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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}}

2 KB (317 words) - 01:22, 19 November 2023
1966 IMO Problems/Problem 6
== Problem == {{IMO box|year=1966|num-b=5|after=Last Problem}}

2 KB (397 words) - 00:37, 17 May 2015
2003 AMC 8 Problems/Problem 6
==Problem== label("$144$",(6,-7.5),N);

845 bytes (116 words) - 00:47, 5 July 2013
2002 AMC 8 Problems/Problem 6
==Problem==

1 KB (201 words) - 17:51, 19 December 2023
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==

1 KB (144 words) - 13:55, 23 October 2016
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>

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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>

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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>

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2011 AMC 8 Problems/Problem 6
==Problem==

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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]

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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]]

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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>

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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>

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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.<

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1968 IMO Problems/Problem 6
== Problem == {{IMO box|year=1968|num-b=5|after=Last Problem}}

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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!)

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Mock AIME 2 2006-2007/Problem 6
#REDIRECT [[Mock AIME 2 2006-2007 Problems/Problem 6]]

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Mock AIME 1 2006-2007/Problem 6
#REDIRECT [[Mock AIME 1 2006-2007 Problems/Problem 6]]

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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}

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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

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1974 AHSME Problems/Problem 6
==Problem==

1 KB (258 words) - 12:42, 5 July 2013
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>

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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]]

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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]]

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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==

2 KB (291 words) - 18:04, 14 July 2021
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>.

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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==

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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==

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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==

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2006 IMO Problems/Problem 6
==Problem== {{IMO box|year=2006|num-b=5|after=Last Problem}}

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2014 AMC 12A Problems/Problem 6
==Problem==

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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 ==

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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 ==

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2012 UNCO Math Contest II Problems/Problem 6
== Problem ==

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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]]

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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]]

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2012 UNC Math Contest II Problems/Problem 6
#REDIRECT [[2012 UNCO Math Contest II Problems/Problem 6]]

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2011 UNCO Math Contest II Problems/Problem 6
== Problem ==

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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

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: 7 Problem A/B, 3/4: 8 Problem A/B, 5/6: 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=Problem 1-2: 3-4 Problem 3-5: 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=Problem 1-5: 1 Problem 6-20: 2 Problem 21-25: 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=Problem 1-10: 2 Problem 11-20: 3 Problem 21-25: 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! ...Problem 6-10: 4 Problem 10-12: 5 Problem 12-15: 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=Problem 1/4: 7 Problem 2/5: 8 Problem 3/6: 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> 

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>. Continuing, <math>9 \equiv 3 \pmod{6}</math>, so <math>\gcd(9,6) = \gcd(6,3)</math>. 

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 Problem SL1-2: 5.5-7 Problem SL3-4: 7-8 Problem SL5+: 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.

15 KB (2,396 words) - 20:24, 21 February 2024
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>

14 KB (2,317 words) - 19:01, 29 October 2021
Right triangle
label("$45^{\circ}$", A, 6*dir(290)); [[2007 AMC 12A Problems/Problem 10 | 2007 AMC 12A Problem 10]]

3 KB (499 words) - 23:41, 11 June 2022
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]]

2 KB (180 words) - 18:06, 6 October 2014
2006 AIME I Problems
== Problem 1 == [[2006 AIME I Problems/Problem 1|Solution]]

7 KB (1,173 words) - 03:31, 4 January 2023
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

3 KB (436 words) - 05:40, 4 November 2022
2006 AIME I Problems/Problem 10
== Problem == draw((6.5,0)--origin--(0,6.5), Arrows(5));

4 KB (731 words) - 17:59, 4 January 2022
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);

4 KB (709 words) - 01:50, 10 January 2022
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}+

3 KB (439 words) - 18:24, 10 March 2015
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]]

2 KB (202 words) - 21:30, 6 October 2014
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]]

2 KB (206 words) - 23:23, 21 June 2021
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>

13 KB (1,957 words) - 12:53, 24 January 2024
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}}

1 KB (213 words) - 15:33, 9 April 2024
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<

3 KB (409 words) - 17:10, 30 April 2024
2006 AMC 12B Problems/Problem 11
== Problem == Joe has 2 ounces of cream, as stated in the problem.

927 bytes (137 words) - 10:45, 4 July 2013
2006 AMC 12B Problems/Problem 13
== Problem == ...qrt {3} \qquad \textrm{(C) } 8 \qquad \textrm{(D) } 9 \qquad \textrm{(E) } 6\sqrt {3}</math>

3 KB (447 words) - 03:49, 16 January 2021
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

4 KB (696 words) - 09:47, 10 August 2015
2006 AMC 12B Problems/Problem 21
== Problem == \mathrm{(D)}\ 6\sqrt {2006}

2 KB (339 words) - 13:15, 12 July 2015
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 ==

1 KB (167 words) - 23:23, 16 December 2021
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>

919 bytes (138 words) - 12:45, 4 August 2017
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 ==

5 KB (908 words) - 19:23, 22 September 2022
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 ==

1 KB (197 words) - 14:16, 14 December 2021
2005 AMC 12B Problems/Problem 5
== Problem == {{AMC12 box|year=2005|ab=B|num-b=4|num-a=6}}

2 KB (223 words) - 14:30, 15 December 2021
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
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

4 KB (607 words) - 21:01, 20 May 2023
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);

2 KB (262 words) - 21:20, 21 December 2020
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>

3 KB (463 words) - 19:28, 6 November 2022
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>

289 bytes (45 words) - 13:14, 16 July 2017
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]]

471 bytes (52 words) - 21:46, 12 August 2014
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));

5 KB (732 words) - 23:19, 19 September 2023
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?

1 KB (191 words) - 17:12, 29 October 2016
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

3 KB (547 words) - 19:15, 4 April 2024
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
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>

2 KB (279 words) - 12:33, 27 October 2019
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]]

45 bytes (5 words) - 10:59, 20 February 2016
2006 AMC 12A Problem 6
#REDIRECT [[2006 AMC 12A Problems/Problem 6]]

45 bytes (5 words) - 11:01, 20 February 2016
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 ==

2 KB (257 words) - 11:20, 2 January 2022
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;

4 KB (729 words) - 01:00, 27 November 2022
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

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