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Create the page "2004 AIME I Problem 15" on this wiki! See also the search results found.
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- == Problem == We approach the problem by [[recursion]]. We [[partition]] the positive integers into the sets9 KB (1,491 words) - 01:23, 26 December 2022
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- This is a problem where constructive counting is not the simplest way to proceed. This next e ...proceed with the construction. If we were to go like before and break the problem down by each box, we'd get a fairly messy solution.13 KB (2,018 words) - 15:31, 10 January 2025
- * [[Mock_AIME_2_2006-2007_Problems#Problem_8 | Mock AIME 2 2006-2007 Problem 8]] ([[number theory]]) *[[1994_AIME_Problems/Problem 9|1994 AIME Problem 9]]2 KB (316 words) - 16:03, 1 January 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME I Problems]]1 KB (135 words) - 18:15, 19 April 2021
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AIME II Problems]]1 KB (135 words) - 12:24, 22 March 2011
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME I Problems]]1 KB (154 words) - 12:30, 22 March 2011
- ...A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member ...amous theorems and formulas and see if there's any way you can make a good problem out of them.51 KB (6,175 words) - 21:41, 27 November 2024
- {{AIME Problems|year=2005|n=I}} == Problem 1 ==6 KB (983 words) - 05:06, 20 February 2019
- == Problem == We approach the problem by [[recursion]]. We [[partition]] the positive integers into the sets9 KB (1,491 words) - 01:23, 26 December 2022
- == Problem == currentprojection = perspective(-2,-50,15); size(200);4 KB (729 words) - 01:00, 27 November 2022
- == Problem == ...ath>\frac{72}{5x^2} - 1 = \frac{27}{x^3} - 1</math> and so <math>x = \frac{15}{8}</math>. Then <math>k = \frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\p5 KB (839 words) - 22:12, 16 December 2015
- == Problem == A [[circle]] of [[radius]] 1 is randomly placed in a 15-by-36 [[rectangle]] <math> ABCD </math> so that the circle lies completely5 KB (836 words) - 07:53, 15 October 2023
- == Problem == Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted quest3 KB (436 words) - 18:31, 9 January 2024
- == Problem == ...of <math>x</math> in <math>P(x)</math> is <math>-1 + 2 - 3 + \ldots + 14 - 15 = -8</math>, so <math>P(x) = 1 -8x + Cx^2 + Q(x)</math>, where <math>Q(x)</7 KB (1,099 words) - 13:41, 30 December 2024
- == Problem == Let <math>S = \{1,2,3,\ldots, 1000\}</math>, and <math>A_{i}= \{i \in S \mid i\, \textrm{ divides }\,1000\}</math>. The number of <math>m</math>'s that ar4 KB (620 words) - 21:26, 5 June 2021
- {{AIME Problems|year=2004|n=I}} == Problem 1 ==9 KB (1,434 words) - 13:34, 29 December 2021
- == Problem == ...n't need to be nearly as rigorous). A more natural manner of attacking the problem is to think of the process in reverse, namely seeing that <math>n \equiv 111 KB (1,857 words) - 12:57, 18 July 2024
- ==Problem== ...math>\frac{250}{800}(60)=\frac{150}{8}</math>. The train then has <math>60-15-\frac{50}{3}-\frac{150}{8}=230/24</math> minutes left to travel 250 miles,4 KB (592 words) - 19:02, 26 September 2020
- {{AIME Problems|year=2004|n=II}} == Problem 1 ==9 KB (1,410 words) - 05:05, 20 February 2019
- {{AIME Problems|year=2003|n=II}} == Problem 1 ==7 KB (1,127 words) - 09:02, 11 July 2023
- == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]6 KB (1,100 words) - 22:35, 9 January 2016
