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Page title matches

• 2010 USAJMO Problems/Problem 6
  #REDIRECT [[2010 USAMO Problems/Problem 4]]
  43 bytes (4 words) - 06:43, 3 June 2010
• 2010 AMC 12B Problems/Problem 6
  {{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #6]] and [[2010 AMC 10B Problems|2010 AMC 10B #12]]}} == Problem ==
  1 KB (166 words) - 16:58, 6 July 2023
• 2019 IMO Problems/Problem 6
  ==Problem== [[File:2019 6 s1.png|450px|right]]
  5 KB (792 words) - 01:52, 19 November 2023
• 2010 IMO Problems/Problem 6
  =2010 IMO Problem 6= == Problem ==
  4 KB (786 words) - 08:46, 12 March 2024
• 2010 AMC 10A Problems/Problem 6
  == Problem 6 ==
  1 KB (140 words) - 18:58, 31 August 2022
• 1966 AHSME Problems/Problem 6
  == Problem ==
  1 KB (164 words) - 12:42, 28 January 2020
• 2011 AMC 12A Problems/Problem 6
  == Problem ==
  2 KB (361 words) - 00:18, 16 July 2024
• 2011 AMC 10A Problems/Problem 6
  ==Problem 6== ...<math>B={6, 7, 8, 9, 10, \cdots , 20}</math>. Then, all the integers <math>6</math> through <math>20</math> would be redundant in <math>A \cup B</math>,
  1 KB (160 words) - 20:19, 21 August 2023
• 2021 AIME II Problems/Problem 6
  ==Problem==
  8 KB (1,364 words) - 01:02, 29 January 2024
• 2011 AMC 12B Problems/Problem 6
  == Problem == In order to solve this problem, use of the tangent-tangent intersection theorem (Angle of intersection bet
  2 KB (370 words) - 13:35, 26 January 2021
• 2001 AMC 10 Problems/Problem 6
  ...1 AMC 12 Problems|2001 AMC 12 #2]] and [[2001 AMC 10 Problems|2001 AMC 10 #6]]}} == Problem ==
  1 KB (183 words) - 16:12, 15 July 2024
• 2011 AIME I Problems/Problem 6
  ==Problem== ...and <cmath>y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}.</cmath> From the problem, we know that the parabola can be expressed in the form <math>y=ax^2+bx+c</
  4 KB (661 words) - 01:18, 11 December 2023
• 1999 IMO Problems/Problem 6
  ==Problem==
  1 KB (199 words) - 07:50, 24 June 2024
• 2011 AIME II Problems/Problem 6
  ==Problem 6== ...he number of ways to partition 6 into 5 non-negative parts is <math>\binom{6+4}4 = \binom{10}4 = 210</math>. The interesting quadruples correspond to pa
  9 KB (1,535 words) - 01:28, 16 January 2023
• 1951 AHSME Problems/Problem 6
  == Problem ==
  831 bytes (141 words) - 12:20, 5 July 2013
• 2011 USAMO Problems/Problem 6
  ==Problem== {{USAMO newbox|year=2011|num-b=5|aftertext=|after=Last Problem}}
  7 KB (1,209 words) - 09:59, 1 August 2024
• 2000 AMC 8 Problems/Problem 6
  ==Problem==
  2 KB (365 words) - 21:02, 28 July 2023
• 2001 AMC 8 Problems/Problem 6
  ==Problem==
  700 bytes (109 words) - 00:38, 5 July 2013
• 2011 AMC 10B Problems/Problem 6
  == Problem==
  1 KB (220 words) - 20:49, 19 July 2024
• 2007 AMC 10B Problems/Problem 6
  ==Problem== ...> points for each incorrect response, and <math>1.5</math> points for each problem left unanswered. After looking over the <math>25</math> problems, Sarah has
  1 KB (184 words) - 21:15, 25 July 2018

Page text matches

• 2006 AMC 12B Problems/Problem 19
  == Problem == \mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8
  4 KB (696 words) - 09:47, 10 August 2015
• 2006 AMC 12B Problems/Problem 21
  == Problem == \mathrm{(D)}\ 6\sqrt {2006}
  2 KB (339 words) - 13:15, 12 July 2015
• 2006 AMC 12B Problems/Problem 23
  == Problem == ...ABC</math>, such that <math>PA=11</math>, <math>PB=7</math>, and <math>PC=6</math>. Legs <math>\overline{AC}</math> and <math>\overline{BC}</math> have
  7 KB (1,169 words) - 14:04, 10 June 2022
• 2006 AMC 12B Problems/Problem 24
  == Problem == \mathrm{(C)}\ \dfrac{\pi^2}{6}
  3 KB (563 words) - 22:45, 24 October 2021
• Talk:Modular arithmetic/Intermediate
  ...h>. What if we aren't this lucky? Suppose we want to solve the following problem: Again, we will solve this problem by computing <math>13^{404}</math> modulo <math>100</math>. The first few
  4 KB (597 words) - 01:41, 19 December 2013
• 2006 AMC 12A Problems\Problem 6
  #REDIRECT [[2006 AMC 12A Problems/Problem 6]]
  45 bytes (5 words) - 14:43, 14 January 2016
• 2006 AMC 12A Problems/Problem 10
  ...ate|[[2006 AMC 12A Problems|2006 AMC 12A #10]] and [[2006 AMC 10A Problems/Problem 10|2006 AMC 10A #10]]}} == Problem ==
  1 KB (167 words) - 23:23, 16 December 2021
• 2006 AMC 12A Problems/Problem 13
  == Problem == <cmath>r_A + r_B + r_C = 6</cmath>
  1 KB (184 words) - 13:57, 19 January 2021
• 2006 AMC 12A Problems/Problem 15
  == Problem == ...{2}\qquad \mathrm{(D) \ } \frac{5\pi}{6} \quad \mathrm{(E) \ } \frac{7\pi}{6}</math>
  919 bytes (138 words) - 12:45, 4 August 2017
• 2006 AMC 12A Problems/Problem 19
  == Problem == ...-axis. The equation of <math>L_1</math> is <math>y=\frac{5}{12}x+\frac{19}{6}</math>, so the coordinate of this point is <math>\left(-\frac{38}{5},0\rig
  2 KB (253 words) - 22:52, 29 December 2021
• 2006 AMC 12A Problems/Problem 20
  ...ate|[[2006 AMC 12A Problems|2006 AMC 12A #20]] and [[2006 AMC 10A Problems/Problem 25|2006 AMC 10A #25]]}} == Problem ==
  6 KB (1,079 words) - 22:48, 12 August 2024
• 2006 AMC 12A Problems/Problem 22
  == Problem == ...rt{6}+\sqrt{3}\qquad \rm{(D) \ } 3\sqrt{2}+\sqrt{6}\qquad \mathrm{(E) \ } 6\sqrt{2}-\sqrt{3}</math>
  2 KB (343 words) - 15:39, 14 June 2023
• 2005 AMC 12A Problems/Problem 25
  == Problem == ...ill actually give an [[octahedron]], not a cube, because it only has <math>6</math> vertices.
  4 KB (495 words) - 01:36, 26 May 2024
• 2005 AMC 12B Problems/Problem 3
  ==Problem== {{AMC10 box|year=2005|ab=B|num-b=4|num-a=6}}
  978 bytes (156 words) - 14:14, 14 December 2021
• 2005 AMC 12B Problems/Problem 4
  ...C 12B Problems|2005 AMC 12B #4]] and [[2005 AMC 10B Problems|2005 AMC 10B #6]]}} == Problem ==
  1 KB (197 words) - 14:16, 14 December 2021
• 2005 AMC 12B Problems/Problem 5
  == Problem == {{AMC12 box|year=2005|ab=B|num-b=4|num-a=6}}
  2 KB (226 words) - 15:09, 23 June 2024
• 2005 AMC 12B Problems/Problem 6
  {{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #6]] and [[2005 AMC 10B Problems|2005 AMC 10B #10]]}} == Problem ==
  2 KB (299 words) - 15:29, 5 July 2022
• 2005 AMC 12B Problems/Problem 7
  == Problem == \mathrm{(A)}\ 6 \qquad
  2 KB (357 words) - 20:15, 27 December 2020
• 2005 AMC 12B Problems/Problem 9
  == Problem == ...h>85</math>. The mean is <math>\dfrac{70\,(2)+80\,(5)+85\,(4)+90\,(3)+95\,(6)}{20}=\dfrac{1720}{20}=86</math>. The difference between the mean and media
  2 KB (280 words) - 15:35, 16 December 2021
• 2005 AMC 12B Problems/Problem 11
  == Problem == ...nties, so you have <math>6</math> bills left. <math>\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15</math> ways. However, you counted the case when you
  4 KB (607 words) - 15:16, 23 June 2024

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