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• 2005 AIME II Problems/Problem 8
  == Problem == ...eft(\sqrt{r_3^2 - O_3T^2}\right) = 2\sqrt{14^2 - \frac{58^2}{7^2}} = \frac{8\sqrt{390}}{7}</cmath>
  4 KB (693 words) - 13:03, 28 December 2021
• 2005 AIME I Problems/Problem 8
  == Problem == {{AIME box|year=2005|n=I|num-b=7|num-a=9}}
  1 KB (161 words) - 17:46, 17 September 2024
• Mock AIME 3 Pre 2005 Problems/Problem 8
  ==Problem== Let <math>N</math> denote the number of <math>8</math>-tuples <math>(a_1, a_2, \dots, a_8)</math> of real numbers such that
  3 KB (520 words) - 12:55, 11 January 2019
• Mock AIME 1 Pre 2005 Problems/Problem 8
  == Problem == Thus, the height of <math>P</math> is <math>\sqrt [3]{8} = 2</math> times the height of <math>P'</math>, and thus the height of eac
  3 KB (446 words) - 00:18, 10 February 2020
• Mock AIME 5 2005-2006 Problems/Problem 8
  == Problem == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=7|num-a=9}}
  511 bytes (79 words) - 21:18, 8 October 2014
• Mock AIME 2 Pre 2005 Problems/Problem 8
  ...03^{2002^{2001}}}\pmod{1000}</math>. The remainder of the RHS modulo <math>8</math> is trivially zero, but the remainder of the RHS modulo <math>125</ma {{Mock AIME box|year=Pre 2005|n=2|num-b=7|num-a=9}}
  1 KB (188 words) - 12:01, 10 August 2020
• Mock AIME 5 Pre 2005 Problems/Problem 8
  454 bytes (80 words) - 00:02, 15 February 2024

Page text matches

• Simon's Favorite Factoring Trick
  ...nd make the equation factorable. It can be used to solve more than algebra problems, sometimes going into other topics such as number theory. == Fun Practice Problems ==
  4 KB (682 words) - 10:25, 18 February 2025
• Principle of Inclusion-Exclusion
  ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ==== Problem ====
  9 KB (1,703 words) - 01:20, 7 December 2024
• Brahmagupta's Formula
  == Problems == ...e integers. 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 ...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
• Ellipse
  defaultpen(fontsize(8)); ==Problems==
  5 KB (892 words) - 21:52, 1 May 2021
• Complex number
  == Problems == *[[2007 AMC 12A Problems/Problem 18]]
  5 KB (860 words) - 15:36, 10 December 2023
• Geometric sequence
  For example, <math>1, 2, 4, 8</math> is a geometric sequence with common ratio <math>2</math> and <math>1 == Problems ==
  4 KB (649 words) - 21:09, 19 July 2024
• 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< == Problems ==
  4 KB (736 words) - 02:00, 7 March 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
• 2005 AIME II Problems/Problem 1
  == Problem == ...nsecutive [[integer]]s. Since 720 is close to <math>9^3=729</math>, we try 8, 9, and 10, which works, so <math>n - 3 = 10</math> and <math>n = \boxed{13
  1 KB (239 words) - 11:54, 31 July 2023
• 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 == <math>15^7 = 3^7\cdot5^7</math> so <math>15^7</math> has <math>8\cdot8 = 64</math> divisors.
  3 KB (377 words) - 18:36, 1 January 2024
• 2005 AIME II Problems/Problem 5
  == Problem == ...og_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </math>
  3 KB (547 words) - 19:15, 4 April 2024
• 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 == Let <math> x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}. </math> Find <math>(x+1)^{48}</math>.
  2 KB (279 words) - 12:33, 27 October 2019

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