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- == Problem == pair A=(0,0),B=(10,0),C=6*expi(pi/3);5 KB (864 words) - 19:55, 2 July 2023
- == 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) - 01:00, 19 June 2019
- == Problem == [[Image:1997_AIME-6.png]]3 KB (497 words) - 00:39, 22 December 2018
- == Problem == [[Image:AIME_1998-6.png|350px]]2 KB (254 words) - 19:38, 4 July 2013
- == Problem == [[Image:1999_AIME-6.png]]2 KB (354 words) - 16:42, 20 July 2021
- == 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</math6 KB (966 words) - 21:48, 29 January 2024
- == 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
- == 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
- == 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. Eac3 KB (477 words) - 18:35, 27 December 2021
- == Problem == G=(6.3333,4);5 KB (787 words) - 17:38, 30 July 2022
- == 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
- == 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 ra4 KB (772 words) - 19:31, 6 December 2023
- == Problem ==3 KB (433 words) - 19:42, 20 December 2021
Page text matches
- ...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 (692 words) - 22:33, 15 February 2021
- ...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."21 KB (3,500 words) - 18:41, 23 April 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 (1,107 words) - 07:35, 26 March 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
- * <math>3! = 6</math> * <math>6! = 720</math>10 KB (809 words) - 16:40, 17 March 2024
- ...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
- 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
- f.p=fontsize(6); f.p=fontsize(6);3 KB (551 words) - 16:22, 13 September 2023
- ...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
- ...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
- ...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
- ...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
- 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/Fo5 KB (768 words) - 20:45, 1 September 2022
- 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
