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Page title matches
- == Problem == label("$3$",(11/2,0),dir(270));8 KB (1,016 words) - 23:17, 30 December 2023
- == Problem == ...ing.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_10 2015 AIME II Problem 10]3 KB (436 words) - 04:40, 4 November 2022
- == Problem == Joe has 2 ounces of cream, as stated in the problem.927 bytes (137 words) - 09:45, 4 July 2013
- ...ms|2006 AMC 12A #11]] and [[2006 AMC 10A Problems/Problem 11|2008 AMC 10A #11]]}} == Problem ==898 bytes (133 words) - 16:52, 3 July 2013
- {{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #11]] and [[2005 AMC 10B Problems|2005 AMC 10B #15]]}} == Problem ==4 KB (607 words) - 14:16, 23 June 2024
- == Problem ==820 bytes (123 words) - 07:05, 17 December 2021
- == Problem == ...</math>, <math>a_{m-1}a_{m-2} = 3</math>; from the recursion given in the problem <math>a_{m-p+1} = a_{m-p-1} - 3/a_{m-p}</math>, so <math>a_{m-p+1} = 3p/a_{3 KB (499 words) - 17:52, 21 November 2022
- == Problem == [[Image:2005 AIME I Problem 11.png]]4 KB (707 words) - 10:11, 16 September 2021
- == Problem ==5 KB (839 words) - 21:12, 16 December 2015
- == Problem ==2 KB (268 words) - 21:20, 23 March 2023
- == Problem ==6 KB (971 words) - 14:35, 27 May 2024
- == Problem == ...cent pair of Birch trees as one separate tree. This then gives <math>\frac{11!}{3! \cdot 3! \cdot 4!}</math> configurations.7 KB (1,115 words) - 23:52, 6 September 2023
- == Problem == ...= Y</math>). Finding the optimal location for <math>X</math> is a classic problem: for any path from <math>F_1</math> to <math>X</math> and then back to <mat5 KB (982 words) - 23:57, 5 December 2024
- == Problem == Let <math>\vec{v} \in V</math> be the polynomial given in the problem, and it is easy to see that <math>[ \vec{v} ]_B = \langle 1, -1, 1, -1, ...6 KB (872 words) - 15:51, 9 June 2023
- == Problem == Find the largest possible value of <math>k</math> for which <math>3^{11}</math> is expressible as the sum of <math>k</math> consecutive [[positive3 KB (418 words) - 17:30, 20 January 2024
- == Problem ==2 KB (422 words) - 23:22, 5 September 2020
- == Problem ==5 KB (851 words) - 17:01, 28 December 2022
- == Problem ==3 KB (519 words) - 08:28, 28 June 2022
- == Problem == ...<math>10!</math> are <math>00</math>, so the last two digits of <math>10!+11!+...+2006!</math> are <math>00</math>.1 KB (170 words) - 13:00, 26 January 2022
- == Problem == In the present problem, <math>n_{}^{}=12</math> and <math>R_{}^{}=1</math>. It follows that <math>4 KB (740 words) - 16:46, 24 May 2024
Page text matches
- <cmath>11-60-61</cmath> * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]6 KB (943 words) - 09:44, 17 January 2025
- == Problem == ...ath>1 + 2 + 3 \cdots + 10 = 55.</math> Knowing this, we can say that <math>11 + 12 \cdots + 20 = 155</math> and <math>21 + \cdots +30 =255</math> and so2 KB (395 words) - 22:29, 3 December 2024
- == Problem == The problem states that <math>x, 2x+6, 2x+14</math> is an arithmetic progression, meani2 KB (335 words) - 04:52, 18 December 2024
- == Problem == label("$3$",(11/2,0),dir(270));8 KB (1,016 words) - 23:17, 30 December 2023
- == Problem == {{AMC10 box|year=2016|ab=A|num-b=11|num-a=13}}2 KB (302 words) - 00:57, 18 October 2024
- == Problem == {{AMC12 box|year=2016|ab=A|num-b=9|num-a=11}}2 KB (402 words) - 13:54, 25 June 2023
- These '''math books''' are recommended by [[Art of Problem Solving]] administrators and members of the [http://aops.com/community AoPS * [http://www.amazon.com/exec/obidos/ASIN/0387982191/artofproblems-20 Problem Solving Strategies] by Arthur Engel contains significant material on inequa24 KB (3,202 words) - 14:33, 13 January 2025
- ==Problem== Find the sum of the infinite series: <center><math>3+\frac{11}4+\frac 94 + \cdots + \frac{n^2+2n+3}{2^n}+\cdots</math>.</center>1 KB (193 words) - 20:13, 18 May 2021
- ...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
- ...rt of Problem Solving]] books, while not devoted specifically to contest [[problem solving]], contain scores of problems from the contests listed below, as we ...ofproblemsolving.com/store/recommendations#contest-1 two excellent contest problem books].2 KB (302 words) - 15:45, 3 October 2019
- ...[[MAML]] (Maine Association of Math Leagues) Meets. Training includes the problem set "Pete's Fabulous 42." ...th Carolina All State Mathematics Team. The qualifying floor this year was 11 out of the 25 questions. After an individual is accepted into the SC All St22 KB (3,532 words) - 10:25, 27 September 2024
- In team rounds, teams of 4 people work fast to answer a problem in 4 minutes. However, the faster a team turns in an answer, the more point A=<insert some kind of math problem here>4 KB (632 words) - 17:21, 21 December 2024
- * <math>11! = 39916800</math> ([[2007 iTest Problems/Problem 6|Source]])10 KB (809 words) - 15:40, 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) - 11:32, 22 March 2011
- ...mpute some examples, for example find <math>3^{31} \pmod{7}, 29^{25} \pmod{11}</math>, and <math>128^{129} \pmod{17}</math>, and check your answers by ca ...hat is the units digit of <math>k^2 + 2^k</math>? ([[2008 AMC 12A Problems/Problem 15]])16 KB (2,660 words) - 22:42, 28 August 2024
- draw(graph(f,1-sqrt(11),1+sqrt(11)),green+linewidth(1)); ...style="text-align:right;">([[2006 AMC 10A Problems/Problem 8|2006 AMC 10A, Problem 8]])</div>3 KB (551 words) - 15:22, 13 September 2023
- ===2023 AIME I Problem 5=== ([[2023 AIME I Problems/Problem 5|Source]])6 KB (922 words) - 16:34, 13 January 2025
- * [[2002 AMC 12A Problems/Problem 11]]1 KB (196 words) - 23:49, 5 January 2021
- === Divisibility Rule for 11 === ...s divisible by 11 if the [[alternating sum]] of the digits is divisible by 11.10 KB (1,572 words) - 21:11, 22 September 2024
- ..., then adding together the totals of each part. Casework is a very general problem-solving approach, and as such has wide applicability. '''Solution''': We divide the problem into cases, based on how long the word is.5 KB (709 words) - 16:40, 24 September 2024