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Page title matches

• 2005 AMC 10B Problems/Problem 6
  #REDIRECT[[2005 AMC 12B Problems/Problem 4]]
  44 bytes (5 words) - 10:51, 29 June 2011
• 2006 USAMO Problems/Problem 6
  == Problem ==
  5 KB (986 words) - 22:46, 18 May 2015
• 2006 AMC 12A Problems/Problem 6
  ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}} == Problem ==
  3 KB (528 words) - 18:29, 7 May 2024
• 2000 AMC 12 Problems/Problem 6
  {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #6]] and [[2000 AMC 10 Problems|2000 AMC 10 #11]]}} ==Problem==
  1 KB (228 words) - 19:31, 29 April 2024
• University of South Carolina High School Math Contest/1993 Exam/Problem 6
  == Problem == ...ty of South Carolina High School Math Contest/1993 Exam/Problem 5|Previous Problem]]
  2 KB (299 words) - 21:06, 5 July 2017
• Mock AIME 1 2006-2007 Problems/Problem 6
  ==Problem== *[[Mock AIME 1 2006-2007 Problems/Problem 5 | Previous Problem]]
  3 KB (460 words) - 15:52, 3 April 2012
• Mock AIME 2 2006-2007 Problems/Problem 6
  == Problem ==
  1 KB (157 words) - 10:51, 4 April 2012
• 1959 IMO Problems/Problem 6
  == Problem ==
  2 KB (383 words) - 05:58, 11 February 2024
• 1969 Canadian MO Problems/Problem 6
  == Problem ==
  2 KB (275 words) - 20:33, 27 November 2023
• 2005 AMC 10A Problems/Problem 6
  ==Problem==
  909 bytes (130 words) - 19:09, 25 December 2022
• 2015 IMO Problems/Problem 6
  ==Problem==
  4 KB (833 words) - 01:33, 31 December 2019
• 1987 IMO Problems/Problem 6
  ==Problem==
  2 KB (430 words) - 13:03, 24 February 2024
• 1981 IMO Problems/Problem 6
  == Problem == ...<math>x=2</math>. <cmath>f(1,0)=2, f(1,1)=3, f(1,2)=4, f(1,3)=5, f(1,4)=6</cmath> This pattern can also be proved using induction. The pattern seems
  2 KB (306 words) - 18:15, 12 April 2024
• 2003 AMC 10A Problems/Problem 6
  #REDIRECT[[2003 AMC 12A Problems/Problem 6]]
  44 bytes (5 words) - 14:44, 30 July 2011
• 2003 AMC 12A Problems/Problem 6
  ...C 12A Problems|2003 AMC 12A #6]] and [[2003 AMC 10A Problems|2003 AMC 10A #6]]}} == Problem ==
  1 KB (210 words) - 15:38, 19 August 2023
• Mock AIME 3 Pre 2005 Problems/Problem 6
  ==Problem==
  3 KB (501 words) - 14:48, 29 November 2019
• 2007 AIME I Problems/Problem 6
  == Problem == ...lid moves, beginning with 0 and ending with 39. For example, <math>0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39</math> is a move sequence. How many move sequences are
  10 KB (1,519 words) - 00:11, 29 November 2023
• 2007 AIME II Problems/Problem 6
  == Problem == ...th> in the second column, we note that <math>3</math> is less than <math>4,6,8</math>, but greater than <math>1</math>, so there are four possible place
  2 KB (338 words) - 15:30, 7 August 2022
• 2007 USAMO Problems/Problem 6
  == Problem == label("$\omega_A$",p_a+x*abs(O-A)*expi(pi/6), (1,1));
  7 KB (1,274 words) - 15:11, 31 August 2017
• 2007 AMC 12B Problems/Problem 6
  ==Problem== Triangle <math>ABC</math> has side lengths <math>AB = 5</math>, <math>BC = 6</math>, and <math>AC = 7</math>. Two bugs start simultaneously from <math>A
  792 bytes (121 words) - 04:21, 15 December 2020

Page text matches

• Constructive counting
  This is a problem where constructive counting is not the simplest way to proceed. This next e ...wer is <math>8 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 = 8 \cdot 7^6 = 941,192</math>, as desired. <math>\square</math>
  12 KB (1,896 words) - 23:55, 27 December 2023
• Euclidean algorithm
  Similarly, <math>42 \equiv 6 \pmod{9}</math>, so <math>\gcd(42,9) = \gcd(9,6)</math>. <br/> Continuing, <math>9 \equiv 3 \pmod{6}</math>, so <math>\gcd(9,6) = \gcd(6,3)</math>. <br/>
  6 KB (924 words) - 21:50, 8 May 2022
• Generating function
  ...multiply the functions together, getting <math>1+3x+6x^2+8x^3+8x^4+6x^5+3x^6+x^7</math>. We want the number of ways to choose 4 eggs, so we just need to
  4 KB (659 words) - 12:54, 7 March 2022
• Binomial Theorem
  ...th>\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots</cmath> for all <math>x</math>. ...[[polynomial]]s with [[binomial coefficient]]s. For example, if a contest problem involved the polynomial <math>x^5+4x^4+6x^3+4x^2+x</math>, one could factor
  5 KB (935 words) - 13:11, 20 February 2024
• Function
  ...function: <math>f(x) = x^2 + 6</math>. The function <math>g(x) = \sqrt{x-6}</math> has the property that <math>f(g(x)) = x</math>. Therefore, <math>g ([[2006 AMC 10A Problems/Problem 2|Source]])
  10 KB (1,761 words) - 03:16, 12 May 2023
• Complementary counting
  ...ng may lead to a quick solution is the phrase "not" or "at least" within a problem statement. ''[[2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least one digit that is
  8 KB (1,192 words) - 17:20, 16 June 2023
• Bijection
  [[2008 AMC 12B Problems/Problem 22]] [[2001 AIME I Problems/Problem 6]]
  1,016 bytes (141 words) - 03:39, 29 November 2021
• Combinatorial identity
  .../www.artofproblemsolving.com/Forum/viewtopic.php?p=394407#394407 1986 AIME Problem 11] ...lving.com/Forum/resources.php?c=182&cid=45&year=2000&p=385891 2000 AIME II Problem 7]
  12 KB (1,996 words) - 12:01, 18 May 2024
• Circle
  ...- 3)^2 + (y + 6)^2 = 25</math> represents the circle with center <math>(3,-6)</math> and radius 5 units. ([[2006 AMC 12A Problems/Problem 16|Source]])
  9 KB (1,581 words) - 18:59, 9 May 2024
• Base numbers
  ...umber can be rewritten as <math>2746_{10}=2\cdot10^3+7\cdot10^2+4\cdot10^1+6\cdot10^0.</math> ...of the decimal place (recall that the decimal place is to the right of the 6, i.e. 2746.0) tells us that there are six <math>10^0</math>'s, the second d
  4 KB (547 words) - 17:23, 30 December 2020
• 2001 AMC 12 Problems/Problem 1
  == Problem == ...d \textbf{(B)}\ 3S + 2\qquad \textbf{(C)}\ 3S + 6 \qquad\textbf{(D)}\ 2S + 6 \qquad \textbf{(E)}\ 2S + 12</math>
  788 bytes (120 words) - 10:32, 8 November 2021
• 2001 AMC 12 Problems/Problem 2
  ...1 AMC 12 Problems|2001 AMC 12 #2]] and [[2001 AMC 10 Problems|2001 AMC 10 #6]]}} == Problem ==
  1,007 bytes (165 words) - 00:28, 30 December 2023
• Math Day at the Beach
  '''Math Day at the Beach''' is a [[mathematical problem solving]] festival for Southern California high school students, hosted by ...oth individual and team competition. Teams represent high schools and have 6 members each. The competition takes place on a Saturday in March.
  4 KB (644 words) - 12:56, 29 March 2017
• Fibonacci sequence
  ...iv style="text-align:right">([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])</div> ...? <div style="text-align:right">([[1998 AIME Problems/Problem 8|1998 AIME, Problem 8]])</div>
  6 KB (957 words) - 23:49, 7 March 2024
• 1988 AJHSME Problems/Problem 5
  ==Problem== label("160",(1.6,.5),NE);
  1 KB (160 words) - 16:53, 17 December 2020
• Fundamental Theorem of Calculus
  Can you do the main problem now? # Here's a slightly different way to think about the main problem, that doesn't use physics. How much does the function <math>f(x)= \frac{x^
  11 KB (2,082 words) - 15:23, 2 January 2022
• Law of Cosines
  \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2
  6 KB (1,003 words) - 00:02, 20 May 2024
• Mathew Crawford
  ...Cameron Matthews. In 2003, Crawford became the first employee of [[Art of Problem Solving]] where he helped to write and teach most of the online classes dur * [[USAMTS]] problem writer and grader (2004-2006)
  2 KB (360 words) - 02:20, 2 December 2010
• Law of Sines
  \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2
  4 KB (658 words) - 16:19, 28 April 2024
• 2005 USAMO Problems/Problem 4
  == Problem == ...2+...+n^2) =</math> <math>\dfrac{(n+1)(n+2)(2n+3)}{6}+\dfrac{n(n+1)(2n+1)}{6}=\boxed{\dfrac{2n^3+6n^2+7n+3}{3}}</math>.
  7 KB (1,276 words) - 20:51, 6 January 2024

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