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- == Problem == {{AIME box|year=2006|n=I|num-b=7|num-a=9}}5 KB (735 words) - 16:15, 3 February 2025
- #REDIRECT [[2006 AIME I Problems/Problem 8]]44 bytes (5 words) - 13:38, 2 July 2009
- == 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 == {{AIME box|year=2005|n=I|num-b=7|num-a=9}}1 KB (161 words) - 17:46, 17 September 2024
- == Problem == {{AIME box|year=2004|n=I|num-b=7|num-a=9}}4 KB (620 words) - 21:26, 5 June 2021
- == Problem == Therefore we have <math>8 \cdot 6</math> normal multiples and <math>3 \cdot 2</math> "half" multiples2 KB (359 words) - 19:58, 24 December 2024
- == Problem == {{AIME box|year=1983|num-b=7|num-a=9}}3 KB (399 words) - 11:53, 30 October 2024
- == Problem == ...<math>\frac{\pi}{2}<\theta<\pi</math> is achieved when <math>\theta=\frac{8\pi}{9}=\boxed{160^\circ}</math>.4 KB (601 words) - 09:11, 23 February 2025
- == Problem == Note: please read the explanation AFTER YOU HAVE TRIED reading the problem but couldn't understand.2 KB (377 words) - 02:17, 16 February 2021
- == Problem == ...quals <math>\log_{10} (abcd...)</math> by sum-product rule of logs, so the problem is reduced to finding the logarithm base 10 of the product of the proper di3 KB (487 words) - 20:52, 16 September 2020
- == Problem == ...> for which there is a unique integer <math>k</math> such that <math>\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}</math>?2 KB (394 words) - 23:16, 14 January 2025
- == Problem == Since this is an AIME problem, there is exactly one correct answer, and thus, exactly one possible value4 KB (538 words) - 13:24, 12 October 2021
- == Problem == 2x_1+2x_2+2x_3+2x_4+2x_5+2x_6+2x_7&=100, &(8) \9 KB (1,312 words) - 12:10, 15 December 2024
- == Problem == [[File:1990 AIME Problem 8.png|center|160px]]3 KB (491 words) - 04:24, 4 November 2022
- == Problem == ...,( \pm 2, \pm 72),( \pm 3, \pm 48),( \pm 4, \pm 36),( \pm 6, \pm 24),( \pm 8, \pm 18),( \pm 9, \pm 16),( \pm 12, \pm 12)</math>; since <math>c</math> is2 KB (309 words) - 23:55, 25 July 2024
- == Problem == {{AIME box|year=1992|num-b=7|num-a=9}}5 KB (778 words) - 21:36, 3 December 2022
- == Problem == ...d back that distinct one case. Or we could just add one, making all of the problem’s different cases counted twice, and divide it all by two, which is <math9 KB (1,400 words) - 14:09, 12 January 2024
- ==Problem== ...the <math>x</math> and <math>y</math> and our displacement is <math>\left[8\sqrt{3}, \frac{(a+b)\sqrt{3}}{6}\right]</math>. (Do you see why we switched5 KB (789 words) - 21:30, 1 January 2025
- == Problem == ...\sum_{y=1}^{99} \left\lfloor\frac{100-y}{y(y+1)} \right\rfloor = 49 + 16 + 8 + 4 + 3 + 2 + 1 + 1 + 1 = \boxed{085}.</cmath>4 KB (646 words) - 12:20, 28 December 2024
- == Problem == {{AIME box|year=1996|num-b=7|num-a=9}}1 KB (155 words) - 19:32, 4 July 2013
Page text matches
- ...e many tools in a good geometer's arsenal. A very large number of geometry problems can be solved by building right triangles and applying the Pythagorean Theo <cmath>8-15-17</cmath>6 KB (978 words) - 12:02, 6 March 2025
- ...idual articles often have sample problems and solutions for many levels of problem solvers. Many also have links to books, websites, and other resources rele * CompetifyHub's Problem Sets [//competifyhub.com/resources/ Free Competition Resources for Grades 114 KB (1,913 words) - 23:52, 6 March 2025
- A more complex example is <math>\frac{x-8}{x+5}+4\ge 3</math>. \frac{x-8}{x+5}+4&\ge 312 KB (1,806 words) - 06:07, 19 June 2024
- ...sed as an abbreviation for American Math Contest, used to refer to the AMC 8, AMC 10, and AMC 12. * [[AMC 8]] — for students grades 8 and under.5 KB (696 words) - 03:47, 24 December 2019
- ...dministered by the [[Mathematical Association of America]] (MAA). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the MAA. ...ficulty=7-9|breakdown=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}}6 KB (836 words) - 04:57, 3 February 2025
- * Performance on problems at practice sessions. * Performance on AMC and AIME, including current and previous years.22 KB (3,532 words) - 11:25, 27 September 2024
- ...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
- ...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
- * <math>8! = 40320</math> ==Problems==10 KB (809 words) - 16:40, 17 March 2024
- ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME II Problems]]1 KB (133 words) - 12:32, 22 March 2011
- <center> <math>3x^2-14x+8</math> (<math>a</math> is positive) return 3x^2-14x+8;3 KB (551 words) - 16:22, 13 September 2023
- ==Problems== ...ast two digits of <math> 7^{81}-3^{81} </math>. ([[Euler's Totient Theorem Problem 1 Solution|Solution]])4 KB (569 words) - 22:34, 30 December 2024
- == 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
- ...uality]]. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures. == Problems ==6 KB (922 words) - 17:34, 13 January 2025
- 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
- ==Problems== ([[2006 AMC 10A Problems/Problem 2|Source]])10 KB (1,761 words) - 03:16, 12 May 2023
- ...ng may lead to a quick solution is the phrase "not" or "at least" within a problem statement. ''[[2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least one digit that is8 KB (1,192 words) - 17:20, 16 June 2023
- defaultpen(fontsize(8)); ==Problems==5 KB (892 words) - 21:52, 1 May 2021
- == Example Problems == *Evaluate <math>\sqrt{61_{8}}</math> as a number in the decimal system.4 KB (547 words) - 17:23, 30 December 2020
- ...sum of the two preceding it. The first few terms are <math>1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...</math>. ...d, too. Then you get something like <math>\dots -8,5,-3,2,-1,1,0,1,1,2,3,5,8\dots</math> . The ratios between successive terms has you continue backward7 KB (1,111 words) - 14:57, 24 June 2024
