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- '''2006 AIME I''' problems and solutions. The first link contains the full set of test pr * [[2006 AIME I Problems]]1 KB (135 words) - 11:31, 22 March 2011
- {{AIME Problems|year=2006|n=I}} [[2006 AIME I Problems/Problem 1|Solution]]7 KB (1,173 words) - 02:31, 4 January 2023
- ...ge 1, </math> find the minimum possible value of <math> |x_1+x_2+\cdots+x_{2006}|. </math> Suppose <math>b_{i} = \frac {x_{i}}3</math>.8 KB (1,334 words) - 16:37, 15 December 2024
- {{AIME box|year=2006|n=I|num-b=13|num-a=15}}6 KB (980 words) - 20:45, 31 March 2020
- ...rsion! Realistically, we would want to prove that the recursion works, but I currently don't know how to prove it. ...h>k</math>. Now we have the sum <math>\sum_{k=1}^{n-1}{(k-1)2^{k-1}}=\sum_{i=1}^{n-2}{k\cdot2^{k}}</math>. However, we did not consider powers of two ye10 KB (1,702 words) - 21:23, 25 July 2024
- {{AIME box|year=2006|n=I|num-b=11|num-a=13}}911 bytes (147 words) - 17:54, 13 December 2017
- {{AIME box|year=2006|n=I|num-b=10|num-a=12}}3 KB (436 words) - 04:40, 4 November 2022
- int i,j; for(i=0; i<3; i=i+1) {4 KB (731 words) - 16:59, 4 January 2022
- ...e integers. Given that <math> \log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006, </math> find the number of possible ordered pairs <math> (a,r). </math> ...th>a, r</math> [[positive integer]]s. <math>a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}</math> so <math>a^{2}r^{11}=2^{1003}</math>.4 KB (651 words) - 17:27, 22 May 2021
- ...are [[congruent (geometry)|congruent]], and each has [[area]] <math>\sqrt{2006}.</math> Let <math>K</math> be the area of rhombus <math>\mathcal{T}</math> ..., D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6);5 KB (735 words) - 15:15, 3 February 2025
- for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray);4 KB (709 words) - 00:31, 5 January 2025
- {{AIME box|year=2006|n=I|num-b=5|num-a=7}}2 KB (237 words) - 18:14, 20 November 2023
- The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </math> whe ...\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</cmath>3 KB (439 words) - 17:24, 10 March 2015
- {{AIME box|year=2006|n=I|num-b=3|num-a=5}}2 KB (353 words) - 21:56, 27 September 2024
- {{AIME box|year=2006|n=I|num-b=2|num-a=4}}4 KB (622 words) - 21:47, 13 October 2024
- {{AIME box|year=2006|n=I|num-b=1|num-a=3}}1 KB (189 words) - 19:05, 4 July 2013
- {{AIME box|year=2006|n=I|before=First Question|num-a=2}}1 KB (174 words) - 15:10, 3 February 2025
Page text matches
- * [[2006_AIME_I_Problems/Problem_1 | 2006 AIME I Problem 1]]6 KB (943 words) - 09:44, 17 January 2025
- ==== AIME I (February 6, 2025)==== ====AIME I (February 1, 2024)====19 KB (2,024 words) - 17:13, 21 February 2025
- ...am progresses. Calculators were permitted on old AMC tests; however, as of 2006 and later, calculators are not permitted for use. ...ints. From 2002 to 2006, unanswered questions were awarded 2.5 points. In 2006 and 2007, unanswered questions were awarded 2 points. Students that are in4 KB (596 words) - 03:53, 3 February 2025
- ...e worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, the number of points for an unanswered question was 2.5 points and before * [[How should I prepare?]]5 KB (646 words) - 03:52, 3 February 2025
- ...tive integer]]s as <math>n!=n \cdot (n-1) \cdots 2 \cdot 1 = \prod_{i=1}^n i</math>. Alternatively, a [[recursion|recursive definition]] for the factor <cmath>\sum_{i=1}^{100}(i!)^{2}</cmath>10 KB (809 words) - 15:40, 17 March 2024
- '''2006 AIME II''' problems and solutions. The first link contains the full set of * [[2006 AIME II Problems]]1 KB (133 words) - 11:32, 22 March 2011
- ...1 </math><div style="text-align:right;">([[2006 AMC 10A Problems/Problem 8|2006 AMC 10A, Problem 8]])</div> ...<div style="text-align:right;">([[2011 AIME I Problems/Problem 6|2011 AIME I, Problem 6]])</div>3 KB (551 words) - 15:22, 13 September 2023
- * [[Mock_AIME_2_2006-2007_Problems#Problem_8 | Mock AIME 2 2006-2007 Problem 8]] ([[number theory]]) ...ombinatorial use of recursion: [[2006_AIME_I_Problems#Problem_11|2006 AIME I Problem 11]]2 KB (316 words) - 15:03, 1 January 2024
- ''[[2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least o ''[[2002 AIME I Problems/Problem 1 | 2002 AIME I Problem 1]] Many states use a sequence of three letters followed by a seque8 KB (1,192 words) - 16:20, 16 June 2023
- [[2001 AIME I Problems/Problem 6]] [[2006 AIME II Problems/Problem 4]]1,016 bytes (141 words) - 02:39, 29 November 2021
- (2018 China Gaokao Syllabus I #17) ([[2006 AIME I Problems/Problem 14|Source]])6 KB (1,003 words) - 23:02, 19 May 2024
- * [[2006_AMC_10A_Problems/Problem_19 | 2006 AMC 10A Problem 19]] * [[2012 AIME I Problems/Problem 2]]4 KB (736 words) - 01:00, 7 March 2024
- ...+ 1} + \ldots + a_n = \left(\sum_{j=1}^n a_j\right) - a_i</math> for <math>i = 1, 2 \ldots, n</math>. Expressing the inequality in this form leads to < * [[2006_AMC_10B_Problems/Problem_10 | 2006 AMC 10B Problem 10]]2 KB (268 words) - 02:02, 3 January 2021
- '''2006 AIME I''' problems and solutions. The first link contains the full set of test pr * [[2006 AIME I Problems]]1 KB (135 words) - 11:31, 22 March 2011
- {{AIME box|year=2005|n=II|before=[[2005 AIME I]]|after=[[2006 AIME I]], [[2006 AIME II|II]]}}1 KB (135 words) - 11:30, 22 March 2011
- ! scope="row" | '''Mock AMC I''' | 200651 KB (6,175 words) - 20:41, 27 November 2024
- * Pre 2006 * 2006 - 20078 KB (901 words) - 19:45, 13 January 2025
- ! Year || Test I || Test II | 2025 || [[2025 AIME I | AIME I]] || [[2025 AIME II | AIME II]]3 KB (400 words) - 19:21, 13 February 2025
- {{AIME Problems|year=2006|n=I}} [[2006 AIME I Problems/Problem 1|Solution]]7 KB (1,173 words) - 02:31, 4 January 2023
- ...ge 1, </math> find the minimum possible value of <math> |x_1+x_2+\cdots+x_{2006}|. </math> Suppose <math>b_{i} = \frac {x_{i}}3</math>.8 KB (1,334 words) - 16:37, 15 December 2024