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- == Problem == {{AIME box|year=2005|n=I|num-b=7|num-a=9}}1 KB (161 words) - 17:46, 17 September 2024
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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
- 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
- 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
- ...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 i5 KB (860 words) - 15:36, 10 December 2023
- ...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
- ...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
- ...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
- {{AIME Problems|year=2006|n=I}} == Problem 1 ==7 KB (1,173 words) - 03:31, 4 January 2023
- {{AIME Problems|year=2005|n=II}} == Problem 1 ==7 KB (1,119 words) - 21:12, 28 February 2020
- == Problem == ...or equal to <math>1000</math> is <math> (\sin t + i \cos t)^n = \sin nt + i \cos nt </math> true for all real <math> t </math>?6 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
- {{AIME Problems|year=2005|n=I}} == Problem 1 ==6 KB (983 words) - 05:06, 20 February 2019
- == Problem == ...th>. When <math>n = 28</math>, this product is <math>980</math>, and since AIME answers are nonnegative integers less than <math>1000</math>, we don't have8 KB (1,249 words) - 21:25, 20 November 2024
- == Problem == ...e to face. Find the number of possible distinguishable arrangements of the 8 coins.5 KB (830 words) - 01:51, 1 March 2023
- == Problem == In [[quadrilateral]] <math> ABCD,\ BC=8,\ CD=12,\ AD=10, </math> and <math> m\angle A= m\angle B = 60^\circ. </math4 KB (567 words) - 20:20, 3 March 2020
- == Problem == ...cubes together contribute a probability of <math>\left(\frac{1}{4}\right)^8 = \frac{1}{2^{16}}</math>4 KB (600 words) - 21:44, 20 November 2023
- == Problem == <center><asy>defaultpen(fontsize(8));5 KB (852 words) - 21:23, 4 October 2023
