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• 2004 AIME I Problems/Problem 8
  == 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 ar
  4 KB (620 words) - 21:26, 5 June 2021

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• Ptolemy's theorem
  ...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> Addit ===2023 AIME I Problem 5===
  6 KB (922 words) - 17:34, 13 January 2025
• Constructive counting
  This is a problem where constructive counting is not the simplest way to proceed. This next e ...the first digit is, we know that it removes one option, so there are <math>8 - 1 = 7</math> options for the second digit.
  13 KB (2,018 words) - 15:31, 10 January 2025
• Recursion
  ...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. * [[Mock_AIME_2_2006-2007_Problems#Problem_8 | Mock AIME 2 2006-2007 Problem 8]] ([[number theory]])
  2 KB (316 words) - 16:03, 1 January 2024
• Fibonacci sequence
  ...sum of the two preceding it. The first few terms are <math>1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...</math>. ...d, too. Then you get something like <math>\dots -8,5,-3,2,-1,1,0,1,1,2,3,5,8\dots</math> . The ratios between successive terms has you continue backward
  7 KB (1,111 words) - 14:57, 24 June 2024
• Complex number
  ...ver, it is possible to define a number, <math> i </math>, such that <math> i = \sqrt{-1} </math>. If we add this new number to the reals, we will have ...rm <math> a + bi </math> where <math> a,b\in \mathbb{R} </math> and <math> i = \sqrt{-1} </math> is the [[imaginary unit]]. The set of complex numbers i
  5 KB (860 words) - 15:36, 10 December 2023
• Arithmetic sequence
  ...91, 83, 75</math> is an arithmetic sequence with common difference <math>-8</math>; however, <math>7, 0, 7, 14</math> and <math>4, 12, 36, 108, \ldots< * [[2005_AMC_10A_Problems/Problem_17 | 2005 AMC 10A Problem 17]]
  4 KB (736 words) - 02:00, 7 March 2024
• 2004 AIME I
  ...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
• 2004 AIME II
  ...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
• 2005 AIME I
  ...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
• Mock AMC
  ...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
• 2005 AIME I Problems
  {{AIME Problems|year=2005|n=I}} == Problem 1 ==
  6 KB (983 words) - 05:06, 20 February 2019
• 2004 AIME I Problems/Problem 15
  == Problem == ...h> block. Thus, our answer is <math>2^9(2^9-1)-8\cdot 2^7 = 2^{18} - 2^9 - 8 \cdot 2^7 = 2^9 \times 509</math>, and the answer is <math>\boxed{511}</mat
  9 KB (1,491 words) - 01:23, 26 December 2022
• 2004 AIME I Problems/Problem 14
  == Problem == ...e of a magician's [[cylinder|cylindrical]] tower whose [[radius]] is <math>8</math> feet. The rope is attached to the tower at ground level and to the u
  4 KB (729 words) - 01:00, 27 November 2022
• 2004 AIME I Problems/Problem 12
  == Problem == <cmath>x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup \cdots
  2 KB (303 words) - 18:43, 16 October 2024
• 2004 AIME I Problems/Problem 11
  == Problem == ...\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}\pi x
  5 KB (839 words) - 22:12, 16 December 2015
• 2004 AIME I Problems/Problem 9
  == Problem == draw((3.5,0)--(3.5,3/8),dashed);
  4 KB (647 words) - 17:43, 23 November 2024
• 2004 AIME I Problems/Problem 5
  == Problem == Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted quest
  3 KB (436 words) - 18:31, 9 January 2024
• 2004 AIME I Problems/Problem 7
  == Problem == ...th>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)</math>
  7 KB (1,099 words) - 13:41, 30 December 2024
• 2004 AIME I Problems
  {{AIME Problems|year=2004|n=I}} == Problem 1 ==
  9 KB (1,434 words) - 13:34, 29 December 2021
• 2004 AIME II Problems/Problem 14
  == Problem == ...and the inequality <math>11b + 111c \le 954 + 2b + 3c \le 954 + 2(10) + 3(8) = 998 < 1000</math>. However, for <math>9k \ge 963 \Longrightarrow n \le 3
  11 KB (1,857 words) - 12:57, 18 July 2024

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