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Create the page "2004 AIME I Problem 9" on this wiki! See also the search results found.
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- == Problem == draw((9*4/14,0)--(9*4/14,5*3/14),dashed);4 KB (647 words) - 17:43, 23 November 2024
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- ...lity, with equality holding when <math>A^*, B^*, C^*</math> are collinear, i.e. when <math>A,B,C</math> lie on a circle containing <math>D.</math> Addit ===2023 AIME I Problem 5===6 KB (922 words) - 17:34, 13 January 2025
- ...e multiply the possibilities for each digit to arrive at our answer: <math>9 \cdot 10 \cdot 10 \cdot 10 = 9000</math>. <math>\square</math> This is a problem where constructive counting is not the simplest way to proceed. This next e13 KB (2,018 words) - 15:31, 10 January 2025
- * [[Mock_AIME_2_2006-2007_Problems#Problem_8 | Mock AIME 2 2006-2007 Problem 8]] ([[number theory]]) *[[1994_AIME_Problems/Problem 9|1994 AIME Problem 9]]2 KB (316 words) - 16:03, 1 January 2024
- ...iv style="text-align:right">([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])</div> ...a triangle. <div style="text-align:right">(Manhattan Mathematical Olympiad 2004)</div>7 KB (1,111 words) - 14:57, 24 June 2024
- ...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
- * [[2005_AMC_10A_Problems/Problem_17 | 2005 AMC 10A Problem 17]] * [[2006_AMC_10A_Problems/Problem_19 | 2006 AMC 10A Problem 19]]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 I Problems]]1 KB (135 words) - 18:15, 19 April 2021
- ...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
- {{AIME Problems|year=2005|n=I}} == Problem 1 ==6 KB (983 words) - 05:06, 20 February 2019
- == Problem == ...ath>. Using [[complementary counting]], we see that there are only <math>2^9 - 1</math> ways.9 KB (1,491 words) - 01:23, 26 December 2022
- == Problem == ...\cdot \frac{5}{6} = \frac{5}{9},</cmath> and the answer is <math>m+n = 5 + 9 = \boxed{014}</math>.2 KB (303 words) - 18:43, 16 October 2024
- == Problem == ...^2}{24\pi - \frac{5}{3}\pi x^2}=\frac{\frac{4}{9}\pi x^3}{12\pi - \frac{4}{9}\pi x^3}</math>. We can take [[reciprocal]]s of both sides to simplify thi5 KB (839 words) - 22:12, 16 December 2015
- == Problem == {{AIME box|year=2004|num-b=9|num-a=11|n=I}}5 KB (836 words) - 07:53, 15 October 2023
- == Problem == draw((9*4/14,0)--(9*4/14,5*3/14),dashed);4 KB (647 words) - 17:43, 23 November 2024
- == Problem == ...dot 37 = 3219</math>. So the [[remainder]]s are all congruent to <math>n - 9 \pmod{37}</math>. However, these numbers are negative for our choices of <m2 KB (250 words) - 23:56, 2 December 2024
- == Problem == ...</math> is [[odd integer | odd]] and <math> a_i>a_{i+1} </math> if <math> i </math> is [[even integer | even]]. How many snakelike integers between 1003 KB (562 words) - 18:12, 4 March 2022
- == Problem == ...um of the product of each pair of distinct terms, or <math>C = \sum_{1 \le i \neq j}^{15} S_iS_j</math>. Also, we know that7 KB (1,099 words) - 13:41, 30 December 2024
- == Problem == Let <math>S = \{1,2,3,\ldots, 1000\}</math>, and <math>A_{i}= \{i \in S \mid i\, \textrm{ divides }\,1000\}</math>. The number of <math>m</math>'s that ar4 KB (620 words) - 21:26, 5 June 2021
- {{AIME Problems|year=2004|n=I}} == Problem 1 ==9 KB (1,434 words) - 13:34, 29 December 2021
