8
  {{AIME box|year=2000|n=I|num-b=7|num-a=9}} [[Category:Intermediate Geometry Problems]]
  4 KB (677 words) - 16:33, 30 December 2023
• 2000 AIME I Problems/Problem 7
  {{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
• 2000 AIME I Problems/Problem 6
  {{AIME box|year=2000|n=I|num-b=5|num-a=7}} [[Category:Intermediate Algebra Problems]]
  6 KB (966 words) - 21:48, 29 January 2024
• 2000 AIME I Problems/Problem 5
  {{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
• 2000 AIME I Problems/Problem 4
  {{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
• 2000 AIME I Problems/Problem 3
  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
• 2000 AIME I Problems/Problem 2
  {{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
• 2000 AIME I Problems/Problem 1
  {{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
• 2001 AIME I Problems/Problem 8
  ...> in base <math>10</math>, it must be equal to <math>2A</math>, so <math>B<2000</math> when <math>B</math> is looked at in base <math>10.</math> If <math>B</math> in base <math>10</math> is less than <math>2000</math>, then <math>B</math> as a number in base <math>7</math> must be less
  3 KB (502 words) - 11:28, 9 December 2023
• 2001 AIME I Problems/Problem 3
  ...{2001}+-x^{2001}=0</math>, so the term with the largest degree is <math>x^{2000}</math>. So we need the coefficient of that term, as well as the coefficien ...{2001}{1} \cdot (-x)^{2000} \cdot \left(\frac{1}{2}\right)^1&=\frac{2001x^{2000}}{2}\\
  2 KB (335 words) - 18:38, 9 February 2023
• 2001 AIME I Problems/Problem 2
  ...begin{align*}\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1} \Longrightarrow \frac{2000}{a+1}=40 \Longrightarrow a=49\end{align*}</cmath> {{AIME box|year=2001|n=I|num-b=1|num-a=3}}
  1 KB (225 words) - 07:57, 4 November 2022
• 2000 AIME II Problems/Problem 15
  {{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
• 2000 AIME II Problems/Problem 14
  ...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
• 2000 AIME II Problems/Problem 13
  {{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
• 2000 AIME II Problems/Problem 12
  {{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
• 2000 AIME II Problems/Problem 11
  {{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
• 2000 AIME II Problems/Problem 10
  {{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
• 2000 AIME II Problems/Problem 9
  ...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</m
  4 KB (675 words) - 13:42, 4 April 2024
• 2000 AIME II Problems/Problem 8
  {{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
• 2000 AIME II Problems/Problem 7
  {{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
• 2000 AIME II Problems/Problem 6
  {{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
• 2000 AIME II Problems/Problem 5
  {{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
• 2000 AIME II Problems/Problem 4
  {{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
• 2000 AIME II Problems/Problem 3
  {{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
• 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
• 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
• 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
• 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) - 21:33, 3 January 2021
• 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 (1,147 words) - 21:58, 23 January 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
• Olympiad Archive
  ...ww.oma.org.ar/enunciados/index.htm ''Full Archive of Mathematical Olympiad Problems''] ** Problems w/o Solutions
  16 KB (1,987 words) - 11:39, 17 February 2024
• 2019 AMC 10C Problems
  Here are the problems from the 2019 AMC 10C, a mock contest created by the AoPS user fidgetboss_4 [[2019 AMC 10C Problems/Problem 1|Solution]]
  12 KB (1,917 words) - 12:14, 29 November 2021
• 2019 AIME II Problems/Problem 9
  ...<math>d(k) = 1</math>, and there is one solution <math>n = 2^4 \cdot 5^3 = 2000</math>. Then we have <math>\frac{S}{20} = \frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \boxed{472}</math>.
  4 KB (609 words) - 14:46, 4 January 2024
• 2020 AIME II Problems
  {{AIME Problems|year=2020|n=II}} [[2020 AIME II Problems/Problem 1 | Solution]]
  7 KB (1,199 words) - 16:17, 12 March 2021
• 2020 AIME II Problems/Problem 3
  ...equation in, we get <math>8^n\cdot3^{20}=3^{2020}</math>, so <math>8^n=3^{2000}</math>. Taking the 100th root, we get <math>8^{\frac{n}{100}}=3^{20}</math {{AIME box|year=2020|n=II|num-b=2|num-a=4}}
  4 KB (502 words) - 04:23, 23 January 2023
• Know the Facts: Gmaas
  - 2000 - 2005: Reconstruction of Newton's Principia Mathematica Gmaasis. (In Engli ...roblem. Therefore, I proved that you cannot use the Gmaas theorem to solve problems. But using theorem 3.141592653589793238462643383 "2. Gmaas can turn things
  88 KB (14,928 words) - 13:54, 29 April 2024
• Mathboy282
  Proudest of: [[2019 AIME II Problems/Problem 15]] Solution 5 [[2023 USAJMO Problems/Problem 6]] Solution 1
  10 KB (1,118 words) - 05:33, 13 January 2024