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  • == Problem == {{AIME box|year=2011|n=II|before=First Problem|num-a=2}}
    1 KB (184 words) - 05:12, 24 November 2024
  • == Problem 2 == {{AIME box|year=2011|n=II|num-b=1|num-a=3}}
    1 KB (214 words) - 16:02, 9 August 2018
  • ==Problem== ...ath> and <math>(160+d)^\circ</math>. Since the step is <math>2d</math> the last term of the sequence is <math>(160 + 17d)^\circ</math>, which must be less
    3 KB (547 words) - 00:13, 31 January 2024
  • == Problem 4 == [[File:2011 AIME II 4.png|400px]]
    6 KB (944 words) - 16:21, 17 November 2024
  • ==Problem== The sum of the first <math>2011</math> terms of a [[geometric sequence]] is <math>200</math>. The sum of th
    3 KB (441 words) - 20:32, 20 January 2024
  • ==Problem 6== ...inations, will the placement of these objects satisfy the condition in the problem. So we know the total number of ordered quadruples is <math>(10*9*8*7/24)=2
    9 KB (1,535 words) - 00:28, 16 January 2023
  • == Problem == {{AIME box|year=2011|n=II|num-b=7|num-a=9}}
    5 KB (839 words) - 20:09, 1 October 2024
  • ==Problem== ...second, etc. up until to the right of the last. This is a stars-and-bars problem, the solution of which can be found as <math>\binom{n+k}{k}</math> where n
    7 KB (1,161 words) - 05:07, 24 November 2024
  • == Problem 12 == ...In my solution I choose to represent people by countries, because all the problem says about distinction among delegates is their country. Now, we can see th
    5 KB (848 words) - 19:53, 15 January 2025
  • ==Problem 14== ...e of these triples are repeated and all are used. By the conditions of the problem, if <math>i</math> is the same in two different triples, then the two numbe
    12 KB (2,005 words) - 04:21, 24 November 2024
  • == Problem 10 == [[Image:2011 AIMEII Problem 10 CASE 2.png|525px]]
    13 KB (1,989 words) - 12:10, 8 December 2024
  • ==Problem== [https://youtu.be/Nufc13M2Rhw?si=qA2ubynUSvE-8cAy 2011 AIME II #13]
    13 KB (2,075 words) - 16:50, 9 November 2024
  • ==Problem== ...the expansionary/recursive definition of determinants (also stated in the problem):
    6 KB (900 words) - 00:16, 25 November 2024
  • ==Problem== ...roblem visually, if that's your thing, since it is a geometric probability problem. Let <math>A = \left\lfloor\sqrt{P(x)}\right\rfloor</math> and <math>B = \s
    8 KB (1,307 words) - 16:05, 20 November 2024
  • ==Problem 9== ...math> and <math>b</math> you receive complex answers (which contradict the problem statement), but the final answer is correct.
    3 KB (466 words) - 14:06, 16 January 2023

Page text matches

  • ...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
  • == Problems == *[[2007 AMC 12A Problems/Problem 18]]
    5 KB (860 words) - 14:36, 10 December 2023
  • ...in the AoPSWiki, please consider adding them. Also, if you notice that a problem in the Wiki differs from the original wording, feel free to correct it. Fi ! Year || Test I || Test II
    3 KB (400 words) - 19:21, 13 February 2025
  • ...eorems concerning [[polygon]]s, and is helpful in solving complex geometry problems involving lengths. In essence, it involves using a local [[coordinate syst ...school students made it popular. The technique greatly simplifies certain problems.
    5 KB (856 words) - 08:41, 4 February 2025
  • ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2010 AIME II Problems|Entire Test]]
    1 KB (160 words) - 17:08, 20 June 2020
  • {{AIME Problems|year=2010|n=II}} == Problem 1 ==
    8 KB (1,246 words) - 20:58, 10 August 2020
  • ...contains the full set of test problems. The rest contain each individual problem and its solution. *[[2011 AIME I Problems]]
    1 KB (157 words) - 19:49, 29 June 2020
  • {{AIME Problems|year=2011|n=I}} == Problem 1 ==
    10 KB (1,634 words) - 21:21, 28 December 2023
  • ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[2011 AIME II Problems]]
    1 KB (156 words) - 18:47, 15 July 2020
  • == Problem == {{AIME box|year=2011|n=II|before=First Problem|num-a=2}}
    1 KB (184 words) - 05:12, 24 November 2024
  • == Problem 2 == {{AIME box|year=2011|n=II|num-b=1|num-a=3}}
    1 KB (214 words) - 16:02, 9 August 2018
  • ==Problem== ...ath> and <math>(160+d)^\circ</math>. Since the step is <math>2d</math> the last term of the sequence is <math>(160 + 17d)^\circ</math>, which must be less
    3 KB (547 words) - 00:13, 31 January 2024
  • == Problem 4 == [[File:2011 AIME II 4.png|400px]]
    6 KB (944 words) - 16:21, 17 November 2024
  • ==Problem== The sum of the first <math>2011</math> terms of a [[geometric sequence]] is <math>200</math>. The sum of th
    3 KB (441 words) - 20:32, 20 January 2024
  • ==Problem 6== ...inations, will the placement of these objects satisfy the condition in the problem. So we know the total number of ordered quadruples is <math>(10*9*8*7/24)=2
    9 KB (1,535 words) - 00:28, 16 January 2023
  • == Problem == {{AIME box|year=2011|n=II|num-b=7|num-a=9}}
    5 KB (839 words) - 20:09, 1 October 2024
  • ==Problem== ...second, etc. up until to the right of the last. This is a stars-and-bars problem, the solution of which can be found as <math>\binom{n+k}{k}</math> where n
    7 KB (1,161 words) - 05:07, 24 November 2024
  • {{AIME Problems|year=2011|n=II}} == Problem 1 ==
    8 KB (1,301 words) - 07:43, 11 October 2020
  • == Problem 12 == ...In my solution I choose to represent people by countries, because all the problem says about distinction among delegates is their country. Now, we can see th
    5 KB (848 words) - 19:53, 15 January 2025
  • ==Problem 14== ...e of these triples are repeated and all are used. By the conditions of the problem, if <math>i</math> is the same in two different triples, then the two numbe
    12 KB (2,005 words) - 04:21, 24 November 2024

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