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• 2016 AMC 10A Problems/Problem 6
  == Problem ==
  967 bytes (143 words) - 03:18, 27 June 2023
• 1970 IMO Problems/Problem 6
  ==Problem==
  1 KB (234 words) - 19:26, 14 July 2017
• 2006 AIME I Problems/Problem 6
  == Problem ==
  2 KB (237 words) - 19:14, 20 November 2023
• 2006 AMC 12B Problems/Problem 6
  == Problem ==
  978 bytes (145 words) - 13:57, 4 December 2015
• 2006 AMC 12A Problems\Problem 6
  #REDIRECT [[2006 AMC 12A Problems/Problem 6]]
  45 bytes (5 words) - 14:43, 14 January 2016
• 2005 AMC 12B Problems/Problem 6
  {{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #6]] and [[2005 AMC 10B Problems|2005 AMC 10B #10]]}} == Problem ==
  2 KB (299 words) - 15:29, 5 July 2022
• 1995 AHSME Problems/Problem 6
  ==Problem==
  1 KB (168 words) - 00:49, 14 October 2013
• 2006 AMC 10A Problems/Problem 6
  == Problem ==
  590 bytes (84 words) - 14:28, 31 May 2023
• 2006 AIME A Problems/Problem 6
  #REDIRECT [[2006 AIME I Problems/Problem 6]]
  44 bytes (5 words) - 12:05, 28 June 2009
• 2005 AIME II Problems/Problem 6
  == Problem == ...xample, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical st
  2 KB (384 words) - 00:31, 26 July 2018
• 2006 AMC 12A Problem 6
  #REDIRECT [[2006 AMC 12A Problems/Problem 6]]
  45 bytes (5 words) - 11:01, 20 February 2016
• 2005 AIME I Problems/Problem 6
  == Problem == ==Solution 6 (De Moivre's Theorem)==
  4 KB (686 words) - 01:55, 5 December 2022
• 2004 AIME I Problems/Problem 6
  == Problem == We divide the problem into two cases: one in which zero is one of the digits and one in which it
  3 KB (562 words) - 18:12, 4 March 2022
• 2004 AIME II Problems/Problem 6
  == Problem == ...al. Also note some other conditions we have picked up in the course of the problem, namely that <math>b_1</math> is divisible by <math>8</math>, <math>b_2</ma
  6 KB (950 words) - 14:18, 15 January 2024
• 1983 AIME Problems/Problem 6
  == Problem == Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math
  3 KB (361 words) - 20:20, 14 January 2023
• 1984 AIME Problems/Problem 6
  == Problem == [[Image:1984_AIME-6.png]]
  6 KB (1,022 words) - 19:29, 22 January 2024
• 1985 AIME Problems/Problem 6
  == Problem == [[Image:AIME 1985 Problem 6.png]]
  5 KB (789 words) - 03:09, 23 January 2023
• 1986 AIME Problems/Problem 6
  == Problem ==
  2 KB (336 words) - 14:13, 6 September 2020
• 1987 AIME Problems/Problem 6
  == Problem ==
  3 KB (530 words) - 07:46, 1 June 2018
• 1988 AIME Problems/Problem 6
  == Problem == [[Image:1988_AIME-6.png]]
  5 KB (878 words) - 23:06, 20 November 2023
• 1989 AIME Problems/Problem 6
  == Problem == pair A=(0,0),B=(10,0),C=6*expi(pi/3);
  5 KB (864 words) - 19:55, 2 July 2023
• 1990 AIME Problems/Problem 6
  == Problem ==
  2 KB (325 words) - 13:16, 26 June 2022
• 2006 AMC 10B Problems/Problem 6
  == Problem ==
  1 KB (181 words) - 13:45, 26 January 2022
• 1991 AIME Problems/Problem 6
  == Problem ==
  3 KB (447 words) - 17:02, 24 November 2023
• 1992 AIME Problems/Problem 6
  == Problem == ...math>h+d\ge 10</math> or <math>c+g\ge 10</math> or <math>b+f\ge 10</math>. 6. Consider <math>c \in \{0, 1, 2, 3, 4\}</math>. <math>1abc + 1ab(c+1)</mat
  3 KB (455 words) - 02:03, 10 July 2021
• 1993 AIME Problems/Problem 6
  == Problem ==
  3 KB (524 words) - 18:06, 9 December 2023
• 1994 AIME Problems/Problem 6
  == Problem == ...}\right)^2 = 100</math>. Thus, the total number of unit triangles is <math>6 \times 100 = 600</math>.
  4 KB (721 words) - 16:14, 8 March 2021
• 1995 AIME Problems/Problem 6
  == Problem == Incorporating this into our problem gives <math>19\times31=\boxed{589}</math>.
  2 KB (407 words) - 08:14, 4 November 2022
• 1996 AIME Problems/Problem 6
  == Problem == ...rical to the first one. Therefore, there are <math>5 \cdot 2^6 + 5 \cdot 2^6</math> ways for an undefeated or winless team.
  3 KB (461 words) - 01:00, 19 June 2019
• 1997 AIME Problems/Problem 6
  == Problem == [[Image:1997_AIME-6.png]]
  3 KB (497 words) - 00:39, 22 December 2018
• 1998 AIME Problems/Problem 6
  == Problem == [[Image:AIME_1998-6.png|350px]]
  2 KB (254 words) - 19:38, 4 July 2013
• 1999 AIME Problems/Problem 6
  == Problem == [[Image:1999_AIME-6.png]]
  2 KB (354 words) - 16:42, 20 July 2021
• 2000 AIME I Problems/Problem 6
  == Problem == ...]s <math>(x,y)</math> of [[integer]]s is it true that <math>0 < x < y < 10^6</math> and that the [[arithmetic mean]] of <math>x</math> and <math>y</math
  6 KB (966 words) - 21:48, 29 January 2024
• 2001 AIME I Problems/Problem 6
  == Problem == Recast the problem entirely as a block-walking problem. Call the respective dice <math>a, b, c, d</math>. In the diagram below,
  11 KB (1,729 words) - 20:50, 28 November 2023
• 2002 AIME I Problems/Problem 6
  == Problem == ...{225}(x_2)=(3+\sqrt{5})+(3-\sqrt{5})=6</math> <math>\Rightarrow x_1x_2=225^6=15^{12}</math>.
  1 KB (194 words) - 19:55, 23 April 2016
• 2003 AIME I Problems/Problem 6
  == Problem == ...h>{4\choose 3} = 4</math> triangles of the first type, and there are <math>6</math> faces, so there are <math>24</math> triangles of the first type. Eac
  3 KB (477 words) - 18:35, 27 December 2021
• 2003 AIME II Problems/Problem 6
  == Problem == G=(6.3333,4);
  5 KB (787 words) - 17:38, 30 July 2022
• 2002 AIME II Problems/Problem 6
  == Problem == ...ve terms are <math>1,2,..,9998</math>, while the negative ones are <math>5,6,...,10002</math>. Hence we are left with <math>1000 \cdot \frac{1}{4} (1 +
  2 KB (330 words) - 05:56, 23 August 2022
• 2001 AIME II Problems/Problem 6
  == Problem == ...a = b</math> would imply <math>m = n</math>, and <math>m < n</math> in the problem, we must use the other factor. We get <math>b = 2/5a</math>, meaning the ra
  4 KB (772 words) - 19:31, 6 December 2023
• 2000 AIME II Problems/Problem 6
  == Problem ==
  3 KB (433 words) - 19:42, 20 December 2021
• 2005 AMC 10B Problems/Problem 6
  #REDIRECT[[2005 AMC 12B Problems/Problem 4]]
  44 bytes (5 words) - 10:51, 29 June 2011
• 2006 USAMO Problems/Problem 6
  == Problem ==
  5 KB (986 words) - 22:46, 18 May 2015
• 2006 AMC 12A Problems/Problem 6
  ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}} == Problem ==
  3 KB (528 words) - 18:29, 7 May 2024
• 2000 AMC 12 Problems/Problem 6
  {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #6]] and [[2000 AMC 10 Problems|2000 AMC 10 #11]]}} ==Problem==
  1 KB (228 words) - 19:31, 29 April 2024
• University of South Carolina High School Math Contest/1993 Exam/Problem 6
  == Problem == ...ty of South Carolina High School Math Contest/1993 Exam/Problem 5|Previous Problem]]
  2 KB (299 words) - 21:06, 5 July 2017
• Mock AIME 1 2006-2007 Problems/Problem 6
  ==Problem== *[[Mock AIME 1 2006-2007 Problems/Problem 5 | Previous Problem]]
  3 KB (460 words) - 15:52, 3 April 2012
• Mock AIME 2 2006-2007 Problems/Problem 6
  == Problem ==
  1 KB (157 words) - 10:51, 4 April 2012
• 1959 IMO Problems/Problem 6
  == Problem ==
  2 KB (383 words) - 05:58, 11 February 2024
• 1969 Canadian MO Problems/Problem 6
  == Problem ==
  2 KB (275 words) - 20:33, 27 November 2023
• 2005 AMC 10A Problems/Problem 6
  ==Problem==
  909 bytes (130 words) - 19:09, 25 December 2022
• 2015 IMO Problems/Problem 6
  ==Problem==
  4 KB (833 words) - 01:33, 31 December 2019
• 1987 IMO Problems/Problem 6
  ==Problem==
  2 KB (430 words) - 13:03, 24 February 2024
• 1981 IMO Problems/Problem 6
  == Problem == ...<math>x=2</math>. <cmath>f(1,0)=2, f(1,1)=3, f(1,2)=4, f(1,3)=5, f(1,4)=6</cmath> This pattern can also be proved using induction. The pattern seems
  2 KB (306 words) - 18:15, 12 April 2024
• 2003 AMC 10A Problems/Problem 6
  #REDIRECT[[2003 AMC 12A Problems/Problem 6]]
  44 bytes (5 words) - 14:44, 30 July 2011
• 2003 AMC 12A Problems/Problem 6
  ...C 12A Problems|2003 AMC 12A #6]] and [[2003 AMC 10A Problems|2003 AMC 10A #6]]}} == Problem ==
  1 KB (210 words) - 15:38, 19 August 2023
• Mock AIME 3 Pre 2005 Problems/Problem 6
  ==Problem==
  3 KB (501 words) - 14:48, 29 November 2019
• 2007 AIME I Problems/Problem 6
  == Problem == ...lid moves, beginning with 0 and ending with 39. For example, <math>0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39</math> is a move sequence. How many move sequences are
  10 KB (1,519 words) - 00:11, 29 November 2023
• 2007 AIME II Problems/Problem 6
  == Problem == ...th> in the second column, we note that <math>3</math> is less than <math>4,6,8</math>, but greater than <math>1</math>, so there are four possible place
  2 KB (338 words) - 15:30, 7 August 2022
• 2007 USAMO Problems/Problem 6
  == Problem == label("$\omega_A$",p_a+x*abs(O-A)*expi(pi/6), (1,1));
  7 KB (1,274 words) - 15:11, 31 August 2017
• 2007 AMC 12B Problems/Problem 6
  ==Problem== Triangle <math>ABC</math> has side lengths <math>AB = 5</math>, <math>BC = 6</math>, and <math>AC = 7</math>. Two bugs start simultaneously from <math>A
  792 bytes (121 words) - 04:21, 15 December 2020
• 2007 AMC 12A Problems/Problem 6
  ...cate|[[2007 AMC 12A Problems|2007 AMC 12A #6]] and [[2007 AMC 10A Problems/Problem 8|2007 AMC 10A #8]]}} ==Problem==
  2 KB (265 words) - 00:20, 30 October 2022
• 2005 AMC 12A Problems/Problem 6
  == Problem == (\mathrm {A}) \ 4 \qquad (\mathrm {B}) \ 5 \qquad (\mathrm {C})\ 6 \qquad (\mathrm {D}) \ 7 \qquad (\mathrm {E})\ 8
  1,006 bytes (166 words) - 21:18, 3 July 2013
• 2006 AIME II Problems/Problem 6
  == Problem == pair Ap = (0, (3 - sqrt(3))/6);
  7 KB (1,067 words) - 12:23, 8 April 2024
• 1988 IMO Problems/Problem 6
  ==Problem==
  4 KB (720 words) - 12:26, 7 April 2024
• 1961 IMO Problems/Problem 6
  ==Problem== ...riangle's vertices, we have <math>G=\frac{1}{3}\left(L+M+N\right)=\frac{1}{6}\left(A+B+C+A^\prime+B^\prime+C^\prime\right)</math>. It is clear now that
  2 KB (301 words) - 23:29, 18 July 2016
• 1960 IMO Problems/Problem 6
  ==Problem== <cmath>W_2 = 6(u^2 - 1)</cmath>
  7 KB (1,214 words) - 18:49, 29 January 2018
• 2004 AMC 12A Problems/Problem 6
  ==Problem==
  1 KB (139 words) - 02:10, 30 December 2020
• 2007 AMC 10A Problems/Problem 6
  == Problem == #<math>\frac{70 - 66}{66} \approx 6\%</math>
  2 KB (211 words) - 22:55, 2 June 2023
• 2002 AMC 12B Problems/Problem 6
  {{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #6]] and [[2002 AMC 10B Problems|2002 AMC 10B #10]]}} == Problem ==
  3 KB (457 words) - 14:53, 17 August 2023
• 1976 IMO Problems/Problem 6
  == Problem ==
  3 KB (531 words) - 16:30, 29 January 2021
• 2001 AMC 12 Problems/Problem 6
  {{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #6]] and [[2001 AMC 10 Problems|2001 AMC 10 #13]]}} == Problem ==
  2 KB (340 words) - 03:02, 28 June 2023
• 2008 AMC 12A Problems/Problem 6
  ...cate|[[2008 AMC 12A Problems|2008 AMC 12A #6]] and [[2008 AMC 10A Problems/Problem 8|2008 AMC 10A #8]]}} ==Problem ==
  2 KB (240 words) - 19:53, 4 June 2021
• 2008 AMC 12B Problems/Problem 6
  ==Problem==
  1 KB (165 words) - 14:18, 16 February 2021
• 2008 AIME I Problems/Problem 6
  == Problem == ...ffect divisibility by <math>67</math>). The second row will be <math>2, 4, 6, \cdots , 98</math>, the third row will be <math>3, 5, \cdots, 97</math>, a
  3 KB (509 words) - 17:21, 22 March 2018
• Mock AIME 1 2007-2008 Problems/Problem 6
  2 KB (288 words) - 17:35, 6 May 2024
• 2008 AIME II Problems/Problem 6
  == Problem ==
  1 KB (223 words) - 23:34, 4 July 2013
• 2006 Alabama ARML TST Problems/Problem 6
  ==Problem== ...loor x \rfloor+\{y\} +z+\{x\}+y +\lfloor z \rfloor=x+x+y+y+z+z=2(x+y+z)=45.6</math>
  862 bytes (139 words) - 12:13, 20 January 2018
• 2008 AMC 10B Problems/Problem 6
  ==Problem==
  2 KB (317 words) - 18:09, 12 April 2024
• 2008 AMC 10A Problems/Problem 6
  ==Problem== ...athrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 5\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 7</math>
  1 KB (209 words) - 22:01, 2 September 2020
• 2008 USAMO Problems/Problem 6
  == Problem == ...ion]] and <math>f(v) = 1</math> otherwise (this corresponds, in the actual problem, to putting a mathematician in the first or second room). Then look at <mat
  13 KB (2,414 words) - 14:37, 11 July 2016
• 2008 Mock ARML 1 Problems/Problem 6
  == Problem == AP &= \frac{8\sqrt{5} \pm \sqrt{20}}{6} = \boxed{\frac{5\sqrt{5}}{3}}, \sqrt{5}.
  2 KB (283 words) - 22:51, 13 April 2015
• 2003 USAMO Problems/Problem 6
  == Problem ==
  5 KB (739 words) - 13:39, 4 July 2013
• 2002 AMC 12A Problems/Problem 6
  {{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #6]] and [[2002 AMC 10A Problems|2002 AMC 10A #4]]}} ==Problem==
  2 KB (258 words) - 04:58, 21 July 2022
• 2001 IMO Problems/Problem 6
  == Problem 6 == {{IMO box|num-b=5|after=Last Problem|year=2001}}
  1 KB (273 words) - 00:23, 19 November 2023
• 2002 USAMO Problems/Problem 6
  == Problem == <cmath>b(n)\geq\frac{1}{6}n^2 - \frac{2}{3}n.</cmath>
  5 KB (940 words) - 17:33, 16 July 2014
• 2001 USAMO Problems/Problem 6
  == Problem ==
  2 KB (302 words) - 18:11, 22 February 2016
• 2002 AMC 10B Problems/Problem 6
  #REDIRECT[[2002 AMC 12B Problems/Problem 3]]
  44 bytes (5 words) - 17:36, 28 July 2011
• 1962 IMO Problems/Problem 6
  ==Problem==
  2 KB (308 words) - 06:29, 16 December 2023
• 1963 IMO Problems/Problem 6
  ==Problem==
  2 KB (249 words) - 14:28, 31 July 2016
• 2004 AMC 12B Problems/Problem 6
  {{duplicate|[[2004 AMC 12B Problems|2004 AMC 12B #6]] and [[2004 AMC 10B Problems|2004 AMC 10B #8]]}} == Problem ==
  1 KB (154 words) - 19:23, 22 October 2022
• 2000 AMC 10 Problems/Problem 6
  #REDIRECT [[2000 AMC 12 Problems/Problem 4]]
  44 bytes (4 words) - 23:34, 26 November 2011
• 1985 AJHSME Problems/Problem 6
  ==Problem== ...eet of paper, and divide into the 2nd equation (which is one way to do the problem), but there are other ways, too.
  1 KB (230 words) - 09:22, 10 January 2023
• 1986 AJHSME Problems/Problem 6
  ==Problem== ...{3} \qquad \text{(C)}\ \frac{2}{3} \qquad \text{(D)}\ 2 \qquad \text{(E)}\ 6</math>
  564 bytes (78 words) - 21:10, 3 July 2013
• 2004 AMC 10B Problems/Problem 6
  ==Problem==
  1 KB (146 words) - 18:48, 26 July 2020
• 1971 IMO Problems/Problem 6
  ==Problem== ...solution was posted and copyrighted by Renan. The original thread for this problem can be found here: [https://aops.com/community/p1074433]
  6 KB (1,192 words) - 14:14, 29 January 2021
• 2009 AMC 12A Problems/Problem 6
  {{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #6]] and [[2009 AMC 10A Problems|2009 AMC 10A #13]]}} == Problem ==
  777 bytes (126 words) - 18:25, 7 August 2020
• 1987 AJHSME Problems/Problem 6
  ==Problem==
  628 bytes (96 words) - 23:52, 4 July 2013
• 2009 AMC 10A Problems/Problem 6
  == Problem == \mathrm{(B)}\ \frac{\pi}{6}
  2 KB (380 words) - 09:21, 8 June 2021
• 2009 AMC 10B Problems/Problem 6
  {{duplicate|[[2009 AMC 10B Problems|2009 AMC 10B #6]] and [[2009 AMC 12B Problems|2009 AMC 12B #5]]}} == Problem ==
  1 KB (159 words) - 08:13, 4 November 2022
• 2009 AIME I Problems/Problem 6
  == Problem == ...2,\ 3^3,\ 4^4,\ \sqrt{5}^{\lfloor\sqrt{5}\rfloor},\ \sqrt{6}^{\lfloor\sqrt{6}\rfloor},\ \sqrt{7}^{\lfloor\sqrt{7}\rfloor},\ \sqrt{8}^{\lfloor\sqrt{8}\rf
  4 KB (595 words) - 16:38, 15 February 2021
• 2005 USAMO Problems/Problem 6
  == Problem ==
  4 KB (725 words) - 23:59, 29 March 2016
• 2009 AIME II Problems/Problem 6
  == Problem == problem is reduced to finding <math>B</math>.
  5 KB (921 words) - 00:15, 11 December 2022
• 2008 IMO Problems/Problem 6
  ==Problem== {{IMO box|year=2008|num-b=5|after=Last Problem}}
  1 KB (283 words) - 01:15, 19 November 2023
• 1988 AJHSME Problems/Problem 6
  ==Problem==
  936 bytes (108 words) - 12:04, 28 July 2020
• 1989 AJHSME Problems/Problem 6
  ==Problem== for(int a=0; a<6; ++a)
  995 bytes (157 words) - 17:07, 29 April 2021
• 1990 AJHSME Problems/Problem 6
  ==Problem==
  792 bytes (115 words) - 00:05, 5 July 2013
• 1964 IMO Problems/Problem 6
  == Problem ==
  4 KB (601 words) - 21:26, 21 November 2023
• 1965 IMO Problems/Problem 6
  == Problem == [https://i.imgur.com/hjGyVyg.png Image of problem Solution]. Credits to user '''awe-sum'''.
  545 bytes (96 words) - 12:52, 29 January 2021
• 2009 USAMO Problems/Problem 6
  == Problem == ...math>t_2</math> for sequence <math>t</math>. Then by the conditions of the problem, we have <math>(s_2 - s_1)(t_2 - t_1)</math> is an integer, or <math>(\frac
  2 KB (463 words) - 12:19, 21 August 2020
• 2009 IMO Problems/Problem 6
  == Problem == {{IMO box|year=2009|num-b=5|after=Last Problem}}
  5 KB (1,055 words) - 01:18, 19 November 2023
• 1991 AJHSME Problems/Problem 6
  ==Problem== 10 & 6 & 4 & 3 & 2 \\
  784 bytes (101 words) - 00:06, 5 July 2013
• 1961 AHSME Problems/Problem 6
  == Problem == <math>\textbf{(A)}\ 6\log{2} \qquad
  697 bytes (95 words) - 20:01, 17 May 2018
• 1989 IMO Problems/Problem 6
  ==Problem==
  11 KB (1,928 words) - 12:26, 26 July 2023
• 1999 AHSME Problems/Problem 6
  ==Problem==
  587 bytes (80 words) - 23:59, 16 March 2020
• 2010 AMC 12A Problems/Problem 6
  {{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #6]] and [[2010 AMC 10A Problems|2010 AMC 10A #9]]}} == Problem ==
  2 KB (320 words) - 04:51, 21 January 2023
• 2007 AMC 8 Problems/Problem 6
  == Problem ==
  764 bytes (115 words) - 12:22, 16 August 2021
• 2010 AIME I Problems/Problem 6
  == Problem == draw(graph(Q,min,max),linetype("6 2")+linewidth(0.7));
  6 KB (1,019 words) - 20:39, 20 November 2023
• 2010 AIME II Problems/Problem 6
  == Problem ==
  2 KB (426 words) - 17:47, 29 June 2022
• 2010 AMC 10B Problems/Problem 6
  ==Problem==
  2 KB (260 words) - 17:00, 1 August 2022
• 2010 USAJMO Problems/Problem 6
  #REDIRECT [[2010 USAMO Problems/Problem 4]]
  43 bytes (4 words) - 06:43, 3 June 2010
• 2010 AMC 12B Problems/Problem 6
  {{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #6]] and [[2010 AMC 10B Problems|2010 AMC 10B #12]]}} == Problem ==
  1 KB (166 words) - 16:58, 6 July 2023
• 2019 IMO Problems/Problem 6
  ==Problem== [[File:2019 6 s1.png|450px|right]]
  5 KB (792 words) - 01:52, 19 November 2023
• 2010 IMO Problems/Problem 6
  =2010 IMO Problem 6= == Problem ==
  4 KB (786 words) - 08:46, 12 March 2024
• 2010 AMC 10A Problems/Problem 6
  == Problem 6 ==
  1 KB (140 words) - 18:58, 31 August 2022
• 1966 AHSME Problems/Problem 6
  == Problem ==
  1 KB (164 words) - 12:42, 28 January 2020
• 2011 AMC 12A Problems/Problem 6
  == Problem ==
  2 KB (361 words) - 20:39, 21 August 2023
• 2011 AMC 10A Problems/Problem 6
  ==Problem 6== ...<math>B={6, 7, 8, 9, 10, \cdots , 20}</math>. Then, all the integers <math>6</math> through <math>20</math> would be redundant in <math>A \cup B</math>,
  1 KB (160 words) - 20:19, 21 August 2023
• 2021 AIME II Problems/Problem 6
  ==Problem==
  8 KB (1,364 words) - 01:02, 29 January 2024
• 2011 AMC 12B Problems/Problem 6
  == Problem == In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection bet
  2 KB (370 words) - 13:35, 26 January 2021
• 2001 AMC 10 Problems/Problem 6
  ...1 AMC 12 Problems|2001 AMC 12 #2]] and [[2001 AMC 10 Problems|2001 AMC 10 #6]]}} == Problem ==
  1,018 bytes (165 words) - 10:33, 8 November 2021
• 2011 AIME I Problems/Problem 6
  ==Problem== ...and <cmath>y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}.</cmath> From the problem, we know that the parabola can be expressed in the form <math>y=ax^2+bx+c</
  4 KB (661 words) - 01:18, 11 December 2023
• 1999 IMO Problems/Problem 6
  ==Problem==
  341 bytes (54 words) - 00:02, 19 November 2023
• 2011 AIME II Problems/Problem 6
  ==Problem 6== ...he number of ways to partition 6 into 5 non-negative parts is <math>\binom{6+4}4 = \binom{10}4 = 210</math>. The interesting quadruples correspond to pa
  9 KB (1,535 words) - 01:28, 16 January 2023
• 1951 AHSME Problems/Problem 6
  == Problem ==
  831 bytes (141 words) - 12:20, 5 July 2013
• 2011 USAMO Problems/Problem 6
  ==Problem== {{USAMO newbox|year=2011|num-b=5|aftertext=|after=Last Problem}}
  7 KB (1,209 words) - 12:50, 25 August 2023
• 2000 AMC 8 Problems/Problem 6
  ==Problem==
  2 KB (365 words) - 21:02, 28 July 2023
• 2001 AMC 8 Problems/Problem 6
  ==Problem==
  700 bytes (109 words) - 00:38, 5 July 2013
• 2011 AMC 10B Problems/Problem 6
  == Problem==
  1 KB (220 words) - 05:34, 25 June 2022
• 2007 AMC 10B Problems/Problem 6
  ==Problem== ...> points for each incorrect response, and <math>1.5</math> points for each problem left unanswered. After looking over the <math>25</math> problems, Sarah has
  1 KB (184 words) - 21:15, 25 July 2018
• 1992 AJHSME Problems/Problem 6
  ==Problem== label("$6$",(2-sqrt(3)/10,0.1),WNW);
  1 KB (174 words) - 00:09, 5 July 2013
• 1958 AHSME Problems/Problem 6
  ==Problem==
  538 bytes (85 words) - 06:09, 3 October 2014
• 2003 AMC 10B Problems/Problem 6
  ...C 12B Problems|2003 AMC 12B #5]] and [[2003 AMC 10B Problems|2003 AMC 10B #6]]}} ==Problem==
  1 KB (220 words) - 14:53, 20 October 2020
• 2010 USAMO Problems/Problem 6
  == Problem == ...s is <math>28 + 5x - {x \choose 2}</math>, which is maximized at x=5 and x=6, and the maximum value is <math>43</math>. Choosing the first 5 numbers as
  3 KB (477 words) - 17:52, 15 January 2022
• 1998 AHSME Problems/Problem 6
  == Problem ==
  1 KB (157 words) - 14:28, 5 July 2013
• 1998 AJHSME Problems/Problem 6
  ==Problem== <math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math>
  2 KB (294 words) - 17:52, 26 October 2020
• 1984 AHSME Problems/Problem 6
  ==Problem==
  1 KB (184 words) - 12:49, 5 July 2013
• 1999 AMC 8 Problems/Problem 6
  ==Problem== Use logic to solve this problem. You don't actually need to use any equations.
  903 bytes (171 words) - 21:17, 30 October 2016
• 1997 AJHSME Problems/Problem 6
  ==Problem==
  1,012 bytes (143 words) - 00:26, 5 July 2013
• 1997 USAMO Problems/Problem 6
  == Problem ==
  6 KB (1,107 words) - 14:12, 12 April 2023
• 2011 IMO Problems/Problem 6
  {{IMO box|year=2011|num-b=5|after=Last Problem}}
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• 1966 IMO Problems/Problem 6
  == Problem == {{IMO box|year=1966|num-b=5|after=Last Problem}}
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• 2003 AMC 8 Problems/Problem 6
  ==Problem== label("$144$",(6,-7.5),N);
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• 2002 AMC 8 Problems/Problem 6
  ==Problem==
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• 1993 AJHSME Problems/Problem 6
  == Problem == <math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10</math>
  664 bytes (103 words) - 00:11, 5 July 2013
• 1996 AJHSME Problems/Problem 6
  ==Problem==
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• 1995 AJHSME Problems/Problem 6
  ==Problem== draw((0,0)--(15,0)--(15,6)--(12,6)--(12,9)--(0,9)--cycle);
  2 KB (277 words) - 11:37, 27 June 2023
• 1997 AHSME Problems/Problem 6
  ==Problem== <math> 1,-2,3,-4,5,-6,\ldots, </math>
  888 bytes (149 words) - 14:12, 5 July 2013
• 2008 AMC 8 Problems/Problem 6
  == Problem == Dividing the gray square into four smaller squares, there are <math>6</math> gray tiles and <math>10</math> white tiles, giving a ratio of <math>
  913 bytes (136 words) - 19:21, 8 August 2021
• 2009 AMC 8 Problems/Problem 6
  ==Problem 6== ...= 6,000</math> gallons of water. At the rate it goes at it will take <math>6,000/2.5 = 2400</math> minutes, or <math>\boxed{\textbf{(A)}\ 40}</math> hou
  1 KB (158 words) - 08:23, 30 May 2023
• 1994 AJHSME Problems/Problem 6
  ==Problem== ...\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math>
  781 bytes (118 words) - 14:25, 12 January 2014
• 1996 AHSME Problems/Problem 6
  ==Problem==
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• 2006 AMC 8 Problems/Problem 6
  == Problem == draw((0,6)--(4,6)--(4,4)--(3,4)--(3,0)--(1,0)--(1,4)--(0,4)--cycle, linewidth(1));</asy>
  809 bytes (117 words) - 21:45, 2 January 2023
• 1985 AHSME Problems/Problem 6
  ==Problem==
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• 1950 AHSME Problems/Problem 6
  == Problem== <cmath>2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1=0</cmath>
  976 bytes (151 words) - 11:57, 5 July 2013
• 2011 AMC 8 Problems/Problem 6
  ==Problem==
  1 KB (194 words) - 00:34, 27 December 2022
• 1985 IMO Problems/Problem 6
  ==Problem== ...solution was posted and copyrighted by DAFR. The original thread for this problem can be found here: [https://aops.com/community/p2751173]
  9 KB (1,788 words) - 00:02, 30 January 2021
• 1971 Canadian MO Problems/Problem 6
  == Problem ==
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• 1970 Canadian MO Problems/Problem 6
  == Problem ==
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• 2012 AMC 10A Problems/Problem 6
  == Problem == ...equal <math>x</math> and <math>y</math>. From the information given in the problem, two equations can be written:
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• 2012 AMC 12A Problems/Problem 6
  #REDIRECT [[2012 AMC 10A Problems/Problem 8]]
  45 bytes (5 words) - 14:32, 12 February 2012
• 1989 AHSME Problems/Problem 6
  == Problem == ...nded by the coordinate axes and the graph of <math>ax+by=6</math> has area 6, then <math>ab=</math>
  911 bytes (141 words) - 21:12, 8 September 2023
• 2004 USAMO Problems/Problem 6
  == Problem == By the given condition in the problem, all the equalities in the above discussion must hold, that is, <math>AI =
  4 KB (700 words) - 23:18, 28 November 2014
• 2012 AMC 12B Problems/Problem 6
  == Problem == ...h> and <math>y</math> are, so we can decide what they are. Let <math>x = 1.6</math> and <math>y = 1.4</math>. We round <math>x</math> to <math>2</math>
  1 KB (246 words) - 07:32, 29 June 2023
• 2012 AMC 10B Problems/Problem 6
  == Problem == The problem statement tells us that Xiaoli performed the following computation:
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• 2004 AMC 8 Problems/Problem 6
  == Problem ==
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• 2012 AIME I Problems/Problem 6
  ==Problem == ...at <math>71</math> is prime and <math>k<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71.<
  1 KB (233 words) - 17:15, 30 July 2022
• 1968 IMO Problems/Problem 6
  == Problem == {{IMO box|year=1968|num-b=5|after=Last Problem}}
  3 KB (427 words) - 12:49, 5 December 2023
• 2012 AIME II Problems/Problem 6
  == Problem 6 == ...o <math>z_{1}^4</math>(if you want to know why, reread what we want in the problem!)
  3 KB (428 words) - 02:34, 31 December 2020
• Mock AIME 2 2006-2007/Problem 6
  #REDIRECT [[Mock AIME 2 2006-2007 Problems/Problem 6]]
  54 bytes (6 words) - 15:29, 3 April 2012
• Mock AIME 1 2006-2007/Problem 6
  #REDIRECT [[Mock AIME 1 2006-2007 Problems/Problem 6]]
  54 bytes (6 words) - 15:49, 3 April 2012
• Mock AIME II 2012 Problems/Problem 6
  ==Problem==
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• Mock AIME I 2012 Problems/Problem 6
  ==Problem==
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• 2000 USAMO Problems/Problem 6
  ==Problem==
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• 2012 USAMO Problems/Problem 6
  == Problem == ...> however, we may use some logic to first layout a plan. Since for <math>n=6,n=4,</math> and <math>n=2</math>, <math>2^{n} - 2^{\frac{n}{2}} = n2^{n-3}
  6 KB (1,081 words) - 13:37, 21 June 2023
• 2008 iTest Problems/Problem 6
  ==Problem== ...he area using the first method: <math>\frac{1}{2}\cdot 5 \cdot 12 = 5\cdot 6 = 30</math>. Therefore, we have <math>\frac{1}{2}L \cdot 13 = 30</math>. Mu
  915 bytes (135 words) - 00:52, 22 June 2018
• 1974 AHSME Problems/Problem 6
  ==Problem==
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• 2006 SMT/Advanced Topics Problems/Problem 6
  ==Problem== ...50\cdot51}{2}=1275 </math>. Therefore, <math> 10n+9\le1275\implies n\le126.6 </math>, so the largest possible winning score is <math> 126 </math>. Notic
  1 KB (221 words) - 21:14, 27 May 2012
• 2006 SMT/Algebra Problems/Problem 6
  ==Problem== <math>c-1 = \tfrac{720}{120} = 6 \leadsto c = 7</math>
  2 KB (386 words) - 01:24, 31 May 2012
• 1980 AHSME Problems/Problem 6
  ==Problem==
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• 2012 USAJMO Problems/Problem 6
  #redirect [[2012 USAMO Problems/Problem 5]]
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• 2010 AMC 8 Problems/Problem 6
  ==Problem==
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• 1998 USAMO Problems/Problem 6
  == Problem ==
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• 1999 USAMO Problems/Problem 6
  == Problem ==
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• 2003 AMC 12B Problems/Problem 6
  #REDIRECT [[2003 AMC 10B Problems/Problem 8]]
  45 bytes (5 words) - 00:17, 5 January 2014
• 2012 AMC 8 Problems/Problem 6
  ==Problem==
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• 2005 AMC 8 Problems/Problem 6
  ==Problem == ...extbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 10 </math>
  518 bytes (78 words) - 18:23, 24 September 2016
• 2013 AMC 10A Problems/Problem 6
  ==Problem==
  2 KB (277 words) - 13:08, 1 July 2023
• 2013 AMC 12A Problems/Problem 6
  == Problem== Since the problem doesn't specify the number of 3-point shots she attempted, it can be assume
  3 KB (419 words) - 11:39, 10 March 2024
• 2013 AMC 10B Problems/Problem 6
  #REDIRECT [[2013 AMC 12B Problems/Problem 5]]
  45 bytes (5 words) - 12:10, 7 April 2013
• 2013 AMC 12B Problems/Problem 6
  {{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #6]] and [[2013 AMC 10B Problems|2013 AMC 10B #11]]}} ==Problem==
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• 2013 AIME I Problems/Problem 6
  ==Problem== ...is equivalent to <math>\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6}{144}=12\cdot11\cdot10\cdot7\cdot3</math>. If all of the math textbooks a
  5 KB (831 words) - 17:20, 9 January 2024
• 1991 AHSME Problems/Problem 6
  == Problem ==
  886 bytes (130 words) - 18:13, 5 September 2021
• 2013 AIME II Problems/Problem 6
  ==Problem 6== ...is small, so <math>10=6x+9</math>. <math>x=1/6\implies \sqrt{10}\approx 19/6</math>. This is 3.16.
  5 KB (826 words) - 09:29, 1 January 2023
• 2013 USAMO Problems/Problem 6
  ==Problem==
  4 KB (691 words) - 18:29, 10 May 2023
• 2005 IMO Problems/Problem 6
  ==Problem== ...more than 2/5 of the contestants. Moreover, no contestant solved all the 6 problems. Show that there are at least 2 contestants who solved exactly 5 p
  394 bytes (56 words) - 01:00, 19 November 2023
• 2014 AIME II Problems/Problem 6
  ==Problem== ...ls a six twice when using the fair die is <math>\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}</math>. The probability that he rolls a six twice using the
  2 KB (356 words) - 09:03, 14 June 2021
• 2013 USAJMO Problems/Problem 6
  ==Problem 6== Again, we experiment. If x = 2, y = 3, and z = 3, then <math>18 > 7 + 4\sqrt{6}</math>.
  3 KB (517 words) - 20:02, 30 April 2014
• 1967 AHSME Problems/Problem 6
  == Problem ==
  530 bytes (88 words) - 01:35, 16 August 2023
• 1967 IMO Problems/Problem 6
  780 bytes (136 words) - 10:02, 3 June 2021
• 1972 IMO Problems/Problem 6
  ==Problem== ...equilateral triangle (it is well-known that this is possible; in fact, the problem here is the <math>3</math>-dimensional version of this), and then complete
  7 KB (1,370 words) - 15:42, 29 January 2021
• 1975 IMO Problems/Problem 6
  ==Problem== ...ion was posted and copyrighted by mathmanman. The original thread for this problem can be found here: [https://aops.com/community/p587109]
  3 KB (563 words) - 16:21, 29 January 2021
• 1962 AHSME Problems/Problem 6
  ==Problem== ...h means that <math>x=6</math>, or the side length of the triangle is <math>6</math>. Thus, the triangle (and the square) have a perimeter of <math>18</m
  1 KB (174 words) - 15:55, 18 April 2018
• 2013 AMC 8 Problems/Problem 6
  ==Problem== ...s in the two boxes that touch it in the row above. For example, <math>30 = 6\times5</math>. What is the missing number in the top row?
  3 KB (477 words) - 00:05, 2 February 2023
• 1952 AHSME Problems/Problem 6
  == Problem==
  649 bytes (95 words) - 21:55, 1 January 2014
• 2014 AMC 10A Problems/Problem 6
  ...C 12A Problems|2014 AMC 12A #4]] and [[2014 AMC 10A Problems|2014 AMC 10A #6]]}} ==Problem==
  4 KB (737 words) - 00:09, 27 June 2023
• 1994 AHSME Problems/Problem 6
  ==Problem==
  685 bytes (102 words) - 17:27, 9 January 2021
• 2015 USAMO Problems/Problem 6
  ===Problem 6===
  3 KB (560 words) - 21:08, 15 December 2018
• 2014 USAJMO Problems/Problem 6
  ==Problem==
  7 KB (1,273 words) - 18:17, 28 August 2021
• 2014 USAMO Problems/Problem 6
  ==Problem==
  2 KB (361 words) - 11:55, 25 June 2020
• 2006 IMO Problems/Problem 6
  ==Problem== {{IMO box|year=2006|num-b=5|after=Last Problem}}
  390 bytes (72 words) - 01:04, 19 November 2023
• 2014 AMC 12A Problems/Problem 6
  ==Problem==
  1 KB (191 words) - 06:08, 23 February 2023
• 2014 AMC 10B Problems/Problem 6
  ==Problem== ...has enough to buy <math>30 \cdot \dfrac{2}{\dfrac{5}{3}} = 30 \cdot \dfrac{6}{5} = \fbox{(C) 36}</math>.
  1 KB (202 words) - 12:03, 2 July 2023
• 2014 AMC 12B Problems/Problem 6
  ==Problem==
  1 KB (223 words) - 10:11, 3 March 2015
• 2014 AIME I Problems/Problem 6
  == Problem 6 == ==Solution 6 (Vieta's solution)==
  7 KB (1,158 words) - 20:50, 8 December 2021
• 1993 AHSME Problems/Problem 6
  == Problem ==
  719 bytes (63 words) - 13:18, 11 August 2023
• 1992 AHSME Problems/Problem 6
  == Problem ==
  760 bytes (138 words) - 22:51, 4 October 2016
• 1990 AHSME Problems/Problem 6
  == Problem == real m = sqrt(6)/12;
  878 bytes (132 words) - 04:39, 4 February 2016
• 1968 AHSME Problems/Problem 6
  == Problem ==
  795 bytes (127 words) - 01:52, 16 August 2023
• 1969 AHSME Problems/Problem 6
  == Problem ==
  1 KB (187 words) - 03:29, 7 June 2018
• 1970 AHSME Problems/Problem 6
  == Problem ==
  925 bytes (159 words) - 03:15, 12 March 2017
• 1977 Canadian MO Problems/Problem 6
  == Problem ==
  259 bytes (45 words) - 14:13, 7 October 2014
• 1973 Canadian MO Problems/Problem 6
  ==Problem==
  445 bytes (76 words) - 17:51, 8 October 2014
• Mock AIME 5 2005-2006 Problems/Problem 6
  == Problem ==
  522 bytes (77 words) - 21:17, 8 October 2014
• 2007 IMO Problems/Problem 6
  == Problem == Now, to solve the problem let <math>H_1, \ldots, H_m</math> be <math>m</math> planes that cover all p
  3 KB (506 words) - 13:38, 23 March 2024
• 2014 IMO Problems/Problem 6
  ==Problem== {{IMO box|year=2014|num-b=5|num-a=6}}
  3 KB (588 words) - 14:50, 27 September 2020
• 2014 UMO Problems/Problem 6
  ==Problem == {{UMO box|year=2014|num-b=5|after=Last Problem}}
  3 KB (628 words) - 19:22, 20 May 2021
• 2013 UMO Problems/Problem 6
  == Problem == <math>n = 4</math>, <math>n = 5</math>, and <math>n = 6</math> are shown below).
  4 KB (466 words) - 15:28, 14 October 2014
• 2013 UNCO Math Contest II Problems/Problem 6
  == Problem ==
  825 bytes (131 words) - 03:29, 13 January 2019
• 2012 UNCO Math Contest II Problems/Problem 6
  == Problem ==
  291 bytes (39 words) - 03:25, 13 January 2019
• 2014 UNCO Math Contest II Problems/Problem 6
  == Problem ==
  1 KB (183 words) - 03:31, 13 January 2019
• 2010 UNCO Math Contest II Problems/Problem 6
  == Problem == What this problem is basically saying is that <math>10000a + 1000b + 100c + 10d + 3 = 3*(2000
  880 bytes (141 words) - 02:56, 13 January 2019
• 2009 UNCO Math Contest II Problems/Problem 6
  == Problem ==
  1 KB (165 words) - 13:46, 12 February 2017
• 2014 UNC Math Contest II Problems/Problem 6
  #REDIRECT [[2014 UNCO Math Contest II Problems/Problem 6]]
  58 bytes (7 words) - 18:15, 19 October 2014
• 2013 UNC Math Contest II Problems/Problem 6
  #REDIRECT [[2013 UNCO Math Contest II Problems/Problem 6]]
  58 bytes (7 words) - 18:22, 19 October 2014
• 2009 UNC Math Contest II Problems/Problem 6
  #REDIRECT [[2009 UNCO Math Contest II Problems/Problem 6]]
  58 bytes (7 words) - 21:14, 19 October 2014
• 2010 UNC Math Contest II Problems/Problem 6
  #REDIRECT [[2010 UNCO Math Contest II Problems/Problem 6]]
  58 bytes (7 words) - 21:17, 19 October 2014
• 2012 UNC Math Contest II Problems/Problem 6
  #REDIRECT [[2012 UNCO Math Contest II Problems/Problem 6]]
  58 bytes (7 words) - 21:21, 19 October 2014
• 2011 UNCO Math Contest II Problems/Problem 6
  == Problem ==
  457 bytes (61 words) - 00:42, 28 October 2015
• 2008 UNCO Math Contest II Problems/Problem 6
  == Problem ==
  2 KB (252 words) - 22:28, 18 March 2018
• 2007 UNCO Math Contest II Problems/Problem 6
  == Problem ==
  1 KB (203 words) - 18:49, 29 January 2018
• 1993 UNCO Math Contest II Problems/Problem 6
  == Problem == 2^2+3^2+6^2 &= 7^2 \\
  1 KB (202 words) - 22:13, 19 August 2021
• 2006 UNCO Math Contest II Problems/Problem 6
  == Problem == ...l of the positive integer divisors of <math>6^2=36</math> is <math>1+2+3+4+6+9+12+18+36=91</math>
  549 bytes (81 words) - 02:33, 13 January 2019
• 2005 CEMC Gauss (Grade 7) Problems/Problem 6
  == Problem ==
  834 bytes (136 words) - 01:45, 23 October 2014
• 1988 AHSME Problems/Problem 6
  ==Problem==
  1 KB (194 words) - 14:31, 31 August 2015
• 1987 AHSME Problems/Problem 6
  ==Problem== label("x", (2.25,6));
  1 KB (175 words) - 12:32, 1 March 2018
• 1986 AHSME Problems/Problem 6
  ==Problem== path table = origin--(1,0)--(1,6)--(6,6)--(6,0)--(7,0)--(7,7)--(0,7)--cycle;
  2 KB (221 words) - 18:11, 1 April 2018
• 2014 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 6
  == Problem ==
  871 bytes (143 words) - 20:57, 25 July 2016
• 2014 AMC 8 Problems/Problem 6
  ==Problem==
  830 bytes (120 words) - 11:15, 2 July 2023
• 2005 Indonesia MO Problems/Problem 6
  ==Problem== 2xy + 2yz + 2xz &= 2x^2 + 2y^2 + 2z^2 - 6 \\
  2 KB (286 words) - 11:56, 17 March 2020
• Mock Geometry AIME 2011 Problems/Problem 6
  ==Problem== draw((6,1+4*sqrt(3))--(4,1));
  3 KB (539 words) - 03:41, 25 December 2022
• 2015 UNCO Math Contest II Problems/Problem 6
  == Problem ==
  926 bytes (128 words) - 09:43, 6 March 2024
• 2015 AMC 10A Problems/Problem 6
  ...C 12A Problems|2015 AMC 12A #4]] and [[2015 AMC 10A Problems|2015 AMC 10A #6]]}} ==Problem==
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• 2015 AMC 12A Problems/Problem 6
  #REDIRECT[[2015 AMC 10A Problems/Problem 8]]
  44 bytes (5 words) - 20:44, 4 February 2015
• 2015 AMC 12B Problems/Problem 6
  ==Problem== ...3, 5, 7, 9, 11,</math> hence the number of odd products is <math>6 \times 6 = 36</math>. Therefore the answer is <math>36/169 \doteq \boxed{\textbf{(A)
  1 KB (200 words) - 20:51, 18 January 2019
• 2015 AMC 10B Problems/Problem 6
  ==Problem==
  1 KB (199 words) - 17:11, 2 August 2022
• 2015 AIME I Problems/Problem 6
  ==Problem== [[File:2015 AIME I 6.png|500px|right]]
  5 KB (782 words) - 16:04, 21 July 2023
• 1998 IMO Problems/Problem 6
  ==Problem==
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• 2015 AIME II Problems/Problem 6
  ==Problem== We get <math>(1, 4, 8)</math>, <math>(3, 6, 6)</math>, and <math>(4, 4, 7)</math>.
  5 KB (946 words) - 14:06, 14 February 2023
• Mock AIME 1 2013 Problems/Problem 6
  ==Problem==
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• 2015 UMO Problems/Problem 6
  ==Problem == {{UMO box|year=2015|num-b=5|after=Last Problem}}
  642 bytes (99 words) - 03:03, 6 November 2015
• 2015 AMC 8 Problems/Problem 6
  ==Problem==
  1 KB (189 words) - 15:55, 21 January 2024
• 2016 AMC 12A Problems/Problem 6
  #REDIRECT [[2016 AMC 10A Problems/Problem 9]]
  45 bytes (5 words) - 12:58, 4 February 2016
• 2018 UNCO Math Contest II Problems/Problem 6
  == Problem == (B) <math>36-48x^2 + 14x^4-x^6</math>
  639 bytes (108 words) - 01:41, 14 January 2019
• 2016 AMC 10B Problems/Problem 6
  ==Problem== ...that <math>5</math> and <math>4</math> work the best as we can't use <math>6</math> and <math>3</math>. Finally, we use <math>2</math> and <math>1</math
  2 KB (365 words) - 12:47, 2 July 2023
• 2016 AMC 12B Problems/Problem 6
  ==Problem== <math>\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16</math>
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• 2016 AIME I Problems/Problem 6
  ==Problem== real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
  14 KB (2,397 words) - 20:04, 27 August 2023
• 2016 AIME II Problems/Problem 6
  ==Problem== For polynomial <math>P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}</math>, define
  2 KB (426 words) - 16:09, 21 July 2023
• 2015 USAJMO Problems/Problem 6
  ==Problem==
  4 KB (769 words) - 20:24, 16 March 2019
• 2011 USAJMO Problems/Problem 6
  #REDIRECT [[2011 USAMO Problems/Problem 4]]
  43 bytes (4 words) - 21:25, 16 April 2016
• 2016 USAJMO Problems/Problem 6
  #REDIRECT [[2016 USAMO Problems/Problem 4]]
  43 bytes (4 words) - 16:41, 21 April 2016
• 2016 USAMO Problems/Problem 6
  ==Problem== {{USAMO newbox|year=2016|num-b=5|aftertext=|after=Last Problem}}
  8 KB (1,516 words) - 10:11, 8 April 2023
• 1954 AHSME Problems/Problem 6
  == Problem 6==
  702 bytes (92 words) - 20:37, 17 February 2020
• 1953 AHSME Problems/Problem 6
  ==Problem 6==
  875 bytes (130 words) - 20:34, 1 April 2017
• 1983 AHSME Problems/Problem 6
  == Problem 6 == \textbf{(C)}\ 6\qquad
  781 bytes (123 words) - 00:43, 20 February 2019
• 1990 IMO Problems/Problem 6
  ==Problem== Then the problem is equivalent to that there exists a way to assign <math>a_1,a_2,\cdots,a_{
  3 KB (522 words) - 13:54, 30 January 2021
• 1996 USAMO Problems/Problem 6
  ==Problem== {{USAMO newbox|year=1996|num-b=5|after=Last Problem}}
  3 KB (584 words) - 07:56, 16 April 2018
• 1975 Canadian MO Problems/Problem 6
  == Problem 6 ==
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• 2016 UNCO Math Contest II Problems/Problem 6
  == Problem ==
  974 bytes (151 words) - 04:03, 13 January 2019
• 1979 AHSME Problems/Problem 6
  == Problem 6 == .../math>, which is <math>\frac{96+80+72+68+66+65}{64}-7 = \frac{447}{64}-7 = 6\frac{63}{64}-7 = \boxed{\textbf{(A) } -\frac{1}{64}}</math>.
  686 bytes (80 words) - 13:05, 5 January 2017
• 2016 AMC 8 Problems/Problem 6
  == Problem == draw((-0.5,6)--(10,6));
  2 KB (279 words) - 14:11, 4 April 2023
• 1975 AHSME Problems/Problem 6
  ==Problem==
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• 2017 AMC 10A Problems/Problem 6
  ==Problem==
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• 2017 AMC 12A Problems/Problem 6
  ==Problem== ...use the <math>19</math> possible integer rod lengths that fall into <math>[6, 24]</math>. However, she has already used the rods of length <math>7</math
  1 KB (183 words) - 13:58, 19 December 2020
• 2017 AMC 10B Problems/Problem 6
  ==Problem== ...extbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7</math>
  989 bytes (157 words) - 20:04, 1 May 2021
• 2017 AMC 12B Problems/Problem 6
  ==Problem== The circle having <math>(0,0)</math> and <math>(8,6)</math> as the endpoints of a diameter intersects the <math>x</math>-axis a
  1 KB (217 words) - 14:19, 26 July 2022
• 2013 Mock AIME I Problems/Problem 6
  ==Problem 6==
  732 bytes (126 words) - 09:22, 23 October 2023
• 2017 AIME I Problems/Problem 6
  ==Problem 6== ...not need to spend time factoring <math>x^2 - 120x + 3024</math>. Since the problem asks for <math>|x_1 - x_2|</math>, where <math>x_1</math> and <math>x_2</ma
  4 KB (726 words) - 16:18, 5 January 2024
• 2017 AIME II Problems/Problem 6
  ==Problem== ==Solution 6==
  7 KB (1,096 words) - 21:03, 12 March 2021
• 1964 AHSME Problems/Problem 6
  == Problem == ...uence starts as <math>-4, -6, -9</math>. The common ratio is <math>\frac{-6}{-4} = \frac{3}{2}</math>. The next term is <math>\frac{3}{2} \cdot -9 = \
  1 KB (223 words) - 04:12, 23 July 2019
• 2017 USAJMO Problems/Problem 6
  ==Problem== {{USAJMO newbox|year=2017|num-b=5|aftertext=|after=Last Problem}}
  8 KB (1,465 words) - 15:30, 12 June 2020
• 2017 USAMO Problems/Problem 6
  ==Problem==
  399 bytes (66 words) - 20:57, 26 December 2017
• 1983 IMO Problems/Problem 6
  ==Problem 6== == See Also == {{IMO box|year=1983|num-b=5|after=Last Problem}}
  5 KB (990 words) - 19:37, 19 December 2021
• 2017 UNCO Math Contest II Problems/Problem 6
  == Problem ==
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• 1978 IMO Problems/Problem 6
  <math>\emph{Problem:}</math> An international society has its members from six different countr ...Then there exists a partition of <math>\{1,2,3,\ldots, 1978\}</math> into 6 difference-free subsets <math>A,B,C,D,E,F</math>. A set S is difference-fre
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• 1969 IMO Problems/Problem 6
  ==Problem==
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• 2017 AMC 8 Problems/Problem 6
  ==Problem==
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• 1981 AHSME Problems/Problem 6
  ==Problem==
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• 1978 AHSME Problems/Problem 6
  == Problem 6 ==
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• 2017 IMO Problems/Problem 6
  ==Problem== {{IMO box|year=2017|num-b=5|after=Last Problem}}
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• MIE 2016/Day 1/Problem 6
  ===Problem 6===
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• MIE 2015/Day 1/Problem 6
  ==Problem 6==
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• 2018 AMC 10A Problems/Problem 6
  == Problem == ...math>y = <math>x</math>. Plugging that in to the original equation, <math> 6/7 y</math> = <math>90</math> and <math> y =105</math>
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• 2018 AMC 12A Problems/Problem 6
  ==Problem== The mean and median are <cmath>\frac{3m+4n+17}{6}=\frac{m+n+11}{2}=n,</cmath>so <math>3m+17=2n</math> and <math>m+11=n</math
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• 2018 AMC 10B Problems/Problem 6
  == Problem == ...\frac{1}{15} \qquad \textbf{(B)} \frac{1}{10} \qquad \textbf{(C)} \frac{1}{6} \qquad \textbf{(D)} \frac{1}{5} \qquad \textbf{(E)} \frac{1}{4}</math>
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• 2018 AMC 12B Problems/Problem 6
  ==Problem==
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• 1974 IMO Problems/Problem 6
  ==Problem== .../math>. So <math>m\leq 6</math>. If <math>n\geq4</math> then <math>m-n\leq 6-n\leq 2</math>. Now assume that <math>n=3</math>. In this case if <math>m\l
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• 1973 IMO Problems/Problem 6
  ==Problem==
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• 2018 USAJMO Problems/Problem 6
  ==Problem 6== {{USAJMO newbox|year=2018|num-b=5|aftertext=|after=Last Problem}}
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• 2018 USAMO Problems/Problem 6
  ==Problem 6==
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• 1960 AHSME Problems/Problem 6
  ==Problem==
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• 2020 USOMO Problems/Problem 6
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• 1963 AHSME Problems/Problem 6
  == Problem ==
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• 1959 AHSME Problems/Problem 6
  == Problem 6==
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• 2018 AIME I Problems/Problem 6
  ==Problem== ...rs <math>z</math> with the properties that <math>|z|=1</math> and <math>z^{6!}-z^{5!}</math> is a real number. Find the remainder when <math>N</math> is
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• 2018 AIME II Problems/Problem 6
  ==Problem==
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• 2012 IMO Problems/Problem 6
  ==Problem== {{IMO box|year=2012|num-b=5|after=Last Problem}}
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• 2013 IMO Problems/Problem 6
  ==Problem== {{IMO box|year=2013|num-b=5|after=Last Problem}}
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• 1973 AHSME Problems/Problem 6
  ==Problem== <math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\text
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• 2018 IMO Problems/Problem 6
  ==Problem== [[File:2018 IMO 6.png|470px|right]]
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• 2002 Indonesia MO Problems/Problem 6
  ==Problem== ...equal <math>5</math> is when <math>p = 1</math> or <math>p = \tfrac{\sqrt{6}}{3}</math>, which are not prime numbers. Thus, the rest of the primes can
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• 2004 Indonesia MO Problems/Problem 6
  ==Problem==
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• 2003 Indonesia MO Problems/Problem 6
  ==Problem== ...f a palace is in a shape of regular hexagon, where the sidelength is <math>6 \text{ m}</math>. The floor of the hall is covered with an equilateral tri
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• 2017 Indonesia MO Problems/Problem 6
  ==Problem==
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• 2009 Indonesia MO Problems/Problem 6
  ==Problem==
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• 2018 AMC 8 Problems/Problem 6
  ==Problem== ...He drives [mathjax]50[/mathjax] miles, so he drives for [mathjax]\dfrac{5}{6}[/mathjax] hours, which is equal to [mathjax]50[/mathjax] minutes (Note tha
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• 1991 IMO Problems/Problem 6
  == Problem == ...<math>i,j</math>; in other words, this sequence simultaneously solves the problem for all <math>a\geq 1</math> simultaneously.
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• 2006 iTest Problems/Problem 6
  ==Problem== ...''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
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• Mock AIME 2 2010 Problems/Problem 6
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• 2019 USAMO Problems/Problem 6
  ==Problem== {{USAMO newbox|year=2019|num-b=5|aftertext=|after=Last Problem}}
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• 1955 AHSME Problems/Problem 6
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• 2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 6
  == Problem ==
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• 2016 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 6
  == Problem == <math>a_2 = 2+0+1+6+2+0+1+6 = 18</math>
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• 2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 6
  == Problem ==
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• 2018 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 6
  == Problem ==
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• 2016 UMO Problems/Problem 6
  ==Problem ==
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• 2019 AMC 10A Problems/Problem 6
  ==Problem==
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• 2019 AMC 12A Problems/Problem 6
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• 2019 AMC 10B Problems/Problem 6
  {{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #6]] and [[2019 AMC 12B Problems|2019 AMC 12B #4]]}} ==Problem==
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• 2019 AMC 12B Problems/Problem 6
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• 2019 AIME I Problems/Problem 6
  ==Problem== ...own above. Note that <math>m\angle KPL = 90^{\circ}</math> as given in the problem. Since <math>\angle KPL \cong \angle KLN</math> and <math>\angle PKL \cong
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• 2019 AIME II Problems/Problem 6
  ==Problem== <cmath>x = 6^{108}</cmath>
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• 2006 Indonesia MO Problems/Problem 6
  ==Problem== ...placement of the two 8s, there are <math>21 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 = 317520</math> phone numbers in this case.
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• 2020 AMC 8 Problems/Problem 6
  ==Problem== ==Video Solution by Math-X (First understand the problem!!!)==
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• 1957 AHSME Problems/Problem 6
  ==Problem== fill((0,4)--(4,4)--(4,0)--(6,0)--(6,4)--(10,4)--(10,10)--(6,10)--(6,14)--(4,14)--(4,10)--(0,10)--cycle,grey);
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• 2006 SMT/Calculus Problems/Problem 6
  ==Problem 6==
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• 2006 SMT/Team Problems/Problem 6
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• 2003 Pan African MO Problems/Problem 6
  ==Problem== |after=Last Problem
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• 1956 AHSME Problems/Problem 6
  == Problem 6==
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• 2020 AIME I Problems/Problem 6
  == Problem == ...then form another right triangle with base <math>7</math>, as given by the problem. The height of this triangle is the difference between the heights of the p
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• 2019 USAJMO Problems/Problem 6
  {{USAJMO newbox|year=2019|num-b=5|aftertext=|after=Last Problem}}
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• 2016 IMO Problems/Problem 6
  ==Problem== {{IMO box|year=2016|num-b=5|after=Last Problem}}
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• 2019 Mock AMC 10B Problems/Problem 6
  ==Problem== ...each die must be distinct, the answer is <math>\frac{2}{5} \cdot \frac{3}{6} = \boxed{\text{(C)} \frac{1}{5}}</math>.
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• 2002 Pan African MO Problems/Problem 6
  ==Problem== |after=Last Problem
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• 2019 AMC 8 Problems/Problem 6
  == Problem 6 == dot((0,6));
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• 1965 AHSME Problems/Problem 6
  == Problem 6==
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• 2020 AMC 10A Problems/Problem 6
  ...C 12A Problems|2020 AMC 12A #4]] and [[2020 AMC 10A Problems|2020 AMC 10A #6]]}} ==Problem==
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• 2020 AMC 12A Problems/Problem 6
  ==Problem== for (int j = 0; j < 6; ++j) {
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• 2020 AMC 10B Problems/Problem 6
  ==Problem==
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• 2020 AMC 12B Problems/Problem 6
  ==Problem==
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• 2007 Indonesia MO Problems/Problem 6
  ==Problem==
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• 1982 AHSME Problems/Problem 6
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• 2021 CMC 12A Problems/Problem 6
  {{duplicate|[[2021 CMC 12A Problems | 2021 CMC 12A #6]] and [[2021 CMC 10A Problems | 2021 CMC 10A #7]]}} ==Problem==
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• 2020 Mock Combo AMC 10 II Problems/Problem 6
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• 1977 IMO Problems/Problem 6
  ==Problem==
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• 1976 AHSME Problems/Problem 6
  =Problem 6=
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• 2020 AIME II Problems/Problem 6
  ==Problem== <cmath>t_6 = \frac{5t_{6-1}+1}{25t_{6-2}} = \frac{5t_{5}+1}{25t_{4}} = \frac{5(\frac{101}{525}) + 1}{25(\frac{103
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• 2020 USOJMO Problems/Problem 6
  ==Problem== {{USAJMO newbox|year=2020|num-b=5|after=Last Problem}}
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• 1995 IMO Problems/Problem 6
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• 2021 AIME I Problems/Problem 6
  ==Problem==
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• 2001 Pan African MO Problems/Problem 6
  ...or our desired choice of <math>R</math> and the value <math>r_R = \frac{1}{6}</math>, so we conclude that <math>OPRQ</math> is rectangular and we are do {{Pan African MO box|year=2001|num-b=5|after=Last Problem}}
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• 2022 AIME I Problems/Problem 6
  == Problem == ...rithmetic progression, <math>a</math> or <math>b</math> can never be <math>6</math>. Since <math>b, 30, 40, 50</math> and <math>a, 30, 40, 50</math> can
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• 2020 CIME I Problems/Problem 6
  ==Problem 6== We reduce the problem to <math>z^{17}+z^7+1</math>, remembering to multiply the final product by
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• 2004 IMO Problems/Problem 6
  ==Problem== ...x\cdot10^{k-1}+a</math> is alternating. We see that either <math>S=\{0,2,4,6,8\}</math> or <math>S=\{1,3,5,7,9\}</math>. In each possible set, each <mat
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• 2021 Fall AMC 10A Problems/Problem 6
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• 2021 JMPSC Accuracy Problems/Problem 6
  ==Problem==
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• 2020 CAMO Problems/Problem 6
  ==Problem 6== ...xt to the squid, then he can just get covered entirely by black squares in 6 moves. In general, if we have a coordinate system with origin at the squid'
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• 2020 IMO Problems/Problem 6
  == Problem ==
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• 2004 Pan African MO Problems/Problem 6
  {{Pan African MO box|year=2004|num-b=5|after=Last Problem}}
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• 2021 AMC 12A Problems/Problem 6
  ==Problem== <math>\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad
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• 2020 SWMC 6 Problems/Problem 1
  ==Problem==
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• 2020 INMO Problems/Problem 6
  ==Problem==
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• 2021 USAJMO Problems/Problem 6
  ==Problem== ...4}}, & a_{4}&=a_{3}+a_{5}, \\ a_{5}&=\frac{1}{a_{4}}+\frac{1}{a_{6}}, & a_{6}&=a_{5}+a_{7}, \\ &\vdots \\ a_{2 n-1}&=\frac{1}{a_{2 n-2}}+\frac{1}{a_{2 n
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• 2021 AMC 10A Problems/Problem 6
  ==Problem== =d\left(\frac{3}{12} + \frac{6}{12} + \frac{4}{12}\right)=\frac{13}{12}d</cmath> hours. At that point, Jea
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• 1972 AHSME Problems/Problem 6
  == Problem 6 ==
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• 1977 AHSME Problems/Problem 6
  == Problem 6 ==
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• 2021 AMC 10B Problems/Problem 6
  #redirect [[2021 AMC 12B Problems/Problem 4]]
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• 2021 AMC 12B Problems/Problem 6
  ...10|2021 AMC 10B #10]] and [[2021 AMC 12B Problems#Problem 6|2021 AMC 12B #6]]}} ==Problem==
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• 1971 AHSME Problems/Problem 6
  ==Problem==
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• 1998 CEMC Gauss (Grade 7) Problems/Problem 6
  == Problem ==
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• 1979 IMO Problems/Problem 6
  ==Problem== ...ution was posted and copyrighted by Myszon11. The original thread for this problem can be found here: [https://aops.com/community/p712390]
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• 1982 IMO Problems/Problem 6
  ==Problem==
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• 1984 IMO Problems/Problem 6
  ==Problem== ...was posted and copyrighted by TheFunkyRabbit. The original thread for this problem can be found here: [https://aops.com/community/p4908989]
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• 1986 IMO Problems/Problem 6
  ==Problem== ...lution was posted and copyrighted by grobber. The original thread for this problem can be found here: [https://aops.com/community/p366620]
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• 1985 AJHSME Problem 6
  == Problem ==
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• 1963 TMTA High School Algebra I Contest Problem 6
  == Problem == |before=[[1963 TMTA High School Algebra I Contest Problem 5| Problem 5]]
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• 2021 USAMO Problems/Problem 6
  ==Problem 6 ==
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• 2021 JMC 10 Problems/Problem 6
  ==Problem 6==
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• 2021 JMPSC Invitationals Problems/Problem 6
  ==Problem==
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• 2021 April MIMC 10 Problems/Problem 6
  ...perimeter of the square. Therefore, the ratio is <math>4\cdot{\sqrt[4]{3}:6=2\cdot \sqrt[4]{3}}:3</math>. <math>2+4+3+3=\fbox{\textbf{(C)} 12}</math>.
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• 2022 USAMO Problems/Problem 6
  ==Problem== {{USAMO newbox|year=2022|num-b=5|after=Last Problem}}
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• 2022 AIME II Problems/Problem 6
  ==Problem==
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• 2021 Fall AMC 12B Problems/Problem 6
  ...AMC 10B #8]] and [[2021 Fall AMC 12B Problems#Problem 6|2021 Fall AMC 12B #6]]}} == Problem ==
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• Mock AIME I 2015 Problems/Problem 6
  ==Problem== ...mber of <math>5</math> digit numbers using only the digits <math>1,2,3,4,5,6,7,8</math> such that every pair of adjacent digits is no more than <math>1<
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• 2021 JMPSC Sprint Problems/Problem 6
  ==Problem== ...lue of <math>60</math> dimes is: <math>60 \cdot 0.10=6.00</math>, or <math>6</math> dollars.
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• Northeastern WOOTers Mock AIME I Problems/Problem 6
  == Problem 6 ==
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• 2021 IMO Problems/Problem 6
  == Problem == {{IMO box|year=2021|num-b=5|after=Last Problem}}
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• 2021 Fall AMC 10B Problems/Problem 6
  ==Problem== ...6^k</math>, where <math>m</math> and <math>k</math> are integers and <math>6</math> is not a divisor of <math>m</math>. What is <math>m+k?</math>
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• 2021 Fall AMC 12A Problems/Problem 6
  ...AMC 10A #7]] and [[2021 Fall AMC 12A Problems/Problem 6|2021 Fall AMC 12A #6]]}} ==Problem==
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• 2021 WSMO Speed Round Problems/Problem 6
  ==Problem==
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• 2022 AMC 8 Problems/Problem 6
  ==Problem== <math>\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8</math>
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• 2022 USAJMO Problems/Problem 6
  ==Problem==
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• Euclid 2020/Problem 6
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• Mock AIME 6 2006-2007 Problems/Problem 3
  ==Problem ==
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• 2021 WSMO Team Round/Problem 6
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• 2023 AMC 8 Problems/Problem 6
  ==Problem==
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• 2022 IMO Problems/Problem 6
  ==Problem== [[File:2022 IMO 6.png|900px|center]]
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• 2021 GMC 10B Problems/Problem 6
  ==Problem==
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• 2020 USAMO Problems/Problem 6
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• ASIA TEAM Problems/Problem 6
  <math>\frac{9}{x}+\frac{4}{y}+\frac{1}{z}=6</math>. ...>x:y:z=3:2:1</math>, or <math>(x,y,z)=(3,2,1)</math>, and <math>xyz=\boxed{6}</math>.
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• 2023 IMO Problems/Problem 6
  ==Problem== [[File:2023 IMO 6.png|300px|right]]
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• 2021 WSMO Accuracy Round Problems/Problem 6
  ==Problem==
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• Mock AIME 6 2006-2007 Problems/Problem 1
  ==Problem==
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• 2022 AMC 10A Problems/Problem 6
  ==Problem==
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• 2022 AMC 12A Problems/Problem 6
  #redirect [[2022 AMC 10A Problems/Problem 8]]
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• 2022 AMC 10B Problems/Problem 6
  ...2 AMC 10B Problems/Problem 6|2022 AMC 10B #6]] and [[2022 AMC 12B Problems/Problem 3|2022 AMC 12B #3]]}} ==Problem==
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• 2022 AMC 12B Problems/Problem 6
  #redirect [[2022 AMC 10B Problems/Problem 8]]
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• 2003 IMO Problems/Problem 6
  == 2003 IMO Problems/Problem 6 == == Problem ==
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• 2020 CIME II Problems/Problem 6
  ==Problem== <cmath>\frac{2 \cdot 63}{64} + \frac{4 \cdot 63}{64^2} + \frac{6 \cdot 63}{64^3} + \ldots</cmath>
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• 2023 AIME I Problems/Problem 6
  == Problem == We break the problem into stages, one for each card revealed, then further into cases based on t
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• 2023 AIME II Problems/Problem 6
  ==Problem == [[File:2023 AIME I 6.png|470px|right]]
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• 2023 AMC 10A Problems/Problem 6
  ==Problem== Note that each vertex is counted <math>2\times 3=6</math> times. Thus, the answer is <math>21\times6=\boxed{\textbf{(D) } 126}
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• 2023 USAMO Problems/Problem 6
  == Problem == A=intersectionpoint(Circle(B,6),Circle(C,9));
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• 2023 SSMO Speed Round Problems/Problem 6
  ==Problem==
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• 2022 SSMO Speed Round Problems/Problem 6
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• 1994 IMO Problems/Problem 6
  ==Problem==
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• 1993 IMO Problems/Problem 6
  ==Problem == {{IMO box|year=1993|num-b=5|after=Last Problem}}
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• 1997 IMO Problems/Problem 6
  ==Problem==
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• 2020 USAJMO Problems/Problem 6
  #redirect [[2020 USOJMO Problems/Problem 6]]
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• 1992 IMO Problems/Problem 6
  ==Problem==
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• 2023 AMC 12A Problems/Problem 6
  ==Problem== ...h>y=\log_{2}x</math>. The midpoint of <math>\overline{AB}</math> is <math>(6, 2)</math>. What is the positive difference between the <math>x</math>-coor
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• 2023 AMC 10B Problems/Problem 6
  ==Problem== ==Video Solution by Math-X (First understand the problem!!!)==
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• 2023 AMC 12B Problems/Problem 6
  ...12|2023 AMC 10B #12]] and [[2023 AMC 12B Problems/Problem 6|2023 AMC 12B #6]]}} ==Problem==
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• 2024 AMC 8 Problems/Problem 6
  ==Problem== ==Video Solution by Math-X (First fully understand the problem!!!)==
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• 2000 IMO Problems/Problem 6
  ==Problem==
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• 2002 IMO Problems/Problem 6
  ==Problem== {{IMO box|year=2002|num-b=5|after=Last Problem}}
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• 1996 IMO Problems/Problem 6
  ==Problem== {{IMO box|year=1996|num-b=5|after=Last Problem}}
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• 2023 USAJMO Problems/Problem 6
  ==Problem== ...names are defined as in the figures below. Figure 1 is the diagram of the problem while Figure 2 is the diagram of the Ratio Lemma. Do note that the point na
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• Mock AIME 6 2006-2007 Problems/Problem 2
  ==Problem ==
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• Mock AIME 6 2006-2007 Problems/Problem 4
  ==Problem==
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• Mock AIME 6 2006-2007 Problems/Problem 5
  ==Problem== At <math>k=6</math>, <math>100k-7=593\;\;86+k^2=202</math>, <math>593>202</math>.
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• Mock AIME 6 2006-2007 Problems/Problem 6
  ==Problem ==
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• Mock AIME 6 2006-2007 Problems/Problem 7
  ==Problem == <math>3, 6, 9, 12, 15,\cdots, 1002</math> gives '''334''' terms
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• Mock AIME 6 2006-2007 Problems/Problem 8
  ==Problem== And the sequence repeats every 6 steps.
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• Mock AIME 6 2006-2007 Problems/Problem 9
  ==Problem== <math>A_{ABC}=\sqrt{(21)(6)(7)(8)}=\sqrt{3^24^27^2}=(3)(7)(4)=\boxed{84}</math>
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• Mock AIME 6 2006-2007 Problems/Problem 10
  ==Problem== For this problem, we're only interested in the <math>y</math>-coordinate. So,
  4 KB (709 words) - 16:20, 25 November 2023
• Mock AIME 6 2006-2007 Problems/Problem 11
  ==Problem== case 6 blues: Total number of octahedrons with the condition = Total from case 2 =
  1 KB (216 words) - 17:40, 26 November 2023
• Mock AIME 6 2006-2007 Problems/Problem 12
  ==Problem== Therefore, <math>S=2007-1=2006 \equiv 6\;(mod\;1000)</math>
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• Mock AIME 6 2006-2007 Problems/Problem 13
  ==Problem==
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• Mock AIME 6 2006-2007 Problems/Problem 14
  ==Problem==
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• Mock AIME 6 2006-2007 Problems/Problem 15
  ==Problem==
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• 1985 OIM Problems/Problem 6
  == Problem ==
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• 1987 OIM Problems/Problem 6
  == Problem ==
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• 1988 OIM Problems/Problem 6
  == Problem ==
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• 1989 OIM Problems/Problem 6
  == Problem ==
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• 1990 OIM Problems/Problem 6
  == Problem ==
  359 bytes (58 words) - 13:44, 13 December 2023
• 1991 OIM Problems/Problem 6
  == Problem ==
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• 1992 OIM Problems/Problem 6
  == Problem == For this problem were going to denote <math>H</math> and <math>T</math> as the area of hexag
  3 KB (626 words) - 09:41, 23 December 2023
• 1993 OIM Problems/Problem 6
  == Problem ==
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• 1994 OIM Problems/Problem 6
  == Problem ==
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• 1995 OIM Problems/Problem 6
  == Problem ==
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• 1996 OIM Problems/Problem 6
  == Problem ==
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• 1997 OIM Problems/Problem 6
  == Problem ==
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• 1998 OIM Problems/Problem 6
  == Problem ==
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• 1999 OIM Problems/Problem 6
  == Problem ==
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• 2000 OIM Problems/Problem 6
  == Problem ==
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• 2002 OIM Problems/Problem 6
  == Problem ==
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• 2023 OIM Problems/Problem 6
  == Problem ==
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• 2022 OIM Problems/Problem 6
  == Problem ==
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• 2021 OIM Problems/Problem 6
  == Problem ==
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• 2001 OIM Problems/Problem 6
  == Problem ==
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• 2003 OIM Problems/Problem 6
  == Problem ==
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• 2022 SSMO Accuracy Round Problems/Problem 6
  ==Problem== ...oportional to <math>7-m</math>. The expected number of rolls until a <math>6</math> is rolled can be expressed as <math>\frac{m}{n}</math>, where <math>
  727 bytes (121 words) - 13:03, 14 December 2023
• 2022 SSMO Team Round Problems/Problem 6
  ==Problem==
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• 2015 OIM Problems/Problem 6
  == Problem ==
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• 2020 OIM Problems/Problem 6
  == Problem ==
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• 2019 OIM Problems/Problem 6
  == Problem ==
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• 2018 OIM Problems/Problem 6
  == Problem ==
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• 2017 OIM Problems/Problem 6
  == Problem ==
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• 2016 OIM Problems/Problem 6
  == Problem ==
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• 2014 OIM Problems/Problem 6
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Page text matches

• Pythagorean Theorem
  <cmath>6-8-10 = (3-4-5)*2</cmath> * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]
  5 KB (886 words) - 21:12, 22 January 2024
• 2005 AMC 12A Problems/Problem 16
  == Problem == <math> (\mathrm {A}) \ 5 \qquad (\mathrm {B}) \ 6 \qquad (\mathrm {C})\ 8 \qquad (\mathrm {D}) \ 9 \qquad (\mathrm {E})\ 10 <
  2 KB (307 words) - 15:30, 30 March 2024
• 2016 AMC 10A Problems/Problem 5
  == Problem == {{AMC10 box|year=2016|ab=A|num-b=4|num-a=6}}
  1 KB (184 words) - 13:58, 22 August 2023
• 2016 AMC 10A Problems/Problem 7
  == Problem == ...using the mean in your reasoning, you can just take the mean of the other 6 numbers, and it'll solve it marginally faster. -[[User:Integralarefun|Integ
  2 KB (268 words) - 18:19, 27 September 2023
• 2016 AMC 10A Problems/Problem 9
  == Problem == <math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf
  2 KB (315 words) - 15:34, 18 June 2022
• 2016 AMC 10A Problems/Problem 10
  == Problem == filldraw(rectangle((1,1),(6,4)),gray(0.75));
  2 KB (337 words) - 14:56, 25 June 2023
• 2016 AMC 10A Problems/Problem 11
  == Problem == ...uad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math>
  8 KB (1,016 words) - 00:17, 31 December 2023
• List of United States elementary school mathematics competitions
  * [[Noetic Learning Math Contest]] - semiannual problem solving contest for elementary and middle school students. [Grades 2-8] ...ills through [https://www.beestar.org/exercise/samples.jsp?adid=105 Weekly Problem Solving Contests] and [https://www.beestar.org/competition/?adid=105 Beesta
  4 KB (473 words) - 16:14, 1 May 2024
• List of United States middle school mathematics competitions
  * [https://www.hardestmathproblem.org Hardest Math Problem] math contest for grades 5-8 with great prizes. * [[Noetic Learning Math Contest]]: a popular problem-solving contest for students in grades 2-8.
  7 KB (792 words) - 10:14, 23 April 2024
• MathILy (serious mathematics infused with levity)
  Class meets for about 7 hours per day, in two shifts (morning and evening), 6 days per week. Each class has a Lead Instructor who is a mathematician with ...ctures and providing proofs. Classes include independent and collaborative problem solving as well as lots of laughter; in this way, students learn creative a
  5 KB (706 words) - 23:49, 29 January 2024
• MATHCOUNTS
  ...cluding Art of Problem Solving, the focus of MATHCOUNTS is on mathematical problem solving. Students are eligible for up to three years, but cannot compete be ...ics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.
  10 KB (1,497 words) - 11:42, 10 March 2024
• ICTM Math Contest
  ...thematics offers two areas of math contests: Grade School (Grades 3, 4, 5, 6, 7, 8 + Algebra 1) and High School (Regional and State Finals). *2024 State FInals - Saturday 4/6/24
  8 KB (1,182 words) - 14:26, 3 April 2024
• Math textbooks
  ...ks''' page is for compiling a list of [[textbook]]s for mathematics -- not problem books, contest books, or general interest books. See [[math books]] for mo * Getting Started is recommended for students grades 6 to 9.
  7 KB (901 words) - 14:11, 6 January 2022
• William Lowell Putnam Mathematical Competition
  ...Problem A/B, 1/2</u>: 7<br><u>Problem A/B, 3/4</u>: 8<br><u>Problem A/B, 5/6</u>: 9}} ...chool olympiads are, although they include more advanced mathematics. Each problem is graded on a scale of 0 to 10. The top five scorers (or more if there are
  4 KB (623 words) - 13:11, 20 February 2024
• Computer science books
  These '''Computer Science books''' are recommended by [[Art of Problem Solving]] administrators and members of the [http://www.artofproblemsolving .../ref_list_smcs.jsp?&mid=1500&div=9 Computer Science Reading for Grade 3-5, 6-8]
  2 KB (251 words) - 00:45, 17 November 2023
• 2005 Alabama ARML TST Problems/Problem 7
  ==Problem== =\frac{2}{2}+\frac{4}{4}+\frac{6}{8}+\frac{8}{16}+\cdots\\
  1 KB (193 words) - 21:13, 18 May 2021
• Inequality
  ...parentheses. For instance, <math>x \in [3,6)</math> means <math>3 \le x < 6</math>. The problem here is that we multiplied by <math>x+5</math> as one of the last steps. W
  12 KB (1,798 words) - 16:20, 14 March 2023
• United States of America Mathematical Talent Search
  The USAMTS is administered by the [[Art of Problem Solving Foundation]] with support and sponsorship by the [[National Securit ...|difficulty=3-6|breakdown=<u>Problem 1-2</u>: 3-4<br><u>Problem 3-5</u>: 5-6}}
  4 KB (613 words) - 13:08, 18 July 2023
• AMC 10
  ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC. ...iculty=1-3|breakdown=<u>Problem 1-5</u>: 1<br><u>Problem 6-20</u>: 2<br><u>Problem 21-25</u>: 3}}
  4 KB (574 words) - 15:28, 22 February 2024
• AMC 12
  ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC! ...ulty=2-4|breakdown=<u>Problem 1-10</u>: 2<br><u>Problem 11-20</u>: 3<br><u>Problem 21-25</u>: 4}}
  4 KB (520 words) - 12:11, 13 March 2024
• American Invitational Mathematics Examination
  ...ministered by the [[Mathematical Association of America]] (MAA). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC! ...<u>Problem 6-10</u>: 4<br><u>Problem 10-12</u>: 5<br><u>Problem 12-15</u>: 6}}
  8 KB (1,057 words) - 12:02, 25 February 2024
• 1954 AHSME Problems/Problem 46
  == Problem 46== draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2));
  3 KB (415 words) - 18:01, 24 May 2020
• 2015 IMO Problems
  ==Problem 1== [[2015 IMO Problems/Problem 1|Solution]]
  4 KB (692 words) - 22:33, 15 February 2021
• United States of America Mathematical Olympiad
  ...administered by the [[American Mathematics Competitions]] (AMC). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC and of the recent expansion ...9|breakdown=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}}
  6 KB (869 words) - 12:52, 20 February 2024
• American Regions Mathematics League
  {{WotWAnnounce|week=June 6-12}} The '''American Regions Math League''' (ARML) is a [[mathematical problem solving]] competition primarily for U.S. high school students.
  2 KB (267 words) - 17:06, 7 March 2020
• Joining an ARML team
  ...tices held during the Spring Semester to determine the team each year. The 6 practices include 3 individual tests to help determine the team and some le ...[[MAML]] (Maine Association of Math Leagues) Meets. Training includes the problem set "Pete's Fabulous 42."
  21 KB (3,500 words) - 18:41, 23 April 2024
• Simon's Favorite Factoring Trick
  ...This yields <math>x(y+5)+6(y+5)=60</math>. Now, we can factor as <math>(x+6)(y+5)=60</math>. ...ecause this keeps showing up in number theory problems. Let's look at this problem below:
  7 KB (1,107 words) - 07:35, 26 March 2024
• Mathematical problem solving
  ...oblem solving]] involves using all the tools at one's disposal to attack a problem in a new way. <math>\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots</math><br>
  2 KB (314 words) - 06:45, 1 May 2014
• Geometric mean
  The geometric mean of the numbers 6, 4, 1 and 2 is <math>\sqrt[4]{6\cdot 4\cdot 1 \cdot 2} = \sqrt[4]{48} = 2\sqrt[4]{3}</math>. * [[1966 AHSME Problems/Problem 3]]
  2 KB (282 words) - 22:04, 11 July 2008
• Principle of Inclusion-Exclusion
  ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ==== Problem ====
  9 KB (1,703 words) - 07:25, 24 March 2024
• Factorial
  * <math>3! = 6</math> * <math>6! = 720</math>
  10 KB (809 words) - 16:40, 17 March 2024
• 2006 AIME II
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME II Problems/Problem 1]]
  1 KB (133 words) - 12:32, 22 March 2011
• Fermat's Little Theorem
  int n = 6; ...),red+linewidth(2));match(2,3,1); </asy>For <math>p=2,3</math> and <math>a=6,4</math>, respectively.</center>
  16 KB (2,658 words) - 16:02, 8 May 2024
• Parabola
  f.p=fontsize(6); f.p=fontsize(6);
  3 KB (551 words) - 16:22, 13 September 2023
• Euler's Totient Theorem
  ...ast two digits of <math> 7^{81}-3^{81} </math>. ([[Euler's Totient Theorem Problem 1 Solution|Solution]]) ...ast two digits of <math> 3^{3^{3^{3}}} </math>. ([[Euler's Totient Theorem Problem 2 Solution|Solution]])
  3 KB (542 words) - 17:45, 21 March 2023
• Number theory/Advanced
  ...uited to studying large-scale properties of prime numbers. The most famous problem in analytic number theory is the [[Riemann Hypothesis]]. ...es <math>G_4</math> and <math>G_6</math> are modular forms of weight 4 and 6 respectively.
  5 KB (849 words) - 16:14, 18 May 2021
• Harmonic mean
  ...instance, if we tried to take the harmonic mean of the set <math>\{-2, 3, 6\}</math> we would be trying to calculate <math>\frac 3{\frac 13 + \frac 16 * [[2002 AMC 12A Problems/Problem 11]]
  1 KB (196 words) - 00:49, 6 January 2021
• Divisibility rules
  ...if its units digit is divisible by 2, i.e. if the number ends in 0, 2, 4, 6 or 8. ...s at the end of it to be divisible by 1,000,000 because <math>1,000,000=10^6</math>.
  8 KB (1,315 words) - 18:18, 2 March 2024
• Induction
  Other, odder inductions are possible. If a problem asks you to prove something for all integers greater than 3, you can use <m ...ernational Mathematics Olympiad | IMO]]. A good example of an upper-level problem that can be solved with induction is [http://www.artofproblemsolving.com/Fo
  5 KB (768 words) - 20:45, 1 September 2022
• Overcounting
  An example of a classic problem is as follows: ...hem twice. A number that is divisible by both 2 and 3 must be divisible by 6, and there are 16 such numbers. Thus, there are <math>50+33-16=\boxed{67}</
  4 KB (635 words) - 12:19, 2 January 2022
• Constructive counting
  This is a problem where constructive counting is not the simplest way to proceed. This next e ...wer is <math>8 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 = 8 \cdot 7^6 = 941,192</math>, as desired. <math>\square</math>
  12 KB (1,896 words) - 23:55, 27 December 2023
• Euclidean algorithm
  Similarly, <math>42 \equiv 6 \pmod{9}</math>, so <math>\gcd(42,9) = \gcd(9,6)</math>. <br/> Continuing, <math>9 \equiv 3 \pmod{6}</math>, so <math>\gcd(9,6) = \gcd(6,3)</math>. <br/>
  6 KB (924 words) - 21:50, 8 May 2022
• Generating function
  ...multiply the functions together, getting <math>1+3x+6x^2+8x^3+8x^4+6x^5+3x^6+x^7</math>. We want the number of ways to choose 4 eggs, so we just need to
  4 KB (659 words) - 12:54, 7 March 2022
• Binomial Theorem
  ...th>\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots</cmath> for all <math>x</math>. ...[[polynomial]]s with [[binomial coefficient]]s. For example, if a contest problem involved the polynomial <math>x^5+4x^4+6x^3+4x^2+x</math>, one could factor
  5 KB (935 words) - 13:11, 20 February 2024
• Function
  ...function: <math>f(x) = x^2 + 6</math>. The function <math>g(x) = \sqrt{x-6}</math> has the property that <math>f(g(x)) = x</math>. Therefore, <math>g ([[2006 AMC 10A Problems/Problem 2|Source]])
  10 KB (1,761 words) - 03:16, 12 May 2023
• Complementary counting
  ...ng may lead to a quick solution is the phrase "not" or "at least" within a problem statement. ''[[2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least one digit that is
  8 KB (1,192 words) - 17:20, 16 June 2023
• Bijection
  [[2008 AMC 12B Problems/Problem 22]] [[2001 AIME I Problems/Problem 6]]
  1,016 bytes (141 words) - 03:39, 29 November 2021
• Combinatorial identity
  .../www.artofproblemsolving.com/Forum/viewtopic.php?p=394407#394407 1986 AIME Problem 11] ...lving.com/Forum/resources.php?c=182&cid=45&year=2000&p=385891 2000 AIME II Problem 7]
  12 KB (1,993 words) - 23:49, 19 April 2024
• Circle
  ...- 3)^2 + (y + 6)^2 = 25</math> represents the circle with center <math>(3,-6)</math> and radius 5 units. ([[2006 AMC 12A Problems/Problem 16|Source]])
  9 KB (1,581 words) - 18:59, 9 May 2024
• Base numbers
  ...umber can be rewritten as <math>2746_{10}=2\cdot10^3+7\cdot10^2+4\cdot10^1+6\cdot10^0.</math> ...of the decimal place (recall that the decimal place is to the right of the 6, i.e. 2746.0) tells us that there are six <math>10^0</math>'s, the second d
  4 KB (547 words) - 17:23, 30 December 2020
• 2001 AMC 12 Problems/Problem 1
  == Problem == ...d \textbf{(B)}\ 3S + 2\qquad \textbf{(C)}\ 3S + 6 \qquad\textbf{(D)}\ 2S + 6 \qquad \textbf{(E)}\ 2S + 12</math>
  788 bytes (120 words) - 10:32, 8 November 2021
• 2001 AMC 12 Problems/Problem 2
  ...1 AMC 12 Problems|2001 AMC 12 #2]] and [[2001 AMC 10 Problems|2001 AMC 10 #6]]}} == Problem ==
  1,007 bytes (165 words) - 00:28, 30 December 2023
• Math Day at the Beach
  '''Math Day at the Beach''' is a [[mathematical problem solving]] festival for Southern California high school students, hosted by ...oth individual and team competition. Teams represent high schools and have 6 members each. The competition takes place on a Saturday in March.
  4 KB (644 words) - 12:56, 29 March 2017
• Fibonacci sequence
  ...iv style="text-align:right">([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])</div> ...? <div style="text-align:right">([[1998 AIME Problems/Problem 8|1998 AIME, Problem 8]])</div>
  6 KB (957 words) - 23:49, 7 March 2024
• 1988 AJHSME Problems/Problem 5
  ==Problem== label("160",(1.6,.5),NE);
  1 KB (160 words) - 16:53, 17 December 2020
• Fundamental Theorem of Calculus
  Can you do the main problem now? # Here's a slightly different way to think about the main problem, that doesn't use physics. How much does the function <math>f(x)= \frac{x^
  11 KB (2,082 words) - 15:23, 2 January 2022
• Law of Cosines
  \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2
  6 KB (1,003 words) - 09:11, 7 June 2023
• Mathew Crawford
  ...Cameron Matthews. In 2003, Crawford became the first employee of [[Art of Problem Solving]] where he helped to write and teach most of the online classes dur * [[USAMTS]] problem writer and grader (2004-2006)
  2 KB (360 words) - 02:20, 2 December 2010
• Law of Sines
  \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2
  4 KB (658 words) - 16:19, 28 April 2024
• 2005 USAMO Problems/Problem 4
  == Problem == ...2+...+n^2) =</math> <math>\dfrac{(n+1)(n+2)(2n+3)}{6}+\dfrac{n(n+1)(2n+1)}{6}=\boxed{\dfrac{2n^3+6n^2+7n+3}{3}}</math>.
  7 KB (1,276 words) - 20:51, 6 January 2024
• Functional equation
  ...function: <math>f(x) = x^2 + 6</math>. The function <math>g(x) = \sqrt{x-6}</math> has the property that <math>f(g(x)) = x</math>. In this case, <mat ...ns can significantly help in solving functional identities. Consider this problem:
  2 KB (361 words) - 14:40, 24 August 2021
• Floor function
  ...math>n</math> satisfy the equation <math>\left[\frac{n}{5}\right]=\frac{n}{6}</math>. [[1985 AIME Problems/Problem 10|(1985 AIME)]]
  3 KB (508 words) - 21:05, 26 February 2024
• 1985 IMO Problems/Problem 5
  == Problem == {{IMO box|year=1985|num-b=4|num-a=6}}
  3 KB (496 words) - 13:35, 18 January 2023
• 2006 AMC 10B
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 10B Problems/Problem 1]]
  2 KB (182 words) - 21:57, 23 January 2021
• Goldbach Conjecture
  <math>6 = 3 + 3</math> Euler, becoming interested in the problem, answered with an equivalent version of the conjecture:
  7 KB (1,201 words) - 16:59, 19 February 2024
• Riemann Hypothesis
  ...nction, it is easy to see that <math>\zeta(s)=0</math> when <math>s=-2,-4,-6,\ldots</math>. These are called the trivial zeros. This hypothesis is one The Riemann Hypothesis is an important problem in the study of [[prime number]]s. Let <math>\pi(x)</math> denote the numbe
  2 KB (425 words) - 12:01, 20 October 2016
• 2004 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME I Problems/Problem 1]]
  1 KB (135 words) - 18:15, 19 April 2021
• 2004 AIME II
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME II Problems/Problem 1]]
  1 KB (135 words) - 12:24, 22 March 2011
• 2005 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME I Problems/Problem 1 | Problem 1]]
  1 KB (154 words) - 12:30, 22 March 2011
• International Mathematical Olympiad
  ...c competitions. Each year, countries from around the world send a team of 6 students to compete in a grueling competition. .../u>: 9.5<br><u>Problem SL1-2</u>: 5.5-7<br><u>Problem SL3-4</u>: 7-8<br><u>Problem SL5+</u>: 8-10}}
  3 KB (490 words) - 03:32, 23 July 2023
• 2006 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME I Problems/Problem 1]]
  1 KB (135 words) - 12:31, 22 March 2011
• 2005 AIME II
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME II Problems/Problem 1]]
  1 KB (135 words) - 12:30, 22 March 2011
• 2006 AMC 12B
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 12B Problems/Problem 1 | Problem 1]]
  2 KB (210 words) - 00:06, 7 October 2014
• Euler's Polyhedral Formula
  ...on &4 &6 & 4 \\ \hline Cube/Hexahedron & 8 & 12 & 6\\ \hline Octahedron & 6 & 12 & 8\\ \hline Dodecahedron & 20 & 30 & 12\\ \hline Icosahedron & 12 & 3 ==Problem==
  1,006 bytes (134 words) - 14:15, 6 March 2022
• Mock AMC
  ...A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member ...amous theorems and formulas and see if there's any way you can make a good problem out of them.
  51 KB (6,175 words) - 20:58, 6 December 2023
• Modular arithmetic/Introduction
  <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> ...\not\equiv 18 \pmod{6}</math> because <math>\frac{91 - 18}{6} = \frac{73}{6}</math>, which is not an integer.
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• Modular arithmetic/Intermediate
  <math>\{ \ldots, -6, -3, 0, 3, 6, \ldots \}</math> <math>\overline{6} \cdot \overline{6} = \overline{6 \cdot 6} = \overline{36} = \overline{1}</math>
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• Right triangle
  label("$45^{\circ}$", A, 6*dir(290)); [[2007 AMC 12A Problems/Problem 10 | 2007 AMC 12A Problem 10]]
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• 2006 AMC 10A
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 10A Problems/Problem 1]]
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• 2006 AIME I Problems
  == Problem 1 == [[2006 AIME I Problems/Problem 1|Solution]]
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• 2006 AIME I Problems/Problem 15
  == Problem == ...sequence. The only thing that will be left will be a sequence <math>\{0,3,6,9,\cdots,3k\}</math> for some even <math>k</math>. Since we started with 20
  6 KB (910 words) - 19:31, 24 October 2023
• 2006 AIME I Problems/Problem 13
  == Problem == ...1000</math>, we have five choices for <math>k</math>, namely <math>k=0,2,4,6,8</math>.
  10 KB (1,702 words) - 00:45, 16 November 2023
• 2006 AIME I Problems/Problem 11
  == Problem == ...e 2 towers which use blocks <math>1, 2</math>, so there are <math>2\cdot 3^6 = 1458</math> towers using blocks <math>1, 2, \ldots, 8</math>, so the answ
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• 2006 AIME I Problems/Problem 10
  == Problem == draw((6.5,0)--origin--(0,6.5), Arrows(5));
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• 2006 AIME I Problems/Problem 9
  == Problem == ...th> [[positive integer]]s. <math>a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}</math> so <math>a^{2}r^{11}=2^{1003}</math>.
  4 KB (651 words) - 18:27, 22 May 2021
• 2006 AIME I Problems/Problem 7
  == Problem == fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray);
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• 2006 AIME I Problems/Problem 5
  == Problem == The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+
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• 2021 JMC 10
  ==Problem 1== [[2021 JMC 10 Problems/Problem 1|Solution]]
  12 KB (1,784 words) - 16:49, 1 April 2021
• 2006 AMC 12B Problems
  == Problem 1 == [[2006 AMC 12B Problems/Problem 1|Solution]]
  13 KB (2,058 words) - 12:36, 4 July 2023
• 2006 AMC 12A
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 12A Problems/Problem 1]]
  1 KB (168 words) - 21:51, 6 October 2014
• 2004 AMC 12A
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AMC 12A Problems/Problem 1]]
  2 KB (186 words) - 17:35, 16 December 2019
• 2004 AMC 12B
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AMC 12B Problems/Problem 1]]
  2 KB (181 words) - 21:40, 6 October 2014
• 2005 AMC 12A
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2005 AMC 12A Problems/Problem 1|Problem 1]]
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• 2005 AMC 12B
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AMC 12B Problems/Problem 1|Problem 1]]
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• 2000 AMC 12
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2000 AMC 12 Problems/Problem 1]]
  1 KB (126 words) - 13:28, 20 February 2020
• 2001 AMC 12
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2001 AMC 12 Problems/Problem 1]]
  1 KB (127 words) - 21:36, 6 October 2014
• 2002 AMC 12A
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2002 AMC 12A Problems/Problem 1]]
  1 KB (158 words) - 21:33, 6 October 2014
• 2003 AMC 12A
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2003 AMC 12A Problems/Problem 1]]
  1 KB (162 words) - 21:52, 6 October 2014
• 2002 AMC 12B
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2002 AMC 12B Problems/Problem 1]]
  1 KB (154 words) - 00:32, 7 October 2014
• 2003 AMC 12B
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2003 AMC 12B Problems/Problem 1]]
  1 KB (160 words) - 20:46, 1 February 2016
• 2006 AMC 12A Problems
  == Problem 1 == [[2006 AMC 12A Problems/Problem 1|Solution]]
  15 KB (2,223 words) - 13:43, 28 December 2020
• 2005 AMC 12A Problems
  == Problem 1 == [[2005 AMC 12A Problems/Problem 1|Solution]]
  13 KB (1,971 words) - 13:03, 19 February 2020
• 2004 AMC 12A Problems
  == Problem 1 == [[2004 AMC 12A Problems/Problem 1|Solution]]
  13 KB (1,953 words) - 00:31, 26 January 2023
• 2003 AMC 12A Problems
  == Problem 1 == [[2003 AMC 12A Problems/Problem 1|Solution]]
  13 KB (1,955 words) - 21:06, 19 August 2023
• 2002 AMC 12A Problems
  == Problem 1 == <math>(2x+3)(x-4)+(2x+3)(x-6)=0 </math>
  12 KB (1,792 words) - 13:06, 19 February 2020
• 2000 AMC 12 Problems
  == Problem 1 == [[2000 AMC 12 Problems/Problem 1|Solution]]
  13 KB (1,948 words) - 12:26, 1 April 2022
• 2001 AMC 12 Problems
  == Problem 1 == ...3\qquad \text{(B)}\ 3S + 2\qquad \text{(C)}\ 3S + 6 \qquad\text{(D)} 2S + 6 \qquad \text{(E)}\ 2S + 12</math>
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• 2002 AMC 12B Problems
  == Problem 1 == \qquad\mathrm{(D)}\ 6
  10 KB (1,547 words) - 04:20, 9 October 2022
• 2003 AMC 12B Problems
  == Problem 1 == <cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath>
  13 KB (1,987 words) - 18:53, 10 December 2022
• 2004 AMC 12B Problems
  == Problem 1 == <math>(\mathrm {A}) 3\qquad (\mathrm {B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm {E}) 15</mat
  13 KB (2,049 words) - 13:03, 19 February 2020
• 2005 AMC 12B Problems
  == Problem 1 == [[2005 AMC 12B Problems/Problem 1|Solution]]
  12 KB (1,781 words) - 12:38, 14 July 2022
• 2006 AMC 12B Problems/Problem 5
  == Problem == {{AMC12 box|year=2006|ab=B|num-b=4|num-a=6}}
  654 bytes (115 words) - 21:47, 1 August 2020
• 2006 AMC 12B Problems/Problem 7
  == Problem == {{AMC12 box|year=2006|ab=B|num-b=6|num-a=8}}
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• 2006 AMC 12B Problems/Problem 9
  == Problem== ...and for each choice there is one acceptable order. Similarly, for <math>c=6</math> and <math>c=8</math> there are, respectively, <math>\binom{5}{2}=10<
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• 2006 AMC 12B Problems/Problem 11
  == Problem == Joe has 2 ounces of cream, as stated in the problem.
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• 2006 AMC 12B Problems/Problem 13
  == Problem == ...qrt {3} \qquad \textrm{(C) } 8 \qquad \textrm{(D) } 9 \qquad \textrm{(E) } 6\sqrt {3}</math>
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• 2006 AMC 12B Problems/Problem 15
  == Problem == // from amc10 problem series
  3 KB (458 words) - 16:40, 6 October 2019
• 2006 AMC 12B Problems/Problem 16
  == Problem == ...n is <math>\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}</math>.
  1 KB (203 words) - 16:36, 18 September 2023
• 2006 AMC 12B Problems/Problem 17
  == Problem == ...h>5</math> and <math>6</math> on each die are in the ratio <math>1:2:3:4:5:6</math>. What is the probability of rolling a total of <math>7</math> on the
  1 KB (188 words) - 22:10, 9 June 2016
• 2006 AMC 12B Problems/Problem 19
  == Problem == \mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8
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• 2006 AMC 12B Problems/Problem 21
  == Problem == \mathrm{(D)}\ 6\sqrt {2006}
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• 2006 AMC 12B Problems/Problem 23
  == Problem == ...ABC</math>, such that <math>PA=11</math>, <math>PB=7</math>, and <math>PC=6</math>. Legs <math>\overline{AC}</math> and <math>\overline{BC}</math> have
  7 KB (1,169 words) - 14:04, 10 June 2022
• 2006 AMC 12B Problems/Problem 24
  == Problem == \mathrm{(C)}\ \dfrac{\pi^2}{6}
  3 KB (563 words) - 22:45, 24 October 2021
• 2006 AMC 12A Problems\Problem 6
  #REDIRECT [[2006 AMC 12A Problems/Problem 6]]
  45 bytes (5 words) - 14:43, 14 January 2016
• 2006 AMC 12A Problems/Problem 10
  ...ate|[[2006 AMC 12A Problems|2006 AMC 12A #10]] and [[2006 AMC 10A Problems/Problem 10|2006 AMC 10A #10]]}} == Problem ==
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• 2006 AMC 12A Problems/Problem 13
  == Problem == <cmath>r_A + r_B + r_C = 6</cmath>
  1 KB (184 words) - 13:57, 19 January 2021
• 2006 AMC 12A Problems/Problem 15
  == Problem == ...{2}\qquad \mathrm{(D) \ } \frac{5\pi}{6} \quad \mathrm{(E) \ } \frac{7\pi}{6}</math>
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• 2006 AMC 12A Problems/Problem 19
  == Problem == ...-axis. The equation of <math>L_1</math> is <math>y=\frac{5}{12}x+\frac{19}{6}</math>, so the coordinate of this point is <math>\left(-\frac{38}{5},0\rig
  2 KB (253 words) - 22:52, 29 December 2021
• 2006 AMC 12A Problems/Problem 20
  ...ate|[[2006 AMC 12A Problems|2006 AMC 12A #20]] and [[2006 AMC 10A Problems/Problem 25|2006 AMC 10A #25]]}} == Problem ==
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• 2006 AMC 12A Problems/Problem 22
  == Problem == ...rt{6}+\sqrt{3}\qquad \rm{(D) \ } 3\sqrt{2}+\sqrt{6}\qquad \mathrm{(E) \ } 6\sqrt{2}-\sqrt{3}</math>
  2 KB (343 words) - 15:39, 14 June 2023
• 2005 AMC 12A Problems/Problem 25
  == Problem == ...ill actually give an [[octahedron]], not a cube, because it only has <math>6</math> vertices.
  4 KB (498 words) - 00:46, 4 August 2023
• 2005 AMC 12B Problems/Problem 3
  ==Problem== {{AMC10 box|year=2005|ab=B|num-b=4|num-a=6}}
  978 bytes (156 words) - 14:14, 14 December 2021
• 2005 AMC 12B Problems/Problem 4
  ...C 12B Problems|2005 AMC 12B #4]] and [[2005 AMC 10B Problems|2005 AMC 10B #6]]}} == Problem ==
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• 2005 AMC 12B Problems/Problem 5
  == Problem == {{AMC12 box|year=2005|ab=B|num-b=4|num-a=6}}
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• 2005 AMC 12B Problems/Problem 6
  {{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #6]] and [[2005 AMC 10B Problems|2005 AMC 10B #10]]}} == Problem ==
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• 2005 AMC 12B Problems/Problem 7
  == Problem == \mathrm{(A)}\ 6 \qquad
  2 KB (357 words) - 20:15, 27 December 2020
• 2005 AMC 12B Problems/Problem 9
  == Problem == ...h>85</math>. The mean is <math>\dfrac{70\,(2)+80\,(5)+85\,(4)+90\,(3)+95\,(6)}{20}=\dfrac{1720}{20}=86</math>. The difference between the mean and media
  2 KB (280 words) - 15:35, 16 December 2021
• 2005 AMC 12B Problems/Problem 11
  == Problem == ...nties, so you have <math>6</math> bills left. <math>\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15</math> ways. However, you counted the case when you
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• 2005 AMC 12B Problems/Problem 13
  == Problem == Suppose that <math>4^{x_1}=5</math>, <math>5^{x_2}=6</math>, <math>6^{x_3}=7</math>, ... , <math>127^{x_{124}}=128</math>. What is <math>x_1x_2
  1 KB (203 words) - 19:57, 24 December 2020
• 2005 AMC 12B Problems/Problem 14
  == Problem == ...>, is tangent to the lines <math>y=x</math>, <math>y=-x</math> and <math>y=6</math>. What is the radius of this circle?
  2 KB (278 words) - 21:12, 24 December 2020
• 2005 AMC 12B Problems/Problem 15
  == Problem == ...h> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</ma
  2 KB (411 words) - 21:02, 21 December 2020
• 2005 AMC 12B Problems/Problem 18
  == Problem == f.p=fontsize(6);
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• 2005 AMC 12B Problems/Problem 19
  == Problem == ...h>. Now we have <math>(a+b)+(a-b)=11+1</math>, <math>2a=12</math>, <math>a=6</math>, then <math>b=5</math>, <math>x=65</math>, <math>y=56</math>, <math>
  2 KB (283 words) - 20:02, 24 December 2020
• 2005 AMC 12B Problems/Problem 20
  == Problem == ...and <math>h</math> be distinct elements in the set <math>\{-7,-5,-3,-2,2,4,6,13\}.</math>
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• 2005 AMC 12B Problems/Problem 24
  == Problem == Our sum is simply <math>2 - 2\cdot\frac{8}{11} = \frac{6}{11}</math>, and thus we can divide by <math>2</math> to obtain <math>\frac
  4 KB (761 words) - 09:10, 1 August 2023
• 2005 AMC 12B Problems/Problem 25
  == Problem == We approach this problem by counting the number of ways ants can do their desired migration, and the
  10 KB (1,840 words) - 21:35, 7 September 2023
• Power of a Point Theorem/Introductory Problem 2
  == Problem == Applying the Power of a Point Theorem gives <math> 6\cdot x = 4\cdot 1 </math>, so <math> x = \frac 23 </math>
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• 2006 AMC 10A Problems
  ==Problem 1== [[2006 AMC 10A Problems/Problem 1|Solution]]
  13 KB (2,028 words) - 16:32, 22 March 2022
• 2006 AMC 10A Problems/Problem 8
  == Problem == <math>b=-6</math>
  2 KB (348 words) - 23:10, 16 December 2021
• 2008 USAMO
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. **[[2008 USAMO Problems/Problem 1]]
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• 2006 AMC 10A Problems/Problem 12
  == Problem == size(150); pathpen = linewidth(0.6); defaultpen(fontsize(10));
  3 KB (424 words) - 10:14, 17 December 2021
• 2006 AMC 10A Problems/Problem 13
  == Problem == ...}{6}</math>. The probability of winning is <math>\frac{1}{2}\cdot \frac{1}{6} =\frac{1}{12}</math>. If the game is to be fair, the amount paid, <math>5<
  1 KB (207 words) - 09:39, 25 July 2023
• 2006 AMC 10A Problems/Problem 15
  == Problem == draw((6,0){up}..{left}(0,6),blue);
  3 KB (532 words) - 17:49, 13 August 2023
• 2006 AMC 10A Problems/Problem 16
  == Problem == D((2*t,2)--(2*t,4)); D((2*t,5)--(2*t,6));
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• 2006 AMC 10A Problems/Problem 18
  == Problem == So, there are <math>6 - 1 = 5</math> choices for the position of the letters.
  2 KB (254 words) - 14:39, 5 April 2024
• 2006 AMC 10A Problems/Problem 20
  == Problem == ...5</math> possible values in <math>\bmod{5}</math> and we are picking <math>6</math> numbers, by the [[Pigeonhole Principle]], two of the numbers must be
  1 KB (187 words) - 08:21, 17 March 2023
• 2006 AMC 10A Problems/Problem 21
  == Problem == ...gements for the non-<math>2</math> or <math>3</math> digits. We have <math>6 \cdot 56</math> = <math>336</math> arrangements for this case. We have <mat
  3 KB (525 words) - 20:25, 30 April 2024
• 2006 AMC 10A Problems/Problem 24
  == Problem == <math>\textbf{(A) } \frac{1}{8}\qquad\textbf{(B) } \frac{1}{6}\qquad\textbf{(C) } \frac{1}{4}\qquad\textbf{(D) } \frac{1}{3}\qquad\textbf
  2 KB (292 words) - 10:19, 19 December 2021
• Set
  ...tiple indistinct elements, such as the following: <math>\{1,4,5,3,24,4,4,5,6,2\}</math> Such an entity is actually called a multiset. ...describe sets should be used with extreme caution. One way to avoid this problem is as follows: given a property <math>P</math>, choose a known set <math>T<
  11 KB (2,021 words) - 00:00, 17 July 2011
• 2006 USAMO
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. **[[2006 USAMO Problems/Problem 1]]
  467 bytes (51 words) - 09:25, 6 August 2014
• 2006 USAMO Problems
  === Problem 1 === [[2006 USAMO Problems/Problem 1 | Solution]]
  3 KB (520 words) - 09:24, 14 May 2021
• 1991 AJHSME Problems
  ==Problem 1== [[1991 AJHSME Problems/Problem 1|Solution]]
  17 KB (2,246 words) - 13:37, 19 February 2020
• 2006 AIME A Problems/Problem 6
  #REDIRECT [[2006 AIME I Problems/Problem 6]]
  44 bytes (5 words) - 12:05, 28 June 2009
• 1990 AJHSME Problems/Problem 1
  ==Problem== ...x digits <math>4,5,6,7,8,9</math> in one of the six boxes in this addition problem?
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• 2022 USAJMO Problems/Problem 1
  ==Problem== ...metric sequence to be <math>\{ g, gr, gr^2, \dots \}</math>. Rewriting the problem based on our new terminology, we want to find all positive integers <math>m
  4 KB (792 words) - 00:29, 13 April 2024
• 2005 AIME II Problems/Problem 1
  == Problem == ...The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find <math>
  1 KB (239 words) - 11:54, 31 July 2023
• 2005 AIME II Problems/Problem 2
  == Problem == *Person 1: <math>\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}</math>
  4 KB (628 words) - 11:28, 14 April 2024
• 2005 AIME II Problems/Problem 5
  == Problem == ...d pair]]s <math> (a,b) </math> of [[integer]]s such that <math> \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </mat
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• 2005 AIME II Problems/Problem 6
  == Problem == ...xample, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical st
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• 2005 AIME II Problems
  == Problem 1 == ...The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find <math>
  7 KB (1,119 words) - 21:12, 28 February 2020
• 2005 AIME II Problems/Problem 7
  == Problem == <cmath>(y^{12}+y^8+y^4+1)(y^2+1)=(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)</cmath>
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• 2005 AIME II Problems/Problem 8
  == Problem == ..._3O_3'</math> is similar to <math>O_1O_2O_2'</math> so <math>O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}{7}</math>. From rectangles, <math>O_3'T=O_1T_1=4</m
  4 KB (693 words) - 13:03, 28 December 2021
• 2005 AIME II Problems/Problem 9
  == Problem == This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</ma
  6 KB (1,154 words) - 03:30, 11 January 2024
• 2005 AIME II Problems/Problem 12
  == Problem == ...e(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F)
  13 KB (2,080 words) - 21:20, 11 December 2022
• 2006 AMC 10A Problem 7
  #REDIRECT [[2006 AMC 12A Problems/Problem 6]]
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• 2006 AMC 12A Problem 6
  #REDIRECT [[2006 AMC 12A Problems/Problem 6]]
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• 2006 AMC 12A Problems/Problem 4
  ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #4]] and [[2006 AMC 10A Problems/Problem 4|2006 AMC 10A #4]]}} == Problem ==
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• 2005 AIME I Problems
  == Problem 1 == [[2005 AIME I Problems/Problem 1|Solution]]
  6 KB (983 words) - 05:06, 20 February 2019
• 2005 AIME I Problems/Problem 1
  == Problem == [[Image:2005 AIME I Problem 1.png]]
  1 KB (213 words) - 13:17, 22 July 2017
• 2005 AIME I Problems/Problem 2
  == Problem == ...th>(6,334)</math>, <math>(12,167)</math>, <math>(167,12)</math>,<math>(334,6)</math>, <math>(501,4)</math>, <math>(668,3)</math>, <math>(1002,2)</math>
  2 KB (303 words) - 01:31, 5 December 2022
• 2005 AIME I Problems/Problem 4
  == Problem == ==Solution 6 (NO ALGEBRA)==
  8 KB (1,248 words) - 11:43, 16 August 2022
• 2005 AIME I Problems/Problem 5
  == Problem == There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.
  5 KB (830 words) - 01:51, 1 March 2023
• 2005 AIME I Problems/Problem 6
  == Problem == ==Solution 6 (De Moivre's Theorem)==
  4 KB (686 words) - 01:55, 5 December 2022
• 2005 AIME I Problems/Problem 7
  == Problem == draw((5,8.66)--(16.87,6.928));
  4 KB (567 words) - 20:20, 3 March 2020
• 2005 AIME I Problems/Problem 9
  == Problem == ...ns, so from these cubes we gain a factor of <math>\left(\frac{2}{3}\right)^6</math>.
  4 KB (600 words) - 21:44, 20 November 2023
• 2005 AIME I Problems/Problem 12
  == Problem == ...d <math> n. </math> For example, <math> \tau (1)=1 </math> and <math> \tau(6) =4. </math> Define <math> S(n) </math> by <math> S(n)=\tau(1)+ \tau(2) + \
  4 KB (647 words) - 02:29, 4 May 2021
• 2005 AIME I Problems/Problem 13
  == Problem == ...>R</math> and <math>U</math> stay together, then there are <math>3 \cdot 2=6</math> ways.
  5 KB (897 words) - 00:21, 29 July 2022
• 2005 AIME II Problems/Problem 15
  == Problem == ...n from the second gives <cmath>20a=240-40x\rightarrow a=12-2x\rightarrow x=6-\frac a2</cmath>
  12 KB (2,000 words) - 13:17, 28 December 2020
• 2005 AIME II Problems/Problem 14
  == Problem == ...> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such tha
  13 KB (2,129 words) - 18:56, 1 January 2024
• 2004 AIME I Problems/Problem 15
  == Problem == We approach the problem by [[recursion]]. We [[partition]] the positive integers into the sets
  9 KB (1,491 words) - 01:23, 26 December 2022
• 2004 AIME I Problems/Problem 14
  == Problem == real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD;
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• 2004 AIME I Problems/Problem 13
  == Problem == ...19}+ \frac3{19}+ \frac 1{17}+ \frac2{17}= \frac6{19} + \frac 3{17} = \frac{6\cdot17 + 3\cdot19}{17\cdot19} = \frac{159}{323}</math> and so the answer is
  2 KB (298 words) - 20:02, 4 July 2013
• 2004 AIME I Problems/Problem 12
  == Problem == ...left(\frac{\frac{4}{5}}{1-\frac{1}{25}}\right)= \frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},</cmath> and the answer is <math>m+n = 5 + 9 = \boxed{014}</
  2 KB (303 words) - 22:28, 11 September 2020
• 2004 AIME I Problems/Problem 10
  == Problem == ...th>r_{ABC} = \frac{[ABC]}{s_{ABC}} = \frac{15 \cdot 36 /2}{(15+36+39)/2} = 6</math>. Thus <math>r_{A'B'C'} = r_{ABC} - 1 = 5</math>, and since the ratio
  5 KB (836 words) - 07:53, 15 October 2023
• 2004 AIME I Problems/Problem 9
  == Problem == ...th> be a [[triangle]] with sides 3, 4, and 5, and <math> DEFG </math> be a 6-by-7 [[rectangle]]. A segment is drawn to divide triangle <math> ABC </math
  4 KB (618 words) - 20:01, 4 July 2013
• 2004 AIME I Problems/Problem 1
  == Problem == ...n + 1) + n = 3210 + 1111n</math>, for <math>n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace</math>.
  2 KB (374 words) - 14:53, 27 December 2019
• 2004 AIME I Problems/Problem 2
  == Problem == The thing about this problem is, you have some "choices" that you can make freely when you get to a cert
  8 KB (1,437 words) - 21:53, 19 May 2023
• 2004 AIME I Problems/Problem 5
  == Problem == Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted quest
  3 KB (436 words) - 18:31, 9 January 2024
• 2004 AIME I Problems/Problem 7
  == Problem == ...\le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*}</cmath>
  5 KB (833 words) - 19:43, 1 October 2023
• 2004 AIME I Problems/Problem 8
  == Problem == There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are similar, but there are two n
  4 KB (620 words) - 21:26, 5 June 2021
• 2004 AIME I Problems
  == Problem 1 == [[2004 AIME I Problems/Problem 1|Solution]]
  9 KB (1,434 words) - 13:34, 29 December 2021
• 2004 AIME II Problems/Problem 15
  == Problem == Now, consider the strip of length <math>1024</math>. The problem asks for <math>s_{941, 10}</math>. We can derive some useful recurrences f
  6 KB (899 words) - 20:58, 12 May 2022
• 2004 AIME II Problems/Problem 14
  == Problem == ...n't need to be nearly as rigorous). A more natural manner of attacking the problem is to think of the process in reverse, namely seeing that <math>n \equiv 1
  11 KB (1,857 words) - 21:55, 19 June 2023
• 2004 AIME II Problems/Problem 12
  == Problem == ...CD </math> be an [[isosceles trapezoid]], whose dimensions are <math> AB = 6, BC=5=DA, </math>and <math> CD=4. </math> Draw [[circle]]s of [[radius]] 3
  3 KB (431 words) - 23:21, 4 July 2013
• 2004 AIME II Problems/Problem 10
  == Problem == ...}</math>. For example, with the binary string 0001001000 <math>y</math> is 6 and <math>x</math> is 3 (note that it is zero indexed).
  8 KB (1,283 words) - 19:19, 8 May 2024
• 2004 AIME II Problems/Problem 8
  == Problem == ...ays and likewise we can partition the 167 in three ways. So we have <math>6\cdot 3\cdot 3 = \boxed{54}</math> as our answer.
  2 KB (353 words) - 18:08, 25 November 2023
• 2004 AIME II Problems/Problem 7
  == Problem == It is clear from the problem setup that <math>0<\theta<\frac\pi2</math>, so the correct value is <math>\
  9 KB (1,501 words) - 05:34, 30 October 2023
• 2004 AIME II Problems/Problem 6
  == Problem == ...al. Also note some other conditions we have picked up in the course of the problem, namely that <math>b_1</math> is divisible by <math>8</math>, <math>b_2</ma
  6 KB (950 words) - 14:18, 15 January 2024
• 2004 AIME II Problems/Problem 5
  ==Problem== From the initial problem statement, we have <math>1000w\cdot\frac{1}{4}t=\frac{1}{4}</math>.
  4 KB (592 words) - 19:02, 26 September 2020
• 2004 AIME II Problems
  == Problem 1 == [[2004 AIME II Problems/Problem 1|Solution]]
  9 KB (1,410 words) - 05:05, 20 February 2019
• 2000 AIME II
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2000 AIME II Problems/Problem 1|Problem 1]]
  1 KB (139 words) - 08:41, 7 September 2011
• 2001 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2001 AIME I Problems/Problem 1|Problem 1]]
  1 KB (139 words) - 08:41, 7 September 2011
• 2000 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2000 AIME I Problems/Problem 1|Problem 1]]
  1 KB (135 words) - 18:05, 30 May 2015
• 1999 AIME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1999 AIME Problems/Problem 1|Problem 1]]
  1 KB (118 words) - 08:41, 7 September 2011
• 1998 AIME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1998 AIME Problems/Problem 1|Problem 1]]
  1 KB (114 words) - 08:39, 7 September 2011
• 1997 AIME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1997 AIME Problems/Problem 1|Problem 1]]
  1 KB (114 words) - 08:39, 7 September 2011
• 1996 AIME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1996 AIME Problems/Problem 1|Problem 1]]
  1 KB (114 words) - 08:39, 7 September 2011
• 1995 AIME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1995 AIME Problems/Problem 1|Problem 1]]
  1 KB (114 words) - 08:38, 7 September 2011
• 1994 AIME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1994 AIME Problems/Problem 1|Problem 1]]
  1 KB (114 words) - 08:43, 7 September 2011
• 1983 AIME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1983 AIME Problems/Problem 1|Problem 1]]
  1 KB (114 words) - 20:35, 31 October 2020
• 1983 AIME Problems
  == Problem 1 == [[1983 AIME Problems/Problem 1|Solution]]
  7 KB (1,104 words) - 12:53, 6 July 2022
• 2019 AIME I Problems/Problem 2
  ==Problem== This problem is essentially asking how many ways there are to choose <math>2</math> dist
  5 KB (830 words) - 22:15, 28 December 2023
• 1984 AIME Problems
  == Problem 1 == [[1984 AIME Problems/Problem 1|Solution]]
  6 KB (933 words) - 01:15, 19 June 2022
• 1986 AIME Problems
  == Problem 1 == [[1986 AIME Problems/Problem 1|Solution]]
  5 KB (847 words) - 15:48, 21 August 2023
• 1987 AIME Problems
  == Problem 1 == [[1987 AIME Problems/Problem 1|Solution]]
  6 KB (869 words) - 15:34, 22 August 2023
• 1988 AIME Problems
  == Problem 1 == ...r -- the correct five buttons. The sample shown below has <math>\{1, 2, 3, 6, 9\}</math> as its combination. Suppose that these locks are redesigned so
  6 KB (902 words) - 08:57, 19 June 2021
• 1989 AIME Problems
  == Problem 1 == [[1989 AIME Problems/Problem 1|Solution]]
  7 KB (1,045 words) - 20:47, 14 December 2023
• 1990 AIME Problems
  == Problem 1 == The [[increasing sequence]] <math>2,3,5,6,7,10,11,\ldots</math> consists of all [[positive integer]]s that are neithe
  6 KB (870 words) - 10:14, 19 June 2021
• 1991 AIME Problems
  == Problem 1 == [[1991 AIME Problems/Problem 1|Solution]]
  7 KB (1,106 words) - 22:05, 7 June 2021
• 1992 AIME Problems
  == Problem 1 == [[1992 AIME Problems/Problem 1|Solution]]
  8 KB (1,117 words) - 05:32, 11 November 2023
• 1993 AIME Problems
  == Problem 1 == [[1993 AIME Problems/Problem 1|Solution]]
  8 KB (1,275 words) - 06:55, 2 September 2021
• 1994 AIME Problems
  == Problem 1 == [[1994 AIME Problems/Problem 1|Solution]]
  7 KB (1,141 words) - 07:37, 7 September 2018
• 1995 AIME Problems
  == Problem 1 == [[Image:AIME 1995 Problem 1.png]]
  6 KB (1,000 words) - 00:25, 27 March 2024
• 1996 AIME Problems
  == Problem 1 == [[1996 AIME Problems/Problem 1|Solution]]
  6 KB (931 words) - 17:49, 21 December 2018
• 1997 AIME Problems
  == Problem 1 == [[1997 AIME Problems/Problem 1|Solution]]
  7 KB (1,098 words) - 17:08, 25 June 2020
• 1998 AIME Problems
  == Problem 1 == ...{12}</math> the [[least common multiple]] of the positive integers <math>6^6</math> and <math>8^8</math>, and <math>k</math>?
  7 KB (1,084 words) - 02:01, 28 November 2023
• 1999 AIME Problems
  == Problem 1 == [[1999 AIME Problems/Problem 1|Solution]]
  7 KB (1,094 words) - 13:39, 16 August 2020
• 2000 AIME I Problems
  == Problem 1 == [[2000 AIME I Problems/Problem 1|Solution]]
  7 KB (1,204 words) - 03:40, 4 January 2023
• 2001 AIME I Problems
  == Problem 1 == [[2001 AIME I Problems/Problem 1|Solution]]
  7 KB (1,212 words) - 22:16, 17 December 2023
• 2002 AIME I Problems
  == Problem 1 == [[2002 AIME I Problems/Problem 1|Solution]]
  8 KB (1,374 words) - 21:09, 27 July 2023
• 2003 AIME I Problems
  == Problem 1 == [[2003 AIME I Problems/Problem 1|Solution]]
  6 KB (965 words) - 16:36, 8 September 2019
• 2000 AIME II Problems
  == Problem 1 == <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
• 2001 AIME II Problems
  == Problem 1 == [[2001 AIME II Problems/Problem 1|Solution]]
  8 KB (1,282 words) - 21:12, 19 February 2019
• 2002 AIME II Problems
  == Problem 1 == [[2002 AIME II Problems/Problem 1|Solution]]
  7 KB (1,177 words) - 15:42, 11 August 2023
• 2003 AIME II Problems
  == Problem 1 == The product <math>N</math> of three positive integers is 6 times their sum, and one of the integers is the sum of the other two. Find
  7 KB (1,127 words) - 09:02, 11 July 2023
• 1983 AIME Problems/Problem 1
  == Problem == == Solution 6 ==
  4 KB (642 words) - 03:14, 17 August 2022
• 1983 AIME Problems/Problem 3
  == Problem == ...non-negative (and moreover, plugging in <math>y=-6</math>, we get <math>-6=6</math>, which is obviously false). Hence we have <math>y=10</math> as the o
  3 KB (532 words) - 05:18, 21 July 2022
• 1983 AIME Problems/Problem 4
  ==Problem== ...ircle is <math>\sqrt{50}</math> cm, the length of <math>AB</math> is <math>6</math> cm and that of <math>BC</math> is <math>2</math> cm. The angle <math
  11 KB (1,741 words) - 22:40, 23 November 2023
• 1983 AIME Problems/Problem 5
  == Problem == One way to solve this problem is by [[substitution]]. We have
  4 KB (672 words) - 10:17, 17 March 2023
• 1983 AIME Problems/Problem 6
  == Problem == Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math
  3 KB (361 words) - 20:20, 14 January 2023
• 1983 AIME Problems/Problem 7
  == Problem == ...in which the trio sits together (as a single entity). There are <math>3! = 6</math> ways to determine their order, and there are <math>(23-1)! = 22!</ma
  9 KB (1,392 words) - 20:37, 19 January 2024
• 1983 AIME Problems/Problem 10
  == Problem == There are <math>6</math> possible sequences.
  5 KB (855 words) - 20:26, 14 January 2023
• 1983 AIME Problems/Problem 11
  == Problem == ...>2s</math>. All other edges have length <math>s</math>. Given that <math>s=6\sqrt{2}</math>, what is the volume of the solid?
  5 KB (865 words) - 21:11, 6 February 2023
• 1983 AIME Problems/Problem 12
  == Problem == draw((-2.6,0)--(-2.6,0.6));
  2 KB (412 words) - 18:23, 1 January 2024
• 1983 AIME Problems/Problem 13
  == Problem == ...example, the alternating sum for <math>\{1, 2, 3, 6,9\}</math> is <math>9-6+3-2+1=5</math> and for <math>\{5\}</math> it is simply <math>5</math>. Find
  5 KB (894 words) - 22:02, 5 April 2024
• 1983 AIME Problems/Problem 14
  == Problem == In the adjoining figure, two circles with radii <math>8</math> and <math>6</math> are drawn with their centers <math>12</math> units apart. At <math>P
  13 KB (2,149 words) - 18:44, 5 February 2024
• 1983 AIME Problems/Problem 15
  == Problem == ...h>. Suppose that the radius of the circle is <math>5</math>, that <math>BC=6</math>, and that <math>AD</math> is bisected by <math>BC</math>. Suppose fu
  19 KB (3,221 words) - 01:05, 7 February 2023
• 1984 AIME Problems/Problem 3
  == Problem == D(A--B--C--cycle); D(A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12);
  4 KB (726 words) - 13:39, 13 August 2023
• 1984 AIME Problems/Problem 5
  == Problem == == Solution 6 ==
  5 KB (782 words) - 14:49, 1 August 2023
• 1984 AIME Problems/Problem 6
  == Problem == [[Image:1984_AIME-6.png]]
  6 KB (1,022 words) - 19:29, 22 January 2024
• 1984 AIME Problems/Problem 7
  == Problem == Open up a coding IDE and use recursion to do this problem. The idea is to define a function (I called it <math>f</math>, you can call
  4 KB (617 words) - 18:01, 9 March 2022
• 1984 AIME Problems/Problem 8
  == Problem == The equation <math>z^6+z^3+1=0</math> has complex roots with argument <math>\theta</math> between
  3 KB (430 words) - 19:05, 7 February 2023
• 1984 AIME Problems/Problem 9
  == Problem == draw(surface(A--B--C--cycle),rgb(0.6,0.7,0.6),nolight);
  6 KB (947 words) - 20:44, 26 November 2021
• 1984 AIME Problems/Problem 11
  == Problem == ...>s, we are placing the extra <math>3</math> <math>n</math>s into the <math>6</math> intervals beside the <math>b</math>s.
  7 KB (1,115 words) - 00:52, 7 September 2023
• 1984 AIME Problems/Problem 14
  == Problem == ...either <math>0 \bmod{6}</math>, <math>2 \bmod{6}</math>, or <math>4 \bmod{6}</math>. Notice that the numbers <math>9</math>, <math>15</math>, <math>21<
  8 KB (1,346 words) - 01:16, 9 January 2024
• 1984 AIME Problems/Problem 15
  == Problem == ...math> \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 </math><br /><math> \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^
  6 KB (1,051 words) - 04:52, 8 May 2024
• 1995 IMO Problems/Problem 2
  == Problem == === Solution 6 ===
  6 KB (1,122 words) - 12:23, 6 January 2022
• 1985 AIME Problems/Problem 15
  == Problem == [[Image:AIME 1985 Problem 15.png]]
  2 KB (245 words) - 22:44, 4 March 2024
• 1985 AIME Problems/Problem 14
  == Problem == ...0 = n^2 + 19n + 90</math> and <math>n^2 -21n + 90 = 0</math> and <math>n = 6</math> or <math>n = 15</math>. Now, note that the top <math>n</math> playe
  5 KB (772 words) - 22:14, 18 June 2020
• 1985 AIME Problems/Problem 12
  == Problem == <cmath>\begin{alignat*}{6}
  17 KB (2,837 words) - 13:34, 4 April 2024
• 1985 AIME Problems/Problem 10
  == Problem == ...mum we can get is <math>1+2+3 = 6</math>, so we only need to try the first 6 numbers.
  12 KB (1,859 words) - 18:16, 28 March 2022
• 1985 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1985|num-b=6|num-a=8}}
  1 KB (222 words) - 11:04, 4 November 2022
• 1985 AIME Problems/Problem 5
  == Problem == ...r this point it must continue to repeat. Thus, in particular <math>a_{j + 6} = a_j</math> for all <math>j</math>, and so repeating this <math>n</math>
  2 KB (410 words) - 13:37, 1 May 2022
• 1985 AIME Problems/Problem 3
  == Problem == ...> (since we know <math>a</math> is positive). Thus <math>c = 6^3 - 3\cdot 6 = \boxed{198}</math>.
  1 KB (205 words) - 18:58, 10 March 2024
• 1986 AIME Problems/Problem 15
  == Problem == f.p=fontsize(6);
  11 KB (1,722 words) - 09:49, 13 September 2023
• 1986 AIME Problems/Problem 12
  == Problem == ...th>\dbinom{6}{0} + \dbinom{6}{1} + \dbinom{6}{2} + \dbinom{6}{3} + \dbinom{6}{4}=57</math> of its subsets have at most four elements (the number of subs
  2 KB (364 words) - 19:41, 1 September 2020
• 1986 AIME Problems/Problem 11
  == Problem == ...first 16 triangular numbers, which evaluates to <math>\frac{(16)(17)(18)}{6} = \boxed{816}</math>.
  6 KB (872 words) - 16:51, 9 June 2023
• 1986 AIME Problems/Problem 9
  == Problem == pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);
  11 KB (1,850 words) - 18:07, 11 October 2023
• 1986 AIME Problems/Problem 8
  == Problem == ...ctorization]] of <math>1000000 = 2^65^6</math>, so there are <math>(6 + 1)(6 + 1) = 49</math> divisors, of which <math>48</math> are proper. The sum of
  3 KB (487 words) - 20:52, 16 September 2020
• 1986 AIME Problems/Problem 7
  == Problem == ...owever, we must change it back to base 10 for the answer, which is <math>3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed {981}</math>.
  5 KB (866 words) - 00:00, 22 December 2022
• 1986 AIME Problems/Problem 5
  == Problem == The key to this problem is to realize that <math>n+10 \mid n^3 +1000</math> for all <math>n</math>.
  2 KB (338 words) - 19:56, 15 October 2023
• 1986 AIME Problems/Problem 4
  == Problem == <center><math>2x_1+x_2+x_3+x_4+x_5=6</math></center>
  1 KB (212 words) - 16:25, 17 November 2019
• 1986 AIME Problems/Problem 3
  == Problem == ...tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}</math>
  3 KB (545 words) - 23:44, 12 October 2023
• 1986 AIME Problems/Problem 2
  == Problem == ...sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).</cmath>
  3 KB (460 words) - 00:44, 5 February 2022
• 1987 AIME Problems/Problem 14
  == Problem == &= (x(x-6) + 18)(x(x+6)+18),
  7 KB (965 words) - 10:42, 12 April 2024
• 1987 AIME Problems/Problem 12
  == Problem == ...that <math>(n + \frac{1}{1000})^3 = n^3 + \frac{3}{10^3} n^2 + \frac{3}{10^6} n + \frac{1}{10^9}</math>. For a given value of <math>n</math>, if <math>(
  4 KB (673 words) - 19:48, 28 December 2023
• 1987 AIME Problems/Problem 11
  == Problem == ...or large factors of <math>2\cdot 3^{11}</math> which are less than <math>3^6</math>. The largest such factor is clearly <math>2\cdot 3^5 = 486</math>;
  3 KB (418 words) - 18:30, 20 January 2024
• 1987 AIME Problems/Problem 9
  == Problem == ...ains a [[point]] <math>P</math> for which <math>PA = 10</math>, <math>PB = 6</math>, and <math>\angle APB = \angle BPC = \angle CPA</math>. Find <math>
  1 KB (200 words) - 18:44, 5 February 2024
• 1987 AIME Problems/Problem 8
  == Problem == Since <math>91n - 104k < n + k</math>, <math>k > \frac{6}{7}n</math>. Also, <math>0 < 91n - 104k</math>, so <math>k < \frac{7n}{8}</
  2 KB (393 words) - 16:59, 16 December 2020
• 1987 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1987|num-b=6|num-a=8}}
  3 KB (547 words) - 22:54, 4 April 2016
• 1987 AIME Problems/Problem 5
  == Problem == {{AIME box|year=1987|num-b=4|num-a=6}}
  1 KB (160 words) - 04:44, 21 January 2023
• 1987 AIME Problems/Problem 3
  == Problem == Thus, listing out the first ten numbers to fit this form, <math>2 \cdot 3 = 6,\ 2^3 = 8,\ 2 \cdot 5 = 10,</math> <math>\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,
  3 KB (511 words) - 09:29, 9 January 2023
• 1988 AIME Problems/Problem 15
  == Problem == ...ring the day, and the boss delivers them in the order <math>1, 2, 3, 4, 5, 6, 7, 8, 9</math>.
  7 KB (1,186 words) - 10:16, 4 June 2023
• 1988 AIME Problems/Problem 13
  == Problem == ...+ 1</math> into the polynomial with a higher degree, as shown in Solution 6.
  10 KB (1,585 words) - 03:58, 1 May 2023
• 1988 AIME Problems/Problem 10
  == Problem == ...[[face]]s 12 [[square]]s, 8 [[regular polygon|regular]] [[hexagon]]s, and 6 regular [[octagon]]s. At each [[vertex]] of the polyhedron one square, one
  5 KB (811 words) - 19:10, 25 January 2021
• 1988 AIME Problems/Problem 9
  == Problem == ...this in to <math>5a^2 + 12a \equiv 1 \pmod{25}</math> gives <math>10a_1 + 6 \equiv 1 \pmod{25}</math>. Obviously, <math>a_1 \equiv 2 \pmod 5</math>, so
  6 KB (893 words) - 08:15, 2 February 2023
• 1988 AIME Problems/Problem 8
  == Problem == ...8}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot f(4,6)\\
  4 KB (538 words) - 13:24, 12 October 2021
• 1988 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1988|num-b=6|num-a=8}}
  1 KB (178 words) - 23:25, 20 November 2023
• 1988 AIME Problems/Problem 6
  == Problem == [[Image:1988_AIME-6.png]]
  5 KB (878 words) - 23:06, 20 November 2023
• 1988 AIME Problems/Problem 5
  == Problem == {{AIME box|year=1988|num-b=4|num-a=6}}
  822 bytes (108 words) - 22:21, 6 November 2016
• 1988 AIME Problems/Problem 1
  == Problem == ...rder -- the correct five buttons. The sample shown below has <math>\{1,2,3,6,9\}</math> as its [[combination]]. Suppose that these locks are redesigned
  1 KB (181 words) - 18:23, 26 August 2019
• 1989 AIME Problems/Problem 15
  == Problem == ...e figure below). Given that <math>AP=6</math>, <math>BP=9</math>, <math>PD=6</math>, <math>PE=3</math>, and <math>CF=20</math>, find the area of <math>\
  13 KB (2,091 words) - 00:20, 26 October 2023
• 1989 AIME Problems/Problem 13
  == Problem == ...pairs: <math>[2,9]</math>, <math>[3,7]</math>, <math>[4,11]</math>, <math>[6,10]</math>. Now, <math>1989 = 180\cdot 11 + 9</math>. Because this isn't an
  2 KB (274 words) - 04:07, 17 December 2023
• 1989 AIME Problems/Problem 10
  == Problem == === Solution 6===
  8 KB (1,401 words) - 21:41, 20 January 2024
• 1989 AIME Problems/Problem 9
  == Problem == &>1.6\cdot1.6\cdot1.3 \\
  6 KB (874 words) - 15:50, 20 January 2024
• 1989 AIME Problems/Problem 8
  == Problem == ...math>f(k)=k^2x_1+(k+1)^2x_2+(k+2)^2x_3+(k+3)^2x_4+(k+4)^2x_5+(k+5)^2x_6+(k+6)^2x_7</cmath> for some <math>k\in\{1,2,3\}.</math>
  8 KB (1,146 words) - 04:15, 20 November 2023
• 1989 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1989|num-b=6|num-a=8}}
  2 KB (239 words) - 16:09, 2 June 2023
• 1989 AIME Problems/Problem 6
  == Problem == pair A=(0,0),B=(10,0),C=6*expi(pi/3);
  5 KB (864 words) - 19:55, 2 July 2023
• 1989 AIME Problems/Problem 5
  == Problem == ...g a heads in one flip of the biased coin as <math>h</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3</math>.
  2 KB (258 words) - 00:07, 25 June 2023
• 1989 AIME Problems/Problem 4
  == Problem == ...> in which plugging into the other expression we know <math>3(25k^3 - 2) + 6 = 75k^3</math> is a perfect square. We know <math>75 = 5^2 \cdot 3</math> s
  3 KB (552 words) - 12:41, 3 March 2024
• 1989 AIME Problems/Problem 1
  == Problem == ...28)+1=755161.</math> Since the alternating sum of the digits <math>7-5+5-1+6-1=11</math> is divisible by <math>11,</math> we conclude that <math>755161<
  4 KB (523 words) - 00:12, 8 October 2021
• 1990 AIME Problems/Problem 14
  == Problem == pair D=origin, A=(13,0), B=(13,12), C=(0,12), P=(6.5, 6);
  7 KB (1,086 words) - 08:16, 29 July 2023
• 1990 AIME Problems/Problem 12
  == Problem == ...gon can be written in the form <math>a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6},</math> where <math>a^{}_{}</math>, <math>b^{}_{}</math>, <math>c^{}_{}</m
  6 KB (906 words) - 13:23, 5 September 2021
• 1990 AIME Problems/Problem 11
  == Problem == Someone observed that <math>6! = 8 \cdot 9 \cdot 10</math>. Find the largest [[positive]] [[integer]] <m
  3 KB (519 words) - 09:28, 28 June 2022
• 1990 AIME Problems/Problem 10
  == Problem == ...^{16}, \ldots n^{144}\}</math> and of set <math>B</math> as <math>\{n^3, n^6, \ldots n^{144}\}</math>. <math>n^x</math> can yield at most <math>144</mat
  3 KB (564 words) - 04:47, 4 August 2023
• 1990 AIME Problems/Problem 9
  == Problem == ...pattern, we find that there are <math>\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}
  3 KB (425 words) - 12:36, 12 May 2024
• 1990 AIME Problems/Problem 7
  == Problem == <!-- replaced: Image:1990 AIME Problem 7.png by I_like_pie -->
  8 KB (1,319 words) - 11:34, 22 November 2023
• 1990 AIME Problems/Problem 5
  == Problem == {{AIME box|year=1990|num-b=4|num-a=6}}
  1 KB (175 words) - 03:45, 21 January 2023
• 1990 AIME Problems/Problem 2
  == Problem == Find the value of <math>(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}</math>.
  5 KB (765 words) - 23:00, 26 August 2023
• 1990 AIME Problems/Problem 1
  == Problem == The [[increasing sequence]] <math>2,3,5,6,7,10,11,\ldots</math> consists of all [[positive integer]]s that are neithe
  2 KB (283 words) - 23:11, 25 June 2023
• 2006 AMC 10B Problems/Problem 4
  == Problem == <math> \textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 6\qquad \textbf{(D) } 8\qquad \textbf{(E) } 9 </math>
  1 KB (172 words) - 10:47, 19 December 2021
• 2006 AMC 10B Problems/Problem 5
  == Problem == {{AMC10 box|year=2006|ab=B|num-b=4|num-a=6}}
  1 KB (242 words) - 18:35, 15 August 2023
• 2006 AMC 10B Problems/Problem 7
  == Problem == {{AMC10 box|year=2006|ab=B|num-b=6|num-a=8}}
  1 KB (179 words) - 10:33, 19 August 2022
• 2006 AMC 10B Problems/Problem 11
  == Problem == ...bf{(A) } 1\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 6\qquad \textbf{(E) } 9 </math>
  1 KB (170 words) - 14:00, 26 January 2022
• 2006 AMC 10B Problems/Problem 13
  == Problem == ...f{(C) }1 \qquad \textbf{(D) \ } \frac{14}{13}\qquad \textbf{(E) } \frac{7}{6} </math>
  1 KB (182 words) - 14:11, 26 January 2022
• 1991 AIME Problems/Problem 13
  == Problem == ...inds that <math>n=44</math> is the largest possible integer satisfying the problem conditions.
  7 KB (1,328 words) - 20:24, 5 February 2024
• 2006 AMC 10B Problems/Problem 15
  == Problem == ...4\sqrt{3}\qquad \textbf{(C) } 8\qquad \textbf{(D) } 9\qquad \textbf{(E) } 6\sqrt{3} </math>
  3 KB (445 words) - 22:01, 20 August 2022
• 1991 AIME Problems/Problem 12
  == Problem == ...o more similar triangles, we find that point <math>E</math> is at <math>(9.6, 7.2)</math>.
  8 KB (1,270 words) - 23:36, 27 August 2023
• 2006 AMC 10B Problems/Problem 16
  == Problem == ...multiples of <math>4</math> (with a few exceptions that don't affect this problem).
  2 KB (336 words) - 10:51, 11 May 2024
• 1991 AIME Problems/Problem 11
  == Problem == dot((cos(i*pi/6), sin(i*pi/6)));
  4 KB (740 words) - 19:33, 28 December 2022
• 2002 Pan African MO
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2002 Pan African MO Problems/Problem 1]]
  581 bytes (73 words) - 13:47, 4 December 2019
• 2006 AMC 10B Problems/Problem 17
  == Problem == <math> \textbf{(A) } \frac{1}{10}\qquad \textbf{(B) } \frac{1}{6}\qquad \textbf{(C) } \frac{1}{5}\qquad \textbf{(D) } \frac{1}{3}\qquad \tex
  1 KB (211 words) - 04:32, 4 November 2022
• 1991 AIME Problems/Problem 9
  == Problem == Note: The problem is much easier computed if we consider what <math>\sec (x)</math> is, then
  10 KB (1,590 words) - 14:04, 20 January 2023
• 2006 AMC 10B Problems/Problem 18
  == Problem == Clearly, the sequence repeats every 6 terms.
  1 KB (158 words) - 01:33, 29 May 2023
• 1991 AIME Problems/Problem 8
  == Problem == ...ath>(\pm1,\pm144),( \pm 2, \pm 72),( \pm 3, \pm 48),( \pm 4, \pm 36),( \pm 6, \pm 24),( \pm 8, \pm 18),( \pm 9, \pm 16),( \pm 12, \pm 12)</math>; since
  2 KB (310 words) - 11:25, 13 June 2023
• 1991 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1991|num-b=6|num-a=8}}
  2 KB (285 words) - 05:15, 13 June 2022
• 1991 AIME Problems/Problem 5
  == Problem == {{AIME box|year=1991|num-b=4|num-a=6}}
  919 bytes (141 words) - 20:00, 4 July 2022
• 1991 AIME Problems/Problem 4
  == Problem == ...<math>-1 \le y \le 1</math>. It is [[periodic function|periodic]] (in this problem) with a period of <math>\frac{2}{5}</math>.
  2 KB (300 words) - 16:01, 26 November 2019
• 1991 AIME Problems/Problem 3
  == Problem == ...st going to be equivalent to multiplying this fraction by <math>\frac{995}{6}</math>. Notice that this fraction's numerator plus denominator is equal to
  5 KB (865 words) - 12:13, 21 May 2020
• 1992 AIME Problems/Problem 12
  == Problem == ...g this inequality may be found by Stars and Bars to be <math>\binom{7+6-1}{6-1} = \boxed{792}</math>.
  2 KB (443 words) - 22:41, 22 December 2021
• 1992 AIME Problems/Problem 9
  == Problem == ...> is irrelevant as long as there still exists a circle as described in the problem.
  5 KB (874 words) - 10:27, 22 August 2021
• 1992 AIME Problems/Problem 8
  == Problem == ...means <math>A</math> looks like <math>(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)</math>. More specifically, <math>A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}</
  5 KB (778 words) - 21:36, 3 December 2022
• 1992 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1992|num-b=6|num-a=8}}
  800 bytes (114 words) - 17:40, 14 March 2017
• 1992 AIME Problems/Problem 6
  == Problem == ...math>h+d\ge 10</math> or <math>c+g\ge 10</math> or <math>b+f\ge 10</math>. 6. Consider <math>c \in \{0, 1, 2, 3, 4\}</math>. <math>1abc + 1ab(c+1)</mat
  3 KB (455 words) - 02:03, 10 July 2021
• 1992 AIME Problems/Problem 5
  == Problem == {{AIME box|year=1992|num-b=4|num-a=6}}
  2 KB (277 words) - 20:45, 4 March 2024
• 1992 AIME Problems/Problem 4
  == Problem == \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt}
  3 KB (476 words) - 14:13, 20 April 2024
• 1992 AIME Problems/Problem 3
  == Problem == ...math>1000n+3000>1006n+2012</math>, so <math>n<\frac{988}{6}=164 \dfrac {4}{6}=164 \dfrac{2}{3}</math>. Thus, the answer is <math>\boxed{164}</math>.
  2 KB (251 words) - 08:05, 2 January 2024
• 1992 AIME Problems/Problem 2
  == Problem == ...ntial ascending numbers, one for each [[subset]] of <math>\{1, 2, 3, 4, 5, 6, 7, 8, 9\}</math>.
  2 KB (336 words) - 05:18, 4 November 2022
• 2006 AMC 10B Problems/Problem 19
  == Problem == \qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3} </math
  5 KB (873 words) - 15:39, 29 May 2023
• 2006 AMC 10B Problems/Problem 20
  == Problem == In rectangle <math>ABCD</math>, we have <math>A=(6,-22)</math>, <math>B=(2006,178)</math>, <math>D=(8,y)</math>, for some inte
  4 KB (594 words) - 15:45, 30 July 2023
• 2006 AMC 10B Problems/Problem 21
  == Problem == ...5</math>, and <math>6</math>, on each die are in the ratio <math>1:2:3:4:5:6</math>. What is the probability of rolling a total of <math>7</math> on the
  3 KB (484 words) - 19:09, 15 October 2023
• 2006 AMC 10B Problems/Problem 23
  == Problem == label("$y$",(1.6,1));
  5 KB (861 words) - 00:53, 25 November 2023
• 2006 AMC 10B Problems/Problem 24
  == Problem == pair X=(-6,0), O=origin, P=(6,0), B=tangent(X, O, 2, 1), A=tangent(X, O, 2, 2), C=tangent(X, P, 4, 1), D=
  4 KB (558 words) - 14:38, 6 April 2024
• 2006 AMC 10B Problems/Problem 25
  == Problem == <math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math>
  5 KB (878 words) - 14:39, 3 December 2023
• 1993 AIME Problems/Problem 14
  == Problem == ...es of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form <math>\sqrt{N}\,</math>
  3 KB (601 words) - 09:25, 19 November 2023
• 1993 AIME Problems/Problem 12
  == Problem == n & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\
  4 KB (611 words) - 13:59, 15 July 2023
• 1993 AIME Problems/Problem 11
  == Problem == In order to begin this problem, we need to calculate the probability that Alfred will win on the first rou
  7 KB (1,058 words) - 20:57, 22 December 2020
• 1993 AIME Problems/Problem 9
  == Problem == ...f <math>16</math>, <math>n \equiv 118 \pmod {125}</math> and <math>n\equiv 6 \pmod {16}</math>. The smallest <math>n</math> for this case is <math>118</
  3 KB (488 words) - 02:06, 22 September 2023
• 1993 AIME Problems/Problem 8
  == Problem == ...th> elements of <math>S.</math> So our final answer is then <math>\frac {3^6 - 1}{2} + 1 = \boxed{365}.</math>
  9 KB (1,400 words) - 14:09, 12 January 2024
• 1993 AIME Problems/Problem 7
  == Problem == There is a total of <math>P(1000,6)</math> possible ordered <math>6</math>-tuples <math>(a_1,a_2,a_3,b_1,b_2,b_3).</math>
  5 KB (772 words) - 09:04, 7 January 2022
• 1993 AIME Problems/Problem 5
  == Problem == {{AIME box|year=1993|num-b=4|num-a=6}}
  2 KB (355 words) - 13:25, 31 December 2018
• 1993 AIME Problems/Problem 4
  == Problem == ...s of <math>186</math>, which are <math>1 \cdot 186, 2\cdot 93, 3 \cdot 62, 6\cdot 31</math>. <math>(d-c)</math> and <math>(d+c-a-b)</math> must be facto
  8 KB (1,343 words) - 16:27, 19 December 2023
• 1993 AIME Problems/Problem 3
  == Problem == :(b) those who caught <math>3</math> or more fish averaged <math>6</math> fish each;
  2 KB (364 words) - 00:05, 9 July 2022
• 1993 AIME Problems/Problem 2
  == Problem == ...)^2}{2} - \sum_{i=0}^9 \frac{(4i+4)^2}{2}\right| = \left|\sum_{i=0}^9 -(8i+6) \right|.</cmath>
  2 KB (241 words) - 11:56, 13 March 2015
• 1993 AIME Problems/Problem 1
  == Problem == The thousands digit is <math>\in \{4,5,6\}</math>.
  3 KB (440 words) - 21:20, 22 July 2021
• 1994 AIME Problems/Problem 15
  == Problem == ...[circumcenter]]s of <math>\triangle PAB, PBC, PCA</math>. According to the problem statement, the circumcenters of the triangles cannot lie within the interio
  4 KB (717 words) - 22:20, 3 June 2021
• 1994 AIME Problems/Problem 12
  == Problem == ...math>\frac{52}{24}n = \frac {13}6n</math> squares in every row. Then <math>6|n</math>, and our goal is to maximize the value of <math>n</math>.
  3 KB (473 words) - 17:06, 1 January 2024
• 1994 AIME Problems/Problem 11
  == Problem == ...'s, <math>6</math>'s, and <math>15</math>'s. Because <math>0</math>, <math>6</math>, and <math>15</math> are all multiples of <math>3</math>, the change
  4 KB (645 words) - 15:12, 15 July 2019
• 1994 AIME Problems/Problem 10
  == Problem == ...- DC^2 = (BC + DC)(BC - DC) = 29^6</math>. Trying out factors of <math>29^6</math>, we can either guess and check or just guess to find that <math>BC +
  3 KB (534 words) - 16:23, 26 August 2018
• 1994 AIME Problems/Problem 9
  == Problem == ...math>P_k = \frac {3}{2k - 1}P_{k - 1}</math>. Iterating this for <math>k = 6,5,4,3,2</math> (obviously <math>P_1 = 1</math>), we get <math>\frac {3^5}{1
  3 KB (589 words) - 14:18, 21 July 2019
• 1994 AIME Problems/Problem 8
  ==Problem== .../math> and our displacement is <math>\left[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}\right]</math>. (Do you see why we switched <math>x</math> and <math>y</mat
  5 KB (788 words) - 13:53, 8 July 2023
• 1994 AIME Problems/Problem 7
  == Problem == ...h>(p,q)</math> and <math>(-p,-q)</math>, for a total of <math>\frac{12}{2}=6</math> lines. Finally, we add the <math>12</math> unique tangent lines to t
  3 KB (442 words) - 19:51, 8 January 2024
• 1994 AIME Problems/Problem 6
  == Problem == ...}\right)^2 = 100</math>. Thus, the total number of unit triangles is <math>6 \times 100 = 600</math>.
  4 KB (721 words) - 16:14, 8 March 2021
• 1994 AIME Problems/Problem 5
  == Problem == <center><math>(1+ 1 +2+3+4+5+6+7+8+9) (1+ 1 +2+3+\cdots+9) (1+ 1 +2+3+\cdots+9)</math></center>
  2 KB (275 words) - 19:27, 4 July 2013
• 1995 AIME Problems/Problem 14
  == Problem == ...48*expi(11*pi/6), D=E+48*expi(7*pi/6), A=E+30*expi(5*pi/6), C=E+30*expi(pi/6), F=foot(O,B,A);
  3 KB (484 words) - 13:11, 14 January 2023
• 1995 AIME Problems/Problem 13
  == Problem == .../math> to <math>k = 6</math>, we get <math>\sum_{k=1}^{6} 4k^2 + 1 = 364 + 6 = 370</math> (either adding or using the [[perfect square|sum of consecutiv
  2 KB (287 words) - 01:25, 12 December 2019
• 1995 AIME Problems/Problem 8
  == Problem == .... This yields the equations <math>x = 2a+1, 6a+2, 12a+3, 20a+4, 30a+5, 42a+6, 56a+7, 72a+8, 90a+9</math>.
  4 KB (646 words) - 17:37, 1 January 2024
• 1995 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1995|num-b=6|num-a=8}}
  3 KB (427 words) - 09:23, 13 December 2023
• 1995 AIME Problems/Problem 5
  == Problem == ...ive <math>3+4i</math>. This gets three equations necessary for solving the problem. <cmath>p+r = 3</cmath> <cmath>pr-qs = 13</cmath> <cmath>-q-s = 4</cmath> S
  3 KB (451 words) - 15:02, 6 September 2021
• 1995 AIME Problems/Problem 4
  == Problem == Circles of radius <math>3</math> and <math>6</math> are externally tangent to each other and are internally tangent to a
  3 KB (605 words) - 11:30, 5 May 2024
• 1995 AIME Problems/Problem 3
  == Problem == ...umber of steps the object may have taken is either <math>4</math> or <math>6</math>.
  3 KB (602 words) - 23:15, 16 June 2019
• 1996 AIME Problems/Problem 15
  == Problem == <cmath>1 - \cos^2 2 \theta = \frac{\cos 4\theta - \cos 6 \theta}{2},</cmath>
  5 KB (710 words) - 21:04, 14 September 2020
• 1996 AIME Problems/Problem 14
  == Problem == <math>\gcd(150,324)=6</math>,
  5 KB (923 words) - 21:21, 22 September 2023
• 1996 AIME Problems/Problem 13
  == Problem == In [[triangle]] <math>ABC</math>, <math>AB=\sqrt{30}</math>, <math>AC=\sqrt{6}</math>, and <math>BC=\sqrt{15}</math>. There is a point <math>D</math> for
  3 KB (521 words) - 01:18, 25 February 2016
• 1996 AIME Problems/Problem 12
  == Problem == &= \frac{9 \cdot 10 + 7 \cdot 9 + 5 \cdot 8 + 3 \cdot 7 + 1 \cdot 6 - 1 \cdot 5 - 3 \cdot 4 - 5 \cdot 3 - 7 \cdot 2 - 9 \cdot 1}{9} \\
  5 KB (879 words) - 11:23, 5 September 2021
• 1996 AIME Problems/Problem 11
  == Problem == Let <math>\mathrm {P}</math> be the product of the [[root]]s of <math>z^6+z^4+z^3+z^2+1=0</math> that have a positive [[imaginary]] part, and suppose
  6 KB (1,022 words) - 20:23, 17 April 2021
• 1996 AIME Problems/Problem 9
  == Problem == ...<math>22 \pmod {32}</math>. He then goes ahead and opens all lockers <math>6 \pmod {32}</math>, leaving <math>22 \pmod {64}</math> or <math>54 \pmod {64
  3 KB (525 words) - 23:51, 6 September 2023
• 1996 AIME Problems/Problem 8
  == Problem == ...> is the harmonic mean of <math>x</math> and <math>y</math> equal to <math>6^{20}</math>?
  1 KB (155 words) - 19:32, 4 July 2013
• 1996 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1996|num-b=6|num-a=8}}
  4 KB (551 words) - 11:44, 26 June 2020
• 1996 AIME Problems/Problem 6
  == Problem == ...rical to the first one. Therefore, there are <math>5 \cdot 2^6 + 5 \cdot 2^6</math> ways for an undefeated or winless team.
  3 KB (461 words) - 01:00, 19 June 2019
• 1996 AIME Problems/Problem 5
  == Problem == {{AIME box|year=1996|num-b=4|num-a=6}}
  3 KB (585 words) - 22:08, 19 November 2022
• 1996 AIME Problems/Problem 4
  == Problem == label("$6$",(unit*(r+1)/2,0,0),N);
  2 KB (257 words) - 17:50, 4 January 2016
• 1997 AIME Problems/Problem 14
  == Problem == ...)\theta) \ge \frac{\sqrt{3}}{2}</cmath>Thus, <cmath>|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \left\lfloor \frac{1997}{12} \right\rfloor =166
  5 KB (874 words) - 22:30, 1 April 2022
• 1997 AIME Problems/Problem 13
  ==Problem== So a total of <math>6</math> doublings = <math>2^6</math> = <math>64</math>, the total length = <math>64 \cdot \sqrt {2} = a\s
  7 KB (1,225 words) - 19:56, 4 August 2021
• 1997 AIME Problems/Problem 12
  == Problem == In our problem <math>f^2(x) = x</math>. It follows that
  11 KB (2,063 words) - 22:59, 21 October 2023
• 1997 AIME Problems/Problem 11
  == Problem 11 == === Solution 6 ===
  10 KB (1,514 words) - 14:35, 29 March 2024
• 1997 AIME Problems/Problem 10
  == Problem == ...reating the set as ordered. The final solution is then <math>\frac{729-27}{6}=\boxed{117}</math>
  3 KB (585 words) - 19:37, 25 April 2022
• 1997 AIME Problems/Problem 9
  == Problem == ...5\cdot 3^4 \cdot 5 + 20\cdot 3^3 \cdot 5\sqrt{5} + 15 \cdot 3^2 \cdot 25 + 6 \cdot 3 \cdot 25\sqrt{5} + 5^3\right)</math> <math>= \frac{1}{64}\left(1030
  4 KB (586 words) - 21:53, 30 December 2023
• 1997 AIME Problems/Problem 8
  == Problem == ...more detailed explanations, see related [[2007 AIME I Problems/Problem 10|problem (AIME I 2007, 10)]].''
  4 KB (638 words) - 16:41, 22 January 2024
• 1997 AIME Problems/Problem 7
  == Problem == <math>\left(\frac{1}{6}t\right)^2 + \left(-110 + \frac{1}{2}t\right)^2 = 51^2</math>.
  4 KB (617 words) - 18:47, 17 July 2022
• 1997 AIME Problems/Problem 6
  == Problem == [[Image:1997_AIME-6.png]]
  3 KB (497 words) - 00:39, 22 December 2018
• 1997 AIME Problems/Problem 5
  == Problem == {{AIME box|year=1997|num-b=4|num-a=6}}
  1 KB (208 words) - 11:46, 4 June 2021
• 1997 AIME Problems/Problem 2
  == Problem == ...mula, that gives us <math>s=1^2+2^2+\cdots+8^2=\dfrac{(8)(8+1)(2\cdot8+1)}{6}=12*17=204</math>.
  3 KB (416 words) - 21:09, 27 October 2022
• 1998 AIME Problems/Problem 15
  == Problem == { }_{2} X_{4},{ }_{4} X_{6}, \ldots,{ }_{n-2} X_{n},(n, n+1)(n+1,1)(1, n+2)(n+2,2)
  9 KB (1,671 words) - 22:10, 15 March 2024
• 1998 AIME Problems/Problem 9
  == Problem == <math>\frac{(60-m)^2}{60^2} = .6</math><br />
  4 KB (624 words) - 18:34, 18 February 2018
• 1998 AIME Problems/Problem 8
  == Problem == | 0 || 1 || 2 || 3 || 4 || 5 || 6
  2 KB (354 words) - 19:37, 24 September 2023
• 1998 AIME Problems/Problem 7
  == Problem == Note that this is an algebraic bijection, we have simplified the problem and essentially removed the odd condition, so now we can finish with plain
  5 KB (684 words) - 11:41, 13 August 2023
• 1998 AIME Problems/Problem 6
  == Problem == [[Image:AIME_1998-6.png|350px]]
  2 KB (254 words) - 19:38, 4 July 2013
• 1998 AIME Problems/Problem 5
  == Problem == Though the problem may appear to be quite daunting, it is actually not that difficult. <math>\
  1 KB (225 words) - 02:20, 16 September 2017
• 1998 AIME Problems/Problem 4
  == Problem == ...m{2}{2} = 6</math> ways. This gives us a total of <math>10 \cdot 2 \cdot 6 = 120</math> possibilities in which all three people get odd sums.
  5 KB (917 words) - 02:37, 12 December 2022
• 1998 AIME Problems/Problem 2
  == Problem == ...count only 1 ordered pair. By doing with 4, we count 4 ordered pairs. With 6, we get 7 pairs. With 8 we get 12. By continuing on, and then finding the d
  6 KB (913 words) - 16:34, 6 August 2020
• 1998 AIME Problems/Problem 1
  == Problem == ...{12}</math> the [[least common multiple]] of the positive integers <math>6^6</math>, <math>8^8</math>, and <math>k</math>?
  2 KB (289 words) - 22:50, 23 April 2024
• 1999 AIME Problems/Problem 15
  == Problem == ...,Q,NE);label("\(R\)",R,SE);</asy><asy>import three; defaultpen(linewidth(0.6));
  7 KB (1,107 words) - 20:34, 27 January 2023
• 1999 AIME Problems/Problem 14
  == Problem == Now <math>[CAP] + [ABP] + [BCP] = [ABC] = \sqrt {(21)(8)(7)(6)} = 84</math>, by [[Heron's formula]].
  7 KB (1,184 words) - 13:25, 22 December 2022
• 1999 AIME Problems/Problem 10
  == Problem == ...endpoints. When we select these segments, we are working with <math>4, 5, 6, 7,</math> or <math>8</math> endpoints in total.
  3 KB (524 words) - 17:25, 17 July 2023
• 1999 AIME Problems/Problem 9
  == Problem == == Solution 6 ==
  6 KB (1,010 words) - 19:01, 24 May 2023
• 1999 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1999|num-b=6|num-a=8}}
  3 KB (475 words) - 13:33, 4 July 2016
• 1999 AIME Problems/Problem 6
  == Problem == [[Image:1999_AIME-6.png]]
  2 KB (354 words) - 16:42, 20 July 2021
• 1999 AIME Problems/Problem 5
  == Problem == {{AIME box|year=1999|num-b=4|num-a=6}}
  1 KB (170 words) - 19:40, 4 July 2013
• 1999 AIME Problems/Problem 1
  == Problem == ...is divisible by <math>3</math>. However, if the common difference is <math>6</math>, we find that <math>5,11,17,23</math>, and <math>29</math> form an [
  2 KB (332 words) - 13:22, 3 August 2020
• 1999 AIME Problems/Problem 8
  == Problem == This problem just requires a good diagram and strong 3D visualization.
  3 KB (445 words) - 19:40, 4 July 2013
• 2000 AIME I Problems/Problem 15
  == Problem == Consider the general problem: with a stack of <math>n</math> cards such that they will be laid out <math
  15 KB (2,673 words) - 19:16, 6 January 2024
• 2000 AIME I Problems/Problem 13
  == Problem == D((-6,0)--(6,0),Arrows(4)); D((0,-6)--(0,6),Arrows(4)); truck((1,0)); truck((2,0)); truck((3,0)); truck((4,0));
  3 KB (571 words) - 00:38, 13 March 2014
• 2000 AIME I Problems/Problem 11
  == Problem == Essentially, the problem asks us to compute <cmath>\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b}</cmat
  4 KB (667 words) - 13:58, 31 July 2020
• 2000 AIME I Problems/Problem 7
  == Problem == note: this is the type of problem that makes you think symmetry, but actually can be solved easily with subst
  5 KB (781 words) - 15:02, 20 April 2024
• 2000 AIME I Problems/Problem 6
  == Problem == ...]s <math>(x,y)</math> of [[integer]]s is it true that <math>0 < x < y < 10^6</math> and that the [[arithmetic mean]] of <math>x</math> and <math>y</math
  6 KB (966 words) - 21:48, 29 January 2024
• 2000 AIME I Problems/Problem 5
  == Problem == If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way
  7 KB (1,011 words) - 20:09, 4 January 2024
• 2001 AIME I Problems/Problem 15
  == Problem == The numbers <math>1, 2, 3, 4, 5, 6, 7,</math> and <math>8</math> are randomly written on the faces of a regula
  11 KB (1,837 words) - 18:53, 22 January 2024
• 2001 AIME I Problems/Problem 14
  == Problem == ...ive mail and <math>1</math> represent a house that does receive mail. This problem is now asking for the number of <math>19</math>-digit strings of <math>0</m
  13 KB (2,298 words) - 19:46, 9 July 2020
• 2001 AIME I Problems/Problem 12
  == Problem == ...sphere]] is inscribed in the [[tetrahedron]] whose vertices are <math>A = (6,0,0), B = (0,4,0), C = (0,0,2),</math> and <math>D = (0,0,0).</math> The [
  6 KB (1,050 words) - 18:44, 27 September 2023
• 2014 USAJMO Problems
  ===Problem 1=== [[2014 USAJMO Problems/Problem 1|Solution]]
  3 KB (600 words) - 16:42, 5 August 2023
• 2005 AMC 10B
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[2005 AMC 10B Problems/Problem 1]]
  1 KB (165 words) - 12:40, 14 August 2020
• 2001 AIME I Problems/Problem 10
  == Problem == ...<math>4</math> OOOs. Doing a sanity check, <math>12 + 8 + 12 + 6 + 8 + 4 + 6 + 4 = 60,</math> which is the total number of points.
  8 KB (1,187 words) - 02:40, 28 November 2020
• 2001 AIME I Problems/Problem 8
  == Problem == ...at most <math>3</math> digits in base-<math>7</math>. Letting <math>a_2 = 6</math>, we find that <math>630_7 = \boxed{315}_{10}</math> is our largest 7
  3 KB (502 words) - 11:28, 9 December 2023
• 2001 AIME I Problems/Problem 7
  == Problem == ...we find that the [[inradius]] is <math>r = \frac{A}{s} = \frac{\sqrt{1311}}6</math>. Since <math>\triangle ADE \sim \triangle ABC</math>, the ratio of t
  9 KB (1,540 words) - 08:31, 1 December 2022
• 2001 AIME I Problems/Problem 6
  == Problem == Recast the problem entirely as a block-walking problem. Call the respective dice <math>a, b, c, d</math>. In the diagram below,
  11 KB (1,729 words) - 20:50, 28 November 2023
• 2001 AIME I Problems/Problem 5
  == Problem == ===Solution 6===
  6 KB (1,043 words) - 10:09, 15 January 2024
• 2001 AIME I Problems/Problem 4
  ==Problem== ...circ}</math> and <math>\angle BAC=60^{\circ}</math>, so <math>BC = 12\sqrt{6}</math>.
  3 KB (534 words) - 03:22, 23 January 2023
• 2002 AIME I Problems/Problem 15
  == Problem == ...el to <math>\overline{GF},</math> <math>BF = AG = 8,</math> and <math>GF = 6;</math> and face <math>CDE</math> has <math>CE = DE = 14.</math> The other
  7 KB (1,181 words) - 20:32, 8 January 2024
• 2002 AIME I Problems/Problem 13
  == Problem == Substituting <math>c=24</math> and solving with algebra now gives <math>a=6\sqrt{31}, b=3\sqrt{70}</math>. Now we can find <math>F</math>. Note that <m
  6 KB (974 words) - 13:01, 29 September 2023
• 2002 AIME I Problems/Problem 9
  == Problem == ...h>2</math> and <math>u</math> must be <math>4</math>, in order for <math>5,6</math> to be paint-able. Thus <math>424</math> is paintable.
  4 KB (749 words) - 19:44, 25 April 2024
• 2002 AIME I Problems/Problem 7
  == Problem == ...6 digits, we can cut out a lot of 6's from <math>858</math> to reduce the problem to finding the first three digits after the decimal of
  2 KB (316 words) - 19:54, 4 July 2013
• 2002 AIME I Problems/Problem 6
  == Problem == ...{225}(x_2)=(3+\sqrt{5})+(3-\sqrt{5})=6</math> <math>\Rightarrow x_1x_2=225^6=15^{12}</math>.
  1 KB (194 words) - 19:55, 23 April 2016
• 2002 AIME I Problems/Problem 5
  == Problem == ...diagonal only form 6 squares. So in total, we have overcounted by <math>9+6=15</math>, and <math>198-15=\fbox{183}</math>.
  1 KB (220 words) - 20:50, 12 November 2022
• 2002 AIME I Problems/Problem 3
  == Problem == ...xtraneous pairs: <math>(1,2), (1,3), (2,3), (1,4), (2,4), (1,5), (2,5), (1,6), (1,7), (1,8),</math> and <math>(1,9)</math>. <math>36-11=\boxed{025}</mat
  2 KB (246 words) - 17:02, 21 May 2023
• 2003 AIME I Problems/Problem 13
  == Problem == ...the sum of these elements is <math>\sum_{i=0}^{5} {2i \choose i} = 1 + 2 +6 + 20 + 70 + 252 = 351</math>.
  4 KB (651 words) - 19:42, 7 October 2023
• 2003 AIME I Problems/Problem 10
  == Problem == ...e); D(A--D(MP("M",M))--B); D(C--M); D(C--D(MP("N",N))--B--N--M,linetype("6 6")+linewidth(0.7));
  7 KB (1,058 words) - 01:41, 6 December 2022
• 2003 AIME I Problems/Problem 9
  == Problem == ...consecutive squares]], namely <math>\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}</math>.
  4 KB (696 words) - 11:55, 10 September 2023
• 2003 AIME I Problems/Problem 7
  == Problem == pair O=(0,0),A=(-15,0),B=(-6,0),C=(15,0),D=(0,8);
  3 KB (490 words) - 18:13, 13 February 2021
• 2003 AIME I Problems/Problem 6
  == Problem == ...h>{4\choose 3} = 4</math> triangles of the first type, and there are <math>6</math> faces, so there are <math>24</math> triangles of the first type. Eac
  3 KB (477 words) - 18:35, 27 December 2021
• 2003 AIME I Problems/Problem 5
  == Problem == ...entprojection = perspective(5,4,3); defaultpen(linetype("8 8")+linewidth(0.6));
  2 KB (288 words) - 19:58, 4 July 2013
• 2003 AIME I Problems/Problem 4
  == Problem == <cmath>\implies \frac{6}{5} = \frac{n}{10}</cmath>
  3 KB (516 words) - 21:59, 22 October 2020
• 2003 AIME I Problems/Problem 3
  == Problem == *<math>21</math> will be the greater number in <math>6</math> subsets.
  2 KB (317 words) - 00:09, 9 January 2024
• 2003 AIME I Problems/Problem 2
  == Problem == *For <math>6</math> circles, the ratio is <math>7/12</math>.
  4 KB (523 words) - 15:49, 8 March 2021
• 2003 AIME I Problems/Problem 1
  == Problem == {{(6!)!}\over{6}}={{720!}\over6}={{720\cdot719!}\over6}=120\cdot719!.</cmath>Because <math>
  756 bytes (104 words) - 07:27, 21 November 2023
• 2024 AMC 8 Problems/Problem 8
  ==Problem== ...extbf{(A) } 3\qquad\textbf{(B) } 4\qquad\textbf{(C) } 5\qquad\textbf{(D) } 6\qquad\textbf{(E) } 7</math>
  2 KB (384 words) - 22:57, 17 February 2024
• 2003 AIME II Problems/Problem 14
  == Problem == ...coordinates of its vertices are distinct elements of the set <math>\{0,2,4,6,8,10\}.</math> The area of the hexagon can be written in the form <math>m\s
  9 KB (1,461 words) - 15:09, 18 August 2023
• 2003 AIME II Problems/Problem 13
  == Problem == Notice that this problem can be converted into a Markov Chain transition matrix.
  15 KB (2,406 words) - 23:56, 23 November 2023
• 2003 AIME II Problems/Problem 10
  == Problem == ...ors even, we let the larger one equal <math>150</math> and the other <math>6</math>, which gives <math>b=48</math>. This checks, so the solution is <mat
  1 KB (218 words) - 14:14, 25 June 2021
• 2003 AIME II Problems/Problem 9
  == Problem == Consider the polynomials <math>P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x</math> and <math>Q(x) = x^{4} - x^{3} - x^{2}
  3 KB (475 words) - 21:53, 6 May 2024
• 2003 AIME II Problems/Problem 8
  ==Problem== ...g terms <math>1,2,3</math>, we use <math>-1,0,1</math> and solve the <math>6</math>th term.
  5 KB (793 words) - 15:18, 14 July 2023
• 2003 AIME II Problems/Problem 7
  == Problem == {{AIME box|year=2003|n=II|num-b=6|num-a=8}}
  2 KB (323 words) - 09:56, 16 September 2022
• 2003 AIME II Problems/Problem 6
  == Problem == G=(6.3333,4);
  5 KB (787 words) - 17:38, 30 July 2022
• 2003 AIME II Problems/Problem 5
  == Problem == ...dentical wedge and sticking it to the existing one). Thus, <math>V=\dfrac{6^2\cdot 12\pi}{2}=216\pi</math>, so <math>n=\boxed{216}</math>.
  1 KB (204 words) - 17:41, 30 July 2022
• 2003 AIME II Problems/Problem 4
  == Problem == ...h>, and calling the foot <math>M</math>, we have <math>WM=XM=\frac{\sqrt3}{6}</math>. Since <math>\cos\angle{WMX}=\cos\angle{AMX}=MX/AM=1/3</math>. By L
  3 KB (563 words) - 17:36, 30 July 2022
• 2003 AIME II Problems/Problem 3
  == Problem == ...restricted. Therefore, the number of seven-letter good words is <math>3*2^6=192</math>
  2 KB (336 words) - 17:29, 30 July 2022
• 2003 AIME II Problems/Problem 1
  == Problem == The [[product]] <math>N</math> of three [[positive integer]]s is <math>6</math> times their [[sum]], and one of the [[integer]]s is the sum of the o
  1 KB (174 words) - 08:56, 11 July 2023
• 2002 AIME II Problems/Problem 15
  == Problem == ...h>\mathcal{C}_{2}</math> intersect at two points, one of which is <math>(9,6)</math>, and the product of the radii is <math>68</math>. The x-axis and th
  7 KB (1,182 words) - 09:56, 7 February 2022
• 2002 AIME II Problems/Problem 13
  == Problem == R=(6.4615,0);
  6 KB (935 words) - 13:23, 3 September 2021
• 2002 AIME II Problems/Problem 12
  == Problem == ...an be represented by the number of paths from <math>(0,0)</math> to <math>(6,4)</math> that always stay below the line <math>y=\frac{2x}{3}</math>. We c
  7 KB (1,127 words) - 13:34, 19 June 2022
• 2002 AIME II Problems/Problem 7
  == Problem == <center><math>1^2+2^2+3^2+\ldots+k^{2}=\frac{k(k+1)(2k+1)}6</math>.</center>
  3 KB (403 words) - 12:10, 9 September 2023
• 2002 AIME II Problems/Problem 6
  == Problem == ...ve terms are <math>1,2,..,9998</math>, while the negative ones are <math>5,6,...,10002</math>. Hence we are left with <math>1000 \cdot \frac{1}{4} (1 +
  2 KB (330 words) - 05:56, 23 August 2022
• 2002 AIME II Problems/Problem 5
  == Problem == ...on-negative integers, for which <math>a^6</math> is not a divisor of <math>6^a</math>.
  3 KB (515 words) - 14:46, 14 February 2021
• 2002 AIME II Problems/Problem 4
  == Problem == [[Image:AIME 2002 II Problem 4.gif]]
  2 KB (268 words) - 07:28, 13 September 2020
• 2002 AIME II Problems/Problem 3
  == Problem == It is given that <math>\log_{6}a + \log_{6}b + \log_{6}c = 6,</math> where <math>a,</math> <math>b,</math> and <math>c</math> are [[posi
  2 KB (263 words) - 22:50, 5 April 2024
• 2001 AIME II Problems/Problem 15
  == Problem == triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8);
  4 KB (518 words) - 15:01, 31 December 2021
• 2001 AIME II Problems/Problem 13
  == Problem == ...gle{BCD}</math>, <math>AB = 8</math>, <math>BD = 10</math>, and <math>BC = 6</math>. The length <math>CD</math> may be written in the form <math>\frac {
  4 KB (743 words) - 03:32, 23 January 2023
• 2001 AIME II Problems/Problem 12
  == Problem == ...h>\frac 18</math>. Thus we have the recursion <math>\Delta P_{i+1} = \frac{6}{8} \Delta P_i</math>, and so <math>\Delta P_i = \frac 12 \cdot \left(\frac
  2 KB (380 words) - 00:28, 5 June 2020
• 2001 AIME II Problems/Problem 11
  == Problem == ...er league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are e
  3 KB (415 words) - 23:25, 20 February 2023
• 2001 AIME II Problems/Problem 10
  == Problem == ...0^{6} - 1 | 10^{6k} - 1</math>, and so any <math>\boxed{j-i \equiv 0 \pmod{6}}</math> will work.
  4 KB (549 words) - 23:16, 19 January 2024
• 2001 AIME II Problems/Problem 9
  == Problem == <math>(6)</math> Five unit squares are blue, <math>4</math> cases in all
  8 KB (1,207 words) - 20:04, 5 September 2023
• 2001 AIME II Problems/Problem 8
  == Problem == ...\frac{2001}{3^k}\right),\ 1 \le \frac{2001}{3^k} \le 3 \Longrightarrow k = 6</math>. Indeed,
  3 KB (545 words) - 23:41, 14 June 2023
• 2001 AIME II Problems/Problem 7
  == Problem == ...(D(MP("A_3",A3,NE)) -- O3 -- foot(O3, A2, O2), linetype("4 4")+linewidth(0.6));
  7 KB (1,112 words) - 02:15, 26 December 2022
• 2001 AIME II Problems/Problem 5
  == Problem == ...ides of a [[triangle]] whose area is positive. Consider sets <math>\{4, 5, 6, \ldots, n\}</math> of consecutive positive integers, all of whose ten-elem
  2 KB (286 words) - 22:32, 5 January 2024
• 2001 AIME II Problems/Problem 4
  == Problem == Let <math>R = (8,6)</math>. The lines whose equations are <math>8y = 15x</math> and <math>10y
  2 KB (240 words) - 20:34, 4 July 2013
• 2000 AIME II Problems/Problem 13
  == Problem == The [[equation]] <math>2000x^6+100x^5+10x^3+x-2=0</math> has exactly two real roots, one of which is <math
  6 KB (1,060 words) - 17:36, 26 April 2024
• 2000 AIME II Problems/Problem 11
  == Problem == pair A=(0,0), D=(1,7), Da = MP("D'",D((-7,1)),N), B=(-8,-6), C=B+Da, F=foot(A,C,D);
  4 KB (750 words) - 22:55, 5 February 2024
• 2000 AIME II Problems/Problem 7
  == Problem == ...><math>\frac 1{2!17!}+\frac 1{3!16!}+\frac 1{4!15!}+\frac 1{5!14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac N{1!18!}</math></ce
  2 KB (281 words) - 12:09, 5 April 2024
• 2000 AIME II Problems/Problem 5
  == Problem == {{AIME box|year=2000|n=II|num-b=4|num-a=6}}
  1 KB (184 words) - 21:13, 12 September 2020
• 2000 AIME II Problems/Problem 4
  == Problem == ...teger with six positive divisors, which indicates that it either is (<math>6 = 2 \cdot 3</math>) a prime raised to the <math>5</math>th power, or two pr
  2 KB (397 words) - 15:55, 11 May 2022
• 2000 AIME II Problems/Problem 2
  == Problem == <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
  == Problem == <center><math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math></center>
  2 KB (292 words) - 13:33, 4 April 2024
• 2005 AMC 10B Problems/Problem 7
  == Problem == label("$\sqrt{2}r$",(-6,0),S);
  2 KB (381 words) - 14:28, 14 December 2021
• 2005 AMC 10B Problems/Problem 10
  #REDIRECT[[2005 AMC 12B Problems/Problem 6]]
  44 bytes (5 words) - 17:37, 30 June 2011
• 2005 AMC 10B Problems/Problem 12
  == Problem == ...rac{5}{2}\left(\frac{1}{6}\right)^{11} \qquad \textbf{(E) } \left(\frac{1}{6}\right)^{10} </math>
  3 KB (385 words) - 14:03, 16 June 2022
• 2005 AMC 10B Problems/Problem 13
  == Problem == ...f <math>3</math> or <math>4</math> but not <math>12</math> are <math>3, 4, 6, 8, </math>and <math>9</math>, a total of five numbers. Since <math>\frac{5
  1 KB (212 words) - 14:44, 15 December 2021
• 2005 AMC 10B Problems/Problem 17
  == Problem == Suppose that <math>4^a = 5</math>, <math>5^b = 6</math>, <math>6^c = 7</math>, and <math>7^d = 8</math>. What is <math>a \cdot b\cdot c \cdo
  2 KB (324 words) - 15:30, 16 December 2021
• 2005 AMC 10B Problems/Problem 18
  == Problem == ...are <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, and <math>9</math>. There are only
  1 KB (195 words) - 15:33, 16 December 2021
• 2005 AMC 10B Problems/Problem 21
  == Problem == ...th>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>7</math>, <math>8</math>, <math>9</math>, or <math>10</math>,
  3 KB (398 words) - 19:17, 17 September 2023
• 2005 AMC 10B Problems/Problem 23
  == Problem == ...(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8 </math>
  3 KB (489 words) - 19:22, 17 September 2023

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