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Create the page "2005 AIME II Problem 8" on this wiki! See also the search results found.
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- == 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
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- ...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
- defaultpen(fontsize(8)); ...latively prime integers, find <math> p+q. </math> ([[2005 AIME II Problems/Problem 15|Source]])5 KB (892 words) - 21:52, 1 May 2021
- *[[2007 AMC 12A Problems/Problem 18]] *[[1984 AIME Problems/Problem 8|1984 AIME Problem 8]]5 KB (860 words) - 15:36, 10 December 2023
- For example, <math>1, 2, 4, 8</math> is a geometric sequence with common ratio <math>2</math> and <math>1 * [[1965 AHSME Problems/Problem 36 | 1965 AHSME Problem 36]]4 KB (649 words) - 21:09, 19 July 2024
- ...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
- ...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
- ...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
- {{AIME Problems|year=2006|n=I}} == Problem 1 ==7 KB (1,173 words) - 03:31, 4 January 2023
- == 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{131 KB (239 words) - 11:54, 31 July 2023
- == 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
- == 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
- == 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
- {{AIME Problems|year=2005|n=II}} == Problem 1 ==7 KB (1,119 words) - 21:12, 28 February 2020
- == 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
- == 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
- == Problem == This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</ma6 KB (1,154 words) - 03:30, 11 January 2024
- == Problem == ...^2</math> equal and solving for <math>x</math> (it is helpful to scale the problem down by a factor of 50 first), we get <math>x = 250\pm 50\sqrt{7}</math>. S13 KB (2,080 words) - 13:14, 23 July 2024
- == Problem == ...eft(\frac 16 \cdot3^3\right) = 36</math>, so the ratio is <math>\frac {36}{8} = \frac 92</math> and so the answer is <math>\boxed{011}</math>.3 KB (436 words) - 20:35, 13 August 2024
- {{AIME Problems|year=2005|n=I}} == Problem 1 ==6 KB (983 words) - 05:06, 20 February 2019
