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Create the page "Problem 6" on this wiki! See also the search results found.
Page title matches
- ==Problem==1 KB (169 words) - 14:07, 5 July 2013
- == Problem == draw((0,6)--(4,6)--(4,4)--(3,4)--(3,0)--(1,0)--(1,4)--(0,4)--cycle, linewidth(1));</asy>809 bytes (117 words) - 21:45, 2 January 2023
- ==Problem==1 KB (185 words) - 18:59, 19 March 2024
- == Problem== <cmath>2(-\dfrac{y+3}{2})^2+6(-\dfrac{y+3}{2})+5y+1=0</cmath>976 bytes (151 words) - 11:57, 5 July 2013
- ==Problem==1 KB (194 words) - 00:34, 27 December 2022
- ==Problem== ...solution was posted and copyrighted by DAFR. The original thread for this problem can be found here: [https://aops.com/community/p2751173]9 KB (1,788 words) - 00:02, 30 January 2021
- == Problem == <math>0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,</math>2 KB (325 words) - 20:05, 12 June 2024
- == Problem ==2 KB (293 words) - 21:24, 21 December 2011
- == Problem == ...equal <math>x</math> and <math>y</math>. From the information given in the problem, two equations can be written:1 KB (161 words) - 13:55, 1 July 2023
- #REDIRECT [[2012 AMC 10A Problems/Problem 8]]45 bytes (5 words) - 14:32, 12 February 2012
- == Problem == ...nded by the coordinate axes and the graph of <math>ax+by=6</math> has area 6, then <math>ab=</math>911 bytes (141 words) - 21:12, 8 September 2023
- == Problem == By the given condition in the problem, all the equalities in the above discussion must hold, that is, <math>AI =4 KB (700 words) - 23:18, 28 November 2014
- == Problem == ...h> and <math>y</math> are, so we can decide what they are. Let <math>x = 1.6</math> and <math>y = 1.4</math>. We round <math>x</math> to <math>2</math>1 KB (246 words) - 07:32, 29 June 2023
- == Problem == The problem statement tells us that Xiaoli performed the following computation:1 KB (187 words) - 16:07, 18 January 2020
- == Problem ==736 bytes (114 words) - 21:31, 24 March 2022
- ==Problem == ...at <math>71</math> is prime and <math>k<71</math> by the conditions of the problem, so the denominator in the argument of this value will always be <math>71.<1 KB (233 words) - 17:15, 30 July 2022
- == Problem == {{IMO box|year=1968|num-b=5|after=Last Problem}}3 KB (427 words) - 12:49, 5 December 2023
- == Problem 6 == ...o <math>z_{1}^4</math>(if you want to know why, reread what we want in the problem!)3 KB (428 words) - 02:34, 31 December 2020
- #REDIRECT [[Mock AIME 2 2006-2007 Problems/Problem 6]]54 bytes (6 words) - 15:29, 3 April 2012
- #REDIRECT [[Mock AIME 1 2006-2007 Problems/Problem 6]]54 bytes (6 words) - 15:49, 3 April 2012
Page text matches
- === Problem 1 === [[2006 USAMO Problems/Problem 1 | Solution]]3 KB (520 words) - 09:24, 14 May 2021
- ==Problem 1== [[1991 AJHSME Problems/Problem 1|Solution]]17 KB (2,246 words) - 13:37, 19 February 2020
- #REDIRECT [[2006 AIME I Problems/Problem 6]]44 bytes (5 words) - 12:05, 28 June 2009
- ==Problem== ...x digits <math>4,5,6,7,8,9</math> in one of the six boxes in this addition problem?1 KB (191 words) - 17:12, 29 October 2016
- ==Problem== ...metric sequence to be <math>\{ g, gr, gr^2, \dots \}</math>. Rewriting the problem based on our new terminology, we want to find all positive integers <math>m5 KB (883 words) - 01:05, 2 June 2024
- == Problem == ...The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find <math>1 KB (239 words) - 11:54, 31 July 2023
- == Problem == *Person 1: <math>\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}</math>4 KB (628 words) - 11:28, 14 April 2024
- == Problem == ...d pair]]s <math> (a,b) </math> of [[integer]]s such that <math> \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </mat3 KB (547 words) - 19:15, 4 April 2024
- == Problem == ...xample, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical st2 KB (384 words) - 00:31, 26 July 2018
- == Problem 1 == ...The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find <math>7 KB (1,119 words) - 21:12, 28 February 2020
- == Problem == <cmath>(y^{12}+y^8+y^4+1)(y^2+1)=(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)</cmath>2 KB (279 words) - 12:33, 27 October 2019
- == Problem == ..._3O_3'</math> is similar to <math>O_1O_2O_2'</math> so <math>O_3O_3'=\frac{6}{14} \cdot 10=\frac{30}{7}</math>. From rectangles, <math>O_3'T=O_1T_1=4</m4 KB (693 words) - 13:03, 28 December 2021
- == Problem == This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</ma6 KB (1,154 words) - 03:30, 11 January 2024
- == Problem == ...e(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F)13 KB (2,080 words) - 21:20, 11 December 2022
- #REDIRECT [[2006 AMC 12A Problems/Problem 6]]45 bytes (5 words) - 10:59, 20 February 2016
- #REDIRECT [[2006 AMC 12A Problems/Problem 6]]45 bytes (5 words) - 11:01, 20 February 2016
- ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #4]] and [[2006 AMC 10A Problems/Problem 4|2006 AMC 10A #4]]}} == Problem ==2 KB (257 words) - 11:20, 2 January 2022
- == Problem 1 == [[2005 AIME I Problems/Problem 1|Solution]]6 KB (983 words) - 05:06, 20 February 2019
- == Problem == [[Image:2005 AIME I Problem 1.png]]1 KB (213 words) - 13:17, 22 July 2017
- == Problem == ...th>(6,334)</math>, <math>(12,167)</math>, <math>(167,12)</math>,<math>(334,6)</math>, <math>(501,4)</math>, <math>(668,3)</math>, <math>(1002,2)</math>2 KB (303 words) - 01:31, 5 December 2022
