Create the page "2005 AIME Problems/Problem 7" on this wiki! See also the search results found.

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• 2005 AIME II Problems/Problem 7
  == Problem == ...y^6+y^4+y^2+1)(y+1) = (y^{15}+y^{14}+y^{13}+y^{12}+y^{11}+y^{10}+y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=\frac{y^{16}-1}{y-1}</cmath>
  2 KB (279 words) - 12:33, 27 October 2019
• 2005 AIME I Problems/Problem 7
  == Problem == label("$C$",(16.87,7),E);
  4 KB (567 words) - 20:20, 3 March 2020
• Mock AIME 3 Pre 2005 Problems/Problem 7
  ==Problem== {{Mock AIME box|year=Pre 2005|n=3|num-b=6|num-a=8}}
  2 KB (330 words) - 10:23, 4 April 2012
• Mock AIME 1 Pre 2005 Problems/Problem 7
  == Problem == {{Mock AIME box|year=Pre 2005|n=1|num-b=6|num-a=8|source=14769}}
  1 KB (221 words) - 17:27, 23 February 2013
• Mock AIME 5 2005-2006 Problems/Problem 7
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=6|num-a=8}}
  853 bytes (134 words) - 21:18, 8 October 2014
• Mock AIME 2 Pre 2005 Problems/Problem 7
  543 bytes (64 words) - 16:34, 24 May 2024

Page text matches

• Brahmagupta's Formula
  == Problems == ...rs. Determine <math>p + q</math>. ([[Mock AIME 3 Pre 2005 Problems/Problem 7|Source]])
  3 KB (543 words) - 19:35, 29 October 2024
• Constructive counting
  This is a problem where constructive counting is not the simplest way to proceed. This next e '''Solution''': We can model this situation as a row of 7 boxes, like this: <cmath>\square \square \square \square \square \square \s
  13 KB (2,018 words) - 15:31, 10 January 2025
• Ellipse
  MC(90,"\mbox{semiminor axis}",7,D((0,0)--(0,3),green+linewidth(1)),E); MC("\mbox{semimajor axis}",7,D((0,0)--(5,0),red+linewidth(1)),S);
  5 KB (892 words) - 21:52, 1 May 2021
• Arithmetic sequence
  ...etic sequence with common difference <math>-8</math>; however, <math>7, 0, 7, 14</math> and <math>4, 12, 36, 108, \ldots</math> are not arithmetic seque == Problems ==
  4 KB (736 words) - 02:00, 7 March 2024
• Diophantine equation
  ...s for any given Diophantine equations. This is known as [[Hilbert's tenth problem]]. The answer, however, is no. ...en, the theorem was finally proven by [[Andrew Wiles]] after he spent over 7 years working on the 200-page proof, and another year fixing an error in th
  9 KB (1,434 words) - 01:15, 4 July 2024
• 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
• 2006 AIME I
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME I Problems]]
  1 KB (135 words) - 12:31, 22 March 2011
• 2005 AIME II
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AIME II Problems]]
  1 KB (135 words) - 12:30, 22 March 2011
• Mock AMC
  ...ake. Sometimes, the administrator may ask other people to sign up to write problems for the contest. * Look at past [[AMC]]/[[AHSME]] tests to get a feel for what kind of problems you should write and what difficulty level they should be.
  51 KB (6,175 words) - 21:41, 27 November 2024
• 2006 AIME I Problems
  {{AIME Problems|year=2006|n=I}} == Problem 1 ==
  7 KB (1,173 words) - 03:31, 4 January 2023
• Set
  ...describe sets should be used with extreme caution. One way to avoid this problem is as follows: given a property <math>P</math>, choose a known set <math>T< ==Problems==
  11 KB (2,019 words) - 17:20, 7 July 2024
• 2005 AIME II Problems/Problem 2
  == Problem == *Person 1: <math>\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}</math>
  4 KB (628 words) - 11:28, 14 April 2024
• 2005 AIME II Problems/Problem 4
  == Problem == ...positive integer]]s that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math>
  3 KB (377 words) - 18:36, 1 January 2024
• 2005 AIME II Problems/Problem 13
  == Problem == ...at it has roots at <math>17</math> and <math>24</math>. Hence <math>P(x)-x+7=A(x-17)(x-24)</math>. In particular, this means that
  4 KB (642 words) - 02:14, 1 June 2024
• 2005 AIME II Problems/Problem 6
  == Problem == {{AIME box|year=2005|n=II|num-b=5|num-a=7}}
  2 KB (384 words) - 00:31, 26 July 2018
• 2005 AIME II Problems
  {{AIME Problems|year=2005|n=II}} == Problem 1 ==
  7 KB (1,119 words) - 21:12, 28 February 2020
• 2005 AIME II Problems/Problem 7
  == Problem == ...y^6+y^4+y^2+1)(y+1) = (y^{15}+y^{14}+y^{13}+y^{12}+y^{11}+y^{10}+y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=\frac{y^{16}-1}{y-1}</cmath>
  2 KB (279 words) - 12:33, 27 October 2019
• 2005 AIME II Problems/Problem 8
  == Problem == pathpen = black + linewidth(0.7);
  4 KB (693 words) - 13:03, 28 December 2021
• 2005 AIME II Problems/Problem 9
  == Problem == This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</ma
  6 KB (1,154 words) - 03:30, 11 January 2024

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