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- == Problem == ...Therefore, the largest element in <math>A</math> is <math>2 + \frac{m-1}{2}</math>.8 KB (1,431 words) - 17:50, 29 December 2024
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- == Proof 2 (inversion) == ...lity, with equality holding when <math>A^*, B^*, C^*</math> are collinear, i.e. when <math>A,B,C</math> lie on a circle containing <math>D.</math> Addit6 KB (922 words) - 17:34, 13 January 2025
- An example of a classic problem is as follows: "How many numbers less than or equal to 100 are divisible by either 2 or 3?"4 KB (635 words) - 12:19, 2 January 2022
- This is a problem where constructive counting is not the simplest way to proceed. This next e === Example 2 ===13 KB (2,018 words) - 15:31, 10 January 2025
- ...alues of the sequence in terms of previous values: <math>F_0=1, F_1=1, F_2=2, F_3=3, F_4=5, F_5=8</math>, and so on. ...th> for <math>n > 0</math> also has the closed-form definition <math>a_n = 2^n</math>.2 KB (316 words) - 16:03, 1 January 2024
- ...m is the sum of the two preceding it. The first few terms are <math>1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...</math>. ...ion|recursively]] as <math>F_1 = F_2 = 1</math> and <math>F_n=F_{n-1}+F_{n-2}</math> for <math>n \geq 3</math>. This is the simplest nontrivial example7 KB (1,111 words) - 14:57, 24 June 2024
- ...mplex numbers''' arise when we try to solve [[equation]]s such as <math> x^2 = -1 </math>. ...from this addition, we are not only able to find the solutions of <math> x^2 = -1 </math> but we can now find ''all'' solutions to ''every'' polynomial.5 KB (860 words) - 15:36, 10 December 2023
- For example, <math>1, 2, 3, 4</math> is an arithmetic sequence with common difference <math>1</math ...-1}, a_n, a_{n+1}</math>. In symbols, <math>a_n = \frac{a_{n-1} + a_{n+1}}{2}</math>. This is mostly used to perform substitutions, though it occasional4 KB (736 words) - 02:00, 7 March 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
- ...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 II3 KB (400 words) - 20:21, 13 February 2025
- {{AIME Problems|year=2005|n=I}} == Problem 1 ==6 KB (983 words) - 05:06, 20 February 2019
- == Problem == ...ath>(501,4)</math>, <math>(668,3)</math>, <math>(1002,2)</math> and <math>(2004,1)</math>, and each of these gives a possible value of <math>k</math>. Thus2 KB (303 words) - 01:31, 5 December 2022
- == Problem == .../math>. Using [[complementary counting]], we see that there are only <math>2^9 - 1</math> ways.9 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 == ...x roots of the form <math> z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34, </math> with <math> 0 < a_1 \le a_2 \le a_3 \le \cdots \le a2 KB (298 words) - 20:02, 4 July 2013
- == Problem == <cmath>x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32}2 KB (303 words) - 18:43, 16 October 2024
- == Problem == ...13 \pi 3^2\cdot 4 = 12 \pi</math> and has [[surface area]] <math>A = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the5 KB (839 words) - 22:12, 16 December 2015
- == Problem == ..._{ABC}}\right)^2 = \frac{15 \times 36}{2} \cdot \frac{25}{36} = \frac{375}{2}</math>.5 KB (836 words) - 07:53, 15 October 2023