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- #REDIRECT [[2006 AMC 12A Problems/Problem 6]]45 bytes (5 words) - 11:01, 20 February 2016
- == Problem == ==Solution 6 (De Moivre's Theorem)==4 KB (686 words) - 12:52, 13 June 2024
- == Problem == We divide the problem into two cases: one in which zero is one of the digits and one in which it3 KB (562 words) - 18:12, 4 March 2022
- == 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</ma6 KB (950 words) - 14:18, 15 January 2024
- == Problem == Let <math>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math3 KB (361 words) - 20:20, 14 January 2023
- == Problem == [[Image:1984_AIME-6.png]]6 KB (1,022 words) - 19:29, 22 January 2024
- == Problem == [[Image:AIME 1985 Problem 6.png]]5 KB (789 words) - 03:09, 23 January 2023
- == Problem ==2 KB (336 words) - 14:13, 6 September 2020
- == Problem ==3 KB (530 words) - 07:46, 1 June 2018
- == Problem == [[Image:1988_AIME-6.png]]5 KB (878 words) - 23:06, 20 November 2023
- == Problem == pair A=(0,0),B=(10,0),C=6*expi(pi/3);6 KB (980 words) - 15:08, 14 May 2024
- == Problem ==2 KB (325 words) - 13:16, 26 June 2022
- == Problem ==1 KB (181 words) - 13:45, 26 January 2022
- == Problem ==3 KB (447 words) - 17:02, 24 November 2023
- == 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)</mat3 KB (455 words) - 02:03, 10 July 2021
- == Problem ==3 KB (524 words) - 18:06, 9 December 2023
- == 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
- == Problem == Incorporating this into our problem gives <math>19\times31=\boxed{589}</math>.2 KB (407 words) - 08:14, 4 November 2022
- == 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) - 00:33, 16 May 2024
- == Problem == [[Image:1997_AIME-6.png]]3 KB (497 words) - 00:39, 22 December 2018
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- ...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 (1,506 words) - 21:31, 14 May 2024
- ...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 4/6/248 KB (1,182 words) - 14:26, 3 April 2024
- ...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
- ...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 are4 KB (623 words) - 13:11, 20 February 2024
- 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
- ==Problem== =\frac{2}{2}+\frac{4}{4}+\frac{6}{8}+\frac{8}{16}+\cdots\\1 KB (193 words) - 21:13, 18 May 2021
- ...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. W12 KB (1,806 words) - 06:07, 19 June 2024
- 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
- ...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
- ...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
- ...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
- == 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
- ==Problem 1== [[2015 IMO Problems/Problem 1|Solution]]4 KB (709 words) - 15:00, 1 June 2024
- ...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
- {{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
- ...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."22 KB (3,533 words) - 20:58, 2 June 2024
- ...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 (1,130 words) - 18:48, 3 June 2024
- ...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
- 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
- ...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
