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• 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) - 12:50, 25 August 2023
• 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) - 05:34, 25 June 2022
• 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
• 1992 AJHSME Problems/Problem 6
  ==Problem== label("$6$",(2-sqrt(3)/10,0.1),WNW);
  1 KB (174 words) - 00:09, 5 July 2013
• 1958 AHSME Problems/Problem 6
  ==Problem==
  538 bytes (85 words) - 06:09, 3 October 2014
• 2003 AMC 10B Problems/Problem 6
  ...C 12B Problems|2003 AMC 12B #5]] and [[2003 AMC 10B Problems|2003 AMC 10B #6]]}} ==Problem==
  1 KB (220 words) - 14:53, 20 October 2020
• 2010 USAMO Problems/Problem 6
  == Problem == ...s is <math>28 + 5x - {x \choose 2}</math>, which is maximized at x=5 and x=6, and the maximum value is <math>43</math>. Choosing the first 5 numbers as
  3 KB (477 words) - 17:52, 15 January 2022
• 1998 AHSME Problems/Problem 6
  == Problem ==
  1 KB (157 words) - 14:28, 5 July 2013
• 1998 AJHSME Problems/Problem 6
  ==Problem== <math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math>
  2 KB (294 words) - 17:52, 26 October 2020
• 1984 AHSME Problems/Problem 6
  ==Problem==
  1 KB (184 words) - 12:49, 5 July 2013
• 1999 AMC 8 Problems/Problem 6
  ==Problem== Use logic to solve this problem. You don't actually need to use any equations.
  903 bytes (171 words) - 21:17, 30 October 2016
• 1997 AJHSME Problems/Problem 6
  ==Problem==
  1,012 bytes (143 words) - 00:26, 5 July 2013
• 1997 USAMO Problems/Problem 6
  == Problem ==
  6 KB (1,107 words) - 14:12, 12 April 2023
• 2011 IMO Problems/Problem 6
  {{IMO box|year=2011|num-b=5|after=Last Problem}}
  2 KB (317 words) - 01:22, 19 November 2023

Page text matches

• 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
• 2005 AMC 12B Problems/Problem 13
  == Problem == Suppose that <math>4^{x_1}=5</math>, <math>5^{x_2}=6</math>, <math>6^{x_3}=7</math>, ... , <math>127^{x_{124}}=128</math>. What is <math>x_1x_2
  1 KB (203 words) - 19:57, 24 December 2020
• 2005 AMC 12B Problems/Problem 14
  == Problem == ...>, is tangent to the lines <math>y=x</math>, <math>y=-x</math> and <math>y=6</math>. What is the radius of this circle?
  2 KB (278 words) - 21:12, 24 December 2020
• 2005 AMC 12B Problems/Problem 15
  == Problem == ...h> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</ma
  2 KB (411 words) - 21:02, 21 December 2020
• 2005 AMC 12B Problems/Problem 18
  == Problem == f.p=fontsize(6);
  2 KB (262 words) - 21:20, 21 December 2020
• 2005 AMC 12B Problems/Problem 19
  == Problem == ...h>. Now we have <math>(a+b)+(a-b)=11+1</math>, <math>2a=12</math>, <math>a=6</math>, then <math>b=5</math>, <math>x=65</math>, <math>y=56</math>, <math>
  2 KB (283 words) - 20:02, 24 December 2020
• 2005 AMC 12B Problems/Problem 20
  == Problem == ...and <math>h</math> be distinct elements in the set <math>\{-7,-5,-3,-2,2,4,6,13\}.</math>
  3 KB (463 words) - 19:28, 6 November 2022
• 2005 AMC 12B Problems/Problem 24
  == Problem == Our sum is simply <math>2 - 2\cdot\frac{8}{11} = \frac{6}{11}</math>, and thus we can divide by <math>2</math> to obtain <math>\frac
  4 KB (761 words) - 09:10, 1 August 2023
• 2005 AMC 12B Problems/Problem 25
  == Problem == We approach this problem by counting the number of ways ants can do their desired migration, and the
  10 KB (1,840 words) - 21:35, 7 September 2023
• Power of a Point Theorem/Introductory Problem 2
  == Problem == Applying the Power of a Point Theorem gives <math> 6\cdot x = 4\cdot 1 </math>, so <math> x = \frac 23 </math>
  289 bytes (45 words) - 13:14, 16 July 2017
• 2006 AMC 10A Problems
  ==Problem 1== [[2006 AMC 10A Problems/Problem 1|Solution]]
  13 KB (2,028 words) - 16:32, 22 March 2022
• 2006 AMC 10A Problems/Problem 8
  == Problem == <math>b=-6</math>
  2 KB (348 words) - 23:10, 16 December 2021
• 2008 USAMO
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. **[[2008 USAMO Problems/Problem 1]]
  471 bytes (52 words) - 21:46, 12 August 2014

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