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- '''2005 AIME I''' problems and solutions. The first link contains the full set of test pr * [[2005 AIME I Problems]]1 KB (154 words) - 11:30, 22 March 2011
- {{AIME Problems|year=2005|n=I}} [[2005 AIME I Problems/Problem 1|Solution]]6 KB (983 words) - 04:06, 20 February 2019
- [[Image:2005 AIME I Problem 1.png]] {{AIME box|year=2005|n=I|before=First Question|num-a=2}}1 KB (213 words) - 12:17, 22 July 2017
- ...values of <math> k </math> does <math> S_k </math> contain the term <math>2005</math>? ...of the sequence <math>S_k</math> is <math>2005</math>. Then <math>1+(n-1)k=2005</math> so <math>k(n-1)=2004=2^2\cdot 3\cdot 167</math>. The [[ordered pair2 KB (303 words) - 00:31, 5 December 2022
- {{AIME box|year=2005|n=I|num-b=2|num-a=4}}2 KB (249 words) - 08:37, 23 January 2024
- {{AIME box|year=2005|n=I|num-b=3|num-a=5}}8 KB (1,249 words) - 20:25, 20 November 2024
- {{AIME box|year=2005|n=I|num-b=4|num-a=6}}5 KB (830 words) - 00:51, 1 March 2023
- ...> P </math> be the product of the nonreal roots of <math> x^4-4x^3+6x^2-4x=2005. </math> Find <math> \lfloor P\rfloor. </math> ...urth root of 2006. Then the roots of the above equation are <math>x = 1 + i^n r</math> for <math>n = 0, 1, 2, 3</math>. The two non-real members of th4 KB (686 words) - 11:52, 13 June 2024
- {{AIME box|year=2005|n=I|num-b=6|num-a=8}}4 KB (567 words) - 19:20, 3 March 2020
- {{AIME box|year=2005|n=I|num-b=7|num-a=9}}1 KB (161 words) - 16:46, 17 September 2024
- {{AIME box|year=2005|n=I|num-b=8|num-a=10}}4 KB (600 words) - 20:44, 20 November 2023
- {{AIME box|year=2005|n=I|num-b=9|num-a=11}}5 KB (852 words) - 20:23, 4 October 2023
- [[Image:2005 AIME I Problem 11.png]] {{AIME box|year=2005|n=I|num-b=10|num-a=12}}4 KB (707 words) - 10:11, 16 September 2021
- ...let <math> b </math> denote the number of positive integers <math> n \leq 2005 </math> with <math> S(n) </math> [[even integer | even]]. Find <math> |a-b| ...only such numbers less than <math>2005</math> (because <math>45^2 = 2025 > 2005</math>).4 KB (647 words) - 01:29, 4 May 2021
- ...e principle]] all of at least one of the two letters must be all together (i.e., stay in a row). ...is correlates with the problem.) Then define for each lattice point <math>(i,j)</math> its triplet thus:5 KB (897 words) - 23:21, 28 July 2022
- {{AIME box|year=2005|n=I|num-b=13|num-a=15}}3 KB (561 words) - 13:11, 18 February 2018
- pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C); {{AIME box|year=2005|n=I|num-b=14|after=Last Question}}5 KB (906 words) - 22:15, 6 January 2024
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- ==== AIME I (February 6, 2025)==== ====AIME I (February 1, 2024)====19 KB (2,028 words) - 03:31, 22 February 2025
- If <math>(A_i)_{1\leq i\leq n}</math> are finite sets, then: <center><math> \left|\bigcup_{i=1}^n A_i\right|=\sum_{i=1}^n\left|A_i\right|9 KB (1,703 words) - 00:20, 7 December 2024
- * <math>\sum_{i=0}^{k}\binom{m}{i}\binom{n}{k-i}=\binom{m+n}{k}</math> * [[2005_AIME_II_Problems/Problem_1 | 2005 AIME II Problem 1]]4 KB (638 words) - 20:55, 5 January 2025
- ...</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 100 * [[2005 AIME I Problems/Problem 5]]13 KB (2,018 words) - 14:31, 10 January 2025
- ...math> q </math> are relatively prime integers, find <math> p+q. </math> ([[2005 AIME II Problems/Problem 15|Source]]) ...re relatively prime positive integers. Find <math>m+n</math>. ([[2001 AIME I Problems/Problem 5|Source]])5 KB (892 words) - 20:52, 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) - 14:36, 10 December 2023
- * [[2005_AMC_10A_Problems/Problem_17 | 2005 AMC 10A Problem 17]] * [[2012 AIME I Problems/Problem 2]]4 KB (736 words) - 01:00, 7 March 2024
- ...re in the form <math>(P_{ia},Q_{ia})</math> for [[positive]] integer <math>i</math>. ...ural to ask whether there is a general solution for Diophantine equations, i.e., an algorithm that will find the solutions for any given Diophantine equ9 KB (1,434 words) - 00:15, 4 July 2024
- ...ot divisible by the square of a prime, find <math>m+n.</math> ([[2005 AIME I Problems/Problem 15|Source]])5 KB (948 words) - 16:04, 21 February 2025
- {{AIME box|year=2004|n=II|before=[[2004 AIME I]]|after=[[2005 AIME I]], [[2005 AIME II|II]]}}1 KB (135 words) - 11:24, 22 March 2011
- '''2005 AIME I''' problems and solutions. The first link contains the full set of test pr * [[2005 AIME I Problems]]1 KB (154 words) - 11:30, 22 March 2011
- '''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
- '''2005 AIME II''' problems and solutions. The first link contains the full set of * [[2005 AIME II Problems]]1 KB (135 words) - 11:30, 22 March 2011
- | 2005 | 200551 KB (6,175 words) - 20:41, 27 November 2024
- ...E 7 Pre 2005 | Mildorf Mock AIME 5 (chronologically, the 7th Mock AIME Pre 2005)]] ** [[Mock AIME I 2011]]8 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
- Suppose <math>b_{i} = \frac {x_{i}}3</math>. ...= \sum_{i = 0}^{2005}(b_{i} + 1)^{2} = \sum_{i = 0}^{2005}(b_{i}^{2} + 2b_{i} + 1)8 KB (1,334 words) - 16:37, 15 December 2024
- ...S'</math> and <math>S''</math> are equal if they include the same objects, i.e., if for every object <math>x</math>, we have <math>x\in S'</math> if and ([[2005 AMC 12A Problems/Problem 13|Source]])11 KB (2,019 words) - 16:20, 7 July 2024
- {{AIME Problems|year=2005|n=II}} [[2005 AIME II Problems/Problem 1|Solution]]7 KB (1,119 words) - 20:12, 28 February 2020