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- == Problem == \mathrm{(B)}\ \frac 3{20}3 KB (485 words) - 13:09, 21 May 2021
- ...ate|[[2006 AMC 12A Problems|2006 AMC 12A #20]] and [[2006 AMC 10A Problems/Problem 25|2006 AMC 10A #25]]}} == Problem ==6 KB (1,083 words) - 13:05, 25 November 2024
- == Problem ==3 KB (437 words) - 22:49, 28 September 2022
- == Problem == <cmath>(0, 2, 4, 5, 9, 11, 13, 20)</cmath>3 KB (584 words) - 00:17, 17 July 2024
- == Problem ==1 KB (187 words) - 07:21, 17 March 2023
- == Problem == <math> AD = \sqrt{ (8-6)^2 + (-42-(-22))^2 } = \sqrt{ (2)^2 + (-20)^2 } = 2\sqrt{101} </math>4 KB (594 words) - 14:45, 30 July 2023
- == Problem ==3 KB (426 words) - 20:48, 1 September 2024
- == Problem == ...<math>64</math> total ways. This makes sense with the construction of the problem. If you are also confused about why we subtracted 1, just think about it th2 KB (263 words) - 17:13, 19 October 2021
- ==Problem==3 KB (437 words) - 13:26, 13 October 2024
- == Problem 20 ==1 KB (200 words) - 23:04, 31 July 2023
- == Problem == [[Image:2007 AMC 12A Problem 20.png|150px]]2 KB (317 words) - 09:10, 16 September 2022
- ==Problem==4 KB (710 words) - 12:59, 1 November 2024
- == Problem == Use areas to deal with this continuous probability problem. Set up a unit square with values of <math>a</math> on x-axis and <math>b</3 KB (552 words) - 22:26, 28 December 2020
- == Problem ==2 KB (244 words) - 06:34, 4 November 2022
- == Problem ==4 KB (663 words) - 06:32, 4 November 2022
- == Problem ==1 KB (177 words) - 17:05, 29 July 2022
- {{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #20]] and [[2002 AMC 10B Problems|2002 AMC 10B #22]]}} == Problem ==4 KB (575 words) - 13:29, 14 January 2025
- == Problem == [[Image:2003_12B_AMC-20.png]]2 KB (382 words) - 17:01, 23 November 2020
- ==Problem==3 KB (429 words) - 13:39, 13 October 2024
- ==Problem== label("\(\frac{20}{7}\)",(B+D)/2,NE);6 KB (951 words) - 15:31, 2 August 2019
Page text matches
- <cmath>20-21-29</cmath> * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]6 KB (943 words) - 09:44, 17 January 2025
- == Problem == <math>\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 40\qquad\te1 KB (169 words) - 13:59, 8 August 2021
- == Problem == ...ots + 10 = 55.</math> Knowing this, we can say that <math>11 + 12 \cdots + 20 = 155</math> and <math>21 + \cdots +30 =255</math> and so on. This is a qui2 KB (395 words) - 22:29, 3 December 2024
- == Problem == ...e largest rectangle minus the second largest rectangle, which is <math>(5x+20) - (3x+6) = 2x + 14</math>.2 KB (335 words) - 04:52, 18 December 2024
- ...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 (1,504 words) - 13:10, 1 December 2024
- ...vidual competition consists of independent student work on 8 questions for 20 minutes. Scores of the top two teams and top ten individuals count toward t ...a contest for junior high schools. This individual competition consists of 20 questions, to be completed in two 30 minute time periods. The top 10 indivi8 KB (1,182 words) - 13:26, 3 April 2024
- ...9|breakdown=<u>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) - 12:11, 20 February 2024
- \x+5-13+4x+20&\ge 3x+15 The problem here is that we multiplied by <math>x+5</math> as one of the last steps. W12 KB (1,806 words) - 05:07, 19 June 2024
- ...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 (636 words) - 21:50, 17 January 2025
- ...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 (529 words) - 08:01, 24 July 2024
- ...y geographical placement. The three top teams usually all place in the top 20, often even in the top 15 or 10. * [http://www.csulb.edu/depts/math/?q=node/20 Math Day at the Beach website.]22 KB (3,532 words) - 10:25, 27 September 2024
- ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ==== Problem ====9 KB (1,703 words) - 00:20, 7 December 2024
- * <math>20! = 2432902008176640000</math> ([[2007 iTest Problems/Problem 6|Source]])10 KB (809 words) - 15:40, 17 March 2024
- ...hat is the units digit of <math>k^2 + 2^k</math>? ([[2008 AMC 12A Problems/Problem 15]]) * Find <math>2^{20} + 3^{30} + 4^{40} + 5^{50} + 6^{60}</math> mod <math>7</math>. ([http://ww16 KB (2,660 words) - 22:42, 28 August 2024
- ...e integers. Determine <math>p + q</math>. ([[Mock AIME 3 Pre 2005 Problems/Problem 7|Source]]) ...common prime factor. What is <math>a+b+c?</math> ([[2022 AMC 10A Problems/Problem 15|Source]])3 KB (543 words) - 18:35, 29 October 2024
- ==Problem== 2. Note the commonality with [[1969 IMO Problems/Problem 5]]. In fact,6 KB (1,054 words) - 17:09, 11 December 2024
- .../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) - 21:22, 15 January 2025
- ...latively prime integers, find <math> p+q. </math> ([[2005 AIME II Problems/Problem 15|Source]]) ...ly prime positive integers. Find <math>m+n</math>. ([[2001 AIME I Problems/Problem 5|Source]])5 KB (892 words) - 20:52, 1 May 2021
- ==Problem== A=3*dir(160); B=origin; C=3*dir(110); D=3*dir(20);1 KB (160 words) - 15:53, 17 December 2020
- ...C = \frac{50}7</math>. We can plug this back in to find <math> AB = \frac{20}7 </math>. ...le ABC, let P be a point on BC and let <math> AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 </math>. Find the value of <math> m\ang3 KB (438 words) - 12:17, 30 November 2024