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• Mock AIME 3 Pre 2005 Problems/Problem 4
  ==Problem== {{Mock AIME box|year=Pre 2005|n=3|num-b=3|num-a=5}}
  2 KB (221 words) - 02:49, 19 March 2015
• Mock AIME 1 Pre 2005 Problems/Problem 4
  == Problem == ...<math>1 + 7 + 7^2 + \cdots + 7^{2004} = \frac{7^{2005}-1}{7-1} = \frac{7^{2005}-1}{6}</math>. Since <math>\varphi(1000) = 400</math>, by [[Fermat's Little
  685 bytes (81 words) - 10:51, 11 June 2013
• Mock AIME 2 Pre 2005 Problems/Problem 4
  == Problem == {{Mock AIME box|year=Pre 2005|n=2|num-b=3|num-a=5|source=14769}}
  817 bytes (114 words) - 17:16, 4 August 2019
• Mock AIME 4 Pre 2005/Problems/Problem 9
  ==Problem== ...h>S = \sum^{\infty}_{n=1} \frac{(7n + 32) \cdot 3^n}{n \cdot (n + 2) \cdot 4^n}</cmath>
  2 KB (319 words) - 01:21, 30 January 2024
• Mock AIME 4 Pre 2005/Problem 1
  2 bytes (1 word) - 14:19, 28 January 2024
• Mock AIME 5 Pre 2005 Problems/Problem 4
  372 bytes (62 words) - 00:05, 15 February 2024

Page text matches

• Brahmagupta's Formula
  <cmath>4[ABCD]^2=\sin^2 B(ab+cd)^2</cmath> <cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath>
  3 KB (543 words) - 19:35, 29 October 2024
• Mock AIME 2 Pre 2005
  The '''Mock AIME 2 Pre 2005''' was written by [[Art of Problem Solving]] community member Mildorf. * [[Mock AIME 2 Pre 2005 Problems|Entire Exam]]
  2 KB (181 words) - 10:58, 18 March 2015
• Mock AIME 7 Pre 2005
  The '''Mock AIME 7 Pre 2005''' was written by [[Art of Problem Solving]] community member Mildorf. * [[Mock AIME 7 Pre 2005 Problems|Entire Exam]]
  1 KB (146 words) - 16:33, 14 October 2022
• Mock AIME 1 Pre 2005 Problems
  == Problem 1 == [[Mock AIME 1 Pre 2005 Problems/Problem 1|Solution]]
  6 KB (1,100 words) - 22:35, 9 January 2016
• Mock AIME 3 Pre 2005 Problems
  ==Problem 1== [[Mock AIME 3 Pre 2005/Problem 1|Solution]]
  7 KB (1,135 words) - 23:53, 24 March 2019
• Mock AIME 3 Pre 2005 Problems/Problem 1
  ==Problem== ..., or <math>84*84=r(10+r)*21</math>, or <math>84*4=r(10+r)</math>. <math>84*4=14*24</math>, so <math>r=14</math>. Thus the area of the circle is <math>\b
  795 bytes (129 words) - 10:22, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 3
  ==Problem== <cmath>2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4</cmath>
  1 KB (191 words) - 10:22, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 5
  ==Problem== ...VC</tt> - the only other combination, two vowels, is impossible due to the problem statement). Then, note that:
  5 KB (795 words) - 16:03, 17 October 2021
• Mock AIME 3 Pre 2005 Problems/Problem 7
  ==Problem== ...ls of <math>ABCD</math> intersect at <math>P</math>. If <math>AB = 1, CD = 4,</math> and <math>BP : DP = 3 : 8,</math> then the area of the inscribed ci
  2 KB (330 words) - 10:23, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 8
  ==Problem== Here are some thoughts on the problem:
  3 KB (520 words) - 12:55, 11 January 2019
• Mock AIME 3 Pre 2005 Problems/Problem 10
  ==Problem== Therefore we have <math>a_n \equiv 6\cdot 16 - 4^2 - 4\cdot 4 - 6 = \boxed{058} \pmod{1000}</math>.
  2 KB (306 words) - 10:36, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 14
  ==Problem== ...th>C</math> and <math>D</math> respectively. If <math>AD = 3, AP = 6, DP = 4,</math> and <math>PQ = 32</math>, then the area of triangle <math>PBC</math
  3 KB (563 words) - 02:05, 25 November 2023
• Mock AIME 3 Pre 2005 Problems/Problem 15
  ==Problem== <math>\sum_{k=1}^{40} \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}}\right)</math>
  2 KB (312 words) - 10:38, 4 April 2012
• Mock AIME 1 Pre 2005 Problems/Problem 1
  == Problem == ...ds digits, then tens digits, then units digits. Every one of <math>\{1,2,3,4,5,6,7,8,9\}</math> may appear as the hundreds digit, and there are <math>9
  1 KB (194 words) - 13:44, 5 September 2012
• Mock AIME 1 Pre 2005 Problems/Problem 3
  == Problem == ...>=r, <math>BP = |r - 1|</math>, <math>CP = |r - 2|</math>, <math>DP = |r - 4|</math>, and <math>EP = |r - 13|.</math> Squaring each of these gives:
  1 KB (217 words) - 06:18, 2 July 2015
• Mock AIME 1 Pre 2005 Problems/Problem 7
  == Problem == ...numbers in the middle (those mentioned in condition [2]). There are <math>4-k</math> <tt>A</tt>s amongst the last six numbers then. Also, there are <ma
  1 KB (221 words) - 17:27, 23 February 2013
• Mock AIME 5 Pre 2005 Problems
  == Problem 1 == [[Mock AIME 5 Pre 2005 Problems/Problem 1|Solution]]
  6 KB (909 words) - 07:27, 12 October 2022
• Mock AIME 5 Pre 2005 Problems/Problem 12
  == Problem == Let <math>m = 101^4 + 256</math>. Find the sum of the digits of <math>m</math>.
  517 bytes (55 words) - 20:01, 23 March 2017
• Ball-and-urn
  ...h> objects in <math>3</math> bins. The number of ways to do such is <math>{4+3-1 \choose 3-1} = {6 \choose 2} = 15</math>. ...ach urn, then there would be <math>{n \choose k}</math> possibilities; the problem is that you can repeat urns, so this does not work.<math>n</math> and then
  5 KB (795 words) - 17:39, 31 December 2024
• Mock AIME 1 Pre 2005 Problems/Problem 12
  == Problem == ...{6}</math>. Thus, the area of <math>ABCD</math> is <math>(10\sqrt{6} + 23)(4\sqrt{6}) = 92\sqrt{6} + 240</math>, and our final answer is <math>92 + 6 +
  2 KB (376 words) - 22:41, 26 December 2016

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