Create the page "Problem 6" on this wiki! See also the search results found.

Page title matches

• 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

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]]
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• 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]]
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• 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.
  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

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